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November 19

Time Dilation? Time speeding?

I think the following section in Special Relativity has a wrong conclusion, “This phenomenon is called time dilation.”, △t’=γ△t should mean time speeding, isn’t it? Please help. Thanks.

Time dilation and length contraction

Writing the Lorentz transformation and its inverse in terms of coordinate differences, where for instance one event has coordinates (x1,t1) and (x'1,t'1), another event has coordinates (x2,t2) and (x'2,t'2), and the differences are defined as Δx = x2 − x1, Δt = t2 − t1, Δx' = x'2 − x'1, Δt' = t'2 − t'1 , we get △t’=γ(△t-(v△x/c^2))， △x’=γ(△x-v△t)， and △t=γ(△t’+(v△x’/c^2)), △x=γ(△x’+v△t’)， Suppose we have a clock at rest in the unprimed system S. Two consecutive ticks of this clock are then characterized by Δx = 0. If we want to know the relation between the times between these ticks as measured in both systems, we can use the first equation and find: △t’=γ△t (for events satisfying Δx = 0) This shows that the time Δt' between the two ticks as seen in the 'moving' frame S' is larger than the time Δt between these ticks as measured in the rest frame of the clock. This phenomenon is called time dilation. Jh17710 (talk) 05:09, 19 November 2010 (UTC) [reply]

lengthy disscussion

I am having a bit of deja vu. Oh, yeah, that's because this question has been asked, by you, before, and answered already. To save everyone the trouble of answering it again, here are the responses from the first time it was answered. If there is something you need clarified from those responses, please feel free to ask that. --Jayron32 05:19, 19 November 2010 (UTC)[reply]

Jayron, Thanks for you link. There were 3 responses and only the 2nd one responded to my explanation, however, there is no name to that 2nd response.

This time, I did not put my explanation and just ask question. If △t’=γ△t and △t'=△t/γ are both for the same moving frame S' and rest frame S, even thay are for different events, △t’ is for the moving clock and △t is for the rest clock. My question is so simple, if we call △t'=△t/γ "time dilation" or "moving clock runs slow" then, we should not call △t’=γ△t the same names, right?

I will respond to your comment and the comment from RedAct when I have time to study it. What I hope is that someone else could help me to understand my simple question.Jh17710 (talk) 06:22, 19 November 2010 (UTC)[reply]

Actually, since neither clock has a privileged frame of reference, when each observer views the other person's clock, he sees it as running slower than his own. There is no time speeding because no observer will ever view another clock as running faster than his own; every observer takes their own frame of reference to be at rest. This is covered at Time_dilation#Relative_velocity_time_dilation. --Jayron32 06:57, 19 November 2010 (UTC)[reply]

You're still making the same conceptual error I pointed out last time, by considering there to be "the moving clock" and "the rest clock". Each observer uses a system of synchronized clocks that are at rest according to that observer, not just one clock. In the example in the article, the clock being examined is a clock in the unprimed system. That clock is "moving", from the perspective of the primed observer. For the two events in the example, △t' is measured with two different clocks in the primed system, so there isn't a clock in the primed system that appears to be running fast according to the unprimed observer. △t' appears "too large" to the unprimed observer, not because there's a clock in the primed system that appears to be running fast, but because the unprimed observer considers the primed system's clocks to not be synchronized properly. Red Act (talk) 12:28, 19 November 2010 (UTC)[reply]

Δt is the elapsed time, not the rate of the clock. These are reciprocals of each other. (The elapsed time might be measured in seconds, the clock rate in Hertz.) Δt' = γΔt means that Δt' is larger (since γ > 1), so the clock is running slower, in some sense, in the primed frame. Does that help? -- BenRG (talk) 00:24, 20 November 2010 (UTC)[reply]

Jayron32, you clearly explained "moving clock runs slow". You also mentioned about one of the key point "If ... each is in communication with the other, ..." and according to Red Act, because the speed of light is finite there having to be a delay for light to travel the distance between the event and the clock, I think the delay is also true for your way of communication. However, my question is for the difference of two "equations".

Red Act, let's have two systems of clocks, not just two clocks with two observers. When you said "...from the perspective of the primed observer." did you mean the primed system is considered the rest system? In that situation Δt' = γΔt is Δt = Δt'/γ , that is the same as Δt' = Δt/γ when S' is considered moving. Then, in your analysis, △t' is measured with two different clocks in the primed system, S'; since all clocks in the S' are synchronized, △t' is the actual time period of two ticks measured by the system of clocks in the S', so that "a" clock in the system of clocks should have same speed as the system of clocks. Finally, your main point is that the unprimed observer considers the primed system's clocks to not be synchronized; I think that could be one way to explain the difference between two systems of clocks if you can explain it in detail. That is why the observer in S considers clocks in S are synchronized and clocks in S' are not synchronized.

Of course there isn't really any such thing as a "rest system" or a "moving system"; all inertial frames of reference can equally validly be said to be "at rest". However, for the purposes of illustrating the principle that "moving clocks run slow", in this particular example it's the unprimed system that needs to be considered to be "moving", and the primed system that needs to be considered to be "at rest". That's because the clock in question isn't moving (is at rest) according to the unprimed system, but it is moving according to the primed system. Unfortunately, the article referred to S' as the "moving" system, which is the opposite of the perspective needed to make the article's analysis mesh with the phrase "moving clocks run slow". I just this morning dropped the word "moving" from the article, because "moving" is basically meaningless when talking about a reference frame, and because it suggests adopting the opposite perspective of what's more appropriate for thinking about time dilation and length contraction. I also changed the article to point out that it's a clock in the unprimed frame that's running slow, as perceived from the primed frame (as opposed to a clock in the primed frame that's running fast).

You said "Δt' = γΔt is Δt = Δt'/γ , that is the same as Δt' = Δt/γ when S' is considered moving". However, it's important to be mindful that Δt' = Δt/γ applies to a different set of circumstances than Δt = Δt'/γ does. It’s not just a matter of a choice as to which system is considered to be moving; the two equations apply to different pairs of events. (In the following, I'm presuming no motion in the y or z directions.) Δt = Δt'/γ only applies to pairs of events such that Δx=0, which is the case in the article's example. Δt' = Δt/γ would only apply to a pair of events such that Δx'=0, which is not the case in the article's example. The only way Δx=0 and Δx'=0 would both be true for a pair of events would be if S and S' were at rest with respect to each other, which gives the trivial case of γ=1, in which case there’s no contradiction between Δt = Δt'/γ and Δt' = Δt/γ.

It isn't possible to explain the differences between the systems of clocks purely in terms of synchronization issues (which is formally known as relativity of simultaneity). Synchronization issues and time dilation both exist, and are distinct. Consider the equation from the article

\Delta t'=\gamma \left(\Delta t-{\frac {v\,\Delta x}{c^{2}}}\right)

. Basically the phrase "time dilation" is referring to the γ factor not being 1, whereas the phrase "relativity of simultaneity" is referring to the

-{\frac {v\,\Delta x}{c^{2}}}

term not being zero.

You are correct in your observation that all clocks in the primed system run at the same speed. Both observers will agree that all clocks in the primed system all run at the same speed (although they will disagree about how fast it is that they are running). However, it still matters which specific clocks in the primed system you use, because according to the unprimed observer, the primed clocks do not all show the same time. "Showing the same time" means to look at all of the clocks simultaneously, and note that at that instant, the clocks have the same value. For example, if you look at all the clocks simultaneously, and they all say "noon", then the clocks all "show the same time". The problem is that the two observers do not agree on what "simultaneous" is. For more information, see relativity of simultaneity. Red Act (talk) 13:13, 20 November 2010 (UTC)[reply]

-{\frac {v\,\Delta x}{c^{2}}}

is always 0 because the v is always 0 in Lorentz Transformation (LT). Why? If you apply the time equation to the spatial equation, you will get the spatial equation of inverse LT, x=γ(x'+vt'). That means, both of LT and inverse LT are coexist at the same time. For v and -v to coexist at the same time the speed v must be 0. Basically, there is a time equation missing in LT, that is t'=γt. Why? Let B(b,0,0) be a point in S. When b<0, let observers in S record the time that the origin point O' of S' moves from B to O by Δt = (0-t) and the distance between B and O by BO, we have BO = |v|Δt; let observers in S' measure the time O' moves from B to O by Δt' = (0-t') then we can calculate the distance between B and O as measured in S' by (BO)' = |-v|Δt'; based on hypothesis of ruler contraction, we have (BO)' = γBO so that |-v|Δt' = γ|v|Δt, then, △t’=γ△t. (0-t')=γ(0-t) so that t’=γt, that means when O' is at B, the time relationship between S' and S is t’=γt. By quite the same procedure, when O' is at B while b>0, t’=γt is also true. When b=0, t'=t=0, so that t’=γt. We have now derived that for any location of O', t’=γt is always true. I think the LT can be written as t’=γt, x'=γ(x-vt), y'=y, and z'=z.Jh17710 (talk) 23:20, 20 November 2010 (UTC)[reply]

You are making the mistake of using (BO)' to mean two different things, and then assuming that the two values of (BO)' are equal. For the equation (BO)'=|-v|Δt' to be valid, (BO)' must be the distance as measured in the primed system between two events such that one event is at B, one event is at O, and the two events are simultaneous as measure in the primed system. But for the equation (BO)' =γBO to be valid, (BO)' must be the distance as measured in the primed system between two events such that one event is at B, one event is at O, and the two events are simultaneous as measure in the unprimed system. You wound up reaching the conclusion that v=0 because it is only when v=0 that those two different meanings of (BO)' have the same value. The problem again boils down to the fact that different observers do not agree on what "simultaneous" is. Please see relativity of simultaneity. Red Act (talk) 04:41, 21 November 2010 (UTC)[reply]

Thanks for your response. I can understand that to measure a moving ruler, we need to snapshot both ends, so that simultaneity is required. However, for a given point B, when B is far away from O, we don't have long ruler in S' to snapshot the length of the moving BO. I think we can use some other way to get it like using the speed |-v| and the measured time period to calculate it. Could you tell me what is wrong to do so? Then, I think you understand that to me, the second (BO)' is also not a measured result. For (BO)' = γBO to be valid, I rely it on the "hypothesis of ruler contraction" but you said that the 2nd {(BO)' must be the distance as measured in the primed system between two events such that one event is at B, one event is at O, and the two events are simultaneous as measure in the unprimed system.}, do you mean, I cannot link the measured BO in S and the calculated (BO)' in S' for fixed points B and O by the "hypothesis of ruler contraction"? Why?Jh17710 (talk) 07:04, 21 November 2010 (UTC)[reply]

Two (BO)'

Yes, the distance between the moving points B and O as measured in S' is (BO)'=vΔt', where Δt' is the time elapsed on a clock stationed at O' as it passes from B to O.

If you take BO to be the distance between B and O as measured in S, then the correct length contraction equation for measuring the distance between B and O as measured in S' is (BO)'=BO/γ.

It does work to equate the two (BO)' if you define the two (BO)' as in this post. With the corrected length contraction equation, both versions of (BO)' are the distance as measured in S' between two events such that one event is at B, one event is at O, and the two events are simultaneous as measured in S'. Red Act (talk) 16:02, 21 November 2010 (UTC)[reply]

Fine, just go by your (BO)'=(BO/γ), so that we have |-v|Δt' = |v|Δt/γ in LT, that means, the time equation in LT could be derived from Δt' = Δt/γ, and we have t'=t/γ follow my reasoning for all posible location of O'. That result still show that the inverse LT could have time equation of t=t'/γ; and the fact that the time equation in LT caused the inverse LT to coexist with LT can put both of t'=t/γ and t=t'/γ to coexist at the same time, and we still have the result that γ=1 and v=0 must be true in LT; mathematically.Jh17710 (talk) 17:29, 25 November 2010 (UTC)[reply]

Δt' = Δt/γ is a correct equation if the Δt' and Δt are between two events which occur at the same location according to S' . Δt = Δt'/γ is a correct equation if the Δt' and Δt are between two events which occur at the same location according to S. If you use both equations, you are using Δt to mean two different things, and using Δt' to mean two different things. Red Act (talk) 08:00, 27 November 2010 (UTC)[reply]

Red Act, this Δt' = Δt/γ means (0-t')=(0-t)/γ in our discussion, after you insisted (BO)'=(BO/γ). Follow (0-t')=(0-t)/γ, we have t'=t/γ, and the main point here is that, besides t'=γ(t-(vx/c^2)) LT also includes the time equation t'=t/γ. I did not put Δt' = Δt/γ and Δt = Δt'/γ together. I just added t'=t/γ into LT; and for the same reasoning, we have t=t'/γ in inverse LT. Then what happened is that within LT we find inverse LT, so that it is LT which puts t'=t/γ and t=t'/γ together. Now, you say that we could not use Δt to mean two different things, you should tell Lorentz that.Jh17710 (talk) 04:19, 28 November 2010 (UTC)[reply]

The LT does not include the equation t'=t/γ. The equation t'=t/γ is incorrect, except for an event such that x=0, or in the case that v=0. Similarly, the equation t=t'/γ is incorrect, except for an event such that x'=0, or in the case that v=0. Red Act (talk) 13:41, 28 November 2010 (UTC)[reply]

Please refer to your comment dated 16:02, 21 November 2010, you said (BO)'=vΔt' is fine and let BO to be the distance between B and O as measured in S, then (BO)'=BO/γ. Since we have BO=vΔt, follow above two (BO)'s, we should have vΔt'=vΔt/γ, then t'=t/γ, for all location of O' in LT. Isn't it? I don't understand the last paragraph in that comment: "With the corrected length contraction equation, both versions of (BO)' are the distance as measured in S' between two events such that one event is at B, one event is at O, and the two events are simultaneous as measured in S'." because the second (BO)' is based on the BO in S and the length contraction, it is not related to the measurement in S', isn't it?Jh17710 (talk) 16:20, 28 November 2010 (UTC)[reply]

Δt'

BenRG, thanks a lot. Your response is the one touched my question. Could you review the mathematics in "Δt' is larger (since γ > 1), so the clock is running slower,.." once again? Normally, Δt'=t'2-t'1, so that, when Δt' is larger, the clock should run faster so that we can count more units between time t'1 and t'2, isn't it? For the same event, Δt' = γΔt means the event is recorded as Δt' of S' units in S' and Δt of S units in S. If Δt' is larger, the clock in S' is quicker. That is why I said, the author of that section should use the clock in S' as the reference clock, same as what Einstein did in year 1905.Jh17710 (talk) 05:43, 20 November 2010 (UTC)[reply]

Δt and Δt' are the time interval between the same two events, just measured with respect to different frames. They are talking about the same clock, not two different clocks. E.g., suppose you're stationary with respect to S, moving uniformly with respect to S'. You note down the times shown on nearby clocks of S and S' (event 1), then time 60 seconds with your stopwatch, then note down the times now shown on the clocks (event 2). Subtracting the unprimed and primed readings gives you Δt and Δt'. "Time dilation" refers to the fact that Δt' > 60 seconds. (Δt = 60 seconds.) The "dilated" clock is your stopwatch, not one of the clocks used to define the systems—although, since you're stationary with respect to S, you could use the nearest clock belonging to S as your stopwatch instead.

The only way to measure the "speed of a clock" is using another clock, so could you say instead that the clocks of S' are running fast? Not exactly, because the two readings you subtracted to get Δt' were taken from two different clocks (you always take the reading from the clock that's nearest at the time of the reading). Thus the difference depends not only on the "speed of the clocks" but also on how they were synchronized. You can't exactly say that any clock is going fast, because which clock would that be? You never looked at any of them more than once. -- BenRG (talk) 08:11, 20 November 2010 (UTC)[reply]

But according to Red Act, all of clocks in one system are running at the same speed, and I think we should assume that to make things easier. I like to hear more detail about your comment "..but also on how they were synchronized.", could you show me some ways they can be synchronized? Thanks.Jh17710 (talk) 23:20, 20 November 2010 (UTC)[reply]

Again, how fast the clocks are running doesn't tell the whole story, as can be trivially illustrated with even a very low speed example. Assume for the purposes of illustration that all timing devices run at the same speed. However, assume that your kitchen wall clock is set to an hour later than your living room wall clock. Leave your living room when your wristwatch and the living room wall clock say 1:00. Walk to the kitchen, and notice that your wristwatch now says 1:01, but the kitchen wall clock says 2:01. If you incorrectly consider all wall clocks to be equivalent when measuring "wall clock elapsed time", then you would incorrectly conclude that wall clocks run 61 times faster than your wristwatch. The solution to avoid that incorrect conclusion is to instead say that wall clocks don't all show the same time (even though they all run at the same speed!), and be careful to specify which wall clock is being used when using a wall clock's time.

You mean the kitchen wall clock says 2:02, because the moving wristwatch runs slow.Jh17710 (talk) 02:13, 21 November 2010 (UTC)[reply]

Nope, I mean the kitchen wall clock says 2:01. I said "assume for the purposes of illustration that all timing devices run at the same speed", and that assumption also applies to the wristwatch. Red Act (talk) 02:59, 21 November 2010 (UTC)[reply]

Then, "...wall clocks run 61 times faster than your wristwatch." should be 60 times faster than or as fast as 61 times of your wristwatch, right?Jh17710 (talk) 07:04, 21 November 2010 (UTC)[reply]

A similar problem occurs in the relativistic situation. If the unprimed observer looks at all clocks in the primed reference frame at time t=0, the primed clocks will not be observed to all show the same time. Instead, at time t=0, the time shown on the primed clocks will show different times depending on the location of the clock, namely, t' = -γvx/c². So you must be careful to consider which primed clock is being used when using a primed clock’s time.

Since you mentioned about the term -γvx/c² again here, could you comment to my reasoning of "v=0 is always true in LT"? After Lorentz assumed ruler contraction, the LT must be t'=γt, x'=γ(x-vt), y'=y, and z'=z; while the inverse LT must be t=γt', x=γ(x'+vt'), y=y', and z=z'. We can still play the same trick Lorentz did to combine spatial equation and inverse spatial equation so that we have t'=γ(t-(vx/c²)). However, that will mean we should combine t'=γt and t=γt' as well, and that will be γ=1 (since for c>|v|>0, γ>1); and that also means v=0. LT is almost a joke.Jh17710 (talk) 02:13, 21 November 2010 (UTC)[reply]

See above for an explanation of why the derivation of v=0 was incorrect. Red Act (talk) 04:41, 21 November 2010 (UTC)[reply]

Thanks a lot. I still have two questions asked above regarding your explanation. Could you help to provide more detail?Jh17710 (talk) 15:56, 21 November 2010 (UTC)[reply]

See above. Red Act (talk) 23:05, 21 November 2010 (UTC)[reply]

In the low speed example, you can avoid the problem by simply setting all wall clocks to have the same time. But that's not possible in the relativistic situation. Clocks at rest in the primed system will all have the same time according to the primed observer, but will not all have the same time according to the unprimed observer.

For information on how clocks are theoretically synchronized in relativistic situations, see Einstein synchronization. Red Act (talk) 01:15, 21 November 2010 (UTC)[reply]

Thanks, I will study it. Hope that is different from the one in his 1905-6-30 paper, because that one is not a complete one, it missed one condition that the distance between the emitting point and reflecting point is the same as the distance between the reflecting point and the ending point. Because of that mistake, Einstein did not prove LT in his paper. The equation in section 3, (t0+t2)/2 = t1 is actually an inequality (t0+t2)/2 < t1 so that follow the correct inequality, Einstein was unable to prove LT. His definition of synchronizatiion is fine if there is only one system so that we can assume that system is rest, but, there are two systems, so that we must add above condition to ensure two clocks are synchronized.Jh17710 (talk) 02:13, 21 November 2010 (UTC)[reply]

You're absolutely right to say that in general, the (τ₀+ τ₂)/2= τ₁ equation is incompatible with a system of synchronized clocks, with the only exception being that it will happen to work as long as the system is at rest. However, the "flaw" that you’re noticing isn't actually a bug, but a feature. Pick an arbitrary inertial system S. From the perspective of S, S is at rest, and all inertial systems are created equal, so when system S applies the Einstein synchronization procedure to its clocks, it will set the clocks to something that appears to be properly "synchronized" according to S. However, those clocks will not appear to be properly synchronized according to anybody else.

Einstein did not "miss" the condition that the distance between the emitting point and reflecting point is the same as the distance between the reflecting point and the ending point. Indeed, that condition is not in general true. When system S is synchronizing its clocks, those two distances will be the same according to how S measures distances. But those distances will not be the same according to any other system. Red Act (talk) 23:05, 21 November 2010 (UTC)[reply]

When we talk about a point, it is a fixed point in the universe. Let me tell you how to define the emitting point based on a book published recently in China. We just need some imagination, let a source of light shine once at point E, then the out going wave front will make a growing but thin out light ball. Do you agree that the center of that growing light ball is a fixed point as in the absolute rest universe? The reason is clear, because the speed of light is independent of the speed of the source of light so that the growing light ball is growing from a fixed point in the whole space, the point E. It is the same for the point R and F, for every event point. The only difficulty is that, unless the source of light stays at that same point while the light ball is growing, otherwise, we have no way to point out where is E, because our technology is unable to show us the actual growing light ball yet. Since now we have the points fixed, we could have the distance ER and RF fixed so that Einstein missed one condition for his two systems.Jh17710 (talk) 17:29, 25 November 2010 (UTC)[reply]

It's incorrect to say "the center of that growing light ball is a fixed point as in the absolute rest universe". The phrase "the center of that growing light ball" has no meaning, without specifying which reference frame you are using to determine the "center" with. The "center" would be a point equidistant from a set of events on the growing light ball, where the events all occur at the same time. However, different inertial frames of reference disagree on what "occur at the same time" means (see relativity of simultaneity, again), which makes them disagree on what the "center" is. Red Act (talk) 06:33, 27 November 2010 (UTC)[reply]

Red Act, if I hold a ball could I say the center point can be decided by the ball without mentioning about other reference frame? Yes, the ball itself is a reference frame. If we can see the expanding ball of light then that will solve our problem here. But, since it is there, even if we can not see it, it is still there, isn't it? Then we check the second issue about the "center". You had made the center point of a ball too complicated. If there is a ball, we can mark four points on the ball with the condition that they are not on one plane, then, draw perpendiculer planes at all middle points of six line segments and the intersection point is the center. The problem with this expanding ball of light is that we have no way to locate it. But it is real and since there must be one and only one center of it, we cannot deny its existence just because we are unable to locate it. Make sense? Jh17710 (talk) 04:56, 28 November 2010 (UTC)[reply]

The ball of light topic is currently being discussed on the current "Absolute Rest Point" topic, so I won't duplicate the whole topic by also discussing it here. Red Act (talk) 13:41, 28 November 2010 (UTC)[reply]

A fresh start with more precise notation

This is really a continuation of the subsection called "Two (BO)'", but with more explicit notation. Up until now, I’ve been trying to stick to your notation as much as possible. However, there are two kinds of imprecision in what you're saying, that not only make it hard for me to follow you at times, but also lead to you making erroneous conclusions. The first problem is using a symbol that pertains to an event or pair of events, without being precisely clear about which event or pair of events the symbol pertains to. The second problem is using equations that are valid for only a subset of events or pairs of events, without being precisely clear as to what subset of events or pairs of event the equations are claimed to be valid for. The solution to the first problem is to from here forward always number all specific events involved, and use subscripts with those event numbers to indicate all coordinates or differences in coordinates. The solution to the second problem is to from here forward always explicitly state what set of events or pairs of events an equation is claimed to be valid for.

I will label events E_p, where p is some integer. I.e., individual events in this problem will be called E₀, E₁, E₂, etc. The coordinates for event E_p are t_p and x_p in the unprimed system, and t'_p and x'_p in the primed system. Given events E_p and E_q, I define Δx_pq=x_q-x_p, Δt_pq=t_q-t_p, Δx'_pq=x'_q-x'_p and Δt'_pq=t'_q-t'_p.

The rigorous versions of all equations used above that I acknowledge to be valid are:

Eq. 1: t'_p=γ(t_p-vx_p/c²), which is valid for all events E_p.

Eq. 2: x'_p=γ(x_p-vt_p), which is valid for all events E_p.

Eq. 3: t_p=γ(t'_p+vx'_p/c²), which is valid for all events E_p.

Eq. 4: x_p=γ(x'_p+vt'_p), which is valid for all events E_p.

Eq. 5: Δt'_pq=γ(Δt_pq-vΔx_pq/c²), which is valid for all pairs of events E_p and E_q.

Eq. 6: Δx'_pq=γ(Δx_pq-vΔt_pq), which is valid for all pairs of events E_p and E_q.

Eq. 7: Δt_pq=γ(Δt'_pq+vΔx'_pq/c²), which is valid for all pairs of events E_p and E_q.

Eq. 8: Δx_pq=γ(Δx'_pq+vΔt'_pq), which is valid for all pairs of events E_p and E_q.

Eq. 9: Δt'_pq=γΔt_pq, which is valid for all pairs of events E_p and E_q satisfying x_p=x_q.

Eq. 10: Δx'_pq=γΔx_pq, which is valid for all pairs of events E_p and E_q satisfying t_p=t_q.

Eq. 11: Δt_pq=γΔt'_pq, which is valid for all pairs of events E_p and E_q satisfying x'_p=x'_q.

Eq. 12: Δx_pq=γΔx'_pq, which is valid for all pairs of events E_p and E_q satisfying t'_p=t'_q.

Eq. 13: t'_p=γt_p, which is valid for all events E_p satisfying x_p=0.

Eq. 14: x'_p=γx_p, which is valid for all events E_p satisfying t_p=0.

Eq. 15: t_p=γt'_p, which is valid for all events E_p satisfying x'_p=0.

Eq. 16: x_p=γx'_p, which is valid for all events E_p satisfying t'_p=0.

Eq. 17: Δx_pq=vΔt_pq, which is valid for all pairs of events E_p and E_q satisfying x'_p=x'_q.

Eq. 18: Δx'_pq=-vΔt'_pq, which is valid for all pairs of events E_p and E_q satisfying x_p=x_q.

I’ll label the events of interest that have been explicitly or implicitly used above as:

E₀: O and O' at the instant they coincide, i.e., t₀=0, t'₀=0, x₀=0, x'₀=0.

E₁: The point B at time t=0, i.e., t₁=0, x₁=-d, where I'm introducing the symbol d to mean what had been called BO. We can use Eq. 1 to calculate t'₁=γvd/c². And because t₁=0, we can use Eq. 14 to calculate x'₁=-γd. This is also a good opportunity to illustrate a length contraction equation being used: Because t₀=t₁, we can use Eq. 10 to calculate Δx'₁₀=γΔx₁₀=γ(0 -(-d))=γd, which is consistent with calculating Δx'₁₀ as Δx'₁₀=x'₀-x'₁=0-(-γd)=γd.

E₂: The points B and O' at the time when they coincide. Being at O' gives x'₂=0. And being at point B gives x₂=-d. And because x'₀=0 also, we have the x'₂= x'₀ criteria needed to be allowed to use Eq. 17 with the events E₂ and E₀. Doing so gives Δx₂₀=vΔt₂₀. Substituting in the definitions for Δx_pq and Δt_pq gives x₀-x₂=v(t₀-t₂). Substituting in the known values x₀=0, t₀=0 and x₂=-d and solving for t₂ gives t₂=-d/v.

If there are any other important events you want to talk about, assign them a number, define them, and derive coordinate values for them using the above equations, being explicit about why the validity conditions for the equations being used are met. Or if there are any other equations you want to talk about, state what the equation is, state the conditions under which you are claiming the equation is valid, and derive the equation from some of the 18 equations above, being careful to use an equation only if the validity conditions for the equation are met.

Based on what you've said so far, it looks like your attempts to find an inconsistency in the Lorentz transformations are or will be along one of the following four similar approaches:

Approach 1: Try to argue that it must be that γ=1 (i.e., v=0) is required in order for equations 13 and 15 to both hold. However, the validity conditions for equations 13 and 15 combined mean that the set of events for which a derivation using both equations is valid are only those events E_p such that both x_p=0 and x'_p=0. Assuming that v≠0, there's exactly one event that satisfies both of those conditions, namely E₀. As proof, substitute x_p=0 and x'_p=0 into equation 2 to get t_p=0. And substitute x_p=0 and x'_p=0 into equation 4 to get t'_p=0. So when both equations 13 and 15 hold, the reason both equations hold isn't because γ=1, but because under the combined validity criteria, t_p=0 and t'_p=0.

Approach 2: Try to argue that it must be that γ=1 (i.e., v=0) in order for equations 14 and 16 to both hold. However, the validity conditions for equations 14 and 16 combined mean that the set of events for which a derivation using both equations is valid are only those events E_p such that both t_p=0 and t'_p=0. Assuming that v≠0, there's exactly one event that satisfies both of those conditions, namely E₀. As proof, substitute t_p=0 and t'_p=0 into equation 1 to get x_p=0. And substitute t_p=0 and t'_p=0 into equation 3 to get x'_p=0. So when both equations 14 and 16 hold, the reason both equations hold isn't because γ=1, but because under the combined validity criteria, x_p=0 and x'_p=0.

Approach 3: Try to argue that it must be that γ=1 (i.e., v=0) in order for equations 9 and 11 to both hold. However, the validity conditions for equations 9 and 11 combined mean that the set of pairs of events for which a derivation using both equations is valid are only those pairs of events E_p and E_q such that both x_p= x_q and x'_p= x'_q. Assuming that v≠0, the only way for both of those conditions to be met is if E_p=E_q. As proof, substitute Δx_pq=0 and Δx'_pq=0 into equation 6 to get Δt_pq=0. And substitute Δx_pq=0 and Δx'_pq=0 into equation 8 to get Δt'_pq=0. So when both equations 9 and 11 hold, the reason both equations hold isn't because γ=1, but because under the combined validity criteria, Δt_pq=0 and Δt'_pq=0.

Approach 4: Try to argue that it must be that γ=1 (i.e., v=0) in order for equations 10 and 12 to both hold. However, the validity conditions for equations 10 and 12 combined mean that the set of pairs of events for which a derivation using both equations is valid are only those pairs of events E_p and E_q such that both t_p= t_q and t'_p= t'_q. Assuming that v≠0, the only way for both of those conditions to be met is if E_p=E_q. As proof, substitute Δt_pq=0 and Δt'_pq=0 into equation 5 to get Δx_pq=0. And substitute Δt_pq=0 and Δt'_pq=0 into equation 7 to get Δx'_pq=0. So when both equations 10 and 12 hold, the reason both equations hold isn't because γ=1, but because under the combined validity criteria, Δx_pq=0 and Δx'_pq=0. Red Act (talk) 00:34, 30 November 2010 (UTC)[reply]

Thanks for your analysis. It clearly listed most possible validity conditions and the results in LT and inverse LT. However, what I tried to do is to establish the time equation based on Galilean Transformation (GT) and "hypothesis of ruler contraction" so that when I started, there is no LT yet. Actually, the time equation I mentioned about is a required condition for the spatial equation of LT to be valid. I will provide more detail next time.Jh17710 (talk) 05:13, 1 December 2010 (UTC)[reply]

The Galilean transformation? In the GT, the time equation is simply t=t', end of story. The GT and the LT are incompatible, except in the trivial case of v=0. You can't derive anything involving the LT in a consistent way, if your starting point involves assuming the GT is valid. To do anything in special relativity, you have to discard the GT as being inaccurate, except as a useful low-speed approximation. Red Act (talk) 07:28, 1 December 2010 (UTC)[reply]

LT is a system modified from GT by assuming ruler contraction. However, he simply took the spatial equation x'=x-vt and assigned γ to the S portion (x-vt) in that equation so that x'=γ(x-vt) became the spatial equation of LT. Actually, that equation is not so simple. We will be able to see it easier if we limit all events in one dimensional space. Then, for t' and t < 0, we have (x'-OO')= x and when t' and t > 0, we have (x'+OO')= x. Since OO'=v|t'| as measured in the moving S', so that (x'+vt')= x for all t'. The actual ruler contraction should be (x'+vt')= γx, that means x' = γx-vt'. It is because of the time equation t'=γt actually exist when we modify GT by assuming ruler contraction, so that the x'=γx-vt' can be written as x'=γx-vγt=γ(x-vt) as well. I will derive the time equation t'=γt in as much detail like you wished above. That equation is for all events under the "ruler contraction version of GT". Not only for the limited condition of x=0 under LT.Jh17710 (talk) 05:05, 2 December 2010 (UTC)[reply]

You say that "LT is a system modified from GT by ruler contraction". However, when you try to derive LT by applying length contraction to GT, you wind up getting t'=γt for all events, which is not LT, because it's inconsistent with LT's Eq. 1 above when x≠0. So when you play around with applying length contraction to GT, you get something different from LT. So according to your own equations, LT is not just a system modified from GT by ruler contraction. Red Act (talk) 06:08, 2 December 2010 (UTC)[reply]

Thanks for your response. Let us go one sentence by one sentence. The first one is what I said. The second one, I am still working on deriving the t'=γt in more detail as you required, event after event; I showed the first step to derive LT as {However, he simply took the spatial equation x'=x-vt and assigned γ to the S portion (x-vt) in that equation so that x'=γ(x-vt) became the spatial equation of LT.} I did not use t'=γt. The second step to derive LT is just combine x'=γ(x-vt) in LT and x=γ(x'+vt') in inverse LT, then we get t'=γ(t-(xv/c^2)) and the derivation of LT by applying ruler contraction to GT is done. The third one, you are corerct. Based on the same starting condition as LT started, I can derive another result. The fourth one, according to my own equation, based on GT and ruler contraction, we can derive LT and other equation; logically speaking, LT is still just a system modified from GT by ruler contraction. If based on E and F we can derive G and H, G is still just based on E and F, isn't it? About the last sentence, could you tell me anything wrong in my two steps of deriving LT? If nothing wrong, then, LT is just a system modified from GT by ruler contraction. Based on LT, if people find something else, that is another issue, isn't it?Jh17710 (talk) 03:50, 3 December 2010 (UTC)[reply]

OK, the statement that "LT is a system modified from GT by ruler contraction" can sort of be viewed as basically being correct, although that statement doesn't have a precise mathematical meaning, if you take x'=γ(x-vt) as expressing what you mean by ruler contraction. However, for clarity, that's quite different from saying that LT can be derived from GT and ruler contraction. A derivation is where you assume that some set of things are true, and logically deduce from those that something else must also be true. Importantly, the original set of things that are assumed to be true at the beginning of the derivation remain assumed to be true at the end of the derivation (I won't get into proof by contradiction here). That's different from what's going on above in arriving at the LT. In the above, what's happening is basically saying "discard all the GT equations as being invalid, and instead assume that x'=γ(x-vt), which happens to look similar to the x'=x-vt equation in the GT." In other words, the GT isn't really being used at all in deriving the LT. Instead, the initial assumptions being used to derive the LT are that x'=γ(x-vt), and that all inertial systems are equally valid (which lets you conclude that x=γ(x'+vt')). So there should be no expectation that the LT would be consistent with anything derived from the GT, or be consistent with anything that's a part of or derived from a different "modified version of the GT".

In particular, note that the t'=γ(t-(xv/c²)) of the LT as derived above is inconsistent with the GT's equation t'=t (except in the trivial case of v=0), so the LT requires that the time portion of the GT must be discarded as well, in addition to discarding the spatial portion. Red Act (talk) 22:37, 3 December 2010 (UTC)[reply]

You have a very good view point of the word "derivation". I should use modify to make things clear, LT is derived from modifying GT by ruler contraction. Does it look better now? THANKS FOR YOUR CORRECTION OF MY MISTAKE. I think LT uses everything of GT including when v=0, t'=t. Yes, you could start LT from x'=γ(x-vt) and x=γ(x'+vt') then combine them for t'=γ(t-(xv/c^2)), however, when you try to compare it with GT, ruler contraction will pop out and we are in the same situation. If you don't compare LT with GT, someone will, isn't it?Jh17710 (talk) 04:20, 4 December 2010 (UTC)[reply]

I don't know what you mean by "LT uses everything of GT". LT disagrees with GT for almost all events and almost all speeds, with the only exceptions being the trivial case of v=0, or for the event for which x=0 and t=0. Red Act (talk) 16:14, 4 December 2010 (UTC)[reply]

Besides the most different one, as you mentioned above, the time equation can fit in when v=0; LT used exactly the same arrangement of GT. And the main uasge is that LT modified the spatial equation in GT directly and inappropriately, but LT is lucky that x'=γ(x-vt) is still accurate after the careless modification as I said above on 12-2-10. I will derive t'=γt in detail later. The equation t'=γt backed up x'=γ(x-vt).Jh17710 (talk) 04:42, 5 December 2010 (UTC)[reply]

Here's a very pertinent homework problem for you: There are three conceptual properties of the LT, that make the LT different from the GT. One is length contraction. One is time dilation. What is the third? Red Act (talk) 07:20, 2 December 2010 (UTC)[reply]

I think there is no FIXED time dilation in LT. As in your equation 13 and 15 {derived from x'=γ(x-vt) and t'=γ(t-(xv/c^2)) in LT, don't need any equation in the inverse LT} the time in S' and S can have dilation or speeding depend on their observers are observing what kind of events. If they are observing events rest in S then the relation is 13, if they are observing events rest in S' then the relation is 15 under the same situation that S is the rest syatem and S' moves.Jh17710 (talk) 03:50, 3 December 2010 (UTC)[reply]

This paragraph doesn't make sense. An event is one specific location at one specific time. Since events only have one time, it's meaningless to talk about an event being "at rest", or "moving", since both of those concepts require a discussion of more than one time, and hence, more than one event. An object is "moving" relative to some reference frame if it is at two different locations as measured in that reference frame at two different times. I.e., if the world line of an object includes two distinct events E_p and E_q such that x_p≠x_q, then the object is "moving" as measured in system S. Something is "at rest" relative to some reference frame if it is at the same location as measured in that reference frame at all times. I.e., if for all distinct pairs of events E_p and E_q on an object's world line, x_p=x_q, then the object is "at rest" as measured in system S. Red Act (talk) 22:37, 3 December 2010 (UTC)[reply]

You are right, however, we can say that a baseball moves from A to B is an event, with a starting event time and ending event time without confusing. But, it is always good to be more precise, let me use Series Of Events (SOE), then there are rest SOE and moving SOE. Actually, in the real world, we are unable to observe an individual event yet. If a rest SOE happens in a very short time period, then we will treat it as an individual event, so that the key issue is if a SOE move or not. When you said "two events", those two events are actually two ends of a SOE, isn't it? Thanks.Jh17710 (talk) 04:20, 4 December 2010 (UTC)[reply]

The standard term for a continuous time-like series of events, such as might be used to describe the path of the center of a baseball through spacetime, is world line.

Note that an event is not unique to one world line. Consider a basketball which goes down through the precise center of a basketball basket's rim at time t=0 in the basket's frame of reference. The center of the basketball traverses a world line that moves, according to the basket's frame of reference. The center of the rim traverses a world line that is at rest, according to the basket's frame of reference. At time t=0, the two world lines intersect at the event whose time is at t=0, and whose location can be described both as being at the center of the basketball, and as being at the center of the rim. It's meaningless to say whether that event is moving or is at rest, since that event is just as much a part of one world line as it is of the other. Red Act (talk) 16:14, 4 December 2010 (UTC)[reply]

I already changed the name to SOE and said moving SOE and rest SOE. Actually, it still make sense if we call an event in a moving SOE a moving event and an event in a rest SOE as rest event. When two events meet like in your case, there are still two events at same location. Before and after t=0, the SOE of the center of basket's rim is rest and the SOE of the center of basketball is moving so that I think it is two events happen at the same time at the same location, not just one event. However, this is a better practice to use SOE for rest or moving.Jh17710 (talk) 06:43, 5 December 2010 (UTC)[reply]

The time dilation is claimed by SR, not LT. In LT, we do have time dilation when x'=0; but, when x=0, t' is not running slow any more. In SR, Einstein claimed that, no matter x'=0 or x=0, or any other situation, so long as S' is moving at constant speed v relative to S, then t'=t/γ is always true for any event. Am I wrong?Jh17710 (talk) 03:50, 3 December 2010 (UTC)[reply]

Sentence 1 is at least inconsistent with Einstein's interpretation of the meaning of the LT equations. For sentence 2, the situation is symmetrical. A clock in S' which has x'=0 at all times will run slow as measured by S. A clock in S which has x=0 at all times will run slow as measured by S'. Sentence 3 is just wrong; Einstein never made any such claim. What he did claim was equation 1. Red Act (talk) 22:37, 3 December 2010 (UTC)[reply]

Sentence 1, Lorentz did not claim time dilation, Einstein calimed it for LT to specify his SR in his 1905-6-30 paper. Sentence 3 is based on the section 4 of Einstein's paper, same paper published 1905-6-30, he said t'=t/γ is always true even if v changed from constant velocity to constant speed. That means, we don't have to care about the direction of v, to v or -v, t'=t/γ is alway valid.Jh17710 (talk) 04:20, 4 December 2010 (UTC)[reply]

Section 4 of Einstein's 1905 paper says "Further, we imagine one of the clocks which are qualified to mark the time t when at rest relatively to the stationary system, and the time τ when at rest relatively to the moving system, to be located at the origin of the co-ordinates of k, and so adjusted that it marks the time τ" (emphasis mine). If the clock discussed in that section is at the origin (of the reference frame that we're calling S), then it is only used to measure the time of events such that x=0, which is exactly the validity criterion specified above for equation 13. Einstein does not claim that equation 13 is valid when x≠0. Red Act (talk) 16:14, 4 December 2010 (UTC)[reply]

Thanks for providing the link. If you read further, Einstein stated "It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide." That is what I mentioned about.Jh17710 (talk) 06:52, 5 December 2010 (UTC)[reply]

The clocks at A and B are only brought up after using (in our notation) t'=t/γ to describe the moving clock that passes through the common origin. The clock that starts at A is a different clock at a different location than the previously discussed clock that passes through the common origin at t=0. Red Act (talk) 06:09, 6 December 2010 (UTC)[reply]

Since we cannot change Einstein's paper any more, we could work out what Einstein meant. On your response dated 12-4-10, the clock is at the origin of S' so that x'=0, not x=0. Equation 15 is valid in LT if x'=0. When Einstein brought up the later issue regarding "any polygonal line" that clock A is the clock with x'=0, except this time, Einstein let v change from constant velocity to constant speed. You can say the clock that starts at A is a different clock at a different location than the previously discussed clock, but, Einstein applied same equation t'=t/γ to that scenario.Jh17710 (talk) 17:15, 6 December 2010 (UTC)[reply]

Yes, I made a mistake in my post of 12/4/10. Instead of saying S, x=0 and equation 13, I should have said S', x'=0 and equation 15.

However, my point still stands that Einstein does not use t'=t/γ (our equation 15) when discussing the clock traveling from A to B. He does use the expression ½ tv²/c², but in that expression, he defines t as "t being the time occupied in the journey from A to B". In other words, in that expression (and not before), he is using t to mean Δt_AB. What he is describing in that paragraph is equation 11, which can be expressed as Δt'_AB=Δt_AB/γ. Equation 11 holds because x'_A=x'_B (even though x'_A≠0 and x'_B≠0). There is no implication for equation 15 in that paragraph. Red Act (talk) 18:45, 6 December 2010 (UTC)[reply]

Right in the paragraph you selected above reasoning, Einstein already change the moving clock from constant velocity v to constant speed on polygonal line. That means, at that point, LT is out of the picture. No more equation 11. Another issue is about no t'=t/γ format in that paragraph. If you read the paragraph right before that paragraph, you will find out Einstein change t'=t/γ to t'=t-(1-(1/γ))t and use the difference amount,(1-(1/γ))t, to describe the difference of two clocks. That means, t'=t/γ is still in the picture.Jh17710 (talk) 05:49, 15 December 2010 (UTC)[reply]

Lorentz transformations are by no means out of the picture in the paragraph where Einstein discusses A and B. A Lorentz transformation in general can be composed of an arbitrary rotation and a Lorentz boost in an arbitrary direction. It can also include a translation, if inhomogeneous Lorentz transformations are part of what's under consideration in a particular situation. It can even include space inversion and/or time reversal, if improper or non-orthochronous Lorentz transformations are under consideration. All of this is explained in the Lorentz transformation article. Rather than LTs being out of the picture in that paragraph, all that's happening is that Einstein is venturing a little bit beyond a standard configuration LT.

The 18 numbered equations above were enumerated assuming that only a standard configuration LT was being dealt with, so most of them (all but 9, 11, 13 and 15) are invalid when considering a LT that involves a rotation. But equation 11 is valid for all orthochronous LTs, even inhomogeneous or improper ones, as long as x'_p=x'_q, y'_p=y'_q and z'_p=z'_q. In the paragraph where Einstein talks about moving a clock along a straight line from A to B, without putting any constraints on the coordinates of A and B, he basically is making the point that equation 11 holds (under the conditions x'_p=x'_q, y'_p=y'_q and z'_p=z'_q) even if the LT is not a standard configuration LT. The physical reason why equation 11 works for more general LTs is basically because two ideal clocks that are stationary with respect to each other will always run at the same speed, no matter where the clocks are positioned, or whether the clocks are oriented differently. Although equations 11 and 15 can both hold for LTs that involve a rotation, equation 15 does not hold as broadly as equation 11 does, because unlike equation 11, equation 15 is not valid for inhomogeneous LTs.

LTs are still being used even in the case where the clock moves from A to B along a polygonal path. For every segment of the path, the clock uses a different coordinate system, so a different LT needs to be used for every segment. But equation 11 is valid along each segment of the path, so it winds up being valid for the whole path.

Yes, Einstein is implicitly continuing to use the equations t'=t/γ and t'=t-(1-(1/γ))t after starting to talk about A and B. However, the meanings of the symbols t and t' (his τ) change at that point. With his original definition of terms, after he said that "…imagine one of the clocks…to be located at the origin of the co-ordinates of k…", t in those equations is taken to mean t_p, and t' is taken to mean t'_p, where the conversation is limited to a standard configuration LP, and limited to an event E_p such that x'_p=0. After he changes his definitions with "…t being the time occupied in the journey from A to B…", he is still implicitly using the equations t'=t/γ and t'=t-(1-(1/γ))t, but t now means Δt_AB and t' now means Δt'_AB, where events E_A and E_B are such that x'_A=x'_B, y'_A=y'_B and z'_A=z'_B. He is not at that point taking t'=t/γ to be valid with t meaning t_p and t' meaning t'_p, for any arbitrary event E_p, as you are attempting to do. Red Act (talk) 20:25, 18 December 2010 (UTC)[reply]

Thanks a lot, Red Act. I feel very comfortable talking with you. A LT in general is just some theoretical reasoning. To make a useful result, let us focus on equation 11 when Δx'=0. The restriction of Δx'=0 means, the equation 11 is good only for events rest in S'. S' is the moving system as observed by observers in S. Now, in so many explanations of SR, how many of them put this restriction in their minds? As a matter of fact, in most physics textbooks, SR is true without any restriction; so long as Δt' is based on the time speed of the moving system and Δt is based on the time speed of the rest system, Δt'=Δt/γ is true for real, not just because of measurement effect. Am I wrong? That is not what I attempted to do, that is what already being done by most physicists. Now, let us think about the equation 11 once again, but this time from the angle of how do we pratically apply it. My question for you is how do observers in S' communicate with observers in S so that they can compare their measurements? What is your suggestion, practically? When physicists claimed that they verified SR or GR, how did they allow above communication to happen? Could you find just one sample, the best sample you know, and explain it briefly so that I can study it? Thanks again.Jh17710 (talk) 15:53, 19 December 2010 (UTC)[reply]

Wheatstone bridge

I'm supposed to find the change in voltage across the central voltmeter as Rx is moved away from equilibrium, assuming the first three resistors have a common resistance R.

However, I am obstructed by several conceptual problems.

Theoretically the voltmeter should have an infinite resistance, right? Yet it permits a finite current. Even for an ideal circuit? I have the relation ΔV / I_v = resistance of the voltmeter = ΔRx + R(I1/I_v - 2). John Riemann Soong (talk) 06:07, 19 November 2010 (UTC)[reply]

The current through an ideal voltmeter is zero. So an ideal voltmeter doesn't affect the circuit it's in at all. You basically analyze the voltages and currents in the circuit as if the voltmeter wasn't there. The voltage measured by the voltmeter is then the difference between the voltages of the two points in the circuit that the voltmeter is connected to. Red Act (talk) 12:51, 19 November 2010 (UTC)[reply]

The galvanometer used in a Wheatstone bridge is not an ideal voltmeter: it has a finite resistance. When the bridge is unbalanced, a current flows through the galvo and deflects it. You then adjust one of the resistors to balance the bridge so that the galvo does not deflect. Once the bridge is balanced, the resistance of the galvo doesn't matter since there is no voltage across it. --Heron (talk) 18:35, 19 November 2010 (UTC)[reply]

I'm presuming the current through the galvanometer is to be neglected in this problem, or else the galvanometer's resistance would have to have been specified as part of the problem. Red Act (talk) 19:33, 19 November 2010 (UTC)[reply]

Yes, I see what you mean. In which case we need to know where John gets his term I_v from, because that implies a non-ideal voltmeter. If I_v is non-zero then of course the equation in our article doesn't apply. --Heron (talk) 19:51, 19 November 2010 (UTC)[reply]

JRS gets a non-zero I_v because he doesn't understand the problem. The problem is completely soluble if I_v is taken to be zero, trivial even, the solution is given in the article. Physchim62 (talk) 20:00, 19 November 2010 (UTC)[reply]

From the article: "The direction of the current indicates whether R2 is too high or too low. Detecting zero current can be done to extremely high accuracy (see galvanometer)." 128.143.181.23 (talk) 20:41, 19 November 2010 (UTC)[reply]

The article is slightly confusing on this point. I've tried to clarify it. --Heron (talk) 11:52, 20 November 2010 (UTC)[reply]

：A galvanometer is not a voltmeter; it's an ammeter. An ideal ammeter has 0 resistance. --140.180.14.145 (talk) 08:01, 21 November 2010 (UTC)[reply]

Exactly. The problem with our article is that it confuses the two. It should make it clear that the original WB used a galvo to obtain zero current at balance, that you can also use a voltmeter to detect balance, and that if you have a high-impedance voltmeter then you can also do useful calculations with

V_{G}

even when the bridge is unbalanced. --Heron (talk) 11:57, 21 November 2010 (UTC)[reply]

why isn't adenosine monophosphate used in alcoholic beverages?

I imagine that as an anti-bitterant it would mask the taste of alcohol really well -- or does it not? John Riemann Soong (talk) 10:14, 19 November 2010 (UTC)[reply]

Why would you want to mask the taste of alcohol? Some people actually enjoy the flavor of their drinks, indeed that is generally the idea. If you really wanted to, you could use a much less expensive alternatives, like fruit juices, simple syrup, colas, etc. A properly trained bartender can mix a drink to suit anyones taste without exotic chemicals. --Jayron32 16:16, 19 November 2010 (UTC)[reply]

Wikipedia:Reference desk/Archives/Science/2010 October 30#does citric acid actively mask the taste of ethanol? Nil Einne (talk) 18:52, 19 November 2010 (UTC)[reply]

I tweaked the wikilink to fix typo in anchor. Hope you don't mind the third-party mod, Nil! DMacks (talk) 19:13, 19 November 2010 (UTC)[reply]

Thanks! Nil Einne (talk) 06:15, 20 November 2010 (UTC)[reply]

I'm not sure I see the logic of using an anti-bitterant to mask the taste of something that most people don't describe as bitter and that is chemically unlike general bitter things or the more specific molecules noted as blocked by AMP. DMacks (talk) 19:08, 19 November 2010 (UTC)[reply]

Indeed. People are more likely to *add* bitter ingredients to drinks (e.g. tonic water of a gin and tonic, or even bitters themselves). -- 140.142.20.229 (talk) —Preceding undated comment added 19:50, 19 November 2010 (UTC).[reply]

The thing is, AMP is apparently a bitter reception antagonist. It's not a mere masker. I would describe the fire of vodka for example, as just really intense bitter. 128.143.181.23 (talk) 20:37, 19 November 2010 (UTC)[reply]

To me, ethanol just tastes like bitter like apple seeds, I find it repulsive. This is the reason for why I dislike strong liquors like spirits. I don't choose my liquors for the taste of the ethanol, ethanol is not the major contributor to the taste of the beverage. Ethanol is not meant to be anything other than a mind altering stimulant. Plasmic Physics (talk) 00:21, 20 November 2010 (UTC)[reply]

You don't like apple seeds? I quite enjoy them. They're a lot of work (you have to take off the endocarp I think it's called) but they have a very nice delicate flavor, reminiscent of almond delight. They're poisonous if you eat a lot of them (cyanogenic glucosides I think), but I only ever eat one or two at a time. --Trovatore (talk) 09:30, 20 November 2010 (UTC)[reply]

See also Apple seed oil. The key safety concern looks like amygdalin according to our apple article. DMacks (talk) 09:34, 20 November 2010 (UTC)[reply]

That is of course, a personal preference. I personally enjoy the taste of ethanol itself, good vodka is basically just water and ethanol. Ethanol is meant to be enjoyed, and for myself I find the flavor effects of ethanol to be more important than the mind altering effects. I actually find those somewhat of an anoyance, as getting drunk severly limits my ability to drink more ethanol. --Jayron32 01:42, 20 November 2010 (UTC)[reply]

OR here but I don't like the taste of most alcoholic drinks. However on those occasions when I've tried them, alcopops have seemed fine, no real different from a random soft drink (hence why I don't drink them a lot). I guess there are some people like JRS who still find alcopops bitter but I suspect they aren't that common and most people who just want to get intoxicated (which I presume is the case if you're trying to completely mask the taste of ethanol) and do find even alcopops bitter probably don't want to spend a lot of money. For those who don't mind alcopops, the fact that it's just masking the bitterness seems irrelevant unless perhaps your diabetic. Nil Einne (talk) 06:29, 20 November 2010 (UTC)[reply]

I would agree that ethanol doesn't seem bitter. But see PMID 12090789 - AMP simply isn't stable at room temperature, even in a pure solution. The organic phosphate bond is just too high in energy and too easily hydrolyzed especially in non-neutral pH. (I assume in the cell it spends most of its time bound to something or other that helps stabilize it, but I'm not recalling just what that might be at the moment) Wnt (talk) 13:17, 20 November 2010 (UTC)[reply]

Species identifcation for File:Sagittaria.jpg

In order to expand on the image description, so the image can be moved to Commons, Is anyone on the Science Reference desk able to provide a more specific species identification?

It appears when the is uploaded, there was some discussion, it may have been mis-identifed.. Sfan00 IMG (talk) 12:46, 16 November 2010 (UTC)[reply]

It looks very much like Sagittaria to me. There are lots of species though. Looie496 (talk) 17:58, 19 November 2010 (UTC)[reply]

shemales

Do shemales really exist? —Preceding unsigned comment added by 59.95.48.140 (talk) 13:28, 19 November 2010 (UTC)[reply]

Yes. See shemale, trans woman and transsexualism. Red Act (talk) 13:38, 19 November 2010 (UTC)[reply]

There are also people born with both genders, see hermaphrodite. --Jayron32 16:14, 19 November 2010 (UTC)[reply]

Also see intersex. Red Act (talk) 18:25, 19 November 2010 (UTC)[reply]

Definitive deletion of a phobia (classical conditioning)

By considering absence of the Renewal effect and the Spontaneous recovery a proof of a definitive deletion of a phobia, can one expect a definitive deletion of a phobia after sufficiently repeated extinctions?--Kooz (talk) 19:27, 19 November 2010 (UTC)[reply]

It's all a matter of probability, not certainty and proof. Any variety of phobias can be conquered, but in some people that means complete extinction and in others it means simply reduction of the response under an acceptable threshold. Time can make things better or worse, depending on the subject and the circumstances. Ideally a therapist will be able to help a patient come up with some exercises the patient can perform by themselves should the problem return. There's nothing exact in psychiatry. Ginger Conspiracy (talk) 04:03, 20 November 2010 (UTC)[reply]

Orgasmic sensation during pull-ups

When doing pull-ups or chin-ups, what is causing the tingling, sexual sensation in the groin area? This question is not looking for medical diagnoses or opinions. Thanks Reflectionsinglass (talk) 20:49, 19 November 2010 (UTC)[reply]

Orgasm#Spontaneous_orgasms covers some of this; it is entirely possible to experience orgasm, or orgasm-like sensations, without any direct stimulation of the genitals. --Jayron32 21:08, 19 November 2010 (UTC)[reply]

Chinups tend to be a very intense physical activity - it's possible that performing them gives you some type of endorphin rush, and combined with the increased peripheral blood flow that results from exercise, your body may be interpreting this as a sign of sexual release. There are suggestions that exercise in general helps to stimulate the libido, so this could be related. It would be interesting to attempt other strenuous exercises to see if it produces similar effects. --jjron (talk) 02:38, 20 November 2010 (UTC)[reply]

Are we allowed to talk about our own experiences in detail? If not, someone can delete this response, but consider this as simply my own experiment. I've been able to reach orgasm by lifting myself and holding myself up, since childhood, four or five years old (no emission, of course). Several years ago, much older and with other means of release, I'd suddenly remembered about this and I thought I would experiment. I used a chin-up bar, held myself up, and was able to experience full orgasm, including emission. It was extremely quick to achieve climax. I doubt I could do it again, I can barely hold myself up for 20 seconds these days. Thanks for the link to spontaneous orgasms. It was too brief a section, unfortunately. I may have to continue looking online. Reflectionsinglass (talk) 07:11, 20 November 2010 (UTC)[reply]

I tell you what, reflections and jjron, if there is anything in what you say, we've beaten the obesity epidemic right here. For my own case, I went to my doctor once and told him that in all my adult life, I felt strong orgasmic sensations when I sneezed. He said "Half your luck...". Myles325a (talk)

November 20

Entropy gun?

Would it be possible to make a weapon that shoots pure entropy? If so, why hasn't it been done? Blobs of entropy would be able to go through any type of protective shielding by scrambling the particles in the matter when it hit. --75.33.217.61 (talk) 00:13, 20 November 2010 (UTC)[reply]

Entropy isn't a substance. It's similarly not possible to make a gun that shoots pure inertia, or viscosity, or prettiness. Paul (Stansifer) 00:16, 20 November 2010 (UTC)[reply]

But would it still be possible to make a weapon that generates large amounts of entropy, even if it isn't pure? The second law of thermodynamics should make that easy. --75.33.217.61 (talk) 00:23, 20 November 2010 (UTC)[reply]

All explosives, in effect, increase the local entropy wherever they go off. That's why it is easier to blow things up than it is to put them together. --Mr.98 (talk) 00:25, 20 November 2010 (UTC)[reply]

Entropy is a measure of disorder. The more disordered something is the larger is its entropy. What you're suggesting is like designing a gun that shoots kilometers (or miles, if you're that way inclined), it simply makes no sense. Plasmic Physics (talk) 00:27, 20 November 2010 (UTC)[reply]

A gun that shoots pure entropy would be like a gun that shoots pure temperature, or pure color, or pure beauty. Abstract concepts can't be shot from a gun. Looie496 (talk) 00:46, 20 November 2010 (UTC)[reply]

I dare say you haven't had much practice. --Trovatore (talk) 11:12, 20 November 2010 (UTC)[reply]

If you say it that way, it actually sounds comical, like something you would find in Hitchiker's Guide to the Galaxy. An abstract concept gun compoared to the point of view gun. Plasmic Physics (talk) 01:10, 20 November 2010 (UTC)[reply]

But again, in all seriousness, all weapons (I've been trying in vain to think of an exception) serve to increase entropy at the local level in their targets. The entire goal of a weapon is to disrupt all of that important and necessary order that is necessary for living organisms or buildings or technologies to stay alive or stay standing or keep functioning. Humans need entropy to be pretty low in order to survive — move their stomach a few inches in the wrong direction, or push down on their neck with a few pounds of force in a small enough time span, and their bodies quickly become disordered, jumbled, rotting, etc. The force of the weapon's ability to increase entropy is usually limited by the ability to keep entropy low — through, say, having a rigid target, or by having a target that will too readily become disordered and dissipate the energy of the entropy-raising projectile (e.g. earthworks or the ceramics used in bullet-proof vests). Entropy change can certainly be a seen as a way to look at how weapons work, but it's a secondary characteristic. --Mr.98 (talk) 14:15, 20 November 2010 (UTC)[reply]

Of course, energy isn't technically matter either, unless you consider energy being carried in photons and packets. Maybe it would qualify if you had some type of antimatter ray gun that would cause small areas to undergo 100% entropic destruction efficiency. Otherwise, a psychological weapon such as staring could increase abstract entropy at a distance for this very abstract concept. ~A H 1^(TCU) 03:33, 21 November 2010 (UTC)[reply]

Parts of the body where cancer can't happen?

Are there any human organs or tissues that haven't been associated with any known cases of cancer? Neon Merlin 00:48, 20 November 2010 (UTC)[reply]

Hair, fingernails and toenails. Maybe teeth? HiLo48 (talk) 01:33, 20 November 2010 (UTC)[reply]

To explain HiLo48's answer, cancer is a disease of cell division; any part of the body where cells can divide (and all living cells can divide) could experience cancer. Hair and fingernails have no living tissue, so they cannot have cancer. --Jayron32 01:38, 20 November 2010 (UTC)[reply]

Cancer can occur within teeth, by the way. The enamel isn't living but inside you have the pulp which is highly vascularised. Regards, --—Cyclonenim | Chat 16:53, 21 November 2010 (UTC)[reply]

Somewhere there should be a table of the frequency of occurrence of cancer in various organs and tissues. It is common to read of cancers of the skin, breast, lung, liver, lymph node, brain, bladder, prostate, pancreas, tongue and cervix. Some of these have specific tests one is supposed to undergo at certain ages, and some even have highly publicized fundraising and awareness-creating events. I have rarely if ever read of cancer of the heart, artery, vein, fat, muscle, sclera, iris, retina, optic nerve, inner ear, cartiledge, appendix, uvula, penis, tonsil, gall bladder or kidney, though I am confident some cases occur. Edison (talk) 20:30, 22 November 2010 (UTC)[reply]

Retinal cancer is most common as retinoblastoma and there are about 40-50 cases in the UK per year (which is pretty rare!). Cancer of the kidney isn't that rare, and cancers of the blood system (i.e. in veins and arteries) can be caused by metastatis into the blood and then into the endothelial lining. Regards, --—Cyclonenim | Chat 14:39, 23 November 2010 (UTC)[reply]

The cancers which metastasize into the circulatory system are more likely to start in the lungs or breasts than in, say, the heart muscle. Why? Isn't cancer more likely to start in the breast than, say, the buttock, the elbow, or the shoulder? The skin gets melanomas or some other cancer when exposed to UV for long periods, just as the lung gets cancer from tobacco. Is there a specific irritant to the breast or testes or bladder or prostate which promotes genesis of cancer, which the elbow or buttock is spared? Where is there a tabulation of cancer incidence of different tissues or organs? Edison (talk) 15:08, 23 November 2010 (UTC)[reply]

The thing here is that we have cancers that are more common because of prevelent carcinogens happening to affect those areas. We have more respiratory tract cancers than other sorts of cancers because of tobacco smoking. Cancer is more likely to start in the breast than the buttocks but that is mostly due to sheer genetics. The genes that cause breast cancer seem to be more suseptible to changes in structure than those coding for other regions, but I don't personally know why that is. This table shows, in the UK, the most common cancers by incidence rate excluding non-melanomic skin cancer. It's important to note that a lot of research is ongoing and not a lot is completed. There may be specific carcinogens affecting things like prostate cancer incidence but they are currently not known. Regards, --—Cyclonenim | Chat 18:31, 23 November 2010 (UTC)[reply]

The name of intermolecular forces acting between alkanes

What is the name of intermolecular forces acting between alkanes, especially large alkanes such as decane? I think they are a special case of van der Waals forces, but van der Waals forces also include H-bonding and dipole-dipole attraction which are not present in alkanes. 220.253.217.130 (talk) 01:22, 20 November 2010 (UTC)[reply]

London dispersion forces. --Jayron32 01:34, 20 November 2010 (UTC)[reply]

Note that our van der Waals force article lists the three major subtypes, including London. DMacks (talk) 06:27, 20 November 2010 (UTC)[reply]

White Dwarf

How close does a white dwarf have to be to another star to "steal" hydrogen? —Preceding unsigned comment added by 76.161.251.228 (talk) 14:26, 18 November 2010 (UTC)[reply]

I don't know what answer to give you, but I can tell you that there is no real cutoff value for the distance. Gravity is responsible for this phenomenon, and it decreases exponentially with radial distance from the white dwarf. You should specify the rate at which mass is gained or lost, and the mass of the white dwarf. Using both these variable, someone can calculate an answer. Plasmic Physics (talk) 03:12, 20 November 2010 (UTC)[reply]

Harrumph. Gravity doesn't decrease exponentially with distance. It decreases as the inverse square of the distance. That is, it varies as 1/r², not 1/a^r. --Anonymous, 06:25 UTC, November 20/10.

Sorry, bad mistake, but the rest of what I said still holds. Plasmic Physics (talk) 06:32, 20 November 2010 (UTC)[reply]

That would be the Roche limit. It depends on the masses and radii of the white dwarf and the star. I would try the formula given in the subsection about a fluid satellite. --Wrongfilter (talk) 11:18, 20 November 2010 (UTC)[reply]

Hmmm ... but the huge difference in densities between the white dwarf and the secondary probably means that the Roche limit is less than the radius of the secondary star, so the stars will collide before the Roche limit is reached. We can get a ballpark upper limit by looking at the period of the cataclysmic variable binaries listed in the Catalog and Atlas of Cataclysmic Variables. Most of these periods are less than 1 day. If the main star has a mass comparable to the Sun, then an orbital period of 1 day means an orbital radius of around 3 million km, or about 4 to 5 times the radius of the Sun. So a ballpark figure is "within 5 solar radii". The exact answer for a given binary will depend on the masses and temperatures of the stars involved, and probably on a lot of more complex factors such as their magnetic fields. Gandalf61 (talk) 11:38, 20 November 2010 (UTC)[reply]

Trapping antihydrogen

The antihydrogen was created back in 1995, as the article says, so why to claim that its atoms are now trapped? It's akin to saying that I baked a bread some time ago and now captured its crumbs. —Preceding unsigned comment added by 89.77.156.31 (talk) 09:09, 20 November 2010 (UTC)[reply]

Antihydrogen is as stable as hydrogen, but it's hard to keep around because it annihilates on contact with ordinary matter. Thus there's a difference between merely producing it and preventing it from annihilating afterwards. -- BenRG (talk) 09:17, 20 November 2010 (UTC)[reply]

I think you have misunderstood the intended meaning of "since 1995" in the antihydrogen article. Each individual atom of antihydrogen only exists for a fraction of second before it is annihilated by contact with "ordinary" matter. The recent breakthrough was to trap atoms of antihydrogen with relatively low energies ("cold" antihydrogen) usaing magnetic fields - but even these atoms could only be kept for one sixth of a second. Antihydrogen atoms are created fresh for each experiment by bombarding a target in a particle accelerator - the antihydrogen atoms that were created back in 1995 are long gone. Gandalf61 (talk) 10:46, 20 November 2010 (UTC)[reply]

Initially, only antihydrogen-nuclei could be studied. Without positrons electrons, the nuclei are negatively charged, allowing them to be captured and contained indefinitely within a Penning trap. The trick was to create neutral atoms, a Penning trap is useless at containing neutral atoms. It is this proble that was recently overcame. Plasmic Physics (talk) 12:30, 20 November 2010 (UTC)[reply]

Space marines/soldiers?

I remember reading a while back about a Navy or Air Force veteran who wanted to start a U.S. Space Marines or something. He planned to send these five guys into space and said they should practice Tai Chi and Meditation techniques to keep them at peak psychological stamina. Anyone know about this?--Editor510 ^{drop us a line, mate} 12:53, 20 November 2010 (UTC)[reply]

I find it unlikely that one would send soldiers to space for any sort of benefit. They could practice Tai Chi and meditation on Earth in quiet rooms if they so desired at much less cost. On top of this, low/zero-gravity is known to affect muscular fitness and with enough time wastage can occur. I don't think it'd be beneficial for soldiers to go to space. Regards, --—Cyclonenim | Chat 15:10, 20 November 2010 (UTC)[reply]

Yea, astronauts who spend more than a week or so in space usually can't even walk when they land. They would hardly strike fear into our nations enemies. APL (talk) 21:10, 20 November 2010 (UTC)[reply]

You mean like Orbital Drop Shock Troopers? Cool. :] WikiDao ☯ (talk) 15:54, 20 November 2010 (UTC)[reply]

Or these space marines? SmartSE (talk) 21:05, 20 November 2010 (UTC)[reply]

Very, very expensive on top of everything else, and they'd have to rely on a certain foreign power of uncertain disposition to get them there and back, now that the Space Shuttle is being retired (unless Virgin Galactic expanded its reach considerably). Clarityfiend (talk) 20:28, 20 November 2010 (UTC)[reply]

Check out Blue Gemini. It was to be a military version of the space program that would have run parallel to the Nasa Gemini program and served as a build-up to a military run space station. It was canceled because it wasn't cost effective to have two space agencies.

Nowadays, of course, by this time next year USA will have no launch vehicles. We couldn't put soldiers up there if we wanted to! APL (talk) 21:10, 20 November 2010 (UTC)[reply]

None of this is it. It was a crazy idea, it was just a concept that some veteran wanted to make real. I think I read about it on Cracked.com? I need to find the relevant Wikipedia article. The men sent up were supposed to be like ambassadors and it was only supposed to be a small team. 'Space Taskforce' might be a name, but I wouldn't swear to it.--Editor510 ^{drop us a line, mate} 23:35, 20 November 2010 (UTC)[reply]

There are two articles on Cracked.com about them neither of which mention it. Is there a chance you've mistaken the role of the United States Marine Corps astronauts? They are afterall the only real "space marines". SmartSE (talk) 12:39, 21 November 2010 (UTC)[reply]

They're not in space, but check out First Earth Battalion for some hippy-dippy military stuff. --Sean 16:23, 22 November 2010 (UTC)[reply]

It was them! Thank you! Unfortunately, I found it out for myself elsewhere before I read this, but I would have been even more grateful if I had come back before. Thanks.--Editor510 ^{drop us a line, mate} 17:14, 27 November 2010 (UTC)[reply]

Understanding inflammation

If inflammation is "part of the complex biological response of vascular tissues to harmful stimuli" and, therefore, as I understand it, a positive state during which the body is attempting to heal tissue damage; why are anti-inflammatories, ice, etc. (with hopes of reducing inflammation) prescribed? By consciously reducing inflammation, aren't we really hampering the body's effort to heal the damage? Or is that we are not really attacking the inflammatory process itself but we're actually speeding it up - and therefore, it goes back to a non-inflammatory state faster, "reducing" the inflammation? --Belchman (talk) 13:56, 20 November 2010 (UTC)[reply]

Surprise: Scientists discover that inflammation helps to heal wounds. So I think this all means that: if you want to 'recover' fast from say a marathon, then a 'cool' but not cold bath gets you back to normal fast, but for training perposes, let nature take it's course. --Aspro (talk) 16:06, 20 November 2010 (UTC)[reply]

Thanks to anyone that helps me with this question, but I'm particularly interested in a physician's opinion if possible - I hope that non-physicians understand this. --Belchman (talk) 17:06, 20 November 2010 (UTC)[reply]

Anti-inflammatory drugs do actually inhibit the inflammatory response, and ice causes local vasoconstriction which impairs the influx of inflammtory mediators. So yes, you're absolutely right in your first hypothesis – we really are putting a damper on the body's natural response to damage. That's why most of the time you're better off just taking a heaped spoonful of Harden-Up instead of reaching for the Brufen every time you get a niggle. The reason we do prescribe anti-inflammatories in some cases is because sometimes the inflammatory repsonse can be worse than the initial injury. Conditions like osteoarthritis and rheumatoid arthritis, and some severe traumas fit this description. I'd say most people use anti-inflammatories just to take the edge off the pain of inflammation, so the reparative effect is still there, just diminished slightly, so the affected person can get on with their lives to some degree. Mattopaedia ^{Say G'Day!} 10:35, 22 November 2010 (UTC)[reply]

Having just recovered from an acute attack I can say gout is another. Nil Einne (talk) 15:52, 22 November 2010 (UTC)[reply]

Indeed. We could make a really big list, but we won't, right? Mattopaedia ^{Say G'Day!} 22:09, 22 November 2010 (UTC)[reply]

Like other immune responses, inflammation may require treatment if the body's reaction spirals out of control such that it interferes with normal activity (see Functio laesa). Fever is the classic example there, where moderate fevers have a variety of host defense effects, but too high or prolonged a fever can put you in a difficult place. Matt Deres (talk) 04:00, 23 November 2010 (UTC)[reply]

Thank you all, those answers were excellent. --Belchman (talk) 22:30, 23 November 2010 (UTC)[reply]

Of note, you're not going to get a physician's answer here. Wikipedia does not give out medical advice. We're all just volunteers providing information we've found through research. — The Hand That Feeds You:^Bite 15:56, 24 November 2010 (UTC)[reply]

Formal classification of fallacies

What are all the formal fallacies the second respondent at <a href="http://theboard.byu.edu/questions/60673/">this</a> site (the one who tries to reverse the Galilean experience into a defense of religion) commits? I was looking at List of fallacies and "appeal to authority" seems to fit the way he's able to determine for himself that "if I feel good after doing X, X must be a good thing." which itself fits under "correlation does not imply causation." Any other/better analyses? Thanks. 76.27.175.80 (talk) 14:04, 20 November 2010 (UTC)[reply]

That's sort of a long post at that link. Can you help narrow down your question by quoting from it exactly where you see the fallacy you are asking about being applied? WikiDao ☯ (talk) 14:46, 20 November 2010 (UTC)[reply]

When he or she says "truth is truth" (which I take as the "Galilean" section — Earth and sun and etc.) it's just tautology. The general argument is axiomatic with a tiny bit of "it makes me happy" thrown into it. But I don't think he or she is committing a strict logical fallacy here. It's just an appeal to faith. He or she is pretty self-consciously aware that this does not constitute scientific evidence and would not be compelling to skeptics. --Mr.98 (talk) 15:03, 20 November 2010 (UTC)[reply]

The "if I feel good after doing X, X must be a good thing" line of reasoning sounds like an argumentum ad consequentiam.

The poster's invocation (at the link given in the OQ) of "I plant the seed and take care of it and it grows" as applying to LDS Church (ie. that it exists as a "major" religion these days; therefore, it must be divinely favored; therefore, God exists and the LDS Church is good and true...) involves Retrospective determinism (and, historically, also involved a certain degree of argumentum ad baculum from what I understand). There are many more fallacies of reasoning in that post. I'd say Wishful thinking may well pertain, too. WikiDao ☯ (talk) 17:48, 20 November 2010 (UTC)[reply]

Here are a few more fallacies being used, to add to the ones already identified above:

The first two paragraphs can be paraphrased as "I'm a scientist, and I know that God exists", which is an appeal to accomplishment.

The fourth paragraph says (with minor paraphrasing) that "the evidence of faith's creation is a confirmation to the individual that comes directly from God, therefore the proof of faith is infinitely more powerful than the logical demonstration of scientific data". That's basically begging the question, which is closely related to circular reasoning. God's existence is the main conclusion to be proven, so it can't logically be used as part of a premise.

The fifth paragraph says (verbatim) "as far as it being verifiable, you can receive a similar witness of the Holy Ghost every time you read the Book of Mormon". The logical content of that statement can be reexpressed as the indicative conditional "if you read the Book of Mormon, then you will receive a witness of the Holy Ghost". Unlike much of the surrounding material, that indicative conditional is actually a falsifiable premise, which is a good thing. However, when I do the experiment, I find that that premise is false. I.e., when I read the Book of Mormon, I just perceive the thing as being utter horseshit, rather than having the sensation that I'm receiving some kind of divine evidence of the Holy Ghost. Not logically supporting a very questionable premise like that amounts to an ipse-dixitism, which is related to a bare assertion fallacy. Red Act (talk) 22:28, 20 November 2010 (UTC)[reply]

Thanks, and something tells me if you told him you didn't hear a "yes" on that last indicative conditional, he'd respond that it's because you didn't read enough or pray hard enough :) 76.27.175.80 (talk) 03:52, 21 November 2010 (UTC)[reply]

Mental disorders

Is bipolar disorder a real disorder or are some people just moody? Is ADD a real disorder or are some people just easily distracted? Is narcissistic personality disorder a real disorder or are some people just full of themselves? --124.254.77.148 (talk) 14:16, 20 November 2010 (UTC)[reply]

Yes. just change all your or's to and's and you're right:) WikiDao ☯ (talk) 14:27, 20 November 2010 (UTC)[reply]

I've known many people who were classic bipolar, and it goes far, far beyond "moody." They swing between completely manic and completely paralyzed depressed. It's clearly some kind of major chemical imbalance and it often completely inhibits their ability to be effective people. I can be "moody" in the sense that sometimes I'm more effective or elevated and sometimes I'm grumpy or sad. But that's not what bipolar is. Now every spectrum disorder (like bipolar, or ADD) has some shades of being similar, at its mildest levels, with what we define as "normal" variation. But when it crosses into the extremes it is pretty apparently the sort of thing which conflicts with how people function and is largely out of their control. That's usually when we class those kinds of extremes as being "disorders". --Mr.98 (talk) 15:07, 20 November 2010 (UTC)[reply]

There is some controversy over the existence of several classified mental disorders, but generally bipolar is characterized by mood states that are persistant—ie. they are nearly impossible to simply "snap out" of, likewise for major depression and various forms of psychosis. ~A H 1^(TCU) 00:54, 21 November 2010 (UTC)[reply]

Vehicle identification

Hello fellows!

Does anybody know who is the manufucturer of these two vehicles and which type is it.

Thanks for helping. Cheers, High Contrast (talk) 16:41, 20 November 2010 (UTC)[reply]

You've also asked at the Miscellaneous desk where I think this question is more appropriate. I suggest the answers should go there. Franamax (talk) 17:03, 20 November 2010 (UTC)[reply]

For the first, the clue is in the name - I searched for "juggernaut tracked vehicle" and found this site, number 32 is a similar vehicle which is a Thiokol juggernaut. Then I searched for "thiokol jugernaut", found this which has a link to this which shows it is a Thiokol Juggernaut 6T that has been adapted with a drill rig. The bulldozer is a lot more difficult though, I'll have a hunt, but if nobody here can help this forum might be the place to ask. SmartSE (talk) 20:49, 20 November 2010 (UTC)[reply]

I've surprised myself, but I think it is a Caterpillar 830M, based on this listing on craigs list and this book which says they were made for the military, which obviously fits with the source and the timing. Here's a better picture of one in another book as well. SmartSE (talk) 21:02, 20 November 2010 (UTC)[reply]

Thanks for helping! Cheers, High Contrast (talk) 08:53, 23 November 2010 (UTC)[reply]

Is bobbing your head to music an instinctual act or a learned one?

Thanks--72.178.134.134 (talk) 17:22, 20 November 2010 (UTC)[reply]

Definitely learnt. People in the tropics obviously got the idea from watching their parrots. [1] --Aspro (talk) 17:40, 20 November 2010 (UTC)[reply]

Aspro's answer was probably a joke. I've seen babies do it, so it's probably instinctual. Ariel. (talk) 23:33, 20 November 2010 (UTC)[reply]

A source to back up Ariel's point, see the procedure of experiment two. Quite why this should be instinctive is another matter... SmartSE (talk) 14:46, 21 November 2010 (UTC)[reply]

A recent BBC news article reported a study that showed chimps using head gestures (up-down and sideways) as humans do. Imagine Reason (talk) 19:24, 22 November 2010 (UTC)[reply]

Composting cooked vegetable matter

I've often seen it said that cooked veg shouldn't be added to compost heaps. Why is this? Thanks. DuncanHill (talk) 17:32, 20 November 2010 (UTC)[reply]

It might be as simple as: cooked veg is sterilised veg and has most of its low temperature enzymes (those that would cause natural lysis) get inactivated. Thus, it would be an ideal culture medium for anaerobic bacteria. Its the thermophilic anaerobics what course smelly drains etc., and there is plenty of heat in a good compost heap. The solution is not to have the heap directly outside kitchen window, but in the traditional place at bottom of the garden. Bones and things in my experience don't cause any problem either. Spread out each layer, rather than dump it on, so that the air can get to it. Any vermin issues gets more than offset by having to buy less cat food.--Aspro (talk) 17:54, 20 November 2010 (UTC)[reply]

Thanks, that makes sense. I suspect that mixed with sufficient raw trimmings, peelings and garden waste it should be fine. DuncanHill (talk) 20:11, 20 November 2010 (UTC)[reply]

Looking at the sites that give this advice not to use cook food, they appear to be for these piddling little green composting drums. They also don't advise wood ash. Thinking back to when I kept two large compost heaps on the go (about 2 foot square a piece), I would put anything organic on them, even chicken heads and their eviscerate, dead rats, pizza shaped roadkill etc., by putting down first, a layer of dirt and then a bit of wood ash over the top of the carcass and offal. The lyle or potassium hydroxide in it, helped I guess, to brake down the tissue in an alkali environment and the soil bacteria stopped any smells arising. Certainly don't remember any smells. If it was organic it when on. What I had left after sieving into the wheel barrow (bones, woody bits and so on), went back on to the new heap for a second pass. Garden waste has plenty of bulk but if one is just using kitchen scraps, then that might need more careful layering as you say. --Aspro (talk) 21:01, 20 November 2010 (UTC)[reply]

You should definitely not put everything organic in the compost, many organic compounds are highly toxic. Some examples of organic things that should not be added to the compost are used motor oil, most types of plastics, left over medication and many chemical weapons. --Gr8xoz (talk) 22:11, 20 November 2010 (UTC)[reply]

The Kim & Bin Laden Chemical Recycling Company Ltd will dispose of any unwanted chemical weapons at very reasonable rates, I understand.

Organic matter...You were saying?....--Aspro (talk) 22:16, 20 November 2010 (UTC)[reply]

We are talking about “garden” compost heaps. Why bring in these other things? Which 'chemical weapons' do you keep at home? The US has spent millions on weapon composting techniques : Bacterial detoxification of organophosphate nerve agents . You might show a light on upstairs but I ask myself “is anybody home?!!”--Aspro (talk) 22:30, 20 November 2010 (UTC)[reply]

If you're using a small compost bin and want to mix the cooked veg with more stuff, you can always add newspaper, tissues, kitchen wipes and cardboard. If the bin starts to smell put plenty of ripped up newspaper in. The one thing I would recommend keeping out is citrus peel, because it does make the compost smell nasty as it rots. Itsmejudith (talk) 00:06, 21 November 2010 (UTC)[reply]

I believe Gr8xoz is making a point about the use of the word "organic". In certain contexts, "organic" simply means "containing carbon". (e.g. organic chemistry) - by that definition, motor oil and chemical weapons count as "organic". On the other hand "organic" in the context of organic farming is a too-narrow restriction on compostables - most conventionally grown fruits and vegetables are perfectly fine for the compost. That said, Aspro is correct in assuming that the context reasonably implies "organic matter". -- 174.24.198.158 (talk) 19:51, 21 November 2010 (UTC)[reply]

Excimer lamp

Hello, what is the overall luminous efficacy (lm/W) of an excimer lamp?, ideal black-body radiator at 7000 K has 49 lm/w, so 180 lm/w like some stupid friend of mine pretends that a self invented excimer lamp should produce would be total trashtalk (his 5w selfinvented excimer lamp produces 900 lm)?TY DST —Preceding unsigned comment added by 86.127.164.33 (talk) 19:10, 20 November 2010 (UTC)[reply]

See Luminous efficacy, 49 lm/W seems to be wrong a ideal black-body radiator at 7000 K has 95 lm/W, the theoretical maximum is 683 lm/W for an ideal source of 555 nm yelow light. Low pressure Sodium-vapor lamp gives up to 200 lm/W. I do not know much about Excimer lamps but 180 lm/W should not be unrealistic. --Gr8xoz (talk) 22:00, 20 November 2010 (UTC)[reply]

Yellow? 555 nm is usually considered green. I agree 180 lm/W seems rather high particularly if we're talking about white light. Even the best commercial cool white LEDs only produce ~140-150 lumens/watt under optimal conditions for example (and the best I've heard from research is 200 lm/W). Nil Einne (talk) 06:44, 21 November 2010 (UTC)[reply]

I see no reason to assume that we talk about white light, but of course 180 lm/W white light is very good and unlikely. Excimer lamps seems to be mostly monochromatic and often in the UV-range, obviously the luminous efficacy is 0 lm/W in that case. I was assuming a monochromatic light in the visible range, and if the claimed 180 lm/W is correct it. will be close to 555 nm. Sorry about the mix-up about colours, I assumed that low pressure Sodium-vapor lamps where close enough to 555 nm that they should have the same colour. Low pressure Sodium-vapor lamps emits 589.3 nm which is yelow. An approximation of the colours are 555 nm and 589.3 nm --Gr8xoz (talk) 12:02, 21 November 2010 (UTC)[reply]

You may be right, I don't really know what an excimer lamp is, I was just surprised the OPs friend would be that excited about a non-white light (although there are obviously some applications like novelty lighting, advertising, traffic lights, aquarium lighting etc depending on the precise colour) as at the moment most interest is generally in white lights. The differences between wavelengths at that area can be rather major, I myself wasn't that aware until I developed an interest in LEDs File:Linear visible spectrum.svg Nil Einne (talk) 13:55, 21 November 2010 (UTC)[reply]

Scott toilet tissue ad touts no inner tube.

But can't that cardboard be recycled? Less waste is good, I suppose, but I'm still curious. Thanks. Imagine Reason (talk) 20:09, 20 November 2010 (UTC)[reply]

So what is the question? Yes a cardboard inner tube can (but may not) be recycled. However a cardboard tube costs money for the producer - a direct overhead. If they have a way of winding paper without the core then once they've the right machines it should cost them less AND they can advertise the cost saving to them as an environmental benefit (and perhaps charge more for it). -- SGBailey (talk) 21:28, 20 November 2010 (UTC)[reply]

A company engaged in greenwashing? No wai! DMacks (talk) 21:34, 20 November 2010 (UTC)[reply]

That's not greenwashing, since it does actually reduce resources. My rule of thumb: If it benefits the environment, but costs more, then it's usually greenwashing. (If things cost more then they use more resources, which means they don't actually benefit the environment. There are some exceptions, but it's a good rule of thumb.) Ariel. (talk) 23:52, 20 November 2010 (UTC)[reply]

There can be some actual green, doesn't mean they get off Scott free (sorry!) on ulterior/"not the whole truth" concerns. Googling various non-extremist news sources and commentaries pretty commonly uses this term for this product, but is fairly split about whether it's "bad" or not--our article on the term doesn't say that there can't be an actual green benefit. They've got a product that costs less to manufacture but they're maybe even going to charge me more for the privlege of their improved process? Who profits more here, company or Nature? The roller is only a small part of the whole product, and the paper itself is only about half post-consumer recycled it appears, so seems like a small improvement (with loud crowing) even if there are much larger strides that could be made. I think I'm going to drop this line of conversation here (I'm not the enviro-wacko this paragaph might make me sound like:) lest it devolve into soapboxing one way or the other. DMacks (talk) 00:14, 21 November 2010 (UTC)[reply]

I'm not sure about that rule of thumb. Some things benefit the environment by not using some cheap but horrible substance. (Mercury, CFCs, DDT, what have you.) APL (talk) 20:22, 21 November 2010 (UTC)[reply]

I did say there were exceptions, and those are exactly it (and pollution in general). But most of the time when you are going to buy something you not faced with such choices. Ariel. (talk) 21:53, 21 November 2010 (UTC)[reply]

Anti matter

Why is antimatter important? I understand trying to understand and produce antimatter is part of understanding what makes up the universe and how it works, but do some people think it would be useful for a new class of weapons, or energy production, or could it affect how we travel or build things. Would it fundamentally change our view of how the universe is structured? Laura —Preceding unsigned comment added by 41.225.41.4 (talk) 21:13, 20 November 2010 (UTC)[reply]

None of the above, which does not mean it doesn't have any uses. For instance, anti-matter is produced inside our bodies in a pet scan machine in order to create a image for medical purposes. 76.123.74.93 (talk) 21:48, 20 November 2010 (UTC)[reply]

The answer to the question "would it fundamentally change our view of how the universe is structured?" isn't really knowable with complete certainty in advance. Some of the experiments scientists hope to do with antimatter include verifying the gravitational interaction of antimatter and testing CPT symmetry. Unexpected results to either of those tests could in principle have major theoretical ramifications. Drastically unexpected results do occur in physics experiments from time to time which lead to major theoretical breakthroughs (although not very frequently). Red Act (talk) 23:04, 20 November 2010 (UTC)[reply]

True, besides, antimatter has already fundamentally changed our view of the universe when it was first discovered decades ago. 76.123.74.93 (talk) 23:12, 20 November 2010 (UTC)[reply]

You might be able to, at some point far in the future, to assemble enough antimatter to make an antimatter bomb but if you had the technology to do that it would probably be easier to just make nuclear fission weapons. It would surely be more unwieldy than the mass/energy ratio would imply, for the same reason nuclear weapons are more wieldy than their fissile cores (it would take some considerable apparatus to keep such a weapon from blowing up before you wanted it to). You can't use antimatter as an energy source as far as I know because it takes more energy to produce it than you'd get as a result, at least with current methods. Any potential technological applications are a long way off as far as anyone can currently guess. --Mr.98 (talk) 23:49, 20 November 2010 (UTC)[reply]

The main value of antimatter lies in the relatively distant future. It is the most efficient possible way to store energy. If you have equal masses of ordinary matter and antimatter, they can be combined to transform the entire total mass into energy. Theoretically there is no way to store the same amount of energy with less mass. Thus antimatter would be the ultimate fuel for something like a starship. Looie496 (talk) 00:18, 21 November 2010 (UTC)[reply]

See W and Z bosons, cosmic inflation, baryogenesis, antineutrino, Kardashev scale#Energy development, antimatter rocket, antimatter tests of Lorentz violation, Higgs boson, black hole starship and Large Hadron Collider. ~A H 1^(TCU) 00:40, 21 November 2010 (UTC)[reply]

http://memory-alpha.org/wiki/Matter-antimatter_reaction_assembly —Preceding unsigned comment added by 142.244.236.20 (talk) 22:10, 21 November 2010 (UTC)[reply]

The answer is mostly Science Fiction. (Star Trek probably has a lot to answer for in this regard.)

The idea that appeals to science fiction writers is the idea that only a small amount of antimatter could produce a phenomenal amount of energy. So if you had a super easy way of storing it, you could a bomb that fits in your pocket but has more power than the Hiroshima bomb. Or you could fuel your car forever on a microscopic amount of fuel. And of course, A serious obstacle in spaceship design is that the fuel is so heavy. Antimatter fuel could go a long way towards solving that.

Of course, for a science fiction writer it's easy enough just to say that we've got a large amount of antimatter. In real life no one's even got a good idea how you would even try to make that much of the stuff. APL (talk) 00:48, 21 November 2010 (UTC)[reply]

Why not just make a gigantic solar-powered anti-mater factory in low-solar orbit and have all of our ships fuel up there? 142.244.236.20 (talk) 23:27, 21 November 2010 (UTC)[reply]

Cost I expect. Googlemeister (talk) 17:00, 22 November 2010 (UTC)[reply]

It is important because it is a fundemental truth about the universe, about which we do not know everything there is to know. There is no greater reason to be human than to learn. --Jayron32 02:25, 21 November 2010 (UTC)[reply]

And a lot of useful applications for discoveries appeared long after they were first found. 142.244.236.20 (talk) 23:27, 21 November 2010 (UTC)[reply]

Rootstocks

Anyone know for sure if "semi vigorous" is more or less vigorous than "moderate vigour". I am comparing Montclaire with St Julien rootstocks and the catalogues are not consistent about which will grow a bigger tree. --BozMo talk 21:43, 20 November 2010 (UTC)[reply]

There may not be any industry standard; each grower or supplier may use its own terminology. --Jayron32 02:23, 21 November 2010 (UTC)[reply]

Tin iodide coloration

Why are the tin iodides colored? Is it impurities or the arrangement of the molecules? --Chemicalinterest (talk) 22:20, 20 November 2010 (UTC)[reply]

Neither. It is due to a ligand-to-metal charge-transfer transition (or LMCT transition for short). In very rough terms, you can think of it as an electron hopping from an iodide ion to a tin ion under the influence of visible light; it doesn't actually go "all the way", because there's a big covalent character to the bond. The effect is greatest for iodides because the ionization energy of the iodide ion is much less than for bormides or chlorides: so iodides are always more like to be coloured, and will always be more strongly coloured and at shades nearer to red, than bromides or chlorides. Physchim62 (talk) 23:29, 20 November 2010 (UTC)[reply]

Our charge-transfer complex article is in pretty good shape I think. DMacks (talk) 00:00, 21 November 2010 (UTC)[reply]

November 21

Full name for invertebrate zoologist named Grube or Grübe, mid C19

Hi all,
Pet peeve is authorities without biographies. "Grube" is the binomial authority for Chirocephalus josephinae, Peripatopsis capensis, as "Grübe" -- and with one "Oersted" -- Themiste alutacea, and so on. From the last mentioned, I'm guessing he or she may well be Danish.
Thank you! --Shirt58 (talk) 09:23, 21 November 2010 (UTC)[reply]

This is where the internet fails! I eventually found it by searching for " Grube nineteenth century taxomony" and found his initials are AE Grube, then google scholared that and found this which says he is Adolph Eduard Grube. The Oersted is Anders Sandøe Ørsted based on the third hit for a search of "A S Ørsted Themiste" (tried posting a link but the syntax gets screwed up as it has square brackets in the URL). SmartSE (talk) 12:28, 21 November 2010 (UTC)[reply]

If you replace the [ and ] with %91 and %93, which are the ASCII escaped equivalents, then it should work. CS Miller (talk) 09:08, 22 November 2010 (UTC)[reply]

Uuh, mmm, yeah, I'll try that ~~the noo~~ the morn --Shirt58 (talk) 13:19, 23 November 2010 (UTC)[reply]

Is autism a human polyphenism similar to those of social insects?

This isn't a topic I know much about, but I was thinking... It is well recognized that locusts switch between a solitary existence and one in which they form vast gregarious migratory swarms. This is dependent on a very simple mechanism involving the foraging gene (a cGMP-dependent protein kinase or PKGI) and increased serotonin levels in the nervous system. This mechanism has a similar function in Drosophila melanogaster, and also controls foraging vs. defender behavior in worker ants.[2]

Now the same sort of PKGI protein controls serotonin uptake by SERT in mammals also.[3] Variants of SERT have been associated with certain autistic traits,[4] Autism can be associated with increased serotonin levels, and various SSRIs have been suggested as treatments.[5] (At first glance the elevated serotonin seems the wrong way, but one could argue that it is in some way compensatory... I'll leave this hanging for now)

Now putting together these things, and if I credit the common ancestor of bilateral animals with some sophistication, I am tempted to suppose that there was some primordial mechanism to use PGKI and serotonin to dictate a different "phase" or behavioral phenotype which was more social in nature. This leads to the idea that at least some forms of autism are a pre-existing, adaptive model of human behavior, rather than a disease. Perhaps it was maintained if as a variation in populations of all preceding ancestors, or at least as a working genetic regulatory system that could be called upon at need; or else it might be that the way the nervous system works allows the same genetic changes to recreate similar changes of behavior even hundreds of millions of years apart from the first instance. (Autism could also result from non-adaptive genetic mutations, just as white skin can arise from albinism rather than an ancient variant of MC1r)

But now to the question. Is it possible for an autistic person to live in a traditional hunter-gatherer society, either Paleolithic or Neolithic (or as best an approximation as can be found) without suffering any selective disadvantage? Apparently autism is still relatively uncharacterized in developing countries, but does exist.[6] Most of the discussion I find about autism and hunter-gatherer lifestyles focuses more on hunter-gatherer diet, said by some to help with the disease, but that's not what I'm looking for. In order for autism to be a true human polyphenism, there has to be some environment in which it is not maladaptive. Actually, that last sentence was quite a stupid thing to say, especially in reference to insect polyphenism! What I mean is that the potential to become autistic would have had to have been preserved by some means, even if the mechanism is prone to be phrased in terms of the ever-controversial group or kin selection. Wnt (talk) 10:06, 21 November 2010 (UTC)[reply]

I don't accept your implicit assumption that autism would automatically be maladaptive in hunter-gatherer societies, at least not in relatively mild forms. You could argue, for example, that mild autism could be a benefit for someone who is engaging in solitary trapping of small animals. Modern hunter-gatherer societies show a high level of individual specialization, and hence can accomodate a range of personality types. Physchim62 (talk) 12:02, 21 November 2010 (UTC)[reply]

Sorry if I was confusing: that's not my assumption but the question. While I would speculate that it is possible that autistic people could do well in such societies (otherwise the answer to the title question is probably "no"), I'd really like to see if there's evidence. Certainly it is believed in conventional circles that autism is a disease and a disability, and I'd like to actually have a concrete example to point to to show otherwise.

Another reason to look for concrete examples is that the rate of autism in society is apparently highly variable (and currently increasing), and only in a society where autism and non-autism are of equal fitness can one see what the "natural" rate of autism among humans might be. (At least, assuming that the developmental decision process is adaptive) I should point out that if it is a polyphenism rather than a disease - in other words, if people are already "pre-loaded" with autistic developmental software and rely only on environmental cues to determine whether to activate it or not) there may be no upper limit on how common it could become - if there is some lifestyle habit, food, chemical, ultrasonic noise, etc. that affects which behavioral phenotype is expressed, you could find yourself in a situation in a few years where 50% or more of children are autistic. It is possible that some unknown trigger (for example, the concentration of carbon dioxide in the atmosphere) could cross a tipping point to cause this, but still be virtually impossible to identify. It is even possible that such stimuli could have a cumulative epigenetic effect in the parents, so that once the change occurs, it is irreversible for a large portion of society. These are of course very remote possibilities, but it's almost an apocalyptic scenario. It would help to rule out very widespread autism in the future if you can identify a society where autistic people do well and show that even there the rate is very low. Wnt (talk) 13:08, 21 November 2010 (UTC)[reply]

This is a very interesting question, but I don't think you're going to get a concrete answer and you're probably asking for opinions, as you've evidently researched this in some depth without finding anything yourself. For starters, I think it's important to question whether autism is maladaptive in today's society, although in severe cases it obviously is, at the less severe end of the spectrum though you'll find many scientists and mathmaticians - [7] by Simon Baron-Cohen who has worked on this e.g [8]. That makes me wonder whether an argument could be made that as today's world is more technical and abstract than in the past, maybe autism is adaptive. This is obviously guesswork, but maybe the industrial revolution selected for genes linked with autism and severe cases of autism are a result of individuals having two recessive alleles or something (sorry, I'm not a geneticist!). This discusses how similar points have been raised regarding schizophrenia and creativity, and reminding us of the thin line between genius and insanity (maybe they are the same). These ideas are linked to neurodiversity - it's pretty obivous that there is no such thing as "normal" when it comes to the way people think and this is a good thing because we are all good at different things. Combined, this has allowed us to go further than if we all thought in the same way and as a society we are stronger as a result. I don't think that anyone has written about this, but in the same way that a more genetically diverse animal population is more likely to survive changes in the environment, the same may well apply to the way our brains work, but in a slightly different way. For example if we need to solve a problem, you may need somebody creative (more schizophrenic) to come up with the idea in the first place, but then someone who pays attention to details (more autistic) to actually design and implement the solution. This links pretty nicely with insect polymorphism, for example in leaf cutter ants where the many different castes perform the jobs they do best and the colony as a whole benefits from that. So, I haven't really answered your question, but I've hopefully provided an alternative viewpoint. SmartSE (talk) 14:36, 21 November 2010 (UTC)[reply]

I'm starting to believe that being Neurotypical is a real live example of the Green-beard effect --Digrpat (talk) 23:46, 21 November 2010 (UTC)[reply]

I believe in neurodiversity, and more generally, that both genetic diversity and other forms of phenotypic plasticity and diversity are maintained within many species. I suspect that there are many of us here on Wikipedia who are some small distance into the autism spectrum; even so, it's been my impression that the average family returning from the pediatrician with a diagnosis of autism is not expecting a great mathematician or even an average student. When autism is accompanied by retardation, I don't know whether that is because it is one symptom of a more widespread problem, or whether autism interferes with learning as it occurs in our society. I suppose I've been assuming the latter when asking about ancient or primitive cultures; if it were the former, I suppose that Aspberger's Syndrome or some otherwise defined subset of the autism spectrum might count as the polyphenism in question above, with more severe autism being something of a red herring. But looking at locusts from an anthropomorphic perspective, I tend to doubt that: a locust who won't follow and eat with the swarm surely must seem profoundly developmentally disabled to its peers.

The "extreme male brain" model of autism seems somewhat offensive, since it's based on the idea (I would be prone to say myth) that women can't do math, science, or technical work. I think that there are (wrong) racist arguments that seem more plausible. To me it would seem more likely that men, lacking a second X chromosome, are somehow more prone to autism on a genetic basis, and a greater than expected number of subtle autistic phenotypes have tinged the studies about differences between male and female. There is also an argument regarding the idea that autism and schizophrenia are opposite ends of a spectrum, and schizophrenics aren't "more female".[9]

vortex launcher : pressure amount area

Hi guys,

recently i have been puzzling over vortex launchers because of the military non lethal one that has been proposed. It is designed to produce a vortex with enough strength to knock someone down. At first i thought that this was folly since the law of equal and opposite reactions meant that if it produced enough force going one way to knock over a person, sureley the person holding it would be knocked down as well. then I began to think, the vortex is caused firstly by an explosion in a 3inch diametre blast chamber (read the report). then the flow is expanded via the nozzle as it travels up it and out and arrives at the target as a 2ft diamtre vortex. Would that mean that even though the pressure in the blast chamber is greater than the vortex pressure on the target, it is over a much smaller area so is not as effective at moving the whole system backwards. A similar example would be a ship on water, if i set off an explosion on a small part of a ship, it may destroy a part of the ship, but the whole thing will not move across the water, but the much smaller pressure of the wind on the sale, which is a significantly larger area, can move the whole ship across the water. Do you think the vortex launcher works something like that, more pressure but in a smaller area at the launcher, so it experiences a jolt, but less pressure but more area at the target so it is moved backwards (knocked over)

Does that make sense or have I missed something.

Many thanks. —Preceding unsigned comment added by 62.3.99.14 (talk) 11:11, 21 November 2010 (UTC)[reply]

It is not necessary that all the momentum comes from the "vortex launcher" since the vortex interacts with the surrounding air. Also the shouter can maybe seek support while the target are unprepared. Even given this I have difficulties in seeing that it could be a practical weapon in most situations. --Gr8xoz (talk) 01:07, 22 November 2010 (UTC)[reply]

Is it safe to drive after drinking cough syrup?

I feel euphoric and other pleasant senses. There's something about cough syrup that just alters the way I feel and etc.

But I have this gut feeling about driving that caused me to ask you here: Is it safe to drive after having had cough syrup? Please let me know ASAP, and don't forget to cite sources. --70.179.178.5 (talk) 13:28, 21 November 2010 (UTC)[reply]

Are you talking about a normal dose, or intentional abuse of dextromethorphan, or something else? There are many different cough syrups with different active ingredients. Do you want to specify one in particular? Wnt (talk) 13:33, 21 November 2010 (UTC)[reply]

Edit conflict - It is policy here not to answer medical advice questions. Please read the package insert of your cough syrup to find out if you can safely drive under medication from its contents, which as user:Wnt pointed out we cannot know. Please ask a qualified Pharmacist or Medical Doctor near your location for their advice regarding this matter. --79.219.104.60 (talk) 13:36, 21 November 2010 (UTC)[reply]

I have added back the above response which was removed by the OP [10] with the claim it was 'unconstructive' Nil Einne (talk) 18:11, 21 November 2010 (UTC)[reply]

Regardless of that, do you really want to trust some randomers on the internet about this? Maybe ask you doctor! SmartSE (talk) 13:56, 21 November 2010 (UTC)[reply]

SmartSE, my doctor is not in on weekends. Otherwise I would've called him already. --70.179.178.5 (talk) 14:46, 21 November 2010 (UTC)[reply]

Wasn't there a leaflet in the box? If so, read it carefully. If not, and you can't get advice anywhere else, be safe and don't drive. No-one here can help you any further because we don't know what is in that particular cough syrup. Don't take medicines for a sense of euphoria, only for the medical conditions they are meant for. Itsmejudith (talk) 17:20, 21 November 2010 (UTC)[reply]

Words of wisdom: when a question says "please let me know ASAP, and don't forget to cite sources", it should be treated as trolling. Looie496 (talk) 17:21, 21 November 2010 (UTC)[reply]

More so when the OP removes helpful answers. Nil Einne (talk) 18:11, 21 November 2010 (UTC)[reply]

Limits on number of channels in MIMO wireless

In my answer to the question Wireless v Fibre on the 16/11 [11] I assumed that the maximum number of independent MIMO- channels in future wireless internet connections is limited to 10. This question got me thinking on what the physical limits really are and how they depends asymptotically to the size of the antenna arrays. I has not been able to find any work on this, maybe be course I do not know the terminology.

In theory it should be possible to place any number N antennas in a arbitrary small area on the transmitter and receiver and get N independent channels but if the antennas are to closely spaced the problem with separating the independent channels will be very ill-conditioned. It will be a hard inverse problem that will get very bad signal to noise ratio.

So the question is: How will the number of possible independent channels depend on the size of the antenna arrays given that the separation of the channels should be a reasonable well-conditioned problem?

This should be the same as to say that the needed power at the sender should scale linearly in the number of channels.

In the case of free space communication my guess based on intuition and calculation of the angular resolution of the antennas are:

$N<C\left({\frac {D_{t}D_{r}}{\lambda L}}\right)^{2}$

Where N is the number of independent channels, C is a small constant depending on how well-conditioned the system should be, $D_{t}$ is the diameter of the transmitter array, $D_{r}$ is the diameter of the receiver array, $\lambda$ is the wave length and L is the distance between the transmitter and the receiver. I think this should be valid for $\lambda <<D_{t}$ , $\lambda <<D_{r}$ , $D_{t}<<L$ and $D_{r}<<L$ .

Is this correct?

An example $D_{t}=3m$ , $D_{r}=3m$ , $L=1000m$ and $\lambda =0.003m$ (100 GHz) would give 9 C channels.

There are much hype around Orbital Angular Momentum (OAM) [12] but I do not think it will make any difference for this limit, is this correct?

In the presence of reflectors the number of channels can be increased and in the extreme case when the transmitter and the receiver are in the focal points of a reflective ellipsoid cover the limit should be something like: $N<C\min \left(\left({\frac {D_{t}}{\lambda }}\right)^{2},\left({\frac {D_{r}}{\lambda }}\right)^{2}\right)$

In a network the capacity can be increased by using more than one base station.

Any ideas of what number of MIMO- channels that can be used in the future for fixed data-links (point to point), stationary roof mounted antennas in a cellular system and in portable devices? --Gr8xoz (talk) 14:19, 21 November 2010 (UTC)[reply]

If you try to pack too many antennas in a small space, the power required will grow exponentially with the number of antennas. Totally negating the benefit, as you could get that by having more bits per symbol anyway. You can try to imagine how many antennas on a portable device. At the wavelengths used you could get about two, and if you allow diverse polarization as well, perhaps you can get 4 independent signals on your handheld device. For larger implementations, it is going to be mostly 2 dimensional on the surface of the earth. In a stationary application, you can have a tradeoff with independent antennas, or getting gain with a directional antenna and increasing the bits per symbol. remember also that the cost will be proportional to the number of antennas, so having 1000 antennas in a circle to get 1000 times the bit rate may not be economical, it may be better to run in an optical fibre. Graeme Bartlett (talk) 09:53, 22 November 2010 (UTC)[reply]

Thank you for the answer. How many antennas is to many in a given space? I assume it is the number of independent channels that is limited, many antennas sending the same signal does not require more power. I also think that about 4 channels on a hand-held device at today's frequencies is around the maximum but the trend are for frequencies to go up over time, from 415 MHz NMT to 2.6 GHz 4G and 5 GHz WiMAX, the question is where does it stop and how many independent channels can be used then.

My impression was that as long as you do not pack the antennas to tight the power rises linearly with the number of channels and the capacity while the power rises exponentially with the number of bits/s/Hz/channel. See Shannon–Hartley theorem.

For the same reason the needed surface-area of a directional antenna will increase exponentially with the number of bits/s/Hz/channel while the needed surface area for a antenna array in a MIMO-system only should increase by the square-root of the needed capacity.

Antenna arrays aligned with the surface of the earth is of curse the way to go for radio astronomy and deep-space communication. I think that in order to get 2D MIMO you will need a projected area perpendicular to the direction of communication, for communication along the surface of the earth it will become a 1D MIMO-system. I was thinking more of integrated antennas on printed circuit-boards in the size of a normal satellite dish operating at high frequencies (10-300 GHz). The current fastest data-link operates at 85 GHz. I do not see the cost of the antennas as important in such an installation but of curse the transceivers and signal processing will cost a loot today but they follow Moore's law.

An optical fibre is of curse a good alternative in many cases.--Gr8xoz (talk) 11:48, 22 November 2010 (UTC)[reply]

Origin of particle diversity

What is the origin of such a diverse world of elementary particles (if you exclude the divine one)? As the vacuum state says, the default vacuum has a zero-point energy, so (if I'm not mistaken) it can support only Higgs boson or photon annihilation at best. How can low-energy Higgs bosons or photons produce so many different particles if they were the original stuff for annihilation? (leaving aside the Big Bang as the high energy origin, because the explosion reason and the origin of pre-Bang ingredients are unclear so far) Twilightchill t 14:33, 21 November 2010 (UTC)[reply]

Nobody knows what the origin of particle diversity is. One of the ambitious goals of string theory and other Theories of Everything is that, if correct, one of these theories will clear up why things are the way they are — it'll turn out that logically they have to be the way they are. (An analogy — in 1913, it was a good question to ask, why are the Bohr atom electron orbits exactly where they are? Doesn't that seem arbitrary? But in 1924 De Broglie showed that their orbits are absolutely necessary according to the idea that the electron was actually a standing wave that could not cancel itself out. What seemed arbitrary at first was really just an imperfect understanding of the nature of the physical universe itself, which really did not allow alternatives.) It's certainly the most ambitious goal of modern physics to figure out why things are as they are, and not some other way. There are some who believe that universes pop into and out of existence all the time (e.g. in the multiverse), and so what we have is a quasi-Darwinian situation where of course the only universe that we happen to have come into existence in happens to be not only stable but contain a mixture of things that are allowable for the formation of suns and planets and water and life (see anthropic principle). For some that's a good enough answer, for others it isn't. --Mr.98 (talk) 14:40, 21 November 2010 (UTC)[reply]

"The default vacuum has a zero-point energy, so it can support only Higgs boson or photon annihilation at best" makes no sense to me, and I think it's simply wrong. In modern physics the vacuum effectively supports everything; we are perturbations of the vacuum. It's also not true that photons and Higgs bosons were the first particles. What determines the particle types is the various ways in which the vacuum can vibrate, so the question is why the vacuum can vibrate in those ways and not others. The various grand unified theories are about finding a simple vacuum "structure" that can vibrate in those ways and not others. For example, in SO(10) grand unification you can imagine there's a 10-dimensional sphere attached to every point in spacetime, and the various Standard Model bosons (except the Higgs) arise as rotations of the spheres, while the different fermions are more or less different spherical harmonics of the spheres. These models predict additional bosons arising from other rotational motions of the spheres, so one has to find a way to explain why those bosons haven't been seen. But that's not too difficult; you can invent a mechanism similar to the one that makes the weak force so weak. Unfortunately, these theories are difficult to test, so we remain ignorant about what's really going on. Hopefully data from the LHC will finally clarify things, but it might not. -- BenRG (talk) 02:22, 22 November 2010 (UTC)[reply]

Electroplating With Lemon Juice?

The other day I decided to clean some extremely corroded and grimy coins that I had laying around my room. One was a United States nickel and the other a US penny. In order to clean the coins, I placed them in a small container of lemon juice in hopes that the acid would help remove some of the grime. After a day, I removed them from the container and to my surprise, the penny was delightfully shiny, while the nickel had turned copper as well! I assumed that some sort of electroplating had occurred, but the container was never exposed to an electrical field. Could someone please help explain to me what has happened to these coins? Thanks! Stripey the crab (talk) 14:51, 21 November 2010 (UTC)[reply]

There was an electrical field, which was set up by the battery you inadvertently created. See Galvanic cell. Going from metallic nickel and copper ions to metallic copper and nickel ions is chemically favorable (see electromotive series and Standard electrode potential (data page)), like going from metallic zinc and copper ions to metallic copper and zinc ions. It's just that, instead of having two separate electrodes with a wire between, you have different parts of the same coin acting as the two electrodes and the wire. -- 174.24.198.158 (talk) 19:26, 21 November 2010 (UTC)[reply]

Waste of lemon juice. --90.219.114.59 (talk) 23:16, 21 November 2010 (UTC)[reply]

I want to repeat this experiment. Will vinegar work for the acid? Will it help to put some table salt in it? Should the coins be touching or not? —Bkell (talk) 05:01, 22 November 2010 (UTC)[reply]

Those are good experiments to try. If we told you what the results would be, they wouldn't be experiments. Cuddlyable3 (talk) 09:08, 22 November 2010 (UTC)[reply]

Heh, okay. :-) —Bkell (talk) 14:23, 22 November 2010 (UTC)[reply]

I don't think that the coins have to touch. The copper(II) oxide and carbonate on the penny coin dissolves in the acid to make copper(II) acetate. The copper(II) acetate reacts with the nickel in the nickel coin to make nickel(II) acetate and copper metal. Sorry for ruining your experiment, do it anyway! --Chemicalinterest (talk) 15:28, 22 November 2010 (UTC)[reply]

Enema washes aren't enough; will taking many laxatives help lose substantial weight?

Hi, the Enema Wash on my BioBidet BB-i3000 ( http://www.BioBidet.com ) isn't doing enough of a job at forcing weight out of my system. It has improved my weight loss, but not by enough in my opinion. I currently weigh 178 lbs. at 5'11.5", but will try to get down to a nice, round 150 lbs. (below 135 is underweight.)

~~So if I take a lot of laxatives, how much will that help me lose more weight? Also, what ill side-effects might it have?~~

~~Besides, what ill side effects will giving myself extra enema-washes from bidet-seats give me? --70.179.178.5 (talk) 14:56, 21 November 2010 (UTC)~~[reply]

Request for medical advice removed. Please consult an appropriate physician for advice on any aggressive weight loss scheme. 71.228.185.250 (talk) 15:05, 21 November 2010 (UTC)[reply]

Mass of energy?

Something is wrong with this set-up, but I'm not sure what. Let's say I turn a kilogram of mass into (1kg)*c^2 energy. If energy is massless, I should be able to lift it up ten metres for free. I then turn it back to matter and run in through a watermill and generate free energy. Is the problem in my assumption that energy is massless? I'm guessing so given that black holes can attract light. 142.244.236.20 (talk) 22:06, 21 November 2010 (UTC)[reply]

If you can turn a Kilogram of mass into energy, why are you bothering with watermills? AndyTheGrump (talk) 23:11, 21 November 2010 (UTC)[reply]

Because I don't want my kilogram consumed in the process; it has sentimental value. 142.244.236.20 (talk) 23:13, 21 November 2010 (UTC)[reply]

And because antimatter is really expensive... Googlemeister (talk) 15:08, 22 November 2010 (UTC)[reply]

It turns out that moving your equivalent-to-one-kilogram-of-matter chunk of energy up a gravity gradient costs exactly the same amount of energy as moving the original one-kilogram mass. For example, if you point a flashlight straight up, the photons will get (ever-so-slightly) redder as they gain altitude. (See gravitational redshift for more details.) Your kilogram mass at ground level will be just a little bit lighter once you've lifted it up ten meters, whether you convert it to energy or leave it as matter during the trip. Each time you repeat the cycle, you lose a little bit of mass/energy — that's where the energy coming out of the watermill in your thought experiment is coming from. TenOfAllTrades(talk) 23:30, 21 November 2010 (UTC)[reply]

What do you mean when you say my kilogram will be lighter after the trip even if I leave it as matter? Does something with more potential energy have less mass? Where did that mass go, did it lose a very small number of atoms? 142.244.236.20 (talk) 02:11, 22 November 2010 (UTC)[reply]

For light to climb out of a gravity well it cannibalizes some of its own energy to do so, so it ends up massing less (compared to other object also out of the gravity well). On the other hand, if you compare that mass to something at the ground you will find no change in the mass because the potential energy of the gravity adds to the mass (so it traded light-energy into gravitational potential energy). Yes, I know it's complicated. If you carry matter up against gravity, you have to add energy externally to do it, since you can't steal energy from the mass itself, this energy you added will, of course, add to the mass of the matter. No atoms are gained or lost, the change in mass (and weight) is entirely due to change in potential energy. Potential energy from gravity is particularly hairy to calculate since it's totally relative to what you are measuring against. There is a reason scientists usually deal only with rest mass when possible, since calculating the true mass of something is not just hard, it's also not a single number - it depends on what you compare to. Ariel. (talk) 02:52, 22 November 2010 (UTC)[reply]

The fundamental assumptions in the question are flawed, mass is identical to energy mass can not be turned in to energy, it is energy and 1 kg will be 1 kg = 9e16 J regardless of what form of energy it is. Energy is not massless, it is mass so the OP:s suspicion of the flaw is correct. Matter is a category of energy that covers almost everything we normally think of as mass, it is not an entirely well defined category. --Gr8xoz (talk) 00:58, 22 November 2010 (UTC)[reply]

TenOfAllTrades gave you the correct answer, but I wanted to add that energy is not massless. Not only does it have mass, it also has weight. So if you had the energy in 1kg all as light (photons), somehow stored in a box, that box would actually weigh 1kg just from the energy. Ariel. (talk) 01:47, 22 November 2010 (UTC)[reply]

If photons have mass, how can they travel at the speed of light? 142.244.236.20 (talk) 02:11, 22 November 2010 (UTC)[reply]

Because although a so-called massless particle like a photon has mass, it has a rest mass of zero. See mass in special relativity. Red Act (talk) 02:35, 22 November 2010 (UTC)[reply]

The fundamental flaw in the question is the assumption that matter can be turned into "pure" energy. You can't. Energy is no substance, it is a quantity that has to be ascribed to a material system depending on its mass and state of motion. Indeed, mass and energy are not the same thing, however often you write down

E=mc^{2}

(that equation is only valid/useful in the rest frame of the mass m - there is the alternative interpretation of "relativistic mass" advocated by Gr8xoz but that is not useful (because it hides the physical difference between mass and energy) and is not used in physics any more). Radiation is not "pure energy" either, it is better viewed as a substance, i.e. matter in a broad sense. A single photon always has zero mass, but a collection of photons (say, a photon gas confined within a box of negligible mass) can have an effective mass. May sound strange, but that's how special relativity works. --Wrongfilter (talk) 10:06, 22 November 2010 (UTC)[reply]

I'm sorry, but this is incorrect. A single photon has mass, and so does a collection of them. Matter CAN be turned into "pure" (as you call it) energy. For example you can convert matter into kinetic energy (assuming you have some anti-matter, but anti-matter is still matter). Mass and energy are in fact the same thing, it's impossible to make a logically consistent distinction between them. You can distinguish between zero rest mass and non zero, but that's all. Ariel. (talk) 11:40, 22 November 2010 (UTC)[reply]

As I said, there are two conventions. The one that is in use in theoretical physics (have a look at a current text book on relativity) uses mass for the invariant length of the 4-momentum vector, and energy for its time component (3-momentum gives the three spatial components). The full equation is

mc^{2}={\sqrt {E^{2}-p^{2}c^{2}}}

. Energy and mass (and 3-momentum) are intricately linked, of course, but they are not the same. If you annihilate matter and anti-matter you don't get pure energy. You get radiation, which has energy. --Wrongfilter (talk) 12:49, 22 November 2010 (UTC)[reply]

Ummm ... if energy and mass are not the same, then perhaps you can demonstrate this by giving an example of a system that has energy without mass, or mass without energy - or a process that changes a system's mass without changing its energy, or vice versa ? Gandalf61 (talk) 13:49, 22 November 2010 (UTC)[reply]

A single photon that has energy

E=h\nu

has momentum

p=h\nu /c

and zero mass. If I switch reference frame, the photon's frequency, and thus its energy and momentum, will change (Doppler effect in SRT) but the mass will remain at 0. A system of two photons (of equal frequency) that are moving in opposite spatial direction has energy 2E, and total momentum p=0, hence mass

2E/c^{2}

. That's why a box containing a gas of photons whose momenta cancel each other (so that the box as a whole stays at rest in my reference frame) will have effective mass which is (numerically) equal to the sum of their energies. There is no system with non-zero mass and zero energy - this is precisely the meaning of

E=mc^{2}

. For a system that is in motion as a whole, the energy will always be larger than its mass (multiplied by c²). Mass is an intrinsic property of a system which is invariant with respect to changes of reference system, energy is not. The mass of a particle tells you what states of motion are possible for it; the energy (and the momentum) tell you what the actual current state of motion of the particle is. --Wrongfilter (talk) 15:14, 22 November 2010 (UTC)[reply]

Locally, incidentally, any process conserves 4-momentum, hence energy and (ordinary) momentum, hence also mass. That's the shortest answer to OP's question. --Wrongfilter (talk) 15:16, 22 November 2010 (UTC)[reply]

So with that definition of mass, two photons of equal frequency travelling in opposite directions have a joint energy of 2E and a joint mass of

2E/c^{2}

- but if you consider each photon individually, it has an energy of E but a mass of 0 ? Gandalf61 (talk) 16:55, 22 November 2010 (UTC)[reply]

That's weird, isn't it? I think it has to do with the fact that the concept of "opposite directions" is not invariant. If you go to another reference system the angle between the directions changes. --Wrongfilter (talk) 17:17, 22 November 2010 (UTC)[reply]

See Mass in special relativity for a discussion of the distinction between "invariant mass" and "relativistic mass" which is what this argument comes down to. Which is the true "mass" is a question of semantics. Rckrone (talk) 18:19, 22 November 2010 (UTC)[reply]

This is one of the reasons that I think the concept of relativistic mass deserves a little more respect than it sometimes gets. Intuitively, the mass of the whole should be the sum of the mass of the parts. That's true for relativistic mass, not true for invariant mass. --Trovatore (talk) 18:17, 22 November 2010 (UTC)[reply]

Pre-mitochondrial eukaryote

How did eukaryotes process energy before merging with mitochondria? 142.244.236.20 (talk) 22:08, 21 November 2010 (UTC)[reply]

See Mitochondrion#Origin and Endosymbiotic_theory. The pre-eukaryotic cells would have probably used biochemical pathways like glycolysis and fermentation (biochemistry). --- Medical geneticist (talk) 22:26, 21 November 2010 (UTC)[reply]

Evolution without Cretaceous–Tertiary extinction event

Would there be major evolutionary disruptions if the Cretaceous–Tertiary extinction didn't occur? I think the evolution would have take much more time at best and the man would have little chances to pop out and survive... —Preceding unsigned comment added by 89.77.158.172 (talk) 23:36, 21 November 2010 (UTC)[reply]

This is very hypothetical. Still, I should point out that mammals filled the ecological niches left by the extinction, for example, by producing megafauna. I think the question here is whether dinosaurs or mammals would have better adapted to the climate changes over the last 65 millions years, such as the ice ages. —Arctic Gnome (talk • contribs) 23:47, 21 November 2010 (UTC)[reply]

Of course the life on earth would be very different if the CT extinction didn't take place. Essentially what you are asking is if life evolved differently, would it be different?.. How different is impossible to say from our perspective. Every evolutionary biologist would love to have a machine that could replay evolution from a chosen point in earth's history, that kind of machine would answer a lot of questions. I would also just point out that "evolution would have taken much more time" is a meaningless statement. Evolution doesn't have a "vector" or purpose except for fitness, the vectors can only be drawn in post hoc. Unlike Star Trek might have you believe, bipedal tetrapods with big brains are not the ultimate goal of evolution. Vespine (talk) 02:38, 22 November 2010 (UTC)[reply]

How many feet do bipedal tetrapods have? --Lgriot (talk) 12:38, 22 November 2010 (UTC)[reply]

Two. WikiDao ☯ (talk) 15:38, 22 November 2010 (UTC)[reply]

It is 100% certain that our species would NOT exist. Cause and effect could not conceivably have led to our species, given such a fundamentally different foundation. It is literally inconceivable. 63.17.93.42 (talk) 04:24, 22 November 2010 (UTC)[reply]

Our article on the extinction (a featured article, no less) mentions that there is not universal support for the exact cause of the extinction, which makes it extremely difficult to figure out how "it" (the cause of the extinction) not happening might affect everything else. In terms of dinosaurs, the article confirms my recollections of the current state of evidence: "The dinosaur fossil record has been interpreted to show both a decline in diversity and no decline in diversity during the last few million years of the Cretaceous..." and "Whether the extinction occurred gradually or very suddenly is debatable, as both views have support in the fossil record." The point being that if non-avian dinosaurs were declining in numbers for whatever reason anyway, their continued existence may not have played much of a part in affecting the survival of birds and mammals that were around at the time. Matt Deres (talk) 14:52, 22 November 2010 (UTC)[reply]

November 22

Human Dissection

Hello. Pretend that I was born yesterday. Why is the identity of a human subject about to go under dissection concealed? Thanks in advance. --Mayfare (talk) 04:21, 22 November 2010 (UTC)[reply]

Well the obvious one is in case someone observing knows the donor. Do you need more reason then that? Vespine (talk) 05:37, 22 November 2010 (UTC)[reply]

There are laws[13] [14] [15] against desecrating corpses. Perhaps the dissector(s) wish to avoid a writ of Habeas corpus.
I congratulate the OP on their first day of speaking English. Cuddlyable3 (talk) 08:57, 22 November 2010 (UTC)[reply]

That's high praise indeed from you, C3! Actually, the rights of the prosector are generally protected by the various Human Tissue Acts against prosecution for desecration of a corpse where the decedent has donated their body. I know C3's comments about habeas corpus represents comedy for the linguist, but I thought I'd just clarify that. The main reason is as per Vespine's response, and to preserve confidentiality for the donor and his or her family. Mattopaedia ^{Say G'Day!} 10:14, 22 November 2010 (UTC)[reply]

engineering survey

What are the importance of surveying to the engineers?

We have articles on Surveying, Civil engineering and Military engineering that may be of interest. Itsmejudith (talk) 13:35, 22 November 2010 (UTC)[reply]

Something interisting: about a year ago a company I used to work for, employed a brand new technology (at least in New Zealand). This technology basically uses sonar scanning to methodically confirm the homogenity of newly cast concrete support columns. Homogenity is important in establishing the strength of concrete. Concrete that is not mixed well, has large zones of differing densities. Even if the bulk of a structure is dense and strong, a single weak spot of low density may cause a catastrophic failure. This may not answer your question, but I think that applying sonar technology in such a way is really neat.

Surveying is very important to engineers, no actually, it is key. Without surveying, engineers would basically be out of a job. Surveying gives crucial information about the terrain, so that engineers know what they're dealing with. It allows them to know how to stop a building from sinking into the ground or toppling over. Plasmic Physics (talk) 13:56, 22 November 2010 (UTC)[reply]

Please note that not all engineers work with buildings. Surveying is not very important to electrical, chemical or aerospace engineers. Googlemeister (talk) 15:06, 22 November 2010 (UTC)[reply]

Tin(IV) sulfate?

From the tin(IV) oxide article:

Similarly, SnO2 dissolves in sulfuric acid to give the sulfate:[3]

SnO2 + 2 H2SO4 → Sn(SO4)2 + 2 H2O

Does tin(IV) sulfate really form like that? If it is so easy to make, it deserves an article. --Chemicalinterest (talk) 15:25, 22 November 2010 (UTC)[reply]

Be WP:BOLD! shoy (reactions) 16:01, 22 November 2010 (UTC)[reply]

But does it exist? Unfortunately, my chemistry experiments have been terminated and I am unable to do some WP:OR to prove that it does or doesn't exist. --Chemicalinterest (talk) 17:49, 22 November 2010 (UTC)[reply]

Stannic sulfate appears to be CAS 19307-28-9. Can also be made by dissolving certain forms of tin metal in concentrated sulfuric acid.(ref: doi:10.1021/ie50259a027) Commercially available from Fluka (maybe as a sulfuric-acid adduct, if you feel like believing their catalog), with a note that it's a reagent used for some standard analytical-chemistry procedure. DMacks (talk) 18:13, 22 November 2010 (UTC)[reply]

Electric heater efficiency

I just got a "Nordik Ceramic Heater" (the kind of electric heater that won't set things on fire if they touch it) and on the box it claims "HIGH EFFICIENCY ceramic elements power consumption / instant settings at High and Low". Am I wrong that (other than radiative losses through windows, sound energy escaping the room, etc.) all electric heaters are ~100% efficient, that pretty much being the definition of a resistive electric load? And that there is essentially no difference in conversion efficiency between a ceramic-element heater and a wire-element heater? And if I'm right, what could be the basis for the claim of "high efficiency"? Thanks! Franamax (talk) 17:44, 22 November 2010 (UTC)[reply]

You're not wrong; the basis might be different meanings of "efficiency". For instance, one could make such a safe heater by wrapping an unsafe heater in a large amount of (non-flammable) insulation, but then it wouldn't heat as quickly (and you would probably have to run the heating element only part of the time to prevent overheating). If they've made a heater with no exposed extreme temperatures but also have some (convective?) system for emitting heat quickly, that could be called "more efficient" in that you wouldn't wait so long for the effects. It might also radiate more rather than conduct; then if you sit by it more of it heats you directly rather than heating a rising plume of air that you're not sitting in. Then you would feel warmer for the same energy spent (or feel as warm for less energy). --Tardis (talk) 18:42, 22 November 2010 (UTC)[reply]

(ec)Yes, that's true from a heat/temperature point of view. But humans don't measure temperature, they measure how fast heat is gained or lost by the skin. So a process that makes the skin feel warm, without actually heating the room can be more than 100% efficient by some measures. That's the idea here - it sends infrared heat to your skin, without (fully) heating the room in between. How effective this is is debatable, because it will only work for the side of you that is facing the machine. Ariel. (talk) 19:33, 22 November 2010 (UTC)[reply]

Can we really digest enzymes "whole," or are they broken down?

So much natural food these days touts the amount of enzymes they contain, and there is so much advice about eating raw or less-cooked foods so that we don't "destroy" the enzymes in the foods.

My question is: don't we digest all proteins into amino acids? And if so, does it matter if we're not consuming enzymes in our newest energy bars or raw foods, so long as we're consuming all the amino acids we need to consume?

I'm tried to research this, but the enzyme page doesn't talk about it, and searching "digest enzymes" on Google comes up with all the enzymes that are used in order to digest things. Is there any published information that deals with the question of whether proteins and enzymes are broken down by the digestive system before they are absorbed, and whether it actually makes any difference to your body whether the proteins have been denatured by cooking?

Thanks! — Sam 63.138.152.135 (talk) 17:51, 22 November 2010 (UTC)[reply]

From here, "The enzymes naturally present in food play an important role in digestion by helping to predigest the ingested food in the upper stomach". I used the Google search term "health benefits enzymes in food". The claim sounds plausible, although I'm not sure whether the enzymes in the food would still function in the acid environment of the stomach. Franamax (talk) 17:59, 22 November 2010 (UTC)[reply]

Hmmm, that's interesting, and is likely what people are talking about when they refer to the benefits of eating enzymes. That said, I can't find anything about "Enzyme University," and they don't have any sources, and they seem to be (passively) pushing supplements, so I'm not certain that they're an authoritative source. — Sam 63.138.152.135 (talk) 18:16, 22 November 2010 (UTC)[reply]

Yes, note I said "sounds" plausible, not "is" plausible. The top Google hits are all sites that think they have something that would be good for you if only you would buy it on a regular basis. They seem to support their claims with anecdotal evidence like "people have more energy" rather than double-blind studies or analysis of undigested nutritive content in the stool. And how would you do a double-blind study anyway? people can usually tell the difference between cooked and raw food. It's also possible that people feel more energy because they get tired of chewing the raw food and don't overstuff themselves as on tasty cooked food. The key measure is how active these enzymes are in the time between mastication of the food and saturation in stomach contents at pH 3.5-4.5, and how active the enzymes are at that pH level. You would likely have to find a much more detailed source to get that information. Franamax (talk) 20:51, 22 November 2010 (UTC)[reply]

I located an "Enzyme University" at [16]. Those responsible for the site clearly have a better knowledge of biology than the average quack, they cite patents that they've developed, and they appear to have some common basis with known products such as the lactase in Lactaid (see [17]). That said, I think that some caution is still required, because I don't think that they are marketing these things as drugs — they aren't mentioning safety and efficacy studies in these pages. The enzymes come from various odd sources, such as fungi, which might not ordinarily be eaten.

I would question the advisability, for example, of trying to replace pepsin activity lost by antacid with an outside enzyme that is active throughout the digestive tract [18], because I would worry that if it works, it might digest signalling proteins on the outside of cells in the gut and end up causing cancers. (Note that chronic use of antacid is definitely a problematic treatment in itself, addicting the user by increasing the stomach's acid production...) Human digestive proteins might have special safety features still unknown to science, and the acid requirement that the product circumvents is a basic safety shut-off that prevents them from damaging the duodenum and other portions of the gut. I should note that the Japanese have been the undisputed masters of coming up with various enzymes and artificially digested foods, but they also have such a high rate of stomach cancer that they have to get tested for it routinely like people get colonoscopies in the U.S.

I think that when you take something as a nutritional supplement rather than a drug, you should know for sure that it is something that has been consumed, either as food or as herbal medicine, for hundreds if not thousands of years. Otherwise you should demand clear, modern evidence of safety. Wnt (talk) 22:18, 22 November 2010 (UTC)[reply]

I agree that that site isn't entire quackery but: "The enzymes naturally present in food play an important role in digestion by helping to predigest the ingested food in the upper stomach. Cooking and processing destroys the natural enzymes found in foods. This places the full digestive burden on the body, which can cause extra stress on the digestive system, leading to incomplete digestion. As a result, vital nutrients may not be released from the food for assimilation by the body." is pretty flawed reasoning in my opinion. The reason we cook food is to break down plant cell walls and denature proteins so that our digestive enzymes can get at the molecules in food and digest them. Cooking will stop enzymes in naturally present in food from functioning, because they aren't enzymes that breakdown food as our enzymes do, since self-digesting isn't a good idea! Also plant have been fighting an evolutionary war against herbivores for millions of years, some of these defences, e.g. tannins and protease inhibitors, decrease the nutritional quality of plants to try and make herbivores eat other plants. It's unlikely that plants contain enzymes that would do the opposite and make themselves more digestible. We've evolved so that we are now fairly dependent on cooking to pre-digest food and I find it difficult to believe that our bodies need any help in digesting food, since the human gastrointestinal tract seems to do a pretty good job. There are of course far more enzymes in our guts than just the ones we produce due to the gut flora which digest many compounds which we can't digest ourselves. I'm always extremely sceptical of a company saying their product is useful, especially when what they are selling is basically just purified mould as they obviously have a conflict of interest. I can't find any independent studies about these supplements and their own research is based on a mechanical model (!) of the digestive system which also "set out to prove" they were effective - not a good way of doing science. SmartSE (talk) 14:57, 23 November 2010 (UTC)[reply]

I fully agree with you that the "we need enzymes in our food to aid digestion" notion is dubious, but I would hesitate to endorse the claim that the (major) purpose of cooking is to break down foods so that our own digestive enzymes can get at them. It's worth remembering that most cells do contain and can manufacture enzymes designed for partial or total self-digestion (see autophagy, apoptosis), but that their function is generally tightly regulated while the organism is still alive. As well, many foodstuffs fresh from the field are going to come with (often-undesirable) hangers-on: microbial contaminants that very much do want to digest their hosts, and secrete all the necessary enzymes to do so. Cooking, smoking, salting, drying, and pickling are all techniques that we use to denature proteins in order to discourage spoilage and inactivate pathogens that would otherwise very quickly pre-digest our foods — whether we wanted them to or not. (It goes without saying that these preservation techniques all also inhibit or inactivate the pathogens responsible for many diseases, which is another perq of a fully-cooked diet.)

Cooking also has the benefit (in many cases) of improving the taste of food. We have evolved to prefer sweeter, fattier foodstuffs (in order to get the calories we need and avoid starvation); cooking foods allows us to simulate these flavors and satisfy that evolutionary imperative (through caramelization and the Maillard reaction for sweetness, and the liquefaction of animal fats for that fatty mouthfeel). Cooking is (evolutionarily speaking) a very recent development; I would be extremely surprised to find that we were in any way dependent on it. TenOfAllTrades(talk) 15:43, 23 November 2010 (UTC)[reply]

Niña

Is there a name for a ship like the Niña that has two square-rigged sails in the front and then a lateen sail at the back? --The High Fin Sperm Whale 19:31, 22 November 2010 (UTC)[reply]

a Carrack. --Dr Dima (talk) 20:14, 22 November 2010 (UTC)[reply]

More recently, a barque. Mikenorton (talk) 23:00, 22 November 2010 (UTC)[reply]

Actually, barques have a gaff rigged sail at the back. Thanks anyway. --The High Fin Sperm Whale 23:31, 22 November 2010 (UTC)[reply]

The Nina and the Pinta were Caravels, originally caravela latinas (lateen sails on all three masts), but re-rigged as caravela redondas to follow the trade winds on the outward leg. And our article on the caravel could really use some work! --Stephan Schulz (talk) 07:52, 23 November 2010 (UTC)[reply]

Soda can exploding

I'm sorry for posting what is probably a trivial question, but why does a can of soda "explode" when opened after shaking? I've checked online, but all of the answers seem either very vague or plain wrong.

Correct me if I'm wrong, but as I understand it, carbonated liquid contains dissolved CO₂, like a solution. But some of the gas can form microbubbles (on their own accord or with the help of nucleating sites? I don't know...). When an unshaken can is opened, the liquid's pressure drops, and the microbubbles are then able to grow in size and leave the liquid. But why would shaking cause so many more bubbles to form? 70.52.44.192 (talk) 21:16, 22 November 2010 (UTC)[reply]

Part of it could be the same reason why bubbles appear in a shaken container of water, a simple mechanical action. But I do wonder if that's not all.

Because when I've shaken a sealed can of soda, it felt like the can became cooler... Wnt (talk) 21:43, 22 November 2010 (UTC)[reply]

There's an article at livescience. The fizz has to do with pressure equilibrium in that when you open the bottle, the pressure inside the bottle equalizes with whatever the pressure is outside the bottle. It may get slightly colder since the diluted carbon dioxide may take some energy to escape. See Solubility, Vapor pressure, partial pressure, Le Chatelier's principle, Pressure, and Boyle's law.Smallman12q (talk) 00:10, 23 November 2010 (UTC)[reply]

That explains the fizz, which I understand, but not why there's a massive increase in bubbles when the soda is shaken (I don't qualify the webpage's explanation, which says that shaking "adds the zing needed to unleash more tiny bubbles and add real splash to a celebration", as a good explanation).

It's very simple: your soda can is not full to the brim with soda. Rather, to keep the soda carbonated during storage, it contains a quantity of compressed gas. When you open a can of unshaken soda, the gas is at the top of the can and can escape freely. When you open a shaken can, the gas is mixed with the soda, and carries a good deal of the soda with it as it escapes. --Carnildo (talk) 00:25, 23 November 2010 (UTC)[reply]

That's certainly part of it, but it can't be all of it. Try it with a clear coke bottle. Shake the bottle vigorously and then open it as soon as the bubbles on the surface break. (Tap the bottle to speed this process.)

In any case, the volume of soda lost from a shaken can often far exceeds the volume of undissolved air in the can, which would not be possible in your theory. APL (talk) 03:38, 23 November 2010 (UTC)[reply]

Do you happen to know the answer to the question then? 70.52.44.192 (talk) 05:01, 23 November 2010 (UTC)[reply]

There is a link at the bottom of the carbonation article which has the answer you are looking for: Whirlpools in a soda pop Explains why shaken soda bottle will spray soda when opened. Vespine (talk) 05:56, 23 November 2010 (UTC)[reply]

Thanks Nil Einne (talk) 11:56, 23 November 2010 (UTC)[reply]

I forgot to mention, but the proper term for "fizzing" would be effervescence.Smallman12q (talk) 13:03, 23 November 2010 (UTC)[reply]

Does the above explain the fountain or geyser you get when you drop a polo mint into a large bottle of cola? Edit: this Diet Coke and Mentos eruption. 92.15.6.86 (talk) 13:10, 23 November 2010 (UTC)[reply]

The article that you linked contains the explanation and links to further research. TenOfAllTrades(talk) 14:52, 23 November 2010 (UTC)[reply]

Thanks. 70.52.44.192 (talk) 15:11, 23 November 2010 (UTC)[reply]

Flaxseed oil and autism

Is there is any real evidence for a connection between taking flaxseed soil, or omega 3 fatty acids and preventing autism? Or if autism has anything to do with dopamine. Thanks in advance. AdbMonkey (talk) 22:23, 22 November 2010 (UTC)[reply]

This is too controversial a subject, to attempt to answer. The best and most informative book I have ever come across is this one. [19] Borrow a copy from your local friendly medical library. You will not ever, get a sensible discussion about this subject here. It is too specialised.--Aspro (talk) 23:21, 22 November 2010 (UTC)[reply]

Ohhh. Thank you. AdbMonkey (talk) 05:34, 23 November 2010 (UTC)[reply]

See Autism#Causes for a general discussion. From the article, it looks like environmental factors are suspected by some to play a role, but genetics is definitely a factor. Paul (Stansifer) 13:27, 23 November 2010 (UTC)[reply]

Chemicals with humorous formulas

Is it possible for Argon Selenide to exist, and if so, would its chemical formula be ArSe? Also, is it possible to manipulate Arsenic sulfide under laboratory conditions so that its ions have equal charges, thereby changing its symbol to AsS?--99.251.211.17 (talk) 23:24, 22 November 2010 (UTC)[reply]

Would Arsole do? We have a subjective list of names at List of chemical compounds with unusual names...though not formulas.Smallman12q (talk) 23:59, 22 November 2010 (UTC)[reply]

That page has been nominated for deletion five times (some Wikipedians do not like articles of a humorous tone). If one of the AFDs ever succeeds, then the article is likely gone forever. If you feel it is an appropriate article, you might wish to watchlist it. I do not feel that this is canvassing, since there is presently no AFD in process. Edison (talk) 14:49, 23 November 2010 (UTC)[reply]

See this for a pretty comprehensive list, with pictures, structure and discussion. Ariel. (talk) 00:10, 23 November 2010 (UTC)[reply]

November 23

Inserting pipes when drilling for oil etc.

When drilling through soft material like mud, clay etc., presumably some sort of pipe has to be inserted into a bore hole as the drilling progresses to prevent the hole from simply collapsing? As the hole becomes longer and longer, how is that achieved? I would have thought it would soon become impossible to ram in new sections of pipe from the top of the borehole due to increasing friction? 86.173.36.159 (talk) 01:03, 23 November 2010 (UTC)[reply]

You'd be surprised. One thing they do is use a type of wet "mud" to lubricate the pipes, but basically, they just keep feeding pipe section after pipe section into the hole. At the end of the last pipe is the drill bit, which leads the way, but as the drill bit clears out enough rock to put in another pipe length, you just add another one. Wikipedia's article on Boring could use some work, but it is really not a complex process. --Jayron32 01:43, 23 November 2010 (UTC)[reply]

Do you know whether it is pressure from above that forces the pipe down or the weight of the pipe above? Intuitively I feel that as the hole gets deeper the gaps in the pipe form somewhere near the top, as the weight of the pipe forces it downward as the hole deepens. If this is this case you are not pushing the pipe all the way to the bottom but just filling a gap. -- Q Chris (talk) 09:13, 23 November 2010 (UTC)[reply]

To expand on Jayron32's explanation, during the drilling of a soft section, the collapse of the hole is prevented by keeping the pressure of the drilling mud high (although not too high or you may hydraulically fracture the well bore). To prevent later problems it is necessary to line the unstable sections of the wellbore using a metal liner known as a casing, after they've been drilled. Once the liner has been set, the borehole is drilled on using a smaller bit, leading to a hole that gets progressively narrower with depth. This starts at the surface by drilling out (or digging out in the case of onshore drilling) the uppermost tophole section and it is into this that the blowout preventer is set once it has been lined. Mikenorton (talk) 11:00, 23 November 2010 (UTC)[reply]

So, please excuse me for adding a related question or two, when the drill bit needs replacing the whole pipe has to be drawn up leaving the bored hole without any lining? With a long pipe, won't the friction become very high and make it difficult to turn the bit? 92.15.6.86 (talk) 13:18, 23 November 2010 (UTC)[reply]

Yes, a broken drill bit (or any other tool breakage in the hole) would be a "minor catastrophe" and could delay drilling operations anywhere from several hours to many days, depending on how long it takes to physically remove the drill string. If there are shards of metal in the bore, then a special grinding tool might need to be put down the hole to chomp up the metal fragments, increasing the costs and delays. And with a very long pipe, friction is very high - so powerful motors (topdrives) are used at the surface. In very "high-tech" drilling, a drillbit can be hydraulically actuated (meaning that instead of turning a drillstring, energy is conveyed by pumping pressurized fluid down the bore, and that pressure provides energy to a mechanical contraption that turns the bit). Alternatively, electric motors can be used down-hole. Neither of these solutions are standard-practice. You might find the Schlumberger Oilfield Glossary a helpful resource - bottomhole assembly has diagrams of the bit and related mechanical parts, and a bunch of links to related terms. Nimur (talk) 14:47, 23 November 2010 (UTC)[reply]

Altering the refractive index of air molecules. —without different gases or high temperatures

I have a question about how one can cause air to become "blurry" —AT ROOM TEMPERATURE— by altering the refractive index of the air molecules in a small area.

I know that a similar effect can be engineered through either raising the temperature tremendously (in a small part of a much larger area), or by pumping a gas with a different refractive index into said area (which is usually a sign to leave IMMEDIATELY!)

I'm curious, however, if one can create a similar effect using a certain frequency of radio waves (or microwaves, etc.), or some other extant technology. To wit, be able to change the refractive index of the air in small area, so that it would appear "blurry" from farther away.

Does any such technology exist? Thank you for reading this! Pine (talk) 01:17, 23 November 2010 (UTC)[reply]

No, because it would be the equivalent of heating the air. Physchim62 (talk) 01:35, 23 November 2010 (UTC)[reply]

What you would have to do is change the density of air in a local area; the only option you may have from the other two you have ruled out is to generate a pressure wave of some sort; unfortunatly in order to be detectable, such a pressure wave would probably be large enough to rip you to shreds; it also could not be localized, so it would travel out causing death and destruction. The problem with all of this is air is a gas; which means that any perterbations you make to density, locally, are VERY quickly dissipated. You just can't do what you are proposing. --Jayron32 01:39, 23 November 2010 (UTC)[reply]

Increasing the pressure in a local area will also inherently increase the temperature in the same local area. Googlemeister (talk) 16:20, 23 November 2010 (UTC)[reply]

Robert Boyle would like to have a word with you. DMacks (talk) 16:56, 23 November 2010 (UTC)[reply]

Just an off-the-cuff thought, but what about using two beams of microwaves of differing frequency which, intersecting at a localised spot, would yield a beat frequency (of, say, 2.54 GHz) that would have a strong dielectric heating effect on the moisture content of the air? 87.81.230.195 (talk) 12:52, 23 November 2010 (UTC)[reply]

Wouldn't the microwave heating cause the air heated to be at a higher temperature from the rest of the room? I suppose a technically correct but unhelpful response would be to have a room where the temp was way different from 70 F but have the "blurry air" be at exactly 70, by heating or cooling. If some air were compressed, then released through a slot in the viewing area, the pressure drop would cause the air to suddenly cool, perhaps with moisture condensing. This is done in some dew point measurement apparatus. Again, there is a temperature change, but the "delivery temperature" of the air could be the required 70 F. Edison (talk) 14:46, 23 November 2010 (UTC)[reply]

What you see when the apparent refractive index of air changes with temperature is an effect of changing air density. (Denser, cooler air contains more molecules per cubic centimeter and has a higher refractive index than less-dense, warmer air.) As suggested, in principle one could use focused infrared or microwave radiation of an appropriate frequency to locally heat a small volume of air from a distance; as it expanded its refractive index would fall, but it would fail your criterion of maintaining 'room temperature' conditions.

You can temporarily increase the density (and refractive index) of air by passing a shock wave through it. (See, for example, this YouTube video; between roughly 0:02 and 0:04 you can see an expanding ring of visible distortion around the exploding automobile.) In principle, I wonder if you might accomplish something similar (and somewhat less destructive) by setting up some kind of standing acoustic wave. It would be awfully loud, however — sound pressures comparable to atmospheric pressure are up around 195 decibels. TenOfAllTrades(talk) 15:06, 23 November 2010 (UTC)[reply]

Or more destructively, with a nuclear weapon. See [20] for some info. DMacks (talk) 15:32, 23 November 2010 (UTC)[reply]

Shock waves (especially nuclear ones) also cause temperature to rise. Googlemeister (talk) 16:21, 23 November 2010 (UTC)[reply]

equilibrium constants units

in formation of ammonia from nitrogen and hydrogen

      N₂ + 3H₂ <--> 2NH₃

units of k_c doesnt cancel. i read about unit of k_c on net and came to know that k_c do have units in the form of (mol/L)⁺ⁿ or(mol/L)^-n.

if it so why dont we write the units , neither it is wrong to not write the units why?

is there any other logic behind not writing units of equilibrium constant except that it is the ratio of same things ie concentrations --Myownid420 (talk) 01:48, 23 November 2010 (UTC)[reply]

Assuming you mean K_c, which is the equilibrium constant (k is the rate constant). Equilibrium constant is ALWAYS a unitless value. When you calculate K_c or K_p, you are using standins for the real value. The equilibrium constant is dependant on a value called activity, which, by a bit of circular logic, is the value which determines how fast a substance reacts in a chemical reaction (and thus, its effect on rate and equilibrium). Activity is unitless, so equilibrium constant is also unitless. It is actually an unmeasurable value, however for most measurements, the ratio of concentrations is close to equal to the ratio of activities so we use concentration as a stand in for it. K_c indicates that concentration is used as the standin for activity, and works for all solutions, and for all gases in a constant pressure environment. In a constant volume environment, we use K_p instead of K_c, which uses pressure values instead of concentrations (normalized to atmospheres). The article Equilibrium constant discusses most of this. In summation, equilibrium constant is always unitless, even though you use "concentration" as an approximation to calculate it; thus if you actually do the dimensional analysis on your calculation, you will often get a "unit" for the equilibrium constant. Ignore that. Equilbrium constant is unitless, period. --Jayron32 04:55, 23 November 2010 (UTC)[reply]

The logic behind the fact that the equilibrium constant is unitless comes from the fact that it is related to the Gibbs free energy change for the reaction by ΔG = −RTlnK. You can't take a logarithm of a quantity that has units. Physchim62 (talk) 12:54, 23 November 2010 (UTC)[reply]

Human-caused / Life-caused

Non-scientist here, so forgive the obliviousness. I know the term used for human-caused (or man-caused) is "Anthropogenic". Two questions: 1) Is there a term used for something caused collectively by life in general? 2) What are some examples of this? I know early plant life had something to do with the early atmosphere changing to something we can breathe today -- could that be considered "life-caused" (or that might be too specific of an example, so it would be limited to "plant-caused" I guess)? If you can point me to a WP article, I would be appreciative. Rgrds. (dynamic IP, may change) --64.85.215.81 (talk) 07:27, 23 November 2010 (UTC)[reply]

Biogenic means "created by life". So, yeah: Biogenic oxygen, biogenic methane, etc. I don't know of an equivalent for "plant-caused". I was thinking floragenic, but that seems to be a company. Someguy1221 (talk) 07:32, 23 November 2010 (UTC)[reply]

I would probably use "phytogenic" for plant-caused. --Jayron32 07:36, 23 November 2010 (UTC)[reply]

But it's sort of a moot point, since the initial oxygenation of the atmosphere was probably caused by cyanobacteria billions of years before there were any plants. Looie496 (talk) 07:46, 23 November 2010 (UTC)[reply]

That would make it "bacteriogenic", which means "bacteria-caused".

Another word pertaining to things caused by life is "zoogenic", which means "animal-caused". Red Act (talk) 08:17, 23 November 2010 (UTC)[reply]

"Biogenic" appears to be what I was looking for and seems to point me in the direction I was hoping, thanks Someguy. ("Zoogenic" is too specific-- I was looking for an all-inclusive term.) Rgrds. --64.85.215.81 (talk) 08:45, 23 November 2010 (UTC)[reply]

Mass flow rate of Refrigerant

what relation can I use to solve mass flow rate problems when given temperatur, pressure and specific volume —Preceding unsigned comment added by 41.221.209.6 (talk) 10:08, 23 November 2010 (UTC)[reply]

what appropriate relations can i use to determine changes in potential and kinetic energies and enthalpy when given temperature, pressure and mass flow rate —Preceding unsigned comment added by 41.221.209.6 (talk) 10:13, 23 November 2010 (UTC)[reply]

In the absence of an empirical equation of state, you can use the Ideal Gas Law or its more accurate form, the Van der Waals equation, to model gas volumes and pressures. Use the assumption that the flow is adiabatic, unless you know there is a pump, heat source, or heat sink. Conserve energy, calculate temperature using a specific heat value for the gas, and assume a phase change at the boiling-point temperature. You can apply the enthalpy of vaporization to conserve heat across the phase-transition. You can assume the fluid is incompressible when in liquid form, unless you have parameters for compressibility. Nimur (talk) 15:39, 23 November 2010 (UTC)[reply]

The Doppler Effect

If one were using the Doppler Effect to determine the speed of an object, how would the angle at which the object is moving in relation to the observer effect the determination? Can the speed still be accurately calculated even if the moving object isn't moving directly away from or at the stationary observer?--160.36.38.212 (talk) 15:41, 23 November 2010 (UTC)[reply]

For the acoustic Doppler effect, only the radial speed matters, so you just add a

\cos \theta

and you're done. For the relativistic Doppler effect, that effect occurs but there is also the transverse Doppler effect (very small at, say, highway speeds). In either case, there's only one value you measure (frequency change), so you can't determine two values (speed and direction). But if you know one of those two, and measure the frequency change, you can get the other. --Tardis (talk) 16:23, 23 November 2010 (UTC)[reply]

What about in the instance of, let's say, determining the speed of a baseball with a radar gun. Since we are trying to determine the speed and can't be sure about the angle of observation, how accurately can the frequency change be used to determine the speed of the baseball?--160.36.38.212 (talk) 16:47, 23 November 2010 (UTC)[reply]

You'll never overestimate its speed (because the error is that if it's moving across your line of fire you'll get 0). The ratio of its true speed to your measurement is the secant of the angle its velocity makes with your line of fire. If you assume that that's, say, no more than 10°, then the true speed is no more than 1.5% above your measurement. But look at the graph of that function to see what happens if the direction isn't known at all. --Tardis (talk) 16:55, 23 November 2010 (UTC)[reply]

I see, not much to worry about as far as error is concerned. Thanks for your help.--160.36.38.212 (talk) 17:03, 23 November 2010 (UTC)[reply]

Does the sun drop faster on the equator?

As headline. —Preceding unsigned comment added by 84.12.125.33 (talk) 16:02, 23 November 2010 (UTC)[reply]

Yes. On visiting Texas from the UK I just couldn't get used to not having a twilight, its light one minute and dark 10 minutes later. The movement of the sun across the sky is also consistent, they can say "15 minutes for each finger-width (at arms length) the sun is above the horizon. In the UK it varies with the time of year but could be hours at midsummer!

The sun moves at that angular velocity all the time and everywhere. But in what direction? If it's moving "horizontally", it's not dropping as fast even though it's moving as fast. The sun typically moves more vertically near the equator and more horizontally near the poles, so there's the effect you're thinking of. However, it varies over the year and day even at one location; clearly at noon the sun is neither rising nor dropping. At the solstices, the sun crosses the horizon closest to one of its stationary points in altitude, so twilight is exaggerated there. During the midnight sun, of course, the sun never (fully) drops even over many days! --Tardis (talk) 16:20, 23 November 2010 (UTC)[reply]