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April 4
Weighing less than nothing
In Don Rosa's comic The Universal Solvent, the Ducks travel to the centre of the Earth in a vertical shaft. At one point, they notice gravity has started working in the opposite direction - it pulls them up instead of down. According to the story, this is because the majority of the Earth's mass is now above the Ducks instead of below them. Assuming travelling so far down were possible in the first place, is this really what would happen? JIP | Talk 11:20, 4 April 2022 (UTC)
- Not really. At any point nearer to the top of the shaft than the centre of the Earth there would be more mass below. Think of the other hemisphere for a start. Only at the centre would gravity cease pulling down. Still, this is science fiction so interposing reality isn't always helpful! Martin of Sheffield (talk) 11:29, 4 April 2022 (UTC)
- Thanks for the answer. It kind of figures. The further down they go, the less they weigh - but their weight stays positive the whole time, as the other hemisphere is still below them. I wonder how Rosa, who has a Bachelor's Degree in civil engineering, failed to spot this. JIP | Talk 11:43, 4 April 2022 (UTC)
- Shell theorem says: "If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell." They would simply feel the gravity of the sphere closer to the centre while everything else would cancel out. I don't think the author actually thought the story was scientifically accurate. It's a Donald Duck story, not hard science fiction. PrimeHunter (talk) 11:56, 4 April 2022 (UTC)
- It's actually quite a bit more complex than that once you factor in the changing density. See Gravity of Earth for the full, unexpurgated, gory details. Martin of Sheffield (talk) 12:05, 4 April 2022 (UTC)
- Density variation as a function of depth is irrelevant to the shell theorem; are you saying density varies significantly in ways not spherically symmetric? —Tamfang (talk) 03:35, 6 April 2022 (UTC)
- But the earth isn't a shell. Using shell theory a body 100 km down a hole experiences no gravitation attraction from the outermost 100 km of the earth's mass. The inner 6,361 km are still pulling down. Repeat the previous two sentences ad nauseum all the way down. What shell theory does tell us is that a problem like this is best handled by modelling the earth as an onion. As you pass through each layer you can discard it and only consider the layers below. Now consider what happens if our onion is composed of differing densities. The outermost layer might comprise only 1% of the mass, so the pull of gravity of the rest is 0.99g (I'm taking serious liberties with the "r2" term of the attraction). Now if the next layer actually held 10% of the onion's mass, then once that was passed through the pull would be 0.89g. Let's assume 5% for the next layer: 0.84% and so forth until we pass through the last layer at the heart of the onion and the pull disappears. I refer you back to Gravity of Earth for a more theoretical and less culinary explanation. Martin of Sheffield (talk) 08:12, 6 April 2022 (UTC)
- I'm still mystified as to what you think you're contradicting. Whatever. —Tamfang (talk) 04:19, 10 April 2022 (UTC)
- This statement: "Density variation as a function of depth is irrelevant to the shell theorem" Martin of Sheffield (talk) 10:45, 10 April 2022 (UTC)
- I'm still mystified as to what you think you're contradicting. Whatever. —Tamfang (talk) 04:19, 10 April 2022 (UTC)
- But the earth isn't a shell. Using shell theory a body 100 km down a hole experiences no gravitation attraction from the outermost 100 km of the earth's mass. The inner 6,361 km are still pulling down. Repeat the previous two sentences ad nauseum all the way down. What shell theory does tell us is that a problem like this is best handled by modelling the earth as an onion. As you pass through each layer you can discard it and only consider the layers below. Now consider what happens if our onion is composed of differing densities. The outermost layer might comprise only 1% of the mass, so the pull of gravity of the rest is 0.99g (I'm taking serious liberties with the "r2" term of the attraction). Now if the next layer actually held 10% of the onion's mass, then once that was passed through the pull would be 0.89g. Let's assume 5% for the next layer: 0.84% and so forth until we pass through the last layer at the heart of the onion and the pull disappears. I refer you back to Gravity of Earth for a more theoretical and less culinary explanation. Martin of Sheffield (talk) 08:12, 6 April 2022 (UTC)
- Density variation as a function of depth is irrelevant to the shell theorem; are you saying density varies significantly in ways not spherically symmetric? —Tamfang (talk) 03:35, 6 April 2022 (UTC)
- Unless they mean they've already passed the center of the Earth within the shaft and was on the "opposite" side already, where "up" instead of "down" is pointing toward the center. GeorgiaDC (talk) 18:44, 4 April 2022 (UTC)
The shell theorem applies to shells (such as a Dyson sphere). However, Earth is a ball. 78.1.172.11 (talk) 21:11, 4 April 2022 (UTC)- Edit: Nvm I've read PrimeHunter's comment again and I see we don't actually disagree. 78.1.172.11 (talk) 21:20, 4 April 2022 (UTC)
- It's actually quite a bit more complex than that once you factor in the changing density. See Gravity of Earth for the full, unexpurgated, gory details. Martin of Sheffield (talk) 12:05, 4 April 2022 (UTC)
- Shell theorem says: "If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell." They would simply feel the gravity of the sphere closer to the centre while everything else would cancel out. I don't think the author actually thought the story was scientifically accurate. It's a Donald Duck story, not hard science fiction. PrimeHunter (talk) 11:56, 4 April 2022 (UTC)
- Thanks for the answer. It kind of figures. The further down they go, the less they weigh - but their weight stays positive the whole time, as the other hemisphere is still below them. I wonder how Rosa, who has a Bachelor's Degree in civil engineering, failed to spot this. JIP | Talk 11:43, 4 April 2022 (UTC)
- Assuming a spherically symmetrical, solid
cowEarth, if you dug a giant whole through the dead center of the earth, and out of the other side, and then jumped in the hole, ignoring air-friction, you would be a harmonic oscillator, barely reaching the other side (your center of mass would clear the surface of the other side of the earth to the exact height your center of mass currently is on this side of the other), bouncing from one side to the other. This video from Science Asylum does a good job explaining the physics of the situation. --Jayron32 12:28, 4 April 2022 (UTC)- If you instantly teleported out a hole and turned real physics back on what would happen? Sagittarian Milky Way (talk) 14:47, 4 April 2022 (UTC)
- If you returned air resistance into the picture, you would behave like a dampened harmonic oscillator, gradually slowing down and your return points would get further and further from either hole, until you came to a gentle rest floating at the exact center of the earth. --Jayron32 15:12, 4 April 2022 (UTC)
- Also, if you returned the Earth's actual gravitational irregularities, I doubt you'd ever make it to the center in one piece. You're going to be moving at terminal velocity pretty quickly, and given the irregularities in gravity on your long trip, you'll probably be pinging off the sides of your shaft rather violently. And, since part of the Earth are molten (liquid) rock and metal, your hole would very quickly fill in, and then exposed to the air above it, would cool down pretty quickly. You wouldn't make it past the Asthenosphere, which means you only would fall for a few kilometers before smacking into hot, partially solidified magma. --Jayron32 16:33, 4 April 2022 (UTC)
- Also, also, as I look at some information in Kola Superdeep Borehole, even within the crust, when you get down past about 10 km, the temperature is hot enough to boil water. You'd probably bake to death before you even hit the bottom. --Jayron32 16:36, 4 April 2022 (UTC)
- Also, also, also, consider that Caisson disease, aka "the Bends", is a real possibility at depths of that level, given the massive increase in air pressure you are likely to experience. --Jayron32 16:52, 4 April 2022 (UTC)
- Also, also, as I look at some information in Kola Superdeep Borehole, even within the crust, when you get down past about 10 km, the temperature is hot enough to boil water. You'd probably bake to death before you even hit the bottom. --Jayron32 16:36, 4 April 2022 (UTC)
- Also, if you returned the Earth's actual gravitational irregularities, I doubt you'd ever make it to the center in one piece. You're going to be moving at terminal velocity pretty quickly, and given the irregularities in gravity on your long trip, you'll probably be pinging off the sides of your shaft rather violently. And, since part of the Earth are molten (liquid) rock and metal, your hole would very quickly fill in, and then exposed to the air above it, would cool down pretty quickly. You wouldn't make it past the Asthenosphere, which means you only would fall for a few kilometers before smacking into hot, partially solidified magma. --Jayron32 16:33, 4 April 2022 (UTC)
- If you returned air resistance into the picture, you would behave like a dampened harmonic oscillator, gradually slowing down and your return points would get further and further from either hole, until you came to a gentle rest floating at the exact center of the earth. --Jayron32 15:12, 4 April 2022 (UTC)
- If you instantly teleported out a hole and turned real physics back on what would happen? Sagittarian Milky Way (talk) 14:47, 4 April 2022 (UTC)
- a decompression accident would be a problem on the way back up, oxygen toxicity due to its high partial pressure would get you on the way down. — Preceding unsigned comment added by 2A01:E34:EF5E:4640:1961:F229:5025:A78 (talk) 11:33, 5 April 2022 (UTC)
- I always thought that (assuming you didn't fill most of it with air compressed to specific gravities >>1) some spectacular implosion at at least c. the speed of sound of each Earth layer would happen, as most of the cylinder surface would have pressure of 1 to 3.6 million atmospheres. How much magma gas would come out of solution from this brief exposure to the hole's lower pressure, and how dramatic of an eruption that could cause I don't know. Would magma not reach the entrance because mantle is denser than continental crust or would it barely leak out or spurt 100 meters high or something stronger and less Hawaiian? How big or small would the earthquake be? How far from the hole would crack? And if it's wide enough for bouncing off the edge to not come first then how far could you fall before things get bad? Sagittarian Milky Way (talk) 23:32, 4 April 2022 (UTC)
- Even worse than gravitational irregularities is the Coriolis force sending you off course. Unless you travel exactly Pole to Pole. PiusImpavidus (talk) 10:15, 5 April 2022 (UTC)
- Ah yes, that too. So many harms from one hole. Sagittarian Milky Way (talk) 13:29, 5 April 2022 (UTC)
- You've obviously met my first girlfriend. --Jayron32 10:55, 6 April 2022 (UTC)
- Ah yes, that too. So many harms from one hole. Sagittarian Milky Way (talk) 13:29, 5 April 2022 (UTC)
- Even worse than gravitational irregularities is the Coriolis force sending you off course. Unless you travel exactly Pole to Pole. PiusImpavidus (talk) 10:15, 5 April 2022 (UTC)
- And that's in a relatively cold section of the crust. In other parts of the world, e.g. the KTB boreholes in Bavaria, the temperature at that depth is so hot that the rock is liquid, and it's not physically possible to drill any further. 51.155.110.141 (talk) 20:36, 5 April 2022 (UTC)
- The comic seems to say that their weight is decreasing (again because of the shell theorem) and that they will soon be weightless (at the center of the Earth), not that gravity is pulling them in the opposite direction. --Amble (talk) 19:18, 4 April 2022 (UTC)
- Did the comic literally say they weigh "less than nothing"? --←Baseball Bugs What's up, Doc? carrots→ 22:19, 4 April 2022 (UTC)
- It's often asserted that a person jumping into a hole more than 35 miles deep would not hit the bottom, but would float at that depth due to the density of the air there. But that assumes air is an ideal gas, as well as neglecting the increasing temperature and the shell theorem. It would be interesting to know the true person-floating depth if it exists. catslash (talk) 22:57, 5 April 2022 (UTC)
- According to the Internet air at the temperature of 35 miles would have to be past the right edge of the chart (10,000 bar) to float a person. Sagittarian Milky Way (talk) 02:50, 6 April 2022 (UTC)
- The effect of changing gravity would cancel out; what matters is the relative densities of the falling body and the medium. —Tamfang (talk) 04:25, 10 April 2022 (UTC)
- It's often asserted that a person jumping into a hole more than 35 miles deep would not hit the bottom, but would float at that depth due to the density of the air there. But that assumes air is an ideal gas, as well as neglecting the increasing temperature and the shell theorem. It would be interesting to know the true person-floating depth if it exists. catslash (talk) 22:57, 5 April 2022 (UTC)
Nitrogen triiodide; is the purple cloud toxic?
I asked this question at Talk:Nitrogen triiodide#Toxicity?, but I thought that it was worth reposting here.
Youtube is full of videos showing Nitrogen triiodide exploding with a purple cloud. Our article on Nitrogen triiodide should say whether that purple cloud is toxic.
The first paragraph of the article says "...releasing a purple cloud of iodine vapor", but later the article says
"The dry material is a contact explosive, decomposing approximately as follows:
8 NI3 · NH3 → 5 N2 + 6 NH4I + 9 I2"
So which is it? "Iodine vapor"? Or is the purple cloud a combination of Ammonia, Ammonium iodide, and Elemental Iodine (I2)? And is the purple cloud an actual vapor or is it a cloud of solid (liquid?) Iodine particles? Iodine boils at 113.7 °C and Ammonium iodide boils at 551 °C so (based upon my extensive education (which consists of getting a C- in high-school chemistry)) it seems that only the Ammonia would be a vapor, while the purple stuff would be an aerosol.
Could someone who understands chemistry please write up a paragraph for that article with citations explaining what is in that purple cloud and whether it is toxic? 2600:1700:D0A0:21B0:5A9:3B45:A28C:DE50 (talk) 18:54, 4 April 2022 (UTC)
- Iodine is a solid under standard conditions, but it sublimes under these conditions to produce some violet iodine vapour - see Iodine#Properties. Mikenorton (talk) 22:01, 4 April 2022 (UTC)
- Ammonia is a colourless gas. I could find no mention of the colour of gaseous ammonium iodide, which suggests it is not remarkable. So the markedly violet colour of the cloud is rather likely purely due to the presence of iodine in the gas state. --Lambiam 22:54, 4 April 2022 (UTC)
- The pictures at [1] look a lot more like a smoke or a dust than a gas to me. Could it be mostly a fine powder of iodine crystals? Of course the crystals create the gas, so there would have to be be gas as well. Either way, could it be that all the article needs is a "for more information see Iodine#Toxicity" link? 2600:1700:D0A0:21B0:5A9:3B45:A28C:DE50 (talk) 02:03, 5 April 2022 (UTC)
- If you watch the video, you can see that the cloud looks like vapour more than dust once it starts to disperse. --Lambiam 09:35, 5 April 2022 (UTC)
- The pictures at [1] look a lot more like a smoke or a dust than a gas to me. Could it be mostly a fine powder of iodine crystals? Of course the crystals create the gas, so there would have to be be gas as well. Either way, could it be that all the article needs is a "for more information see Iodine#Toxicity" link? 2600:1700:D0A0:21B0:5A9:3B45:A28C:DE50 (talk) 02:03, 5 April 2022 (UTC)
- Ammonia is a colourless gas. I could find no mention of the colour of gaseous ammonium iodide, which suggests it is not remarkable. So the markedly violet colour of the cloud is rather likely purely due to the presence of iodine in the gas state. --Lambiam 22:54, 4 April 2022 (UTC)
- What do we do when we see a chemical explosion and a cloud and we're not sure how hazardous it is?
- Well, it turns out that this is very common!
- I pull out my handy paper copy of the Department of Transportation Emergency Response Guidebook (2020 edition, available for free in PDF format and probably available for free at your local US fire department!)
- This is a guidebook for first responders "during the initial phase of a transportation incident involving hazardous materials/dangerous goods". How often do we find a weird chemical with unknown safety parameters? Too often.
- Page 1 of the book is the decision-making flow-chart. Follow along - the flow chart will direct you to Guide 111: "Unidentified Cargo" - and lists all the potential hazards. "Evacuation: Immediate precautionary measure - Isolate spill or leak area for at least 100 meters (330 feet) in all directions."
- If you say the dry material is a "contact explosive": That'd be ... Guide 112 - "EVACUATION - Immediate precautionary measure. Isolate spill or leak area immediately for at least 500 meters (1/3 mile) in all directions. MAY EXPLODE AND THROW FRAGMENTS 1600 METERS (1 MILE) OR MORE IF FIRE REACHES CARGO."
- Later when we have time for it, we can pull out the MSDS and figure out how to comply with the necessary local and national chemical handling guidelines. In the short term, don't play with chemicals if you aren't trained to deal with them and their hazards.
- Meanwhile - if you think people who are skilled and trained in chemistry have the time to micro-analyze the safety concerns of every possible type of chemical they might encounter, and evaluate whether this particular purple cloud matches that particular textbook example, and then clearly cite the details on a free encyclopedia, you are grossly misinformed. Don't mess around with chemicals, don't mess around with explosives, and if you're not sure if it's hazardous, assume it is hazardous and don't mess around with it.
- Nimur (talk) 23:18, 4 April 2022 (UTC)
- Got it. You have chosen to criticize someone for daring to ask that a Wikipedia article on Nitrogen triiodide not give the reader two different answers for what it turns into when it explodes.
- And we shouldn't have information such as Iodine#Toxicity, Phosphorus#Precautions, Organic mercury#Toxicity and safety, Thermite#Hazards, or any of the other thousands of places where Wikipedia talks about the hazards of a substance. Got it.
- Please go away. Your answer was not helpful. 2600:1700:D0A0:21B0:5A9:3B45:A28C:DE50 (talk) 02:03, 5 April 2022 (UTC)
- Telling an editor with over 16 years experience here to "go away" is not likely to get you anywhere. --←Baseball Bugs What's up, Doc? carrots→ 02:59, 5 April 2022 (UTC)
- Let me apologize for my harsh tone.
- My key point remains valid: even if we have reputable sources that describe the textbook-example chemistry reaction, there is no reasonable way to know if the particular instance in a particular demonstration is safe. The correct procedure - a procedure that I didn't invent, but cited from a reliable source!: assume the material is hazardous until other facts are known. Does this procedure apply in every context? Perhaps not. But, when we discuss hazards with anonymous participants on the internet - I feel that it is great to start from a position of zero assumptions, and aim for maximum safety. Please keep in mind that these discussions are archived and searchable - somebody in the future (who is not presently a participant) may find this discussion via a web-search; they might perform that search at some future time when they are seeing a strange and unidentified purple cloud and seeking information about it (...Imagine, under what circumstances, this Wikipedia archive might come up on some future human's screen - perhaps when a user types "is the purple cloud toxic" into a general-purpose internet search engine - or, speaks to some future voice-activated artificially-intelligent software-agent that was trained on a free database of information - and magically finds the answer on Wikipedia...! Ask yourself, profoundly, whether I am speaking in the hypothetical.) Other readers might have less experience than the person who asked the original question. This is just one motivating reason I write from such a cautious perspective: "assume it's a hazard" is fine advice, and probably more future-proof than a blanket assurance derived from one particular case.
- It doesn't hurt to throw in a few "don't play with matches" every now and then - and I never mean to do so as an affront or insult to any particular individual - I only say these things for completeness.
- Nimur (talk) 17:24, 5 April 2022 (UTC)
- Telling an editor with over 16 years experience here to "go away" is not likely to get you anywhere. --←Baseball Bugs What's up, Doc? carrots→ 02:59, 5 April 2022 (UTC)
[2] says: "Although it is possible to make pure nitrogen triiodide (which is a red solid) by reacting boron nitride with iodine fluoride, the substance that is actually produced by the reaction I used to perform is more complex. It is an adduct of nitrogen triiodide and ammonia. An adduct is a not-quite compound. The two molecules, nitrogen triiodine and ammonia, are attracted to each other to form a new molecule that is represented by the usual NI3 and NH3 with a dot in between. Initially on formation there are five ammonias to each nitrogen triiodide, but as the substance dries it gives off ammonia to end up with a one to one adduct. What has formed in the solid are chains of nitrogen triiodide with ammonia molecules linking them."
I found the idea of an Adduct interesting. Is this worth covering in the article? 2600:1700:D0A0:21B0:5A9:3B45:A28C:DE50 (talk) 02:03, 5 April 2022 (UTC)
- It's just iodine, from personal experience making the stuff without it being held in ammonia. Abductive (reasoning) 09:36, 5 April 2022 (UTC)
- The ammonia and iodine gases that are released are toxic. The ammonia air level reported to be considered immediately dangerous to life or health is 300 ppm.[3] The iodine air level reported to be considered immediately dangerous to life or health is 2 ppm.[4][5] --Lambiam 10:01, 5 April 2022 (UTC)
Special relativity : 3.4 Relativity without the second postulate
With only the principle of relativity [1], what about the CERN collider [2] in which proton beams traveling in opposite directions collide at 99.9999991% of c? If you consider each beam as a frame of reference, for each, the other arrives at a speed close to 2c. And in case each beam is 50% of c you get a Lorentz factor = 1/0, so you can't apply the Lorentz transformation, only the Galilean transformation, right? In fact Einstein's relativity and the Lorentz transformation apply only to electromagnetic rays, not to inertial objects themselves with not invariant speed , but only to the reflection of electromagnetic rays on inertial objects! Didn't Einstein confuse what we see with what is? Malypaet (talk) 23:21, 4 April 2022 (UTC)
- Hasn't the constancy of the speed of light been consistently experimentally confirmed? --Lambiam 09:25, 5 April 2022 (UTC)
- It is important not note that speeds do not add the same way as you think they do. Which is to say that if you have two objects moving relative to each other, the velocity of object 2 from the frame of reference of object 1 is not simply to add the velocities. It is approximately that at sufficiently low speeds (which is to say, that WELL within significant figures, you get the same answer whether you use the Newtonian calculation or the Einsteinian one). At speeds that close to the speed of light, you need to use the correct equation to calculate relative speeds, which is the Velocity-addition formula,
- where c=1 (a common way to write these equations to simplify the numbers. This formula will simplify to simple addition of velocities for sufficiently small values of V. Your mistake is assuming that two particles colliding will have a combined velocity of 2c. They don't. They have a combined velocity of a very very very small amount less than c, using the Einsteinian velocity addition formula I cited above. Also, we know they do this because we did the experiments many many many times, and the results match the predictions of the theory.--Jayron32 12:14, 5 April 2022 (UTC)
- There is an error in this formula. The correct one is
- I corrected the error in the article as well. Ruslik_Zero 20:02, 6 April 2022 (UTC)
- There is an error in this formula. The correct one is
- The Einsteinian calculation is based on the Lorentz transformation, itself based on the constancy of the speed of light, for the displacement of light. It therefore makes no sense to use this calculation for objects with non-constant speed. You explain how the addition of kinetic energies? Malypaet (talk) 12:33, 5 April 2022 (UTC)
- I don't care if it doesn't make sense to you. Your ability to understand it doesn't matter here. The math derives naturally from the postulates, and is confirmed by experiment. There are formulations one could do that do not rely on the inability of massed particles from exceeding the constant speed of light, but those formulations are bonkers... It's the equivalent of the silly epicycle nonsense used to maintain a geocentric frame of reference in astronomy. The math works, but putting the Earth in orbit around the sun works better. In the same way, the constant speed of light, and the fact that objects with mass cannot exceed it works better than the alternative. It is the "heliocentric model" for relativity. --Jayron32 13:11, 5 April 2022 (UTC)
- You haven't answered my question about adding kinetic energies. At the LHC, a proton at a speed close to c collides with an immobile target with an energy of 7Tev, it collides another proton in the opposite direction with an energy of 14Tev. So if in these two cases we have a relative speed close to c (and not 2c for the collision), how to explain the difference in energy of the impact? Malypaet (talk) 21:04, 5 April 2022 (UTC)
- Imagine it getting heavier the closer it is to c so pushes have diminishing returns. Sagittarian Milky Way (talk) 02:42, 6 April 2022 (UTC)
- "getting heavier the closer it is to c", but with which formula ? with newton's mechanics it's simple and here the predictions give the right result, the addition of the speeds gives the right value for the energy, right? Malypaet (talk) 12:18, 7 April 2022 (UTC)
- Imagine it getting heavier the closer it is to c so pushes have diminishing returns. Sagittarian Milky Way (talk) 02:42, 6 April 2022 (UTC)
- You haven't answered my question about adding kinetic energies. At the LHC, a proton at a speed close to c collides with an immobile target with an energy of 7Tev, it collides another proton in the opposite direction with an energy of 14Tev. So if in these two cases we have a relative speed close to c (and not 2c for the collision), how to explain the difference in energy of the impact? Malypaet (talk) 21:04, 5 April 2022 (UTC)
- I don't care if it doesn't make sense to you. Your ability to understand it doesn't matter here. The math derives naturally from the postulates, and is confirmed by experiment. There are formulations one could do that do not rely on the inability of massed particles from exceeding the constant speed of light, but those formulations are bonkers... It's the equivalent of the silly epicycle nonsense used to maintain a geocentric frame of reference in astronomy. The math works, but putting the Earth in orbit around the sun works better. In the same way, the constant speed of light, and the fact that objects with mass cannot exceed it works better than the alternative. It is the "heliocentric model" for relativity. --Jayron32 13:11, 5 April 2022 (UTC)
- I do not dispute the constancy of the speed of light, but how one can extrapolate the Lorentz transformations applied to a displacement of a light ray between mirrors and based on the constancy of the speed of light, whatever the referential , therefore to an inertial object whose speed of movement is not invariable, whatever the frame of reference? A mirage in the desert doesn't show reality, does it? As according to Einstein's principle of relativity, there is no absolute frame of reference, with the colliders of Protons, we have very close to 2c for the shock of the protons, moreover it is reported there that during the shock the kinetic energies add up. So what do you have as a more serious explanation? Malypaet (talk) 12:19, 5 April 2022 (UTC)
- I don't know what else to do than to refer you again to the article Velocity-addition formula, which does ALL of the math, in full three dimensions, in ALL of the gory details. If you start with the constancy of the speed of light, and that nothing with mass can exceed that speed, then you do the math, this is the results you get. Experiments confirm that math. If you, as an observer standing next to the particle accelerator measure the two particles velocity to each be at 99.9999991% of c, they do NOT collide with a combined speed of twice that value; they collide with a speed of much closer to but not over c. The measured energies from the collision confirm this predicted calculation. I don't know what else to say except that the math is in that article and experiments do confirm it. Einstein started with the postulates of invariant speed of light and that nothing with a mass could reach that speed, and this is the equation that falls out of that math. --Jayron32 12:28, 5 April 2022 (UTC)
- All this became apparent 300 years before the formulae were developed to explain it. When the telescope was invented, astronomers knew the exact positions of Jupiter's satellites from occultation observations. Knowing their velocities and direction of travel they checked the observed times of the occultations against the times they would have been seen allowing for the effect the velocities would have had on the time taken for the light to arrive. They were astonished to find that they had no effect whatsoever - the perceived speed was always the same. 79.73.134.90 (talk) 14:17, 5 April 2022 (UTC)
- Yes, Rømer's determination of the speed of light is a famous experiment. However, people still held on to old beliefs (people will often shoehorn experimental results that conflict with existing theories in bizarre ways, see the aforementioned epicycles), and it really wasn't until the negative results of the Michelson–Morley experiment that would lead special relativity, completely rewriting how we understand motion. Science is not a singular process, and the growth of scientific knowledge happens gradually, over centuries. After all, while we credit calculus to Newton and Leibniz, the basics of calculus actually date as far back as The Method of Mechanical Theorems by Aristotle, but were largely ignored as mathematical "tricks" until given formal form some two millenia later. Just like no one knew what Aristotle was on to, no one knew what Rømer was on to for centuries. --Jayron32 16:18, 5 April 2022 (UTC)
- All this became apparent 300 years before the formulae were developed to explain it. When the telescope was invented, astronomers knew the exact positions of Jupiter's satellites from occultation observations. Knowing their velocities and direction of travel they checked the observed times of the occultations against the times they would have been seen allowing for the effect the velocities would have had on the time taken for the light to arrive. They were astonished to find that they had no effect whatsoever - the perceived speed was always the same. 79.73.134.90 (talk) 14:17, 5 April 2022 (UTC)
- I don't know what else to do than to refer you again to the article Velocity-addition formula, which does ALL of the math, in full three dimensions, in ALL of the gory details. If you start with the constancy of the speed of light, and that nothing with mass can exceed that speed, then you do the math, this is the results you get. Experiments confirm that math. If you, as an observer standing next to the particle accelerator measure the two particles velocity to each be at 99.9999991% of c, they do NOT collide with a combined speed of twice that value; they collide with a speed of much closer to but not over c. The measured energies from the collision confirm this predicted calculation. I don't know what else to say except that the math is in that article and experiments do confirm it. Einstein started with the postulates of invariant speed of light and that nothing with a mass could reach that speed, and this is the equation that falls out of that math. --Jayron32 12:28, 5 April 2022 (UTC)
- It is important not note that speeds do not add the same way as you think they do. Which is to say that if you have two objects moving relative to each other, the velocity of object 2 from the frame of reference of object 1 is not simply to add the velocities. It is approximately that at sufficiently low speeds (which is to say, that WELL within significant figures, you get the same answer whether you use the Newtonian calculation or the Einsteinian one). At speeds that close to the speed of light, you need to use the correct equation to calculate relative speeds, which is the Velocity-addition formula,
- Far from affecting only electromagnetic rays (e.g. light) and not inertial objects as the OP supposes, Lorentz contraction affects any geometric quantity related to lengths, so from the perspective of a moving observer, areas and volumes will also appear to shrink along the direction of motion. See Lorentz_transformation#Physical_implications. A proton has mass and is in no way immunized from Lorentz transformation just by the OP describing it as "an inertial object whose speed of movement is not invariable, whatever the frame of reference". The double negative "not invariable" confuses me. Philvoids (talk) 17:19, 5 April 2022 (UTC)
- When you write "will appear to shrink", it is by the different vision that the radar echo of the light on a volume in the 2 reference frames gives you, isn't it? But if you are blind, the light and its speed no longer affect the size of the volume and yet it still exists. Malypaet (talk) 20:40, 5 April 2022 (UTC)
- OP, the above repetitious appeal to authority is a bit depressing; its kind of like watching a novice chess player being hammered by other more experienced players. On the other hand, cows like myself do prefer chewing on cud, before digesting it. Anyway, you wrote "In fact Einstein's relativity and the Lorentz transformation apply only to electromagnetic rays, not to inertial objects...". That is not correct, reference frame transformations apply to all coordinates of space and time and therefore to all matter waves. With respect to relativity, your velocity addition of 2c that does not preserve c, a Galilean transformation, as you acknowledged, and is thus misapplied and not within the scope of the theory. --Modocc (talk) 18:05, 5 April 2022 (UTC)
- So you criticize my correction of the OP's misconceptions by then... correcting the OP's misconceptions... I'm not sure how to respond to that. I guess just say "I'm sorry, I was wrong" and move on? What modifications would you like me to make to my answers to make them less bad? --Jayron32 18:10, 5 April 2022 (UTC)
- Your corrections are fine. Modocc (talk) 19:02, 7 April 2022 (UTC)
- So you criticize my correction of the OP's misconceptions by then... correcting the OP's misconceptions... I'm not sure how to respond to that. I guess just say "I'm sorry, I was wrong" and move on? What modifications would you like me to make to my answers to make them less bad? --Jayron32 18:10, 5 April 2022 (UTC)
It appears that a primary confusion of the OP is why the (seemingly empirical) constancy that applies to c also affects everything else. Another confusion might be why we so confidently postulate the invariance of c if it's empirically derived. GeorgiaDC (talk) 18:50, 5 April 2022 (UTC)
- Very briefly, on both points, c can also be derived from the Maxwell equations and ends up depending only on the properties of vacuum, namely ε0 and μ0. Since there's no special treatment for any inertial (constant velocity) observer, the Maxwell equations and properties of vacuum should be the same; hence c should also be the same. Once c is fixed, you can conclude from spacetime diagrams that observers with different velocities end up with different time and space measurements even for the same space-time interval (i.e. the same starting and end events). GeorgiaDC (talk) 19:11, 5 April 2022 (UTC)
- In parallel, you don't necessarily need constancy of c to begin with. Here are some derivations (https://iopscience.iop.org/article/10.1088/0143-0807/24/3/312/meta https://aapt.scitation.org/doi/abs/10.1119/1.10490 http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf) that start off with the properties of spacetime (homogeneity, isotropy, etc) and/or the group property of applying transformations. The result would not be tied to the "speed of light" or speed of anything per se. Anyhow, the transformation function eventually ends up depending on a parameter k (or k2). If it's infinite, we live in a universe with Galilean transformation; if finite, we have Lorentz. GeorgiaDC (talk) 19:23, 5 April 2022 (UTC)
- What I understand with your answers, it is that nothing explains how one associates a specific formula with the characteristics of the light for objects not having these characteristics. You always use equations from the characteristics of light (Lorentz transformation) to justify their suitability for other objects. Your explanations look like a self-justifying loop. In colliders, the predictions of Newton's mechanics give the right energy result, even at speeds close to c (or c2?), this would tend to dissociate the applications of the laws, Lorentz transformation for light and Galilean transformation for inertial objects, right? Malypaet (talk) 12:29, 7 April 2022 (UTC)
- AFAIK, Newtonian calculations do not give correct values. Also, Lorentz transformations are not used just "for light". They are used because of the speed limit set by light, not as a means of measuring light. Lorentz transformations are necessary because the limit of the speed of light introduce an asymptote into the geometry of objects moving in spacetime. That means relative calculations one would do in Euclidean geometry, things like the pythagorean theorem and the basic trigonometric functions, don't work in the same way; so the Galilean transformations, which use Euclidean math, don't work. You need to use Hyperbolic functions. Lorentz transformations are just what you get when you do hyperbolic functions in 4D Minkowski spacetime. But, and I cannot stress this enough, your misconception is that Lorentz transformations were derived to deal with light alone. They were not. They were invented to deal with the speed limit of light, which is to say that objects with mass can approach but not equal the speed of light. The do depend on the speed of light, because the speed of light defines the asymptote that controls the hyperbolic functions in question, so that's the connection there. --Jayron32 12:40, 7 April 2022 (UTC)
- "the limit of the speed of light introduces an asymptote in the geometry of objects moving through spacetime", ok, I understand very well that using light reflection as a measuring instrument to describe this geometry in a space-time, this is the basis of special relativity stated by Einstein in chapter 2. But I insist, because if instead of this reflection, to measure this geometry in space-time it is these objects which directly emit regular bursts of light (chapter 2, burst of light reflected by mirror, replaced by object at speed v, burst of light in A, then B and again in A). In this last exercise of thought, you see that in the different frames of reference, the speed of light does not intervene, the clocks are synchronous, we remain in the transformations of Galileo. Moreover, the speed limit observed in accelerators is not surprising, since the means of propulsion by electric or magnetic fields are components of light, therefore logically with the same limitations, hence the colliders to exceed this limit, in all relativity. There is some logic there, right? Malypaet (talk) 22:10, 7 April 2022 (UTC)
- Chapter 2 of what book? Unless we're all reading the same book, we can't possibly know what diagrams you are looking at or what text you are reading? --Jayron32 11:40, 8 April 2022 (UTC)
- Translation:"On the Electrodynamics of Moving Bodies" by A Einstein.[6] Malypaet (talk) 21:18, 8 April 2022 (UTC)
- Chapter 2 of what book? Unless we're all reading the same book, we can't possibly know what diagrams you are looking at or what text you are reading? --Jayron32 11:40, 8 April 2022 (UTC)
- "the limit of the speed of light introduces an asymptote in the geometry of objects moving through spacetime", ok, I understand very well that using light reflection as a measuring instrument to describe this geometry in a space-time, this is the basis of special relativity stated by Einstein in chapter 2. But I insist, because if instead of this reflection, to measure this geometry in space-time it is these objects which directly emit regular bursts of light (chapter 2, burst of light reflected by mirror, replaced by object at speed v, burst of light in A, then B and again in A). In this last exercise of thought, you see that in the different frames of reference, the speed of light does not intervene, the clocks are synchronous, we remain in the transformations of Galileo. Moreover, the speed limit observed in accelerators is not surprising, since the means of propulsion by electric or magnetic fields are components of light, therefore logically with the same limitations, hence the colliders to exceed this limit, in all relativity. There is some logic there, right? Malypaet (talk) 22:10, 7 April 2022 (UTC)
- Correct: constancy of c applies to light. Incorrect: constancy of c does not affect non-light inertial objects (I assume you mean objects with non-zero rest mass). Briefly, because c is fixed, you would notice that an observer (say B) with a different velocity will have a different definition of simultaneity from your own when manipulating light. Now, if principle of relativity still holds, then as far as B is concerned, they're at rest and you're moving. That means they can't tell, and that whatever definition of simultaneity derived from manipulating light must match the results from manipulating even massive objects! Unless of course principle of relativity is broken, but that is not and should not be the case for multitude of other reasons. The alternative derivations linked above (they might be a bit dense but I do recommend that you try to understand them) do not start from the speed of light in particular. As long as there's some evidence whether transformations are Lorentz xor Galilean, we would know whether k2 is finite or infinite respectively. GeorgiaDC (talk) 20:42, 11 April 2022 (UTC)
- AFAIK, Newtonian calculations do not give correct values. Also, Lorentz transformations are not used just "for light". They are used because of the speed limit set by light, not as a means of measuring light. Lorentz transformations are necessary because the limit of the speed of light introduce an asymptote into the geometry of objects moving in spacetime. That means relative calculations one would do in Euclidean geometry, things like the pythagorean theorem and the basic trigonometric functions, don't work in the same way; so the Galilean transformations, which use Euclidean math, don't work. You need to use Hyperbolic functions. Lorentz transformations are just what you get when you do hyperbolic functions in 4D Minkowski spacetime. But, and I cannot stress this enough, your misconception is that Lorentz transformations were derived to deal with light alone. They were not. They were invented to deal with the speed limit of light, which is to say that objects with mass can approach but not equal the speed of light. The do depend on the speed of light, because the speed of light defines the asymptote that controls the hyperbolic functions in question, so that's the connection there. --Jayron32 12:40, 7 April 2022 (UTC)
- What I understand with your answers, it is that nothing explains how one associates a specific formula with the characteristics of the light for objects not having these characteristics. You always use equations from the characteristics of light (Lorentz transformation) to justify their suitability for other objects. Your explanations look like a self-justifying loop. In colliders, the predictions of Newton's mechanics give the right energy result, even at speeds close to c (or c2?), this would tend to dissociate the applications of the laws, Lorentz transformation for light and Galilean transformation for inertial objects, right? Malypaet (talk) 12:29, 7 April 2022 (UTC)
References
April 6
"Hyraxes are sometimes described as being the closest living relative of the elephant". Can i harvest ivory from the gophers in my garden? Zindor (talk) 00:34, 6 April 2022 (UTC)
- Our article Ivory is written in a confusing way. It says that "certain" teeth are traded as ivory, but doesn't clearly explain which ones. It also mentions the two canine-like teeth of elk as "ivory teeth", implying that the other teeth are not ivory. In fact those two teeth lack enamel, which is the crucial difference. Hippopotamus teeth, on the other hand, have enamel on one surface, but are still traded as ivory. I'm unclear on whether the enamel is normally removed from hippo ivory. Similarly, the tusks of hyraxes have enamel on the front, but not the back, and as such they are just like a rodent's incisors, including those of a gopher. In principle if you could remove the enamel from a gopher's front teeth you would have produced two tiny scraps of ivory. Possibly any teeth count as ivory, even with a coating of enamel. However, size seems important to commercial value. Card Zero (talk) 03:50, 6 April 2022 (UTC)
- Other than being superficially similar looking (perhaps due to convergent evolution), hyraxes are not particularly closely related to gophers. Hyraxes and Gophers only belong to the same infraclass, being Placentalia, hyraxes belong to superorder Afrotheria of magnorder Atlantogenata, while gophers belong to superorder Euarchontoglires of magnorder Boreoeutheria. Which is to say that gophers are more closely related to humans (having a more recent common ancestor) than they are to hyraxes. --Jayron32 10:52, 6 April 2022 (UTC)
Osprey hunting behavior
There’s a video currently trending on Insta showing an Osprey diving into the water, grabbing a fish, and then coming out of the water and flying away. One would think their wet feathers would make it difficult to fly back out of the water again. How do they do this? Viriditas (talk) 01:22, 6 April 2022 (UTC)
- According to birdwatchingpro.com and Osprey Tales, they are oily, producing oil for their feathers from preen glands. Though apparently this effect is removed if an osprey is caught in an oil spill. Feather#Functions says that powder down can provide waterproofing, although I don't see that mentioned in relation to ospreys. Not sure why the article on down feathers doesn't mention this function. The Feather article also claims that "The feathers of cormorants soak up water and help to reduce buoyancy, thereby allowing the birds to swim submerged", but Cormorant disagrees with this, pointing out that, like ospreys, they have preen glands, and the truth about cormorants seems mysterious. (Fortunately you didn't ask.) Card Zero (talk) 02:31, 6 April 2022 (UTC)
- The swimming action of the fish obviously propels the Osprey back out of the water. The action doesn't occur in Cormorants because they don't hold the fish with talons. Zindor (talk) 02:42, 6 April 2022 (UTC)
- Obviously? What if the fish is swimming in the other direction when it's caught? Bazza (talk) 10:48, 6 April 2022 (UTC)
- The articles sarcasm and joke should explain the physics of the situation sufficiently, I believe. --Jayron32 11:04, 6 April 2022 (UTC)
- I understand both, but didn't notice the body language or inflection in the comment. Stranger things have been written here as truth. Bazza (talk) 11:07, 6 April 2022 (UTC)
- Bazza, i would suggest that the osprey only pounces on fish that are surfacing, so facing in the right direction. Zindor (talk) 12:44, 6 April 2022 (UTC)
- But by what means do waterlogged cormorants take to the air? Card Zero (talk) 13:31, 6 April 2022 (UTC)
- If you observe cormorants you will see they leave the water onto an object (small island etc) and stretch their wings out. This dries them off. If they were out at sea i suspect that there is a fair chance they could still take off, having enough water saturation to decrease buoyancy doesn't necessarily mean they couldn't fly, but it would certainly hinder it. Zindor (talk) 14:20, 6 April 2022 (UTC)
- But by what means do waterlogged cormorants take to the air? Card Zero (talk) 13:31, 6 April 2022 (UTC)
- Bazza, i would suggest that the osprey only pounces on fish that are surfacing, so facing in the right direction. Zindor (talk) 12:44, 6 April 2022 (UTC)
- I understand both, but didn't notice the body language or inflection in the comment. Stranger things have been written here as truth. Bazza (talk) 11:07, 6 April 2022 (UTC)
- The articles sarcasm and joke should explain the physics of the situation sufficiently, I believe. --Jayron32 11:04, 6 April 2022 (UTC)
- Obviously? What if the fish is swimming in the other direction when it's caught? Bazza (talk) 10:48, 6 April 2022 (UTC)
- The swimming action of the fish obviously propels the Osprey back out of the water. The action doesn't occur in Cormorants because they don't hold the fish with talons. Zindor (talk) 02:42, 6 April 2022 (UTC)
_drying_wings.jpg)
- Flightless cormorants do this too, for some reason. Or for no reason? Card Zero (talk) 14:40, 6 April 2022 (UTC)
- Probably just a vestigial or re-purposed fixed action pattern. The wing drying is either a beneficial trait or has no negative effect and they haven't evolved out of it. Zindor (talk) 15:15, 6 April 2022 (UTC)
- Flightless cormorants do this too, for some reason. Or for no reason? Card Zero (talk) 14:40, 6 April 2022 (UTC)
- FYI, sometimes their feathers *do* get too wet to get out of the water again, or the fish they caught is too heavy to fly off with - however ospreys can actually swim if the need arises. They do the butterfly stroke with their wings. --Iloveparrots (talk) 11:19, 6 April 2022 (UTC)
How would purple rain originate on an alien planet
What would need to happen for this to happen? — Preceding unsigned comment added by Nexxotic (talk • contribs) 03:17, 6 April 2022 (UTC)
- You can ask how it would happen on Earth too. It could be due to animal excrement after eating purple grapes. There are not many purple minerals or purple gases that might make purple liquids. My second guess would be on a movie set with artificially dyed rain. Graeme Bartlett (talk) 08:32, 6 April 2022 (UTC)
- Until he passed, the famous atmospheric scientist Prince Rogers Nelson knew how to do it, but sadly, he took his knowledge to the grave. --Jayron32 10:43, 6 April 2022 (UTC)
- In sufficiently high concentration, permanganate solutions (in water) are intensily pink to purple, a colour known as Fuchsia: . But what natural process would produce the permanganate, and how would it get into the drops? Several manganese minerals such as pyrolusite are common on Earth and therefore also plausibly on planet Prince. Are there naturally occurring carbonates or hydroxides with which it could fuse? --Lambiam 12:40, 6 April 2022 (UTC)
- Some.Category:Carbonate minerals, Category:Hydroxide minerals. But the with which it could fuse part is beyond my ken. Since permanganates are oxidizing agents, a permanganate thunderstorm could be fun. Card Zero (talk) 13:44, 6 April 2022 (UTC)
- Not permanganate, but Mars has a not-insignificant-amount of perchlorate in the soils, perchlorate is also a strong oxidizing agent, and dealing with it is part of the discussion on any sort of serious Mars visit by humans. If Mars had rain, it would not be hard to envision it having some perchlorate in it. --Jayron32 15:54, 6 April 2022 (UTC)
- Some.Category:Carbonate minerals, Category:Hydroxide minerals. But the with which it could fuse part is beyond my ken. Since permanganates are oxidizing agents, a permanganate thunderstorm could be fun. Card Zero (talk) 13:44, 6 April 2022 (UTC)
- Purple alien bacteria. On Earth, see "Schiermeier, Q. 'Rain-making' bacteria found around the world. Nature (2008). Viriditas (talk) 21:55, 6 April 2022 (UTC)
- Blood rain is typically due to algae spores. So, yeah, some purple alien life could fall from the alien sky, if we assume there is alien life. Card Zero (talk) 23:37, 6 April 2022 (UTC)
- Wait it can really rain "blood"? Truth in heavy metal. Sagittarian Milky Way (talk) 01:35, 7 April 2022 (UTC)
- Charles Fort documented a large number of cases of anomalies; raining blood was one of them. -- Jack of Oz [pleasantries] 20:01, 7 April 2022 (UTC)
- Wait it can really rain "blood"? Truth in heavy metal. Sagittarian Milky Way (talk) 01:35, 7 April 2022 (UTC)
- Blood rain is typically due to algae spores. So, yeah, some purple alien life could fall from the alien sky, if we assume there is alien life. Card Zero (talk) 23:37, 6 April 2022 (UTC)
- Saharan dust episodes intense enough to colour the sky - and occasionally the rain - an orange-brown color are rare enough to make the news but far from unheard of. I can't reallt think of anything which might transform that orange to purple though. — Preceding unsigned comment added by 2A01:E34:EF5E:4640:995C:79C2:9A9B:C3C2 (talk) 11:00, 7 April 2022 (UTC)
April 7
Total gear reduction ratio
I'm trying to work out whether any generalizations can be made about the total gear reduction ratios of various speed-shift gearboxes (aka 'transmissions') used in tracked or half-tracked military vehicles. I'm aware that some people simply use the product of all the ratios, but in my sources the authors multiply the 'jumps' between each ratio (e.g. 9.21 / 4.56 = 2.02 as below), or divide the highest ratio by the lowest (9.21 / 0.69). The result is pretty much the same, I'm not concerned about the exact figures. I think the letters i and q are used in the formulae, but again I'm not too bothered about them.
Thus a certain gearbox is stated to have a total (or overall) reduction ratio of 1:13.4 - the same figure also can be reached by dividing the max. road speed in top gear at max. revs by the speed in 1st gear.
Gear Ratio Jump
1:x
1 9.21 2.02
2 4.56 1.59
3 2.87 1.56
4 1.83 1.45
5 1.27 1.41
6 0.90 1.31
7 0.69
Overall reduction ratio = 13.4
Using the above methods, here are nine gearboxes in order of overall reduction ratio:
a 8 gears, 11.2 / 0.66 i = 1:17 b 8 gears, 11.0 / 0.68 i = 1:16 c 10 gears, 8.00 / 0.55 i = 1:14.55 d 7 gears, 9.21 / 0.69 i = 1:13.4 e 7 gears, 70.0 / 5.50 i = 1:12.7 (km/h) f 6 gears 40.0 / 5.5 i = 1:7.27 (km/h) g 6 gears, 42.0 / 5.8 i = 1:7.24 (km/h) h 5 gears, 32.4 / 5.1 i = 1:6.35 (km/h) j 4 gears, 5.58 / 1.00 i = 1:5.58
Letters b and d used the same 23 litre engine, c, f, g & h used a 12 litre engine, and e and j used a 4.2 litre engine. Although a high top speed is desirable, cross-country tractability is probably to be preferred.
Is there any purpose to comparing these ratios? What, if anything, can be deduced from these figures? Cheers, MinorProphet (talk) 12:10, 7 April 2022 (UTC)
- It is unclear what you are asking. The road speed of a vehicle is given as distance per time unit, such as expressed in km/h. The number of revolutions of a gear is a dimensionless number; the rotation speed is given as number per time unit, which has the same dimension as the hertz. You cannot directly derive the gear ratio from a combination of the two, but you can convert distance to number of revolutions of the wheels of the car if you use the wheel diameter. Is there any purpose to comparing the weights of two bullfrogs? Yes, if you are the judge in a competition of "whose bullfrog is the heaviest". Probably not if you are just on a leisurely stroll and happen to spot two bullfrogs. Whether a comparison has a purpose depends on what you want to achieve. --Lambiam 14:02, 7 April 2022 (UTC)
- Thanks for your reply. What is the purpose of stating the total gear reduction ratio? It's often given in statistics about transmissions, in the way that the power and torque figures of an engine are also given and can be fruitfully compared. Take the gearboxes e and j in the list above. The first is semi-automatic, the other manual. They were both used with the same engine in two prime movers of similar capabilities. What advantage might one have over the other? Cheers, MinorProphet (talk) 19:09, 7 April 2022 (UTC)
- The performance of the combination of engine and transmission depends on many things, not the least of which is what aspects you want to grade this performance on. Is it sheer power? And if so, at which speed? An engineer doing the calculations may need to know the applicable mechanical advantage and thereby the torque ratio. I cannot readily think of a plausible advantage of knowing the gear ratio for a typical car buyer, who should be more interested in the fuel consumption given the envisaged use. I'm not into cars, though. --Lambiam 07:28, 8 April 2022 (UTC)
- My opening sentence probably wasn't clear enough, but I'm talking [torqueing?] about tanks weighing up to 40 tons, which have somewhat different design criteria. T[h]anks anyway. :) MinorProphet (talk) 20:44, 8 April 2022 (UTC)
- Doesn't this tell you the range of total torque available as output? A vehicle with 1:17 is able to vary the output torque of its engine + gearbox more, compared to 1:5.58. GeorgiaDC (talk) 20:01, 11 April 2022 (UTC)
- My opening sentence probably wasn't clear enough, but I'm talking [torqueing?] about tanks weighing up to 40 tons, which have somewhat different design criteria. T[h]anks anyway. :) MinorProphet (talk) 20:44, 8 April 2022 (UTC)
- The performance of the combination of engine and transmission depends on many things, not the least of which is what aspects you want to grade this performance on. Is it sheer power? And if so, at which speed? An engineer doing the calculations may need to know the applicable mechanical advantage and thereby the torque ratio. I cannot readily think of a plausible advantage of knowing the gear ratio for a typical car buyer, who should be more interested in the fuel consumption given the envisaged use. I'm not into cars, though. --Lambiam 07:28, 8 April 2022 (UTC)
- Thanks for your reply. What is the purpose of stating the total gear reduction ratio? It's often given in statistics about transmissions, in the way that the power and torque figures of an engine are also given and can be fruitfully compared. Take the gearboxes e and j in the list above. The first is semi-automatic, the other manual. They were both used with the same engine in two prime movers of similar capabilities. What advantage might one have over the other? Cheers, MinorProphet (talk) 19:09, 7 April 2022 (UTC)
Why might a chewing gum brand glue to fingers easier than another?
One used-up gum can be rolled into a ball and easily pulled off, try that with the other and it always bonds to finger pads strong and never gets pulled off 100% like the other and what didn't get pulled off takes extensive rubbing to fully remove. Sagittarian Milky Way (talk) 17:46, 7 April 2022 (UTC)
- Chewing gum is generally composed of a few different things, usually some kind of resilient plastic substance (called gum base), either an artificial polymer, or natural ones such as latex or chicle, along with sweeteners and flavorings. What in there is sticky? Could be any of them. Some gum bases are going to be stickier than others. Wet sugar is pretty much always sticky; though depending on how long you chewed it and how much you digested out of the gum with your saliva would affect that. Without knowing the specific ingredients in the specific gum, it would be hard to say what is making it sticky. --Jayron32 17:59, 7 April 2022 (UTC)
- See Chewing_gum#Stickiness. Philvoids (talk) 21:14, 7 April 2022 (UTC)
- So, just after you have washed your hands with soap, the wad won't stock to your pads. But when they are dry and greasy it bonds. This effect does not depend on the brand. --Lambiam 22:40, 7 April 2022 (UTC)
- It absolutely is some composition difference like I dunno polymer length, not skin hydrophobicity. Maybe it'd take longer if I didn't use soapwater to reduce thumb-to-forefinger friction but it still takes a lot of trying to use soap to unfuck residues that have been fucked into thin gluey films from trying to rub the residue off without soap. Sagittarian Milky Way (talk) 00:31, 8 April 2022 (UTC)
- I trust you have done a double-blind experiment to determine that the bonding is brand-dependent. The adhesion with which the goo clings to a surface and the cohesion with which it sticks together are different things, though. Cleaning with (soapy) water only works if the water can creep between the goo and the surface it is clinging to. However, because the goo is hydrophobic, the water is goo-phobic. If it could speak, it would say, uh-uh, I'm not gonna go into that mess. --Lambiam 07:04, 8 April 2022 (UTC)
- It isn't the-less-sticky-one-was-always-soon-after-obliterating-skin-oil-with-soap-dependant. Both cohesive and adhesive proprieties are different. Yes I know that's why it takes so long, the solvent needs to get between the hydrophobic oleophilic goo and the skin oil it loves so much and that takes a lot of trying to force the contact side to roll over so it can break contact. The water of soaps and detergent should not be completely oleophobic, as letting water dissolve oleophilic things (maybe not all gums) is the whole point of soaps and detergents. Sagittarian Milky Way (talk) 14:42, 8 April 2022 (UTC)
- I trust you have done a double-blind experiment to determine that the bonding is brand-dependent. The adhesion with which the goo clings to a surface and the cohesion with which it sticks together are different things, though. Cleaning with (soapy) water only works if the water can creep between the goo and the surface it is clinging to. However, because the goo is hydrophobic, the water is goo-phobic. If it could speak, it would say, uh-uh, I'm not gonna go into that mess. --Lambiam 07:04, 8 April 2022 (UTC)
- It absolutely is some composition difference like I dunno polymer length, not skin hydrophobicity. Maybe it'd take longer if I didn't use soapwater to reduce thumb-to-forefinger friction but it still takes a lot of trying to use soap to unfuck residues that have been fucked into thin gluey films from trying to rub the residue off without soap. Sagittarian Milky Way (talk) 00:31, 8 April 2022 (UTC)
- So, just after you have washed your hands with soap, the wad won't stock to your pads. But when they are dry and greasy it bonds. This effect does not depend on the brand. --Lambiam 22:40, 7 April 2022 (UTC)
- See Chewing_gum#Stickiness. Philvoids (talk) 21:14, 7 April 2022 (UTC)
Double-slit experiment and not wave but only input stream?
In the experiments of double slits [1], originally "Young's slits" with light which is supposed to have confirmed its wave nature, we now also have flows of electrons, atoms and molecules. Outside I noticed that a flow of water molecules meeting an obstacle, produced waves, the same on the bow of a boat. We therefore have a phenomenon of resonance, which could be found with electrons, atoms and molecules. Would this not be an argument in favor of the corpuscular hypothesis of Newton's light, knowing that a flux can also be considered as a train of impulse with common characteristics with a wave: frequency, period and length of wave ? Malypaet (talk) 23:16, 7 April 2022 (UTC)
- Newton's corpuscular theory argued that the straight lines of reflection demonstrated light's particle nature as only particles could travel in such straight lines. Refraction could be explained if particles of light accelerated laterally upon entering a denser medium. Newton's contemporaries Hooke and Huygens refined the wave viewpoint, showing that if light traveled at different speeds in different media, refraction was easily explained as the medium-dependent propagation of light waves. See Wave–particle duality. Philvoids (talk) 00:53, 8 April 2022 (UTC)
- When I quote Newton's hypothesis, a photon is a simple particle like an electron, an atom, etc... and which has nothing to do with Einstein's. I can then interpret a flux of photons as n elementary rays of photons coming from "Planck resonators" (pulse trains). In this case we are indeed in a fluid mechanics as for water and this flow of photons will then be able to carry a resultant wave created by a disturbance, like the edges of a slit. In this case there is no duality, always flows, sometimes carriers of waves. Has anyone considered this possibility? Malypaet (talk) 22:36, 8 April 2022 (UTC)
- Isaac Newton 1642 - 1727 never used the term "photon" which was coined in 1926 by Gilbert Newton Lewis for what Einstein had called a light quantum. Please see Photon#Historical_development and consider whether you question the awards of Nobel Prizes to Einstein in 1921 (energy quantization explains the photoelectric effect) and to Compton in 1927 (photons have momentum). Philvoids (talk) 13:13, 9 April 2022 (UTC)
- I know that. But today we can also very well use the term photon for Newton's corpuscle, as an alternative hypothesis to Einstein's theory of relativity. In this case for a frequency "ν" we have "ν" photons corpuscles in an elementary flow of corpuscles. Malypaet (talk) 21:20, 10 April 2022 (UTC)
- Isaac Newton 1642 - 1727 never used the term "photon" which was coined in 1926 by Gilbert Newton Lewis for what Einstein had called a light quantum. Please see Photon#Historical_development and consider whether you question the awards of Nobel Prizes to Einstein in 1921 (energy quantization explains the photoelectric effect) and to Compton in 1927 (photons have momentum). Philvoids (talk) 13:13, 9 April 2022 (UTC)
- When I quote Newton's hypothesis, a photon is a simple particle like an electron, an atom, etc... and which has nothing to do with Einstein's. I can then interpret a flux of photons as n elementary rays of photons coming from "Planck resonators" (pulse trains). In this case we are indeed in a fluid mechanics as for water and this flow of photons will then be able to carry a resultant wave created by a disturbance, like the edges of a slit. In this case there is no duality, always flows, sometimes carriers of waves. Has anyone considered this possibility? Malypaet (talk) 22:36, 8 April 2022 (UTC)
- So, it looks like you're confusing two different things here. The wave-like nature of light is not a vibration of photons. Photons are not the medium of the vibration in the way that, say, water is the medium of vibration for the wake of the boat you describe. Likewise, with matter waves, it is not the vibration of matter; it is that matter is the vibration itself. The Young's slit experiment works for single particles sent one-at-a-time, not just for groups of particles, so it's not like a group of electrons are carrying a wave as a group, like a water wave. And the experiment works in a vacuum, so it's not like the electron (or whatever) is traveling through air and generating a wake, which is might be the source of wave-like behavior. You get interference patterns even if single electrons are sent through one at a time, and even if the experiment is conducted in a pure vacuum. So we're left with two major questions: with what does a single electron interfere with? Itself? Other electrons that will be sent in the future? How does a point particle, which is indivisible, go through two slits at once? These are not simple questions to answer, and if you aren't struggling at a core philosophical level with any answer given to these questions, and don't see inherent paradoxes any answer to these questions give, you aren't doing Quantum mechanics correctly. Neils Bohr supposedly said "Anyone who is not shocked by quantum theory has not understood it.", and Richard Feynman similarly said "I think I can safely say that nobody understands quantum mechanics". What you have there is 1) a scientist who has as good a claim as any to inventing the field of QM, and 2) A scientist who has as good a claim as any to advancing the field further than anyone since its invention (specifically with QED), and they aren't comfortable with the conclusions that QM leads one to. And yet, these are the laws and theories, and the experiments confirm them, no matter how nonsensical they seem, we have a theory that makes predictions that experiments have borne out to shocking accuracy millions of time. --Jayron32 12:10, 8 April 2022 (UTC)
- Even more shocking spooky action at a distance. 80.43.81.32 (talk) 12:35, 8 April 2022 (UTC)
- Even weirder than Young's original experiment are variations such as the Delayed-choice quantum eraser experiment. --Jayron32 12:44, 8 April 2022 (UTC)
- Sorry, but in the original Young's experiment there are 2 halves of an original flow that each aim for their slot, one part goes through it either straight or by hitting the edges (like water on a rock) and the rest is reflected on the outside of the slits. So 2 input streams which can create 2 waves carried by these output streams and which will intersect.
- When there is only one particle at a time, you have to be a very bad shooter with a trembling hand not to know which slit will be crossed and then you enter into a problem of probability, which has nothing left to do with the original experience. A photon, an electron, etc... is necessarily smaller than a slit, isn't it?
- Figure 2 [7] shocks me for the size of the photon at the entrance to the 2 slits! Malypaet (talk) 22:43, 8 April 2022 (UTC)
- It's not like a stream of water hitting some rocks. It's more like waves in a pool of otherwise stationary water. When the photon is detected somewhere, it has arrived at that exact location and has taken the path leading there. If there's more than one such path, it has taken all of them at the same time. But as long as the photon propagates, it follows every possible path, is everywhere at the same time (although not at the same probability). The single photon goes through the left slit, through the right slit, is reflected in the mirror next to the slit, all at the same time. There's no hitting the edges though. What makes it weird is that it even works with the smallest possible wave, i.e. a single quantum of energy, which can only be detected at one place, still taking every possible path as a wave. The size of a photon is like the size of a wave at sea (and I'm not talking about the height): huge, but it still fits through a small hole a quayside. If you take a laser beam, the length of a photon is something like the inverse of the bandwidth of the laser and the width of the photon is the width of the beam. PiusImpavidus (talk) 10:12, 9 April 2022 (UTC)
- So i can understand that your photon is not an elementary object, but a group object where some are passing by the fisrst slit, some by the second and some reflected between the 2 slit. So , this group is everywhere in the same time ! But for an electron, its size is much smaller than a slit. It cannot aim and precisely cross a slit? Malypaet (talk) 22:09, 9 April 2022 (UTC)
- No, the photon is an elementary object, just like the electron, and it can hit only a single detector at a time. The weird thing is, even when it's a single object, it behaves like a wave going everywhere at the same time. The same is true for electrons. Quantummechanics is weird, much weirder than you imagined. PiusImpavidus (talk) 10:56, 10 April 2022 (UTC)
- An electron has a mass, a supposed size, unlike the photon and if it turns out that the photon has a mass, the wave/particle duality disappears, right? Malypaet (talk) 22:24, 10 April 2022 (UTC)
- An electron has mass, but no size. A photon has supposedly no mass (but we can't really prove that) and no size. Wave/particle duality exists for both. PiusImpavidus (talk) 09:20, 11 April 2022 (UTC)
- Furthermore, when a particle, like a photon or an electron, travels between the emitter and detector (whether in the Young's experiment, or really any path between where it is emitted and where it interacts with some bit of matter (an "observation" in QM-speak)), it takes every available path to get there, even fantastically complicated and strange paths. The chance that it hits any one point of the detector is the weighted average of the probability of each of near-infinite number of possible paths. If you want to really know why Feynman is famous, it's not just for being a charismatic, bohemian personality (though he was that too), it's that he worked out the mathematical shortcuts to calculate the averages between all of those near-infinite number of paths. This was the path integral formulation. --Jayron32 11:53, 11 April 2022 (UTC)
- And the trap penning giving an electron radius ? Malypaet (talk) 20:56, 11 April 2022 (UTC)
- An electron has mass, but no size. A photon has supposedly no mass (but we can't really prove that) and no size. Wave/particle duality exists for both. PiusImpavidus (talk) 09:20, 11 April 2022 (UTC)
- An electron has a mass, a supposed size, unlike the photon and if it turns out that the photon has a mass, the wave/particle duality disappears, right? Malypaet (talk) 22:24, 10 April 2022 (UTC)
- No, the photon is an elementary object, just like the electron, and it can hit only a single detector at a time. The weird thing is, even when it's a single object, it behaves like a wave going everywhere at the same time. The same is true for electrons. Quantummechanics is weird, much weirder than you imagined. PiusImpavidus (talk) 10:56, 10 April 2022 (UTC)
- So i can understand that your photon is not an elementary object, but a group object where some are passing by the fisrst slit, some by the second and some reflected between the 2 slit. So , this group is everywhere in the same time ! But for an electron, its size is much smaller than a slit. It cannot aim and precisely cross a slit? Malypaet (talk) 22:09, 9 April 2022 (UTC)
- It's not like a stream of water hitting some rocks. It's more like waves in a pool of otherwise stationary water. When the photon is detected somewhere, it has arrived at that exact location and has taken the path leading there. If there's more than one such path, it has taken all of them at the same time. But as long as the photon propagates, it follows every possible path, is everywhere at the same time (although not at the same probability). The single photon goes through the left slit, through the right slit, is reflected in the mirror next to the slit, all at the same time. There's no hitting the edges though. What makes it weird is that it even works with the smallest possible wave, i.e. a single quantum of energy, which can only be detected at one place, still taking every possible path as a wave. The size of a photon is like the size of a wave at sea (and I'm not talking about the height): huge, but it still fits through a small hole a quayside. If you take a laser beam, the length of a photon is something like the inverse of the bandwidth of the laser and the width of the photon is the width of the beam. PiusImpavidus (talk) 10:12, 9 April 2022 (UTC)
- Even weirder than Young's original experiment are variations such as the Delayed-choice quantum eraser experiment. --Jayron32 12:44, 8 April 2022 (UTC)
- Malypaet, You are trying to visualise and conceptualise quantum phenomena in terms of the everyday "Classical" macroscopic physics that creates our everyday experiences, and which consequently has molded all our thought processes.
- This cannot be done! No way of doing so has been devised in over a century of effort by the best minds on the planet. One just has to accept the logically correct mathematics and consistently demonstrated experimental measurements of the "quantum world" as proof of the existence of aspects of our universe outside the scope of our human sensorium, that do not work in the same way as those aspects within it. {The poster formerly known as 87.81.230.195} 90.200.65.249 (talk) 17:34, 10 April 2022 (UTC)
- "This cannot be done! ... One just has to accept..." is antithetical to dealing with any dogma. All of it. Models, classical, modern or speculative are only as good as the bars or hurdles they have managed to pass over and no better. Times change, people change and often because someone becomes aware of even better models, and sometimes they can even upright the upside down. The education of current thought is paramount, but there are limits to what can be taught. Modocc (talk) 21:48, 10 April 2022 (UTC)
- A theory is valid only as long as it is not invalidated, even if it can take several centuries. You are taking an argument from authority which has nothing scientific about it. Take Planck's formula "E=hν" which you call Einstein's "quantum of light", there you have the energy of a photon which depends on its frequency. Now consider Newton's hypothesis, if "ν" is a number of cycles per second of a wave (hertz), it can also be the number of corpuscles per second of an elementary flux coming from Planck's "resonator". In this case, while keeping a notion of group of objects for the "quantum of light" you find the kinetic energy of a corpuscle (photon of Newton) "E=h=1/2 mc2" and its mass “ mγ=2h/c2”, i.e.:
- ”mγ=1.474 499 464 763 88 x 10-50 kg “
- Value often approached experimentally as by Coulomb's law test [8] and more recently with FRBs [9]. Why is it forbidden to use my classical vision, if it gives simple, coherent results that can be found experimentally, as here for the mass of the photon? Malypaet (talk) 21:50, 10 April 2022 (UTC)
- I am not arguing from authority in favour of (or against) any theory; I am suggesting that it is impossible for any macroscopic (classically immersed) being (such as ourselves) to meaningfully describe quantum phenomena in macroscopic terms, because our minds have no suitable referents. This doesn't mean that we cannot describe such models mathematically, and describe what results from them, just that we can't visualise them. {The poster formerly known as 87.81.230.195} 90.200.65.249 (talk) 22:17, 10 April 2022 (UTC)
- You went further than simply lack of comprehension when you said "...as proof of the existence of aspects of our universe outside the scope of our human sensorium..." emphasis in bold added. None of it is proof. Sometimes the earliest modelers lack enough data to make accurate maps. Modocc (talk) 22:40, 10 April 2022 (UTC)
- Just to be clear, I meant proof that certain things are outside the scope of the human sensorium, (and therefore beyond our ability to conceptualise and describe them in words) not proof of exactly what those things are. However, I do not think we are on the same wavelength, so there is little point in continuing this aspect of the discussion. {The poster formerly known as 87.81.230.195} 90.200.65.249 (talk) 16:09, 11 April 2022 (UTC)
- You went further than simply lack of comprehension when you said "...as proof of the existence of aspects of our universe outside the scope of our human sensorium..." emphasis in bold added. None of it is proof. Sometimes the earliest modelers lack enough data to make accurate maps. Modocc (talk) 22:40, 10 April 2022 (UTC)
- I am not arguing from authority in favour of (or against) any theory; I am suggesting that it is impossible for any macroscopic (classically immersed) being (such as ourselves) to meaningfully describe quantum phenomena in macroscopic terms, because our minds have no suitable referents. This doesn't mean that we cannot describe such models mathematically, and describe what results from them, just that we can't visualise them. {The poster formerly known as 87.81.230.195} 90.200.65.249 (talk) 22:17, 10 April 2022 (UTC)
- Newton's classical treatment cannot work as such. Read the history. Modocc (talk) 21:59, 10 April 2022 (UTC)
- I read Newton's story and he describes light well as a flow of corpuscles having mass, with the knowledge of the time. After reading Planck's story on black body radiation, we can very well represent an elementary ray as the shot of corpuscle photons by a machine gun, with such and such a frequency "ν". Over one second you have the power of the shot by adding the kinetic energy of each corpuscle and if you multiply by one second you find the energy corresponding to the frequency, is that clear? Malypaet (talk) 22:50, 10 April 2022 (UTC)
- Yes, but you may have missed my reply 3 days ago regarding the KE and mass that is measured. "See https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass" It differs. Also see https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation#Observations_conflicting_with_Newton's_formula. Photons have twice the expected Newtonian deflection due to gravity, but they also have twice the expected KE for their velocity. Thus you are not wrong nor is it forbidden to take that view, but there is a fair bit of mathematics involved when modeling kinetics and kinematics, one of the most important ones is the relativistic Doppler effect. Modocc (talk) 23:55, 10 April 2022 (UTC)
- you refer me to a formula resulting from an invalidated theory if the photon has a mass, to demonstrate that the photon has no mass. It's a vicious circle and the only solution to this kind of problem is to get out of the circle.
- As for the deflection of a mass photon by the gravity of an observed half, in the vicinity of the sun or any other star, I would like to know if the atmosphere or other characteristics due to the proximity would not have they also their share for the other half?
- Many other observations can be explained by the mass of the photon/Newton's corpuscle:
- Gravity black holes with escape velocity > c.
- Dark matter = ocean of photons/mass corpuscles in permanent movement in all directions and bathing the universe, which explains well in spiral galaxies, the orbit of stars around their centers, contrary to the theory of relativity of Einstein.
- These photons/particles which leave the visible universe with their mass can also explain the acceleration of the expansion of the universe. Malypaet (talk) 22:48, 11 April 2022 (UTC)
- You may want to pick up an older textbook that does a better job of introducing classical and the relativistic Doppler relations. Its the symmetrical Doppler relation that is supported by measurements and must be accurately modeled to have any merit. Modocc (talk) 01:35, 11 April 2022 (UTC)
- If you say that the frequency of light is the number of photons per second, then what is the intensity? What happens if you add two identical light sources together? Experimentally, it doubles the intensity, not the frequency. What happens when you split a beam in two? Experimentally, it halves the intensity, not the frequency. How do you want to explain the photo-electric effect? That was crucial to the development of quantummechanics. How do you create a black-body spectrum, with many frequencies at the same time? No, if the theory uses photons, the frequency must be proportional to the energy per photon and the intensity to the number of photons per second.
- If you want to understand physics, you're welcome. If you just want to publish your own theories, write a (soft) sci-fi novel. I might like it. PiusImpavidus (talk) 09:20, 11 April 2022 (UTC)
- That the OP conflated intensity with wave number is unremarkable, especially if the wave packet is a spherical short pulse of light with uniform density. BTW, you wrote " the width of the photon is the width of the beam". No, as a particle it behaves as a point particle with no width that is modeled as part of plane wave that is unsupported by a medium.. Modocc (talk) 12:11, 11 April 2022 (UTC)
- Depends on how you see “width” and the photon. The photon has no width of its own, I agree with that (obviously), but the photon (as a wave) is present over the full width of the beam. The width of the beam is the width of the wave and the photon is the wave. Or the wave is a superposition of photons, but it works for a single photon as well. So I think it's fair to say that the width of a photon is the width of the beam. A photon doesn't even have to be continuous. Did I switch too easily to a wave description? PiusImpavidus (talk) 17:32, 11 April 2022 (UTC)
- It depends. Forty-some years ago one of my teachers naively thought that there is some astronomically small chance that all the air in the room could exit the door and leave us breathless; of course ignoring all the thermal wave interactions that prevent that from happening. In the same vein, the photon is exactly wherever it hits a target. In addition, its interactions with matter's near-fields form the group waves that we are able to measure/calculate with virtual particles as it remains a particle with its invariant power determined by its nonzero Poynting vector. The origin and extent of randomness in those processes like with what the teacher had asserted are model/interpretation dependent and last I checked there are at least seventeen interpretations of the quantum wave functions and they don't even have to be physical. Normally, instead of conflating particles with waves I would just say that particles contribute to waves of different sizes such as the laser beam depending on the particles' characteristics and leave it at that. Modocc (talk) 20:13, 11 April 2022 (UTC)
- Depends on how you see “width” and the photon. The photon has no width of its own, I agree with that (obviously), but the photon (as a wave) is present over the full width of the beam. The width of the beam is the width of the wave and the photon is the wave. Or the wave is a superposition of photons, but it works for a single photon as well. So I think it's fair to say that the width of a photon is the width of the beam. A photon doesn't even have to be continuous. Did I switch too easily to a wave description? PiusImpavidus (talk) 17:32, 11 April 2022 (UTC)
- I have described the frequency of an elementary flux of corpuscles, whose elementary energy quantum depends on this frequency, which is given by Planck's formula. For the intensity, it's simple, you multiply/parallelize these elementary flows and this will give you one or more beams of greater or lesser intensity, it's still easy to imagine, it's the model of the electric current.
- With respect to the blackbody, the higher the thermal radiation energy entering it, the higher the dominant frequency will be, and vice versa. All these light beams of greater or lesser intensity/frequency intersect (there are a lot), have elastic shocks on the material which maintain and contribute to thermal agitation. Malypaet (talk) 21:40, 11 April 2022 (UTC)
- For the photoelectric effect, no change, an elementary flux will eject an electron struck by the number of corpuscles corresponding to the energetic quantum of Planck/Einstein. Malypaet (talk) 21:51, 11 April 2022 (UTC)
- That the OP conflated intensity with wave number is unremarkable, especially if the wave packet is a spherical short pulse of light with uniform density. BTW, you wrote " the width of the photon is the width of the beam". No, as a particle it behaves as a point particle with no width that is modeled as part of plane wave that is unsupported by a medium.. Modocc (talk) 12:11, 11 April 2022 (UTC)
- Yes, but you may have missed my reply 3 days ago regarding the KE and mass that is measured. "See https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass" It differs. Also see https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation#Observations_conflicting_with_Newton's_formula. Photons have twice the expected Newtonian deflection due to gravity, but they also have twice the expected KE for their velocity. Thus you are not wrong nor is it forbidden to take that view, but there is a fair bit of mathematics involved when modeling kinetics and kinematics, one of the most important ones is the relativistic Doppler effect. Modocc (talk) 23:55, 10 April 2022 (UTC)
- I read Newton's story and he describes light well as a flow of corpuscles having mass, with the knowledge of the time. After reading Planck's story on black body radiation, we can very well represent an elementary ray as the shot of corpuscle photons by a machine gun, with such and such a frequency "ν". Over one second you have the power of the shot by adding the kinetic energy of each corpuscle and if you multiply by one second you find the energy corresponding to the frequency, is that clear? Malypaet (talk) 22:50, 10 April 2022 (UTC)
April 8
Literature about effects of different speakers on a person
At first I thought it was just me having bad days, but then after years I noticed that if I hear some audio (podcast/music) from certain speakers - say the one integrated in a good laptop - I get relatively quickly headache (not even after 1 hour I have to stop it). If instead I hear from speakers that are still good but not great but a bit away from me (say 2-3 meters), I have no problems. Conversely with headphones I can endure audio for hours, without major problems. It is just my body behaving in a weird way or is there literature about this that shows that some audio sources are less pleasant than others for some people? --Pier4r (talk) 15:00, 8 April 2022 (UTC)
- That reminds me of this oldie:
- Patient: It hurts when I do this!
- Doctor: So don't do it!
- --←Baseball Bugs What's up, Doc? carrots→ 18:06, 8 April 2022 (UTC)
- It probably relates to signal/noise ratio, or distortion. At some point, the distortion level makes processing the "signal" (music, speech, etc.) taxing on the brain.citation needed [edit: 2603:6081:1C00:1187:4C16:31AD:E403:46E4 (talk) 21:19, 8 April 2022 (UTC)]
- Further thought: this reminds me of an old Marantz ad with the tagline "Don't turn down the music, turn down the distortion". 2603:6081:1C00:1187:4C16:31AD:E403:46E4 (talk) 22:42, 8 April 2022 (UTC)
- Even further thoughts, probably off-topic - Marantz (in my not-so-welcome HO) had a fantastic marketing campaign: I don't think I've ever bought anything, ever, just on the strength of the advertising alone, except their early '80s hi-fi separates - 'champagne gold' brushed aluminium finish. They never said it's a pig to keep clean, tho' ... MinorProphet (talk) 09:49, 9 April 2022 (UTC)
- Further thought: this reminds me of an old Marantz ad with the tagline "Don't turn down the music, turn down the distortion". 2603:6081:1C00:1187:4C16:31AD:E403:46E4 (talk) 22:42, 8 April 2022 (UTC)
- I had a brief hunt for some decent literature about this, but failed to come up with anything useful. It's a bit like asking why driving down the motorway at 90mph is more relaxing in a V-12 Jaguar than in a 700 cc Smart. I expect that frequency response, sound pressure and equalization (EQ) have some bearing on the matter. Some well-known speaker manufacturers have put some effort into the problem - this review mentions some of the better-sounding integrated laptop speakers. I imagine that the sort of literature you're looking for would include something along the lines of the charts in this article. As it says, "Apple’s official tech spec page doesn’t list the frequency range of the speakers. Booo! Are they trying to hide something? Try googling “Frequency range” and “MacBook Pro”. You won’t find any easy results." There may be blogs from sound engineers from the speaker makers mentioned, eg JBL, Harmon Kardon, Bang and Olufsen - or even contact the firms directly. Sorry not to be more helpful. MinorProphet (talk) 21:52, 8 April 2022 (UTC)
- Further thought: some speakers may be optimized for reproducing MP3s - if you are are used to listening to full-range sound sources, eg CD, DVD, your brain might be struggling to fill in the missing frequencies. Some audio-oriented laptops include sound-processing technologies to deal with the issue. Some people™ have particularly sensitive hearing, others just don't care or can't hear the difference. MinorProphet (talk) 22:16, 8 April 2022 (UTC)
- Thank you! Pier4r (talk) 07:39, 11 April 2022 (UTC)
- Further thought: some speakers may be optimized for reproducing MP3s - if you are are used to listening to full-range sound sources, eg CD, DVD, your brain might be struggling to fill in the missing frequencies. Some audio-oriented laptops include sound-processing technologies to deal with the issue. Some people™ have particularly sensitive hearing, others just don't care or can't hear the difference. MinorProphet (talk) 22:16, 8 April 2022 (UTC)
- I had a brief hunt for some decent literature about this, but failed to come up with anything useful. It's a bit like asking why driving down the motorway at 90mph is more relaxing in a V-12 Jaguar than in a 700 cc Smart. I expect that frequency response, sound pressure and equalization (EQ) have some bearing on the matter. Some well-known speaker manufacturers have put some effort into the problem - this review mentions some of the better-sounding integrated laptop speakers. I imagine that the sort of literature you're looking for would include something along the lines of the charts in this article. As it says, "Apple’s official tech spec page doesn’t list the frequency range of the speakers. Booo! Are they trying to hide something? Try googling “Frequency range” and “MacBook Pro”. You won’t find any easy results." There may be blogs from sound engineers from the speaker makers mentioned, eg JBL, Harmon Kardon, Bang and Olufsen - or even contact the firms directly. Sorry not to be more helpful. MinorProphet (talk) 21:52, 8 April 2022 (UTC)
- Pier, when i was younger i used to throw up a lot when in my dad's car. It turned out he had the CD player set to max bass and max treble, to make it worse he would listen to the Archers on BBC Radio 4. I adjusted the settings right down one day and i was perfectly fine from then onwards! Zindor (talk) 00:44, 9 April 2022 (UTC)
- It's difficult to adjust for all the variables in your scenario: the disparate speaker equipment, the variant types of sound processioning (including both purposeful equalization functions and artifacts from software and signal processing), proximity, and any additional idiosyncratic factors related to you yourself; I honestly wouldn't even hazard a guess, and the list of possible explanations run a gamut from the technological to the biophysical. However, all of that said, there is one thing that kinda-sorta occurs to me most immediately here, and Zindor's post immediately above touches upon it: you'll want to remember that not all sound which affects you physiologically is in the audible range. There are in particular some frequencies just on fringes of perceptible sound to which the sound detecting organs in the inner ear are sensitive. And sub-audible low frequencies in particular can, to some extent, effect the organs with sound waves which are even completely outside the ability to simulate perception of sound through the normal mechanisms, but still irritate the structures, particularly over an extended period of time. Unfortunately, any number of the technological and physical factors described above could cause such noise in your overall sound, so it doesn't narrow you in much, but a little experimentation such as young Zindor's may nevertheless bear fruit. SnowRise let's rap 19:18, 9 April 2022 (UTC)
April 10
What is a 1833 Boston soda factory?
I'm reading a biography on a person (Daniel) with a time period around 1833. It says, His first work in that city was pumping soda for Mr. Darling, who then had a large soda factory on Franklin Ave in Boston. What was Daniel doing for his job? What was a 1833 soda factory? What did it make, in today's terms?--Doug Coldwell (talk) 14:02, 10 April 2022 (UTC)
- Fizzy pop. I have found a Thomas Darling, and a Darling and Pollard, listed as soft drink bollters. See also [10] DuncanHill (talk) 14:14, 10 April 2022 (UTC)
- The company was Darling & Ireland, listed here. Bottles from the company are shown on this page and on that, and apparently they go for several hundred dollars these days. Primarily that would have been soda water I guess, not sure whether they were (always?) flavoured. Addendum: DuncanHill's last link explains how Pollard turned into Ireland. --Wrongfilter (talk) 14:25, 10 April 2022 (UTC)
- Very good. I do believe that clears this up with the source above found by User: DuncanHill that says further, In 1845 Mr. Ireland returned to Boston, and with his brother George, purchased the business of Darling & Pollard, manufacturers of soda-water on Franklin avenue, which they subsequently sold out to Scripture & Parker, on Court square. Thanks for clearing this up.--Doug Coldwell (talk) 14:58, 10 April 2022 (UTC)
- See also Soda jerk and Soda fountain#History. Alansplodge (talk) 12:17, 11 April 2022 (UTC)
- Very good. I do believe that clears this up with the source above found by User: DuncanHill that says further, In 1845 Mr. Ireland returned to Boston, and with his brother George, purchased the business of Darling & Pollard, manufacturers of soda-water on Franklin avenue, which they subsequently sold out to Scripture & Parker, on Court square. Thanks for clearing this up.--Doug Coldwell (talk) 14:58, 10 April 2022 (UTC)
For a F-4 Phantom II at Lancaster, California
On this page, I saw a F-4 Phantom II is at Lancaster, California, who can find its serial number? --Great Brightstar (talk) 17:47, 10 April 2022 (UTC)
- Seems to be here. --OuroborosCobra (talk) 18:50, 10 April 2022 (UTC)
- Last Military Serial: 64-0952 USAF / Construction Number: 1364 / Last Civil Registration: N401AV [11] Alansplodge (talk) 12:13, 11 April 2022 (UTC)
April 11
Electrotyping of mid-19th century
Even after I looked up electrotyping I still don't get it. Can someone explain what it is in simple layman's terms. I'm reading a biography from a 1915 book and it says, He invented the art of electrotyping in copper, wood-cuts and type such as are used today. In simple terms what does this mean? The book goes on to say, In 1846 he electrotyped in copper the arm of a child. What exactly did this inventor do here?--Doug Coldwell (talk) 10:50, 11 April 2022 (UTC)
- It's the same as electroplating, basically, just using a shaped mold which gets filled in with the material being plated (typed in this case, I guess). The way it works is that you have an electrically conductive mold (the cathode), and some kind of metal ions in solution, along with an anode made of the metal whose ions are in the solution, which you are going to plate. If you drive an electric current through the solution, the metal ions pick up the electrons at the cathode, and become reduced to the native metal; at the same time, electrons are taken from the anode, oxidizing the metal to provide more ions to the solution. So long as an electromotive force (i.e. voltage) is correctly applied, the process will continue until the anode runs out. --Jayron32 11:17, 11 April 2022 (UTC)
- To respond to your italicised sentence specifically. First a negative mould is made of the object, whether it be an child's arm or more commonly a wood-cut. The mould can be made of a low-temperature setting substance such as plaster of Paris. The mould is coated with a very thin layer of a conductive material, often graphite. From memory (but unfortunately without citations) the conductive surface can be applied directly to the original and can then be taken up by the mould. Then, as jayron32 explains copper is deposited onto this layer creating a negative of a negative = a positive! Once a sufficient layer has been built up for the particular application, it can be parted from the mould and a backing applied as required. Martin of Sheffield (talk) 11:35, 11 April 2022 (UTC)
- There was also a papier mache version of the process. Here's a book about it: Stereotyping and electrotyping. ...In fact there are plenty of other books linked to at the foot of the article. Card Zero (talk) 12:08, 11 April 2022 (UTC)
- Be careful with that book. The first part contrasts and compares the plaster and papier maché methods, but that is for stereotyping a quite different process. Starting on page 112 it only mentions wax as the mould material for electotyping. Martin of Sheffield (talk) 13:53, 11 April 2022 (UTC)
- Oh, dammit. Perhaps I'm being confidently wrong. Card Zero (talk) 15:49, 11 April 2022 (UTC)
- Be careful with that book. The first part contrasts and compares the plaster and papier maché methods, but that is for stereotyping a quite different process. Starting on page 112 it only mentions wax as the mould material for electotyping. Martin of Sheffield (talk) 13:53, 11 April 2022 (UTC)
- There was also a papier mache version of the process. Here's a book about it: Stereotyping and electrotyping. ...In fact there are plenty of other books linked to at the foot of the article. Card Zero (talk) 12:08, 11 April 2022 (UTC)
- To respond to your italicised sentence specifically. First a negative mould is made of the object, whether it be an child's arm or more commonly a wood-cut. The mould can be made of a low-temperature setting substance such as plaster of Paris. The mould is coated with a very thin layer of a conductive material, often graphite. From memory (but unfortunately without citations) the conductive surface can be applied directly to the original and can then be taken up by the mould. Then, as jayron32 explains copper is deposited onto this layer creating a negative of a negative = a positive! Once a sufficient layer has been built up for the particular application, it can be parted from the mould and a backing applied as required. Martin of Sheffield (talk) 11:35, 11 April 2022 (UTC)