User talk:JarahE

Welcome

Hello, JarahE, and welcome to Wikipedia. Thank you for your contributions. I hope you like the place and decide to stay. If you are stuck, and looking for help, please come to the Wikipedia Boot Camp, where experienced Wikipedians can answer any queries you have! Or, you can just type {{helpme}} on your user page, and someone will show up shortly to answer your questions. Here are a few good links for newcomers:

I hope you enjoy editing here and being a Wikipedian! By the way, you can sign your name on Talk and vote pages using three tildes, like this: ~~~. Four tildes (~~~~) produces your name and the current date. If you have any questions, see the help pages, add a question to the village pump or ask me on my talk page. Again, welcome! Kukini 16:08, 6 March 2006 (UTC)[reply]

Supergravity

Thanks for the comments on supergravity, I replied there. Also: I invite you to join the conversations on the talk page of Wikipedia:WikiProject Physics. You might find Wikipedia:WikiProject Mathematics interesting as well. linas 01:36, 14 March 2006 (UTC)[reply]

Remarks

One note, it is good to use an edit summary; it helps others see what you changed. Thanks. :)

Thank you for the examples at unitary operator. I have a suggestion, the example of unitary projective operator is rather big, and I would think it may be better off forked to its own article unitary projective operator, while keeping at unitary operator just a short blurb, more inline with the other examples. Wonder what you think, you can reply here, thanks. Oleg Alexandrov (talk) 02:20, 15 March 2006 (UTC)[reply]

Yeah, I thought it was too big too. Feel free to split it off, but the other should be Projective Unitary Operator. Really I feel more comfortable writing about the Projective Unitary Group, since I don't know much about operator theory and writing about the group I can talk about its topology. JarahE 16:16, 15 March 2006 (UTC)[reply]

You are putting too many uppercase letters, sir. :) Must be "projective unitary group" that is. :) Oleg Alexandrov (talk) 02:32, 16 March 2006 (UTC)[reply]

Dear sir, in the correction you've italicized U(N)'s, SU(N)'s and PU(N)'s (although only some of them). I was under the impression, me Lord, that these groups aren't supposed to be italicized (although admittedly I've italicized them myself many times in papers). JarahE 13:10, 16 March 2006 (UTC)[reply]

My Lord, you may be right, as I saw in other math articles, SU may no need italic, however, the N inside of them may need. But note that when in math tags, the SU gets italicized, so maybe it should be \mathrm{SU}. Don't know, I leave it up to you. Oleg Alexandrov (talk) 16:05, 16 March 2006 (UTC)[reply]

Enemy number one

The Poincaré symmetry ISO(3,1), except for ISO(3) which is ok

Our universe, which is more or less the Freedman-Lemaître-Robertson-Walker universe, is not invariant under Lorentz boosts. Except for local perturbations like galaxies, it does respect the remaining ISO(3) symmetry. This is a robust statement that survives homogenous dark energy, inflationary periods, etc. Thus the ISO(3,1) symmetry is at best a spontaneously broken symmetry, Higgsed by the VEV of the graviton, i.e. by the metric. Ordinarily when a symmetry is spontaneously broken, it is restored at high energy/temperature, when the offending VEV, the order parameter of the symmetry breaking, vanishes. However the boost symmetry is broken by the big bang itself, I would be surprised if the bang bang becomes negligible in the ultraviolet. Thus I don't think that, in this instance, there is any meaningful distinction between a sponaneous breaking and an explicit breaking of the boost symmetry.

The boosts do not commute with the Hamiltonian.

The boost generators are skew-Hermitian, but to be conserved charges (they do not include time reversal) they would need to be Hermitian matrices. It is well known, of course, that the boosts are not conserved charges, I just want to emphasize that in quantum field theory I don't know of any other examples in which we impose a continuous symmetry that is not generated by a conserved charge.

Our universe possesses a prefered reference frame, the reference frame of the cosmic microwave background (CMB). Our velocity with respect to this frame has been accurately measured. This is the frame of the last scattering surface, and so it is most likely the frame of the big bang itself.

No finite number of Lorentz-invariant counterterms is sufficient to cancel the divergences in Einstein gravity. Fortunately counterterms preserving only the ISO(3) symmetry will be dynamically generated in the effective action as a result of the forementioned Higgsing of the boosts. Plenty of finite combinations of ISO(3)-invariant counterterms suffice to renormalize gravity sacrificing neither locality nor unitarity ... perhaps too many.

We have no experimental evidence to suggest the existence, nor the consistency of a Lorentz-invariant spacetime, such as Minkowski space

Boosts do not preserve the area, in Planck units, of the event horizon of a black hole. This area is commonly believed to obey the Bekenstein-Hawking entropy formula which states that it is proportional to the entropy of the black hoke, and is therefore the logarithm of a natural number. Lorentz boosts transform lengths and therefore areas by arbitrary scalings, and so generically take the logarithm of natural number to a real number which is not the logarithm of a natural number. Moreover, this natural number is usually associated with the number of physical microstates, as has been demonstrated explicitly for some very supersymmetric black holes in string theory, and therefore should be Lorentz-invariant. While the Planck area is not invariant under the usual Lorentz boosts, it may be invariant for example under the Lorentz boosts in doubly-special relativity.

ISO(3) preserves a Hamiltonian, in particular it preserves the Hamiltonian of observers at rest with respect to the CMB. Given a choice of Hamiltonian, one can define a vacuum in quantum field theory. In a gauge theory one can then impose a physical state condition, eliminating undesirable, negative-normed states like temporal photons. This cannot be done in a generally covariant manner, a non-inertial observer witnesses a vacuum related by a Bogoliubov transformation in which the physical state condition is no longer satisfied. The physical state condition therefore choses a set of inertial observers, not a velocity, but an acceleration (so really this is more a point against general covariance). JarahE 16:34, 4 April 2006 (UTC)[reply]

Hey, that was one of the more interesting things I've read in a while. Are you planning on moving this to an article at some point? linas 04:11, 25 April 2006 (UTC)[reply]

Re: Black holes

Jarah -

First of all, as long as you keep it out of the article space, I do not mind your using Wikipedia as a soapbox.

I see your point about Hawking radiation and free scalars. One of the surprises that I got was when I compujted the amount of Hawking radiation the would be emitted under my distorted Schwarzschild metric, and got zero for an answer. Later, when I was taking GR courses at U. Md., I learned that Hawking radiation was uniquely associated with the event horizon as a null Killing surface. As the black hole is unreachable in my theory, the loss of Hawking radiation is to be expected.

If you like, you can see if the metric $ds^{2}=(1-2m/r)dt^{2}-[(r-m)^{2}/(r-2m)^{2}]dr^{2}-r^{2}d\theta ^{2}-(r\sin \theta )^{2}d\phi ^{2}$ has any the other signs of a "sick" theory that you noted. I would be curious about your findings.

I will look at your arXiv article soon. I have just resubmitted by article, and should have some breathing room for the next few weeks. --EMS | Talk 22:23, 22 April 2006 (UTC)[reply]

EMS -

Thanks for your message. Your solution still appears to have a horizon, so unless you've changed the dispersion relations for electromagnetic waves I'd still suspect that there will be a problem. However, if there's no Hawking radiation, there's no problem. The problem is essentially that ghosts get Hawking radiated.

r=2m

does still exist, and still is singular, but if you cannot pass it. If you should reach it (which is theoretucally possible), you would be send back from whence you came. However, to reach it there needs to be a black hole, and that inability to pass

r=2m

prevents the formation of such a beast. As for Hawking radiation, I guess I need to recreate my exercise for you, but that is not a priority for me at this time. In the meantime, I can send you a copy of the current version of my article if you like. (I have just resubmitted it, but have not yet gotten around to placing the new version on my web site.) --EMS | Talk 00:17, 23 April 2006 (UTC)[reply]

But now that you've mentioned that you're at U.Md., I should say that I just spent the last few hours reading Eling and Jacobson's paper from yesterday on black holes in the Einstein-Aether theory. They seem to have some static solutions, but they're singular and so I think there's a reasonable chance that they can't form by the collapse of a spherical shell. It's true that I could raise this objection for Schwarzschild and I'd be wrong, but here there's a new repulsive force coming from the potential term for the ether and so maybe here it can beat gravity and prevent a horizon from forming? In particular their transition from collapsing stars to black holes which they say on page 4 needs to be accompanied by some kind of strange burst seems pretty suspicious to me.

I'm printing their paper from March on stars now so I can see if when you try to shrink them to black holes they bounce back. Anyway, this Einstein-Ether theory seems to me like a very concrete setting in which you can challenge black holes, since they could be ruled out even classically (I say classically, but really I guess there are hidden hbars in the coupling of the ether, but they're already in the Lagrangian so they're easy to deal with).

Have you had a chance to talk to these guys about the Einstein-ether project and black holes? JarahE 23:50, 22 April 2006 (UTC)[reply]

You misunderstand me. I took GR course at U. Md in '99 and '00, but am not affiliated with U. Md now. I did get some encouragement from Tad Jacobson to try to publish my work, and am grateful to him for that, but have not been in contact with him recently. Perhaps it is time to drop him a line. --EMS | Talk 00:17, 23 April 2006 (UTC)[reply]

You read my mind. Yeah, I don't know Ted. I just discovered his Einstein-Aether theory Friday when a student in a class I was teaching told me about the paper (I don't read gr-qc). But as you'll see in my preprint I was suggesting something similar, Ted's beat me to it but I'm interested to learn all that I can about it now. His black hole paper from yesterday is full of open problems, including whether they're even stable.

The problem of whether collapsing matter can form a black hole in their theory isn't so well defined, since you have to put in by hand how the matter is coupled. But since I'm already predisposed, like you, to conclude that black holes can't form I can reverse engineer the coupling that I need.

In summary, yeah, don't loose your connection to Ted, I think he's on to something very important, even if his theory isn't quite right. But before going to talk to him it'd probably be a good idea to familiarize yourself with what he's been doing lately.JarahE 00:28, 23 April 2006 (UTC)[reply]

Jarah -

I have looked over your arXiv paper. I was particularly intrigued by the following:

Principle: The universe as seen by any observer obeys the usual laws of quantum mechanics.

In my own theory, I state that gravitational time dilation and gravitational length contraction occur with respect to spacetime as an observer views it. So we both are keying on the direct observations of observers, which is intriguing.

OTOH, I cannot help but wonder if you are missing something: I find it hard to credit a view in which "ghosts" appear for the continuously accelerating observer but not for the intertial one. Let me put it to you this way: My motion/acceleration should not cause photons to appear elsewhere in spacetime. Similarly there is something wrong if I must see ghosts but the inertial observer whom I am instantaneously co-moving with does not. So I think that the generic horizon problem that you are citing is a red herring. However, that is not to contradict your assertion that Hawking radiation is evidence of trouble. (You may find that you can zero out the Rindler ghosts but not the event-horizon/Hawking-radiation ghosts. I would be much more comfortable with your work if you could demonstrate that.) --EMS | Talk 15:46, 23 April 2006 (UTC)[reply]

This is the point, I think that the particles that you observe should be independent of your acceleration. But the Unruh effect, which everyone believes but me (and apparently you), states that your acceleration tells you how many particles you see. So I'm using the Unruh effect to create a contradiction. I can't make the argument without the Unruh effect, because the whole point is to argue that the Unruh effect is incorrect.

For Hawking ghosts I think it'd be even more serious because the ghosts would have nontrivial correlators with the physical sector. This already happens in a paper of Strominger's from the early 90s that I cite. JarahE 16:52, 23 April 2006 (UTC)[reply]

I was not aware of the Unruh effect. I can believe it, as acceleration can for instance generate Cherenkov radiation, but that obviously is not quite the same mechanism. I do see your point above even so. So I would guess that either the Unruh effect is indeed incorrect, or spacetime is odder than you are giving it credit for. Just be advised that I cannot exclude the later possibility given my own research. --EMS | Talk 18:21, 23 April 2006 (UTC)[reply]

It's not just you that can't exclude it, really pretty much everyone accepts it although Jacobson seems to put in qualifiers about how it has built in assumptions about transPlanckian physics. So far as I know my article is the only argument against the Unruh effect, but it's an argument, density matrices with negative eigenvalues mean negative probabilities and I think this could be a big problem with the Unruh scenario. JarahE 18:25, 23 April 2006 (UTC)[reply]

I must admit that I have some difficultly with purely philosophical arguments: It's all fine and dandy not to like something, but what are you going to do about it? That is part of the reason I came up with FBGR. In your case, I am loathe to declare Minkowski spacetime to be corrupt. Most likely, once you work things out properly, you will find that the positive and negative eigenvalues cancel, and that Unruh radiation has a zero probablity of existing. Otherwise, there may be a subtle interaction with virtual photons for accelerating observers. I won't put the later beyond QM to produce, but I agree that this is an odd effect. However, it is also one that I don't intend to lose sleep over. --EMS | Talk 23:55, 23 April 2006 (UTC)[reply]

Calculating a probability and getting a negative answer isn't a philosophical objection, it means that one of the underlying assumptions is wrong (or the calculation, but I've checked that over with lots of people and there's no mistake there).

Of course you're right that the big open question is to find a theory, either in the literature or by constructing it, that doesn't have this problem. There's no need to eliminate Minkowski space as a solution, it suffices to eliminate Rindler space (the thing seen by the accelerating observer in GR) as a solution. Rindler space is a solution to Einstein's equations, but it's a common situation these days that Einstein's equations have solutions that are sick and these sicknesses are cured in the quantum theory, where the solution gets some quantum corrections. For example Horava and friends have shown that this is the case with closed timelike curves in Godel universes. So the solution I think is that the accelerating observer in Minkowski space doesn't see the Rindler metric, but instead the Rindler metric with some quantum corrections that make the horizon go away. This wouldn't be a particularly new phenomenon, in the fuzzball story quantum corrections eliminate the event horizons in extremal black holes in string theory.

As for working things out properly, I didn't make any mistake in the calculations, really there's nothing to them, its the framework that's the problem. Also there's no missing QM effect, the theory is free and I know exactly how it works, it's only got one physical state after all and all of the loop diagrams vanish so there are no quantum corrections. So calculation is not a classical approximation.

Your comment about cancellations of course is relevant. The key is that while there are cancellations in the correlators, there is no cancellation in individual terms in the density matrix. For example, there are elements with only temporal photons and nothing else (no ghosts) to cancel them. If you measure the system there's a nontrivial chance that you project to this state, which then has nothing to cancel its pathologies.

Summary: I'm not claiming that Minkowski is sick, just that the accelerating observer doesn't see Rindler space. But there's no danger of a calculational error, the ghosts are free scalars and free scalars are in a thermal state according to Unruh, that's all you need. JarahE 00:33, 24 April 2006 (UTC)[reply]

JarahE wrote:

I'm not claiming that Minkowski is sick, just that the accelerating observer doesn't see Rindler space.

To me, you are now arguing that Minkowski spacetime isn't Minknowski spacetime. The Rindler wedge is a part of Minkowski spacetime, and under special relativity, that is what will be observed.

At this point, I would like to to take a different tact. You also wrote above:

For Hawking [radiation,] ... the ghosts would have nontrivial correlators with the physical sector.

What do you mean by a "trivial correlator"? Am I correct in inferring from this that the Rindler wedge has a trivial correlator while the Schwarzschild solution's event horizon has a nontrivail one? --EMS | Talk 19:07, 24 April 2006 (UTC)[reply]

I don't understand your reference to special relativity. For one, I don't see why we're using special relativity to understand acceleration, special relativity just tells me how to boost from one frame to another. Minkowski space is invariant under boosts, you don't get Rindler space no matter how many times you boost it. So, I disagree that I'm objecting to Minkowski space (I'm agnostic on whether quantum gravity should make sense in Minkowski space). Instead I'm making the weaker claim that no matter how much you accelerate, every event will eventually be in your past. This is in contrast with general relativity, but this doesn't bother me as general relativity isn't a good approximation at the arbitrarily high energies involved.

Your comment on the correlators is important. The Rindler and Minkowski observer agree on the subset of correlation functions in which all points lie inside the Rindler wedge. This has historically been taken as evidence that everything is ok in the Rindler picture. However the correlators don't tell you the whole story, they don't even seem to know about the temperature (not without some more information like a decomposition into which trajectories give which correlators). One critical piece of extra information that the density matrix has but the correlators don't, which is really the source of the problem, is that there are elements with nonzero eigenvalue but with negative normed-states. This means that there is a finite probability of measuring a negative-normed state ... which can never happen. The correlators don't know this because the Minkowski observer, who has the same correlators, has a density matrix with a single nonzero eigenvalue corresponding to the physical state and so he is ok.JarahE 19:23, 24 April 2006 (UTC)[reply]

Minkowski spacetime is the spacetime of special relativity (SR), and Rindler's metric is just a diffeomorphism of the Minkowski metric. So SR itself calls for Rindler horizons to form. Certainly that is a nuisance to me in arguing against event horizons and black holes, but I stepped around it by distinguishing between a "dynamic" horizon which exists only for the observer and the "static" horizon of the black hole which exists at the same events for all external observers.

As for this discussion of Unruh radiation, I must admit I am coming up against a limit. I studied eigenvalues and eigenstates when I did quantum mechanics in college, but that stuff is 30 years in my past, and I need to refresh myself on it. I see your concern about measuring something with a negative probablitity of happening, but I can offer little more than towards resolving the problem than the thoughts that I have already expressed. Perhaps if there was some was of measuring the Unruh radiation we could then see whether you or everyone else is correct. In either case, the result would redefine the problem. --EMS | Talk 20:50, 24 April 2006 (UTC)[reply]

I can't follow your argument on SR at all. Minkowski spacetime is the spacetime of SR, good. Now you want to consider a diffeomorphism??? In SR you only get to do Lorentz transformations, not diffeomorphisms, maybe you're thinking of GR? Rindler space has a different metric from Minkowski space, in particular Minkowski space is geodesically complete and Rindler space isn't. So if you've constructed a diffeomorphism between them you must not be transforming the metric covariantly, or else you'd find that they'd both have to have the same geodesic completeness property. Thus I don't see what SR has to do with diffeomorphisms, nor what diffeomorphisms have to do with the relation between Minkowski and Rindler space.

As for measuring the Unruh radiation, this is a good point. What is known about measurements of Unruh radiation? Is there some argument that we it's too small to have any experimental signatures that we could pick up? Say when binary neutron stars are falling into each other or in supernovae or some other events that could have big accelerations ... this is definitely something I should know ... JarahE 21:02, 24 April 2006 (UTC)[reply]

SR and diffeomorphisms

I think I need to appraise you on how I view the relationship between special relativity and general relativity. In a strict sense, you are correct that only the Minkowski metric is used in SR itself. Even so, Einstein in 1907 inferred the existance of gravitational time dilation using SR, and I similarly infer gravitational length contraction using SR. So through the use of the Lorentz transformation, fundamental acceleration effects are manifested. Of course, life gets easier if you can move away from the SR paradigm of only using the Minkowski metric and its related coordinate system. So you can create another view of spacetime for an accelerated observer: $ds^{2}=(1+gz)^{2}dt^{2}-dx^{2}-dy^{2}-dz^{2}$ in gemetrized coordinates (such that $c=1$ ). This is still the Minkowski spacetime, just mapped differently, and with the "trouble" that $z<-1/g$ is not a "real" part of the map. Of course, I see that as the fault of the map, not of the spacetime.

In any case, while I am using GR tools here this is still IMO the SR spacetime. That the horizon at $z=-1/g$ should generate photons I find to be odd, but as I said above I dislike the idea of saying a-priori than quantum mechanics cannot cause such a thing to happen. (It is possible that as this issue gets researched more, it may come to be put on a sounder theoretical basis wherein the effect is due to the local acceleration rather than the distant horizon.) --EMS | Talk 03:03, 25 April 2006 (UTC)[reply]

Your metric is not a complete specification of the spacetime, you need to also include a range for the variables. One choice will give you the full Minkowski space, then we're back where we started and are still considering the inertial observer. Another choice will give you the Rindler wedge. The Rindler wedge isn't just a change of coordinates from Minkowski, because it has a smaller range for the variables. SR never tells you to shrink the range for the variables, it just says to transform everything covariantly. When Einstein started thinking about how to extend SR to include acceleration, we started on a train of thought which led him to GR, which, of course, isn't valid at the high energies that come into play here due to the infinite red-shift of light from the horizon as seen by the Rindler observer (which means he sees things that were very blue, and so not described by GR).

As for particles being created by the horizon. The way that Unruh works is that virtual pairs of particles are created everywhere, but sometimes one falls behind the horizon so it seems like a single particle was created. This is all fine and good, until you do something crazy like trace out the states behind the horizon. This is because the fact that the partner existed was important, since the pairs cancel each others pathologies. Thus forgetting about stuff behind the horizon is what's killing the positive-definiteness of the Hilbert space and therefore the positivity of probabilities. This is why I argue against forgetting about (tracing over) stuff behind the horizon, but this means that stuff behind the horizon has a physical effect, which means it's still in causal contact, which means that in the quantum theory there is no horizon. JarahE 15:32, 25 April 2006 (UTC)[reply]

Oddness on top of oddness. The whole exercise of the Unruh radiation now seems corrupt to me since the virtual photon is not out of causal contact with the other virtual photon. Instead, the loss of causal contact is between the accelerated observer and the event of the virtual photons recombining. That the accelerated observer never sees that event does not mean that the virtual photon is not still heading towards it.

This is a very different situation from Hawking radiation, where the virtual photons lose causal contact with each other. --EMS | Talk 19:13, 25 April 2006 (UTC)[reply]

Since the virtual photon on the inside of the Rindler wedge is heading away from it's origin at the speed of light, no message can get to it from its partner on the other side unless it stops or turns. So it is out of contact, although it could get into contact if it wanted to. Similarly the Rindler observer could always get into contact with the whole Minkowski universe if he'd just stop accelerating. So the Rindler observer and virtual photon are both out of causal contact in the same sense. In the black hole case it's different, the guy outside can never see his friend who has fallen in without jumping in himself (and even then maybe not). JarahE 19:22, 25 April 2006 (UTC)[reply]

That argument misses the point. Either the virtual photon rejoins its partner, or it does not. What the accelerating observer is doing does not change that. In fact the observer's acceleration is in the future for the photons, and so cannot exert a causal influence on them. So either the photons disassociated or they did not, and that result is the same for all observers, inertial or not. The Rindler horizon is a coordinate horizon, not a physical one. The fact that you cannot see the end of the process does not mean that it did not end. After all, the trouble with Zeno's Paradox is not that Achilles never passes the tortise but instead that the exercise never does. --EMS | Talk 19:37, 25 April 2006 (UTC)[reply]

The pair-created photons that are creating the problems here aren't the ones that annihilate, but those that make it out to infinity and so contribute to the S-matrix. Of these, all but a measure zero subset are really moving away at the speed of light along a straight line, and so never hear from their partner since no message could catch up. The ones that annihilate, since the theory is free, don't couple to anything and so don't contribute to the Unruh effect. JarahE 19:46, 25 April 2006 (UTC)[reply]

Then maybe the CMB is Unruh radiation! All that I am saying is that the Rindler horizon cannot cause Unruh radiation because it is not a physical horizon such that all observers will agree that it exists at a given set of events. If the photons recombine for the inertial observer, then they also recombine for the accelerated observer. The difference is that the event of the recombination is inaccessible to the accelerated observer, but that does not destroy its existance. So either we are always surrounded by a sea of thermal photons (like the CMB) or we are not. It's that simple. --EMS | Talk 20:13, 25 April 2006 (UTC)[reply]

The whole point of Unruh radiation is that it's observer dependent. Personally I think that the number of particles should be observer independent, but I'm the only one. And Unruh's argument indicates that I'm wrong, but it has a hidden assumption about high-energy physics ... but it is consistent for the system he considers.

As for the CMB, we don't live in Minkowski space so it's not quite the same, but if we did, then you'd conclude that observers in free fall (inertial frames) don't see Unruh radiation, which includes us. I once calculated the rate at which we'd be accelerating if the CMB were Unruh radiation, and I got a really huge/implausible number, we'd feel a force that big. Plus you'd expect the Unruh radiation to just come from one direction. JarahE 20:18, 25 April 2006 (UTC)[reply]

First of all we are not inertial observers. Under the equivalence principle of general relativity we are accelerated observers (unless you happen to live in free fall). In any case, the rest of your points with respect to my facetious suggestion are on the mark. I gather that you are playing the devil's advocate here, but are teaching me about Unruh radiation in the process, which I appreciate. At this point, I can only see a falacy in taking a piece of Minkowski spacetime, and treating the particle pairs that cross the border of the wedge as producing free thermal photons a-la Hawking radiation. It is like expecting a road to end at the edge of a map. --EMS | Talk 02:15, 26 April 2006 (UTC)[reply]

I'm not in free-fall. But my planet (and I presume yours) is, to a very good approximation (neglecting solar wind) in free-fall all the time. The surface isn't, but the center is and the satellites that measure the CMB are in free-fall too. Experimentally, a person on the surface of the Earth (who therefore isn't in free-fall) measures the same CMB (up to absorption by the atmosphere) as a satellite, which is in free-fall. So the Earth's gravitation isn't an important effect here. Really the CMB seems to be quite well explained by primordial plasma condensing into atoms, this is one of the big successes of cosmology (and one of the only points where I agree with the party line).

I agree with you on the dislike for geodesically incomplete spacetimes. I don't like them either. Which is why I'm trying to prove that they're inconsistent. JarahE 02:27, 26 April 2006 (UTC)[reply]

I agree with you on the CMB.
The issue with Unruh radiation for me is not geodesic incompleteness. The Rindler wedge exists on Minkowski spacetime, which is goedesically complete. What is incomplete is the mapping between the Rindler wedge and Minkwoski spacetime. So the issue involves the boundary condition at the horizon, not the spacetime itself.

--EMS | Talk 13:22, 26 April 2006 (UTC)[reply]

Ok, it is something like a boundary condition, certainly it's something related to the boundary. However SR doesn't tell you anything about boundary conditions, and GR doesn't apply at the energies involved. If you pretend that it does apply and then try to quantize electromagnetism as usual you get Unruh radiation. I want to demonstrate that this is the wrong thing to do (which is harder than just arguing that it's unjustified, which is a consequence of the fact that GR doesn't apply to the high energies involved). To argue that it's the wrong thing to do, I show that the resulting quantum theory has negative probabilities. JarahE 17:37, 26 April 2006 (UTC)[reply]

I think that it is time to wrap this up. I accept that there is a flaw with Unruh radiation. However, I advise trying to be imaginative about where the flaw is. In principle, GR applies in this case, but at the same time flat spacetime is a very special case, lacking the features which instigated the creation of GR in the first place. I really think that the key is that Rindler wedge is just a piece of the Minkowski spacetime. Thus either the intensity or the energy of the Unruh radiation should be zero IMO. --EMS | Talk 23:18, 26 April 2006 (UTC)[reply]

Unruh effect

Interesting conversation above. For experimental measurements of the Unruh effect, I suppose one could just google. But, pretending Google doesn't exist, I note that the Casimir effect is a pure zero-point-energy effect. So the question is, does the Casimir effect change when the apparatus is accelerated (presumably because some of those virtual photons are becoming real due to the Unruh effect and are getting lost somehow)? Next, I notice that while traditionally, the Casimir effect is hard to measure, the latest MEMS (micro-electronic mechanical sensors) have become small enough that the Casimir effect is no longer ignoreable in their design. So if the Unruh effect changed the Casimir effect, and it wasn't totally miniscule for <100G accelerations, then its measureable in today's labs.

My current understanding of the Casimir effect is that it is essentially one and the same as the van der Waals attraction. (see Talk:Casimir effect). That is, one should consider the total possible spectrum of eigenvalues of the Hamiltonian, when one has a Hamiltonian with a free parameter (e.g. the separation between a pair of atoms, or between a pair of metal plates, etc.) The van der Waals force is just the derivative w.r.t. the parameter, of the sum-total of the spectrum. So, is the van der Waals force, the thing that keeps certain liquids and solids together, invariant under accelaration? Or are some of the photons that are helping atoms stick together being radiated away? Actually, I'm not even sure how to treat a Hamiltonian in an accelerated reference frame. Hmmm.. I guess that's the point ... way past my bed-time ... that's my excuse for being incoherent. linas 04:52, 25 April 2006 (UTC)[reply]

I don't have much time now, but please free feel to google experimental signatures for Unruh, I am curious. As for the Casimir effect, yeah I'd expect some shift, but it'd be supressed by something like the ratio of the acceleration to the Planck acceleration, so probably you couldn't measure this in a laboratory. I think the only way to get a high enough acceleration is to look at an astrophysical source, something with ultra high energy cosmic rays maybe. But I've been wrong before. JarahE 15:36, 25 April 2006 (UTC)[reply]

Large Hadron Collider is Science Collaboration of the Week

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Topological string theory

Thank you for your contributions to topological string theory and related articles. I think they are very useful in helping mathematicians and physicists understand each other. Joshua Davis 19:43, 31 August 2006 (UTC)[reply]

No problem. Coming from Duke, which in my small experience is the university with the best communication between mathematicians and physicists, this is quite a compliment. I don't know the math side so well (as you, for example) so I've been making these articles a bit physics-heavy. Feel free to fill this gap whenever you feel the urge. For example I wasn't brave enough to describe how instantons measure Gromov-Witten invariants and why they're not integral, and in Chern-Simons theory I didn't talk about how surgery lets you reduce calculations to knot invariants.JarahE 10:54, 1 September 2006 (UTC)[reply]

Total re-write of the main Physics page is in progess

You might like to join us at Physics/wip where a total re-write of the main Physics page is in progess. At present we're discussing the lead paragraphs for the new version, and how Physics should be defined. I've posted here because you are on the Physics Project participant list. --MichaelMaggs 08:04, 11 October 2006 (UTC)[reply]

Peer review for the list of baryons

Hi,

I noticed that you had an interest in particle physics, so I wondered you could head over the List of baryons and Talk:List of baryons pages a give some feedback. I'm currently trying to bring that article to Featured List status, but I'm not a particle physicist so I probably made half a dozen mistakes. Any help will be appreciated. Thanks in advance. Headbomb (talk · contribs) 21:37, 23 April 2008 (UTC)[reply]

WikiProject Physics participation

You received this message because your were on the old list of WikiProject Physics participants.

On 2008-06-25, the WikiProject Physics participant list was rewritten from scratch as a way to remove all inactive participants, and to facilitate the coordination of WikiProject Physics efforts. The list now contains more information, is easier to browse, is visually more appealing, and will be maintained up to date.

If you still are an active participant of WikiProject Physics, please add yourself to the current list of WikiProject Physics participants. Headbomb {ταλκ – WP Physics: PotW} 14:57, 25 June 2008 (UTC)[reply]

October 2008

Welcome to Wikipedia. Although everyone is welcome to contribute constructively to the encyclopedia, adding content without citing a reliable source, as you did to Alpha Magnetic Spectrometer, is not consistent with our policy of verifiability. Take a look at the welcome page to learn more about contributing to this encyclopedia. If you are familiar with Wikipedia:Citing sources, please take this opportunity to add references to the article. Thank you. -MBK₀₀₄ 19:49, 31 October 2008 (UTC)[reply]

I didn't add a reference, because the reference that I used is the sole reference which is already listed on the wikipedia page. It is the official AMS website. - Jarah

ArbCom elections are now open!

Hi,
You appear to be eligible to vote in the current Arbitration Committee election. The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to enact binding solutions for disputes between editors, primarily related to serious behavioural issues that the community has been unable to resolve. This includes the ability to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail. If you wish to participate, you are welcome to review the candidates' statements and submit your choices on the voting page. For the Election committee, MediaWiki message delivery (talk) 13:45, 23 November 2015 (UTC)[reply]