tag:blogger.com,1999:blog-8883034.post1898083589409158647..comments2024-10-02T09:09:09.121+02:00Comments on atdotde: Relativistic transformation of temperatureRoberthttp://www.blogger.com/profile/06634377111195468947noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-8883034.post-7543221399288943852017-06-10T05:40:51.323+02:002017-06-10T05:40:51.323+02:00>> I do not agree that it is out of the que...>> I do not agree that it is out of the question to have a thermometer with a relative velocity in thermal equilibrium with a heat bath at rest.<br /><br />Assume thermometer and system are made from the same type of atoms (to keep the argument simple).<br />The distribution of velocities of those atoms has to be different, otherwise they would not be moving relative to each other in any reasonable sense.<br />But if the velocity distributions are different, they are not in equilibrium.<br /><br />wolfgangnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-54921120596256085512017-06-09T14:59:12.580+02:002017-06-09T14:59:12.580+02:00Rather than commenting here, I updated the origina...Rather than commenting here, I updated the original article in order to be able to write formulas. See above. Roberthttps://www.blogger.com/profile/06634377111195468947noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-70206314611275330972017-06-09T11:54:57.331+02:002017-06-09T11:54:57.331+02:00This whole story doesn't make sense but in gen...This whole story doesn't make sense but in general, not special, relativity. <br /><br />Another way of realizing what's going on is that, when the temperature is fixed, the energy isn't-it fluctuates. Which means that time translations are not well-defined, therefore Poincaré invariance isn't a global symmetry and it doesn't make sense to ask any question that depends on this assumption. While all these issues may have been discussed in many ways in the past, it's the intuition that's come from Unruh's work, building on Hawking's, that's relevant and clarifies the picture. Once one works within the context of non-inertial frames, there's nothing more to discuss-beyond what could happen if backreaction can't be ignored. While this is an issue for the applications that Hawking has in mind, it isn't for Unruh, where the accelerating system is a probe of the spacetime. Anonymoushttps://www.blogger.com/profile/06379940898372024773noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-60532755256351557502017-06-09T09:19:31.709+02:002017-06-09T09:19:31.709+02:00You want to know how temperature transforms under ...You want to know how temperature transforms under Lorentz transformations?<br /><br />Let's consider an example. Take an ideal gas. In the rest-frame of the gas (i.e., the frame in which the mean velocity of the gas molecules is zero), the pressure and density are proportional to each other. And the constant of proportionality is (up to a constant factor) the temperature.<br /><br />Now, in that frame, the stress tensor T_{\mu\nu} is diagonal, T_{\mu\nu} = diag(ρ,p,p,p). Hence, the temperature is (up to the aforementioned factor) the ratio of the distinct eigenvalues of T_{\mu\nu}.<br /><br />In any other Lorentz frame, T_{\mu\nu} is, of course, no longer diagonal. But it's still a symmetric tensor. The temperature is defined to transform as a scalar field, which is the ratio of the distinct eigenvalues of T_{\mu\nu} in the local rest-frame of the gas.<br /><br />The generalization to other fluids with other equations of state is straightforward. And the whole story generalizes to General Relativity. See any textbook on cosmology.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-16547271638047848072017-06-09T07:49:37.121+02:002017-06-09T07:49:37.121+02:00I think that one should define the temperature in ...I think that one should define the temperature in equilibrium as the inverse length of the Euclidean time circle. If you start your experiment with a line element<br /><br />ds^2 = -dt^2 + dx^2<br /><br />and posit, in addition, that \tau=it (or is it -it? I always forget) has periodicity<br /><br />\tau \sim \tau+\beta<br /><br />then you can define the temperature as the inverse length of the time circle.<br /><br />You can carry out whatever coordinate transformation you want. It may be that the parametric length of the time circle will change as a result, but the temperature which is defined as the actual length of the time circle will remain the same since it is defined in a coordinate invariant way.<br /><br />The entire discussion can be stated more covariantly. Given a line element with a timelike Killing vector we can define an equilibrated configuration as one which is periodic in Euclidean time. The temperature would be the inverse length of the thermal circle which is coordinate invariant. <br /><br />So this is a notion of temperature which is well defined. You might wonder what your thermometer will measure when you do the actual experiment. There's some discussion that can be carried out here but before that, let me point out that you'd be using the thermometer in a setting which it wasn't designed for, so it's not clear why one would do that (apart from saying that it's an interesting academic question) and it's not clear why one would want to call the result of the measurement a temperature. <br /><br />If you are in equilibrium and do want to measure the temperature I would suggest measuring the average energy density instead, which is well defined. With that information (and perhaps the energy flux) one can obtain the temperature.Amosnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-86759102920856201672017-06-08T21:11:49.828+02:002017-06-08T21:11:49.828+02:00@Robert
as you already stated there is a lot of l...@Robert<br /><br />as you already stated there is a lot of literature on this, e.g. arxiv.org/abs/1606.02127<br /><br />In order for a "thermometer" to measure the temperature of a system (the rail) both need to be in equilibrium, but moving relative to each other they cannot be, so the measurement is just meaningless.<br /><br />Btw you dont really have to go to the relativistic case, the non-relativistic Doppler effect is sufficient for your "paradox" imho.<br /> wolfganghttp://tsm2.blogspot.comnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-23068083089539925122017-06-08T20:00:44.392+02:002017-06-08T20:00:44.392+02:00In the non-relativistic approximation, temperature...In the non-relativistic approximation, temperature is related to the average kinetic energy-only the latter isn't a Lorentz invariant quantity.That's why it isn't useful, in the relativistic context. <br /><br />This presentation: http://www.hartmanhep.net/topics2015/3-Rindler.pdf might, also, be useful.Anonymoushttps://www.blogger.com/profile/06379940898372024773noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-8424607452169265612017-06-08T15:23:25.057+02:002017-06-08T15:23:25.057+02:00Isn't temperature still going to be a measure ...Isn't temperature still going to be a measure of the average kinetic energy of the constituent atoms of the rail? If they're Boltzmann distributed with mean kT in the rest frame then why wouldn't the temp as seen by the conductor be the mean of the new velocity distribution in which each probability bin has the same probability mass but is now associated with the boosted velocity?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8883034.post-74278889859305622942017-06-08T13:42:01.069+02:002017-06-08T13:42:01.069+02:00Temperature, in a relativistic context, does mean ...Temperature, in a relativistic context, does mean acceleration. The relation was found by Unruh. So an observer, at equilibrium in T, is uniformly accelerating with an acceleration, a, related to the temperature by T = hbar a/(2π). http://www.scholarpedia.org/article/Unruh_effect<br />One notices that it is a quantum effect-so for a classical object the limit is quite tricky. <br />Anonymoushttps://www.blogger.com/profile/06379940898372024773noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-60663734846628043432017-06-08T13:32:57.997+02:002017-06-08T13:32:57.997+02:00Stam, I am not sure I understand your point. I don...Stam, I am not sure I understand your point. I don't think temperature implies acceleration, at best it's the other way round. Just for the fun of it, please compute the acceleration corresponding to 300K that I roughly have here in my room. I am sure, this would exceed the 5-7g that my body might be willing to tolerate by many orders of magnitude.Roberthttps://www.blogger.com/profile/06634377111195468947noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-65125567678043743082017-06-08T13:32:24.503+02:002017-06-08T13:32:24.503+02:00Finite temperature, in a special relativistic cont...Finite temperature, in a special relativistic context, means acceleration. This was the point of Unruh's observation: a uniformly accelerating observer, in Minkowski spacetime, is in equilibrium at a specific temperature; conversely, a relativistic observer, in equilibrium at a fixed temperature, is uniformly accelerating.<br /><br />Therefore the resolution of the apparent paradox regarding the Doppler shift is the usual one: a uniformly accelerating frame isn't an inertial frame. Anonymoushttps://www.blogger.com/profile/06379940898372024773noreply@blogger.comtag:blogger.com,1999:blog-8883034.post-69980108019209331132017-06-08T13:28:58.863+02:002017-06-08T13:28:58.863+02:00This comment has been removed by the author.Anonymoushttps://www.blogger.com/profile/06379940898372024773noreply@blogger.com