Imagine a long rail of a railroad track. It is uniformly heated to a temperature T and is in thermodynamic equilibrium (if you like a mathematical language: it is in a KMS state). On this railroad track travels Einstein's relativistic train at velocity v. From the perspective of the conductor, the track in front of the train is approaching the train with velocity v, so one might expect that the temperature T appears blue shifted while behind the train, the track is moving away with v and so the temperature appears red-shifted.
Following this line of thought, one would conclude that the conductor thinks the rail has different temperatures in different places and thus is out of equilibrium.
On the other hand, the question of equilibrium should be independent of the observer. So, is the assumption of the Doppler shift wrong?
A few remarks: If you are worried that Doppler shifts should apply to radiation then you are free to assume that both in front and in the back, there are black bodies in thermal contact with the rail and thus exhibiting a photon gas at the same temperature as the rail.
You could probably also make the case for the temperature transforming like the time component of a four vector (since it is essentially an energy). Then the transformed temperature would be independent of the sign of v. This you could for example argue for by assuming the temperature is so high that in your black body photon gas you also create electron-positron pairs which would be heavier due to their relativistic speed relative to the train and thus requiring more energy (and thus temperature) for their creation.
A final remark is about an operational definition of temperature at relativistic speeds: It might be difficult to bring a relativistic thermometer in equilibrium with a system if there is a large relative velocity (when we define temperature as the criterium for two systems in contact to be in equilibrium). Or to operate a heat engine between he front part of the rail and the back while moving along at relativistic speed and then arguing about the efficiency (and defining the temperature that way).
Update one day later:
Thanks for all your comments. We also had some further discussions here and I would like to share my conclusions:
1) It probably boils down to what exactly you mean when you say ("temperature"). Of course, you want that his at least in familiar situations agrees with what thermometers of this type or another measure. (In the original text I had hinted at two possible definitions that I learned about from a very interesting paper by Buchholz and Solveen discussing the Unruh effect and what would actually be observed there: Either you define temperature that the property that characterises equilibrium states of systems such there is no heat exchange when you bring in contact two systems of the same temperature. This is for example close to what a mercury thermometer measures. Alternatively, you operate a perfect heat engine between two reservoirs and define your temperatures via
$$\eta = \frac{T_h - T_c}{T_h}.$$
This is for example hinted at in the Feynamn lectures on physics.
One of the commentators suggested using the ratio of eigenvalues of the energy momentum tensor as definition of temperature. Even though this might give the usual thing for a perfect fluid I am not really convinced that this generalises in the right way.
2) I would rather define the temperature as the parameter in the Gibbs (or rather KMS) state (it should only exist in equilibrium, anyway). So if your state is described by density matrix $\rho$, and it can be written as
$$\rho = \frac{e^{-\beta H}}{tr(e^{-\beta H})}$$
then $1/\beta$ is the temperature. Obviously, this requires the a priori knowledge of what the Hamiltonian is.
For such states, under mild assumptions, you can prove nice things: Energy-entropy inequalities ("minimisation of free energy"), stability, return to equilibrium and most important here: passivity, i.e. the fact you cannot extract mechanical work out of this state in a cyclic process.
2) 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. You could for example imagine a mirror fixed next to the track and in thermal equilibrium with the track. A second mirror is glued to the train (and again in thermal equilibrium, this time with a thermometer). Between the mirrors is is a photon gas (black body) that you could imagine equilibrating with the mirrors on both ends. The question is if that is the case.
3) Maybe rails and trains a a bit too non-spherical cows, so lets better look at an infinitely extended free quantum gas (bosons or fermions, your pick). You put it in a thermal state at rest, i.e. up to normalisation, its density matrix is given by
$$\rho = e^{-\beta P^0}.$$
Here $P^0$ is the Poincaré generator of time translations.
Now, the question above can be rephrased as: Is there a $\beta'$ such that also
$$\rho = e^{-\beta' (\cosh\alpha P^0 + \sinh \alpha P^1)}?$$
And to the question formulated this way, the answer is pretty clearly "No". A thermal state singles out a rest frame and that's it. It is not thermal in the moving frame and thus there is no temperature.
It's also pretty easy to see this state is not passive (in the above sense): You could operate a windmill in the slipstream of particles coming more likely from the front than the back. So in particular, this state is not KMS (this argument I learned from Sven Bachmann).
4) Another question would be about gravitational redshift: Let's take some curve space-time and for simplicity assume it has no horizons (for example, let the far field be Schwarzschild but in the center, far outside the Schwarzschild radius, you smooth it out. Like the space-time created by the sun). Make it static, so it contains a timeline Killing vector (otherwise no hope for a thermal state). Now prepare a scalar field in the thermal state with temperature T. Couple to it a harmonic oscillator via
$$ H_{int}(r) = a^\dagger a + \phi(t, r) (a^\dagger + a).$$
You could now compute a "local temperature" by computing the probability that the harmonic oscillator is in the first excited state. Then, how does this depend on $r$?
Update one day later:
Thanks for all your comments. We also had some further discussions here and I would like to share my conclusions:
1) It probably boils down to what exactly you mean when you say ("temperature"). Of course, you want that his at least in familiar situations agrees with what thermometers of this type or another measure. (In the original text I had hinted at two possible definitions that I learned about from a very interesting paper by Buchholz and Solveen discussing the Unruh effect and what would actually be observed there: Either you define temperature that the property that characterises equilibrium states of systems such there is no heat exchange when you bring in contact two systems of the same temperature. This is for example close to what a mercury thermometer measures. Alternatively, you operate a perfect heat engine between two reservoirs and define your temperatures via
$$\eta = \frac{T_h - T_c}{T_h}.$$
This is for example hinted at in the Feynamn lectures on physics.
One of the commentators suggested using the ratio of eigenvalues of the energy momentum tensor as definition of temperature. Even though this might give the usual thing for a perfect fluid I am not really convinced that this generalises in the right way.
2) I would rather define the temperature as the parameter in the Gibbs (or rather KMS) state (it should only exist in equilibrium, anyway). So if your state is described by density matrix $\rho$, and it can be written as
$$\rho = \frac{e^{-\beta H}}{tr(e^{-\beta H})}$$
then $1/\beta$ is the temperature. Obviously, this requires the a priori knowledge of what the Hamiltonian is.
For such states, under mild assumptions, you can prove nice things: Energy-entropy inequalities ("minimisation of free energy"), stability, return to equilibrium and most important here: passivity, i.e. the fact you cannot extract mechanical work out of this state in a cyclic process.
2) 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. You could for example imagine a mirror fixed next to the track and in thermal equilibrium with the track. A second mirror is glued to the train (and again in thermal equilibrium, this time with a thermometer). Between the mirrors is is a photon gas (black body) that you could imagine equilibrating with the mirrors on both ends. The question is if that is the case.
3) Maybe rails and trains a a bit too non-spherical cows, so lets better look at an infinitely extended free quantum gas (bosons or fermions, your pick). You put it in a thermal state at rest, i.e. up to normalisation, its density matrix is given by
$$\rho = e^{-\beta P^0}.$$
Here $P^0$ is the Poincaré generator of time translations.
Now, the question above can be rephrased as: Is there a $\beta'$ such that also
$$\rho = e^{-\beta' (\cosh\alpha P^0 + \sinh \alpha P^1)}?$$
And to the question formulated this way, the answer is pretty clearly "No". A thermal state singles out a rest frame and that's it. It is not thermal in the moving frame and thus there is no temperature.
It's also pretty easy to see this state is not passive (in the above sense): You could operate a windmill in the slipstream of particles coming more likely from the front than the back. So in particular, this state is not KMS (this argument I learned from Sven Bachmann).
4) Another question would be about gravitational redshift: Let's take some curve space-time and for simplicity assume it has no horizons (for example, let the far field be Schwarzschild but in the center, far outside the Schwarzschild radius, you smooth it out. Like the space-time created by the sun). Make it static, so it contains a timeline Killing vector (otherwise no hope for a thermal state). Now prepare a scalar field in the thermal state with temperature T. Couple to it a harmonic oscillator via
$$ H_{int}(r) = a^\dagger a + \phi(t, r) (a^\dagger + a).$$
You could now compute a "local temperature" by computing the probability that the harmonic oscillator is in the first excited state. Then, how does this depend on $r$?
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12 comments:
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.
Therefore the resolution of the apparent paradox regarding the Doppler shift is the usual one: a uniformly accelerating frame isn't an inertial frame.
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.
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
One notices that it is a quantum effect-so for a classical object the limit is quite tricky.
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?
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.
This presentation: http://www.hartmanhep.net/topics2015/3-Rindler.pdf might, also, be useful.
@Robert
as you already stated there is a lot of literature on this, e.g. arxiv.org/abs/1606.02127
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.
Btw you dont really have to go to the relativistic case, the non-relativistic Doppler effect is sufficient for your "paradox" imho.
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
ds^2 = -dt^2 + dx^2
and posit, in addition, that \tau=it (or is it -it? I always forget) has periodicity
\tau \sim \tau+\beta
then you can define the temperature as the inverse length of the time circle.
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.
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.
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.
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.
You want to know how temperature transforms under Lorentz transformations?
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.
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}.
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.
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.
This whole story doesn't make sense but in general, not special, relativity.
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.
Rather than commenting here, I updated the original article in order to be able to write formulas. See above.
>> 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.
Assume thermometer and system are made from the same type of atoms (to keep the argument simple).
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.
But if the velocity distributions are different, they are not in equilibrium.
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