# Is there a relativistic (quantum) thermodynamics?

Does a relativistic version of quantum thermodynamics exist? I.e. in a non-inertial frame of reference, can I, an external observer, calculate quantities like magnetisation within the non-inertial frame?

I'd be interested to know if there's a difference between how to treat thermodynamics in a uniformly accelerated reference frame and in a non-uniformly accelerated reference frame.

Thanks!

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We have an Unruh temperature, and Unruh radiation. –  QGR Feb 10 '11 at 15:28
I assume that when you say thermodynamics, you really want statistical mechanics (the two are similar but not the same!). The basis of statistical mechanics is that all states which satisfy macroscopic constraints are equiprobable. In a non-relativistic setting, usually energy is a good constraint, which introduces $\beta$ (reciprocal temperature) as a Lagrange multiplier. In relativistic settings, we need to constraint 4-momentum $p_\mu$, so we get a sort of 4-vector temperature $\beta^\mu$. Other constraints such as magnetisation (in some particular reference frame) can be applied similarly. –  genneth Feb 10 '11 at 15:33
@genneth Thanks! Yes I do mean statistical mechanics. Do Do you know of a paper that describes what you say in your comment? –  Jane Feb 10 '11 at 16:12
@QGR Yes, I understand that you can't have thermal equilibrium for nonuniform acceleration, but could some sort of nonequilibrium version of thermodynamics/statistical mechanics be used? –  Jane Feb 10 '11 at 16:14
@Jane: I'm afraid nothing comes to mind --- Roy's suggestion of MTW "Gravitation" is great, but admittedly might be a bit too heavy-handed. I note in passing though, that there are quite subtle issues to do with relativity and thermodynamics; for example, a system in thermal equilibrium defines a preferred direction of time, and indeed of energy --- you can invert the expression $\rho = 1/Z \exp(-\beta H)$ to obtain $H = -ln(\rho Z)/\beta$. There is in fact quite general theorem (due to Connes?) which yields this (something about automorphisms of the observables algebra). –  genneth Feb 10 '11 at 16:55
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## 3 Answers

There is a classic treatise on "Relativity, Thermodynamics and Cosmology" from R. Tolmann from the 1930s - it is still referenced in papers today. This generalises Thermodynamics to Special Relativity and then General Relativity. As a simple example the transformation law for Temperature is stated as: $T=\sqrt(1-v^2/c^2)T_0$ when changing to a Lorentz moving frame.

Another example is that "entropy density" $\phi$ is introduced, which is also subject to a Lorentz transformation. Finally this becomes a scalar with an associated "entropy 4-vector" in GR. The Second Law is expressed using these constructs by Tolmann.

There is some discussion in Misner, Thorne and Wheeler too.

Of course both these texts also include lots of regular General Relativity Theory which you may not need.

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Thanks, I'll have a look at those books. –  Jane Feb 10 '11 at 16:52
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Jane, you had to mean quantum statistical mechanics. Thermodynamics only emerges in the thermodynamic limit of statistical mechanics, which requires a large number of degrees of freedom, and when you have a large number of degrees of freedom, the relevant quantities automatically behave classically, too. In this sense, thermodynamics is always classical (non-quantum).

Quantum statistical physics is easily promoted to special relativity. But one must understand that a thermal ensemble is given by $$\rho\approx \exp(-\beta H)$$ where $H$ is the energy, the generator of translations in time. You may extend it in a relativistically invariant way, $$\rho\approx \exp(-\beta H - \vec\beta_p \cdot \vec p)$$ where $\vec\beta_p$ are spatial components of inverse temperature. However, the density matrix above is equivalent - by a Lorentz transformation - to the previous one. Only the norm of the 4-vector $(\beta,\beta_p)$ has an invariant physical meaning, and you may always go to the frame where the spatial components vanish. In this frame, you get an ordinary thermal ensemble.

(Of course, another question is whether one knows how to calculate the thermal properties of objects in a particular relativistic theory. That involves new mathematical challenges. But the definition of the objects - e.g. the thermal density matrix - is totally unaffected by having a relativistic theory.)

So the formula for the thermal state is actually identical for relativistic quantum statistical mechanics - and obviously, it's also identical in the classical limit of it, the relativistic classical statistical mechanics. The 4-vector of the inverse temperature had to be time-like, otherwise the density matrix would be divergent (the momentum-dominated "energy" wouldn't be bounded from below).

In general relativity, you may define the temperature (and its preferred reference frame) locally, and locally, the physics reduces to that of special relativity which was explained above. However, there's no real ensemble with a fixed global temperature globally, except for geometries that are static - that have a time-like Killing vector field. That's because to define the thermal ensemble, you need a periodic Euclidean time (the energy generating time translations enters the exponential in the density matrix), and if the original solution nontrivially depended on time, you couldn't make the time periodic in the imaginary direction.

There are many fancy thermal effects in curved spaces - such as particle production; Unruh effect; and Hawking radiation. The most exciting portion is black hole thermodynamics. However, to get there, one must first understand why the extension of statistical mechanics to the special relativistic context isn't really conceptually new. It may be useful if you asked a more advanced question about the references when you figure out what is actually out there so that your question becomes somewhat more well-defined.

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Yes, the Unruh effect is strange. An inertial observer and an accelerating observer going by see different vacua, and the accelerated one sees thermal black body radiation where there is none for the other. Pisin Chen proposed a lab test of this-Testing Unruh Radiation with Ultra-intense Lasers. Pisin Chen and Toshi Tajima. Physical Review Letters 83, 256-259 (1999), but I don't believe as yet there has been any clearcut demonstration of the radiation (?Pioneer anomaly--unlikely...) –  Gordon Feb 11 '11 at 6:27
š So if I wanted to calculate, for example, the internal energy of a quantum system in a non-inertial frame, I could do it in one of two ways? - 1. Calculate the internal energy "normally" and then Lorentz transform (or whatever the relevant transformation is for internal energy) as @Roy referred to. Or 2. Treat the density matrix relativistically but transform it as you say. And then calculate internal energy buy calculating $U = tr(\rho H)$. –  Jane Feb 12 '11 at 10:00
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There's always a relativistic version of any theory in a weak sense that we can construct a system that is Lorentz *co*variant. The question is then what Lorentz non-invariant entities have to be added. As Luboš implies, the additional structure in the case of quantum statistical fields is a time-like 4-vector. In particular, the quantum vacuum state is invariant under the action of the Poincare group, whereas a thermal state is invariant only under the subgroup of the Poincare group that leaves a particular time-like 4-vector invariant. If we look at the $n$-point functions of the Klein-Gordon free field in the vacuum state and in a thermal state, taking $T^\mu$ to be invariant, both states are Gaussian, so the 2-point function gives full information, but the Fourier transforms of the 2-point functions are given by $2\pi\delta(k^2-m^2)\theta(k_0)$ and by $\coth{\!\left(\!\frac{\hbar \sqrt{|\mathbf{k}|^2+m^2}}{2\mathsf{kT}}\!\right)} 2\pi\delta(k^2-m^2)\theta(k_0)$, respectively. The $|\mathbf{k}|^2$ is the length of the 3-vector component of the wave-number relative to the time-like 4-vector $T^\mu$. This deforms the measure on the mass-shell, but it is still on the mass-shell. You can find a derivation of this, more-or-less, in my arXiv:quant-ph/0411156, which is also published in Phys. Lett. A 338, 8-12(2005).

Hence, we have a quantum statistical free field theory. Interacting QFT only exists perturbatively in 1+3 dimensions, and hence questionably, and the situation for quantum statistical field theory is the same.

Edit: A small change of $\tanh{}$ to $\coth{}$! The wrong way up for the way I stated it here. The $\coth{\!\left(\!\frac{\hbar \sqrt{|\mathbf{k}|^2+m^2}}{2\mathsf{kT}}\!\right)}$ factor behaves more-or-less as $1/\sqrt{|\mathbf{k}|^2+m^2}$ when $|\mathbf{k}|$ is small, low frequency, and $m\hbar$ is small relative to $\mathsf{kT}$, making the measure on the mass-shell effectively the same as the classical thermal state in that case, but behaves as a constant, $1$, when $|\mathbf{k}|$ is large, high frequency, making the measure on the mass-shell effectively the same as the quantum vacuum state in that case.

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