# Which temperature transformation does QFT allow?

## Background

Taken from here Is temperature a Lorentz invariant in relativity?:

Einstein himself, in a 1907 review (available in translation as Am. J. Phys. 45, 512 (1977), e.g. here), and Planck, one year later, assumed the first and second law of thermodynamics to be covariant, and derived from that the following transformation rule for the temperature: $$T' = T/\gamma, \quad > \gamma = \sqrt{1/(1-v^2/c^2)}.$$ So, an observer would see a system in relativistic motion "cooler" than if he were in its rest frame.

However, in 1963 Ott (Z. Phys. 175 no. 1 (1963) 70) proposed as the appropriate transformation $$T' = \gamma T$$ suggesting that a moving body appears "relatively" warmer.

Later on Landsberg (Nature 213 (1966) 571 and 214 (1967) 903) argued that the thermodynamic quantities that are statistical in nature, such as temperature, entropy and internal energy, should not be expected to change for an observer who sees the center of mass of the system moving uniformly. This approach, leads to the conclusion that some thermodynamic relationships such as the second law are not covariant and results in the transformation rule: $$T' = T$$

## Question

What position does QFT take? (assuming it takes a unique one?) I feel QFT is the right theory to describe this because one can talk about temperature in the context of a field. Also QFT is has a nice way of talking about co-ordinate transformations too as symmetries.

• I've been unclear on this for a while as well. When I learned about the "headlight effect" I was puzzled about how a moving observer carrying a bright light could agree about the temperature gradient of their spaceship with a stationary observer who sees the light concentrated in the direction of motion-- shouldn't the stationary observer see the bow and stern of the ship heating up faster then the port and starboard on account of the light? While the moving observer sees the whole ship heat uniformly. – Diffycue Sep 14 '19 at 17:37
• @Diffycue The headlight effect doesn't seem to include QFT ... But I'm sure your comment would be better here? physics.stackexchange.com/questions/83488/… – More Anonymous Sep 14 '19 at 17:39
• it's not clear to me that "Effect X doesn't include theory Y" is a coherent statement. But at any rate, modelling the operation of the lightbulb (a source of photons, field quanta) in a way that is consistent with SR is not something I could do without QFT. – Diffycue Sep 14 '19 at 17:51
• @Diffycue I said that because of this: physics.stackexchange.com/questions/359886/… – More Anonymous Sep 14 '19 at 17:55
• Right; the headlight effect on-its-own follows from a straightforward application of SR to light rays emanating from a point, but the resolution to the situation I describe might require QFT to model properly the way in which the bulb produces photons or statistical field theory to model properly the way in which the room heats up. – Diffycue Sep 14 '19 at 18:18

How light is a feather? Light = color, or light = weight? Those are different questions, so of course they have different answers. When I skim through the literature about how Lorentz trnasformations affect temperature, I see people arguing about how light a feather is.

We can debate how the words should be used, but the physics is straightforward.

Consider a Lorentz-symmetric QFT. To reduce clutter, I'll use units in which the speed of light $$c$$ and Boltzmann's constant $$k$$ are both equal to $$1$$. Let $$P_0,P_1,P_2,P_3$$ be the operators representing the system's total energy $$(P_0)$$ and total momentum ($$P_1,P_2,P_3$$). Start with the density matrix $$\rho\propto \exp(-\beta P_0) \tag{1}$$ for some positive constant $$\beta$$. If we apply a boost in the $$1$$-direction with rapidity $$\theta$$, then the resulting density matrix is $$\rho'\propto \exp\big( -\beta P_0 \cosh\theta -\beta P_1 \sinh\theta \big) \tag{2}$$ with the same constant $$\beta$$ as before. This is straightforward and uncontroversial. But does it answer the question?

• If the question is "How does the state (1) transform under a boost?" then the answer is (2). That's the QFT content of this post.

• If the question is "What state should I use?" then the answer is another question: what situation are you trying to model, and which details do you want to assume are known?

• If the question is "When physicists say 'temperature', what do they mean?" then I'll list a couple of possibilities below...

If the state is (1), then the quantity $$1/\beta$$ is typically called "temperature." More generally, if the state is $$\exp(-\beta_0 P_0-\beta_1 P_1)$$, then the quantity $$1/\beta_0$$ is a reasonable guess about what the author/speaker might mean by "temperature." Applied to (2), this would give $$1/(\beta\cosh\theta)$$ for the temperature. Or the author/speaker might use "temperature" to refer instead to the appearance of the radiation emitted by the thermal body, as perceived by an observer in some specific location. Then the temperature depends on the observer's location: the radiation in the boosted frame is not isotropic, because it depends on the boost direction. We could also consider other definitions, which may give other answers.

Words are not always used consistently, and nothing I say here is going to fix that eternal problem.

As the excellent answer by Chiral Anomaly already states, QFT takes no particular stance on the issue unless you specify more precisely what you mean by "temperature". I'll just elaborate on why this would be a hard thing to do.

A very good definition of temperature is that it's whatever a thermometer measures, just like how time is whatever a clock measures in relativity. Now, forgetting about boosts for the moment, why should we expect such a definition to work at all? Why is it that in everyday life, all properly calibrated thermometers and clocks precisely agree with each other, even if they're made of completely different materials?

For time, it's because the time elapsed for any clock can be calculated geometrically, as the length of its path through spacetime. And for temperature, it's because thermal equilibrium has the deeper meaning of maximizing entropy when energy exchange is allowed. When a thermometer is placed in a hot system, whatever that system is, the system will transfer energy to the thermometer, until the rate it loses entropy equals the rate the thermometer gains entropy. At the end, you get a temperature reading $$\frac{1}{T} = \frac{\partial S}{\partial E}.$$ The result only depends on the relationship between entropy and energy of the system you're measuring. If you put a thermometer in your oven, it'll give the same reading in thermal equilibrium whether it's shiny, dirty, painted white, or painted red, even though it'll interact completely differently with the radiation field in each case.

More generally, there is an associated temperature-like quantity for every conserved quantity that can be exchanged to reach equilibrium. If two systems can exchange particle number, then in equilibrium they have equal $$\partial S / \partial N$$, and we call this quantity $$\mu/T$$. If two systems can exchange volume, then they match their $$\partial S / \partial V$$, and we call this quantity $$P/T$$. The reason we don't have to worry about $$\mu$$ and $$P$$ for thermometers is because most thermometers we use, to a good approximation, don't exchange particles with their environment, or change in volume. (And if we had a thermometer that did, we probably wouldn't even call it a thermometer.)

So what about boosts? The problem is that once you boost systems, they pick up another conserved quantity: the overall momentum, which leads to three more quantities $$\mathbf{R}/T = \partial S / \partial \mathbf{P}$$. And while it's easy to build a thermometer that exchanges energy but not particles, it's basically impossible to imagine one that exchanges energy but not momentum. (It would have to somehow never experience any net force, yet also not just be transparent.)

The result is that different thermometer designs can give completely different results, depending on how they couple to momentum, making it meaningless to speak of a unique temperature at all. Quantum field theory doesn't change this, because the problem has nothing to do with describing the thermodynamic system, and everything to do with how it's measured.