# Is temperature an invariant in Special Relativity?

1. Jan 4, 2014

### yuiop

In another current thread on the possible invariance of pressure, I mentioned:
What is the current consensus on how temperature transforms in relativity?

Here is another simple thought experiment. Consider two very long rectangular objects A and B, that are moving relative to each other, while remaining in thermal contact along their longest sides. In a particular reference frame (S), A and B are moving at the same speed wrt S, and have equal temperatures.

If we take the Ott view that temperature increases with relative motion, then in the rest frame of object A, the temperature of object B is greater than that of A and heat flows from B to A.

The converse would then be true in the rest frame of B, so that heat flows in the opposite direction from A to B. This would appear to be a physical contradiction to what is measured in the rest frame of A.
A similar contradiction appears if we imagine temperature gets colder with relative motion.

If temperature is not a Lorentz invariant, it would appear that one of the consequences is that entropy is not a Lorentz invariant either.

Another approach is to consider phase transitions, (as in the pressure thread). If temperature is not Lorentz invariant, then the absolute temperature of the triple point of water for example, would be an observer dependent quantity.

2. Jan 4, 2014

Staff Emeritus
Non-relativistically, there are several equivalent definitions of temperature. Relativistically, they are no longer equivalent. So the answer to questions like yours can only be "it depends on what you call temperature".

Note that this is largely academic. This only matters when a) you have many objects moving at relativistic speeds with respect to each other (otherwise relativity is unnecessary) and b) the objects are in thermal equilibrium with each other (otherwise temperature is undefined). This is a very rare situation.

3. Jan 5, 2014

### Jonathan Scott

What does "thermal contact" mean in this case? I think the answer will depend on your definition, as well as on the definition of temperature.

4. Jan 5, 2014

### yuiop

I originally meant in physical contact so that heat is transferred by conduction with the additional assumption of ideal frictionless sliding contact surfaces. If possible if you could give an answer for surfaces in very close proximity but with constant separation distance so heat transfer id by radiation, then that would also be of interest.

I am assuming that whatever the process of heat transfer, that heat naturally flows from hot bodies to cold bodies, but maybe that is not a safe assumption in relativity. *

* It occurred to me that a possible exception might be a black hole which can have negative heat capacity. Given that, a more general rule might be that heat naturally flows in the direction that increases the entropy of the system.

Last edited: Jan 5, 2014
5. Jan 5, 2014

### Jonathan Scott

Temperature and heat are macroscopic terms relating to the average bulk distribution of internal kinetic energy over a set of particles in static or slow-moving situations, and the rules which apply to them are basically statistical results which cover the motion of large enough numbers of particles. What actually happens in a given case can be calculated by setting up a microscopic model of what the particles are doing and applying the appropriate SR transforms, but it is not clear whether the bulk properties of the transformed system can similarly be described in terms of "temperature".

The amount of heat energy in an object can be treated like part of the rest energy for transformation purposes. However, I don't personally know of a standard method of extending the definition of temperature to include the case where the object is moving at high speed relative to the reference frame; the whole concept of temperature as a scalar quantity assumes that the internal motion is the same in all directions. Both of these terms relate only to internal energy.

Heat conduction requires contact in order to propagate microscopic motion, but if there is significant relative motion, the transferred energy between the particles making up the surfaces due to heat energy will be much less significant than the transferred energy due to the relative velocity. Even if the interface is hypothetically "frictionless", any vibrations on one side cannot be seen as a plane vibration on the other because of differences in simultaneity, so there will necessarily be components of force which are not perpendicular to the interface.

As for radiation, I would presume you should be able to calculate heat transfer due to radiation by looking at the radiation emitted in the rest frame of one object and transforming it to the frame of the other. This radiation would of course be subject to frequency shifting and direction aberration as seen from the other frame.

6. Jan 6, 2014

### tom.stoer

CMB is a nice example ;-)

7. Jan 6, 2014

### vanhees71

Well, this is of course an "academic question" but it's by no means just theoretical speculation. The entire field of ultrarelativistic heavy-ion collisions that is under vigorous investigation at the Relativistic Heavy Ion Collider (RHIC) at the Brookhaven National Laboratory (BNL) and the Large Hadron Collider (LHC) at CERN.

There heavy nuclei like gold or lead nuclei are accelerated to very high energies and hit each other head on. The result is the creation of a very hot and dense fireball that collectively expands in a way that can be described astonishingly well with nearly ideal relativistic hydrodynamics.

The debate about the nature of the thermodynamic quantities in relativistic physics dates back already to the very beginnings of special relativity, and a lot of confusion in this matter has been created. Nowadays we aim at the use of covariant definitions of quantities, and temperature is always defined as a scalar field. This is achieved by defining the temperature as being measured with a thermometer that is at rest relative to the fluid cell under consideration. The corresponding quantity is the average energy-momentum tensor of the fluid, which transforms as a 2nd rank tensor. The temperature and pressure are defined in the (local) restframe of the fluid and thus are invariants.

Of course, the general relativistic case is interesting in astrophysics and cosmology. The CMBR is indeed a nice example.

8. Jan 6, 2014

### tom.stoer

We had this discussion several times - does anybody know a reference paper?

9. Jan 6, 2014

### vanhees71

A classical paper is

C. Eckart, The Thermodynamics of Irreversible Processes. III. Relativistic Theory of the Simple Fluid, Phys. Rev. 58, 919 (1940)

Another one from the more phenomenological-thermodynamics point of view is

N. G. van Kampen. Relativistic thermodynamics of moving systems. Phys. Rev., 173(1):295, 1968.

From a kinetic-theory perspective, you may look at my lecture notes:

http://fias.uni-frankfurt.de/~hees/publ/roorkee.pdf

10. Jan 6, 2014

### yuiop

... and presumably nearly ideal relativistic thermodynamics? If the nuclei coming from opposite directions have momentum of equal magnitude in the 'lab' frame, then the lab is in the rest frame of the zero momentum frame of the fireball and we do not need to do a Lorentz transform. The relativistic aspect is only required to analyse the very high velocities of the individual particles in the fireball rather than the motion of the fireball itself relative to the lab.
As I understand it, a scalar field can have a gradient but is invariant under any Lorentz transformation. You describe the "corresponding quantity" as transforming as a second rank tensor which is not necessarily invariant under transformation. Could you clarify?

My motivating interest is how we apply the gas law PV=nRT in relativity. If we define temperature and pressure as invariants, then we have:

$V' = \frac{n_0R_0T_0}{P_0}$

This is a contradiction, because the volume term on the LHS is not invariant while all the quantities on the right are constants, proper quantities or Lorentz invariants (signified by zero subscripts) so the RHS must be invariant. This would require us to redefine the gas law as:

$V' = \frac{1}{\gamma}\frac{n_0R_0T_0}{P_0}$

which suggests the universal gas constant is not a Lorentz invariant.

11. Jan 6, 2014

### vanhees71

In the heavy ion collisions in the center-mass frame the created particles build a fireball that collectively expands like a fluid. See the Wikipedia:

https://en.wikipedia.org/wiki/Quark-gluon_plasma

The usual ideal-gas law holds true in the relativistic case and takes the usual form in the restframe of the fluid cell under consideration (see Sect. 1.6 in my lecture notes):

http://fias.uni-frankfurt.de/~hees/publ/roorkee.pdf

For an ideal fluid the energy-momentum tensor reads

$$T^{\mu \nu}=(P + \epsilon) u^{\mu} u^{\nu} - P g^{\mu \nu},$$
where $u^{\mu}$ is the covariant fluid-flow vector and $P$ and $\epsilon$ are the pressure and energy density measured in the fluid restframe, respectively.

12. Jan 6, 2014

### yuiop

Even in non relativistic thermodynamics, we have to subtract the kinetic energy due to the motion of the body as a whole before calculating temperature in terms of kinetic energy of the particles themselves. If we do this for a moving object in relativity, the average kinetic energy of the constituent particles is reduced by a factor of gamma. This correlates with the notion that energy is proportional to frequency and frequency of the vibrating particles is reduced by a factor of gamma due to time dilation.
This also agrees with the assertion by Einstein and Planck that the temperature of a moving body is colder than its proper temperature $(T' = T_0/\gamma)$.

The Einstein notion of relativistic temperature also seems to make sense in the context of chemical reactions. Chemical reaction rate appears to slow down when it occurs in a body moving relative to us. This must be so due to time dilation. Chemical reaction rate is also know to be proportional to temperature so there appears to be a correspondence here.

I see the difficulties you raise here and I am not sure how to resolve them at this time.

In a very heuristic analysis, length contraction of the emitting surface area of the moving body intensifies the effective radiation per unit area of the emitter relative to the receiving body by a factor of gamma. The relativistic transverse Doppler equation reduces the received radiation frequency by a factor of gamma. This reduction is due to the time dilation of the emitting frequency of the emitting body and supports Einstein's assertion that the temperature of a moving body is colder, because it effectively radiates less. The two effects, (increase in intensity due to length contraction and reduction of frequency due the relativistic Doppler effect) cancel each other out. The radiation received equals the radiation emitted if the two bodies have equal proper temperature. This is a very desirable result as it avoids the contradiction of the direction of the flow of heat depending on the reference frame. However, this heuristic analysis has ignored the aberration effect and it is not clear if that effect is effectively self cancelling or not.

If we go along with Einstein (and ignore Ott, Landsberg, etc) and assume $(T' = T_0/\gamma)$ then it will require relativistic reformulation of a lot of the classical thermodynamic equations.

For example the energy transferred from body a to body b per unit time by radiation (Stefan-Boltzman law) is classically:

$E/t = \sigma \epsilon A(T^4_a-T^4_b)$

In order that no net energy is transferred from one body to the other when their proper temperatures are equal the equation would have to be reformulated to something like:

$E/t = \sigma \epsilon (A'_a T_a^{'4 }\gamma_a^2- A'_b T_b^{'4} \gamma_b^2)$

While this is not pretty, the gas law when we assume $T' = T_0/\gamma$ becomes:

$V' = \frac{nR(T_0/\gamma)}{P_0} = \frac{nRT'}{P_0}$

where I am also assuming pressure is an invariant, which seems reasonable in the context of the gas laws where the scalar isotropic static pressure is relevant form of pressure.

Last edited: Jan 6, 2014
13. Jan 6, 2014

### yuiop

We are observing this fireball in its rest frame so it does not directly tell us anything about its volume, temperature and pressure under transformation.

I am more interested in the unusual form when the fluid cell is not in the rest frame of the observer.

Last edited: Jan 6, 2014