## The imagined scenario:

Part A:

From special relativity we know that velocity is a relative physical quantity, that is, it is dependent on the frame of reference of choice. This means that kinetic energy is also relative, but this does not undermine the law of conservation of energy as long as we are consistent with our choice of frame. So far so good.

Part B:

On the other hand, from statistical mechanics, we know that the average kinetic energy of a system and its temperature are directly related by the Boltzmann constant $$\langle E_k \rangle = \frac{1}{2}m\langle v^2 \rangle = \frac{3}{2} k_B T$$ which leads to conclude that when the notion of temperature in physics is expressed in terms of a system's kinetic energy, then it too ought to be a relative quantity, which is a bit mind-boggling, because I had always thought of temperature as absolute.

Part C:

Furthermore, we know that all objects at non-zero temperature, radiate electromagnetic energy with a wavelength given as a function of the body/object's temperature, this is the Blackbody radiation. Thus in principle, I am able to infer an object's temperature (i.e. the temperature in its own rest-frame of reference) by measuring its emitted radiation, regardless of the frame I find myself in. But this seems to violate the previously expected relativity of temperature as defined by average kinetic energy.

## Proposed resolutions:

The resolutions that I imagine to this paradox are:

• a) Depending on the frame of reference from which I measure the emitted blackbody radiation of the object, the radiation will undergo different Doppler blue/red-shifts. Thus the relativity of the temperature in the context of blackbody radiation, is preserved due to the Doppler effect.

• b) I suspect that treating temperature as nothing but an average kinetic energy does not in general hold true, and to resolve this paradox, one should work with a more general definition of temperature (which I admit I do not know how in general temperature ought to be defined, if not in terms of state of motion of a system's particles).

Case a) resolves this hypothetical paradox by including the Doppler effect, but does not contradict the relativity of temperature.

Case b) on the other hand, resolves the problem by challenging the definition that was used for temperature, which in the case that we define temperature more generally, without relating to kinetic energy, may leave temperature as an absolute quantity and not relative to a frame.

## Main question:

• But surely only one can be correct here. Which begs to ask: what was the logical mistake(s) committed in the above scenario? In case there was no mistake, which of the two proposed resolutions are correct? If none, what is then the answer here? Very curious to read your input.
• There is no reference frame where the relative motion of the molecules are zero. They are also constantly changing direction and experiencing acceleration which is not an intertial reference frame. – Peter R Jun 13 '16 at 16:13
• @JohnRennie Dear Mr Rennie, thanks a lot for the linked post, I had not come across it yet. Regarding the duplicate mark, although there's overlap between the two posts, namely the discussion on whether temperature is relative or not, but my post is rather presenting a concrete scenario and asking questions about it, including how temperature is correctly defined in such circumstances. Moreover, I still don't see the bottom line, so am I to go with case a or b? or both are wrong? Would you mind reconsidering the duplicate mark? – user929304 Jun 13 '16 at 16:20
• OK,I have reopened your question, though I still feel that the information in the other question makes this one redundant. – John Rennie Jun 13 '16 at 16:55
• @PeterR very interesting, maybe that s where im going wrong, although i was only referring to the average kinetic energy. Would you expand on your point as an answer if time allows? – user929304 Jun 13 '16 at 17:54
• The average kinetic energy is just that. it describes the energy content of the heated molecules. As the post below explained, it's best to look at the heated body in its own reference frame and apply relativistic doppler shift for the change in the radiation due to relativistic effects. If you tried to analyse the relativistic effect of the heated body traveling relative to you, all you would be doing is applying the same velocity vector to each molecule and arrive at the same place anyway. It would fall out as a constant. – Peter R Jun 13 '16 at 19:05

Temperature is related to kinetic energy in the rest frame of the fluid/gas. In non-relatvistic kinetic theory the distribution function is $$f(p) \sim \exp\left(-\frac{(\vec{p}-m\vec{u})^2}{2mT}\right)$$ where $\vec{u}$ is the local fluid velocity. The velocity can be found by demanding that the mean momentum in the local rest frame is zero. Then $\vec{u}$ transform as a vector under Galilean transformations, and $T$ is a scalar.

In relativistic kinetic theory $$f(p) \sim \exp\left(-\frac{p\cdot u}{T}\right)$$ where $p$ is the four-momentum, $u$ is the four-velocity, and $T$ is the temperature scalar. The rest frame is defined by $\vec{u}=0$, and in the rest frame $f\sim \exp(-E_p/T)$, as expected.

The relativistic result is known as the Juttner distribution (Juttner, 1911), and is discussed in standard texts on relativistic kinetic theory, for example Cercignani and Kremer , equ. (2.124), and de Groot et al , equ (ch4)(25). See also (2.120) in Rezzolla and Zanotti. For an intro available online see equ. (55-58) of Romatschke's review. Neumaier notes that some (like Beccatini ) advocate defining a four-vector field $\beta_\mu=u_\mu/T$, and then define a frame dependent temperature $T'\equiv 1/\beta_0$. I fail to see the advantage of this procedure, and it is not what is done in relativistic kinetic theory, hydrodynamics, numerical GR, or AdS/CFT.

Ultimately, the most general definition of $T$ comes from local thermodynamics (fluid dynamics), not kinetic theory, because strongly correlated fluids (classical or quantum) are not described in kinetic theory. The standard form of relativistic fluid dynamics (developed by Landau, and explained in his book on fluid dynamics) also introduces a relarivistic 4-velocity $u_\mu$ (with $u^2=1$), and a scalar temperature $T$, defined by thermodynamic identities, $dP=sdT+nd\mu$. The ideal fluid stress tensor is $$T_{\mu\nu}=({\cal E}+P) u_\mu u_\nu -Pg_{\mu\nu}$$ where ${\cal E}$ is the energy density and $P$ is the pressure. Note that for a kinetic system the parameter $u_\mu$ in the Juttner distribution is the fluid velocity, as one would expect. More generally, the fluid velocity can be defined by $u^\mu T_{\mu\nu}={\cal E}u_\nu$, which is valid even if dissipative corrections are taken into account.

Regarding the paradox'': Temperature is not relative, it is a scalar. The relation in B is only correct in the rest frame. The Doppler effect is of course a real physical effect. The spectrum seen by an observer moving vith relative velocity $v$ is $f\sim\exp(-p\cdot v/T)$, which exhibits a red/blue shift. The spectrum only depends on the relative velocity, as it should. Measuring the spectrum can be used to determine both the relative velocity and the temperature. However, if you look at a distant star you only measure light coming off in one direction. Then, in order to disentangle $u$ and $T$, you need either a spectral line, or information on the absolute luminosity.

• Thanks for your answer. Would you be so kind and add a few words as to how we can relate these facts to resolving the paradox? – user929304 Jun 17 '16 at 14:27
• Please add references that confirm your assumed starting point in the relativistic case. Note that the subject is controversial, so simply stating your point of view without references is not OK. – Arnold Neumaier Jun 18 '16 at 13:43
• The definition of temperature implied by your setting is not the canonical one. Planck and Einstein used the noncovariant canonical temperature (changing nontrivially under a Lorentz transform), which differs by your covariant temperature by a factor $u^0$. See the details in my revised answer. – Arnold Neumaier Jun 19 '16 at 12:43
• I think the papers of Planck and Einstein are mainly of historical interest. There is a vast literature on relativistic fluids and plasmas (heavy ions, relativistic astrophysics) that uses the conventions mentioned above. – Thomas Jun 19 '16 at 15:23
• Thus what temperature is and how it transforms is a matter of convention, and the conventions changed some time after 1968. – Arnold Neumaier Jun 23 '16 at 5:08

[June 19,2016: thoroughly revised, giving a more detailed, comparative presentation and better references]

General case. In relativistic thermodynamics, inverse temperature $\beta^\mu$ is a vector field, namely the multipliers of the 4-momentum density in the exponent of the density operator specifying the system in terms of statistical mechanics, using the maximum entropy method, where $\beta^\mu p_\mu$ (in units where $c=1$) replaces the term $\beta H$ of the nonrelativistic canonical ensemble. This is done in

C.G. van Weert, Maximum entropy principle and relativistic hydrodynamics, Annals of Physics 140 (1982), 133-162.

for classical statistical mechanics and for quantum statistical mechanics in

T. Hayata et al., Relativistic hydrodynamics from quantum field theory on the basis of the generalized Gibbs ensemble method, Phys. Rev. D 92 (2015), 065008. https://arxiv.org/abs/1503.04535

F. Becattini, Covariant statistical mechanics and the stress-energy tensor, Phys. Rev. Lett 108 (2012), 244502. https://arxiv.org/abs/1511.05439

Conservative case. One can define a scalar temperature $T:=1/k_B\sqrt{\beta^\mu\beta_\mu}$ and a velocity field $u^\mu:=k_BT\beta^\mu$ for the fluid; then $\beta^\mu=u^\mu/k_BT$, and the distribution function for an ideal fluid takes the form of a Jüttner distribution $e^{-u\cdot p/k_BT}$.

For an ideal fluid (i.e., assuming no dissipation, so that all conservation laws hold exacly), one obtains the format commonly used in relativistic hydrodynamics (see Chapter 22 in the book Misner, Thorne, Wheeler, Gravitation). It amounts to treating the thermodynamics nonrelativistically in the rest frame of the fluid.

Note that the definition of temperature consistent with the canonical ensemble needs a distribution of the form $e^{-\beta H - terms~ linear~ in~ p}$, conforming with the identification of the noncovariant $\beta^0$ as the inverse canonical temperature. Essentially, this is due to the frame dependence of the volume that enters the thermodynamics. This is in agreement with the noncovariant definition of temperature used by Planck and Einstein and was the generally agreed upon convention until at least 1968; cf. the discussion in

R. Balescu, Relativistic statistical thermodynamics, Physica 40 (1968), 309-338.

In contrast, the covariant Jüttner distribution has the form $e^{-u_0 H/k_BT - terms~ linear~ in~ p}$. Therefore the covariant scalar temperature differs from the canonical one by a velocity-dependent factor $u_0$. This explains the different transformation law. The covariant scalar temperature is simply the canonical temperature in the rest frame, turned covariant by redefinition.

Quantum general relativity. In quantum general relativity, accelerated observers interpret temperature differently. This is demonstrated for the vacuum state in Minkowski space by the Unruh effect, which is part of the thermodynamics of black holes. This seems inconsistent with the assumption of a covariant temperature.

Dissipative case. The situation is more complicated in the more realistic dissipative case. Once one allows for dissipation, amounting to going from Euler to Navier-Stokes in the nonrelativistic case, trying to generalize this simple formulation runs into problems. Thus it cannot be completely correct. In a gradient expansion at low order, the velocity field defined above from $\beta^\mu$ can be identified in the Landau-Lifschitz frame with the velocity field proportional to the energy current; see (86) in Hayata et al.. However, in general, this identification involves an approximation as there is no reason for these velocity fields to be exactly parallel; see, e.g.,

P. Van and T.S. Biró, First order and stable relativistic dissipative hydrodynamics, Physics Letters B 709 (2012), 106-110. https://arxiv.org/abs/1109.0985

There are various ways to patch the situation, starting from a kinetic description (valid for dilute gases only): The first reasonable formulation by Israel and Stewart based on a first order gradient expansion turned out to exhibit acausal behavior and not to be thermodynamically consistent. Extensions to second order (by Romatschke, e.g., https://arxiv.org/abs/0902.3663) or third order (by El et al., https://arxiv.org/abs/0907.4500) remedy the problems at low density, but shift the difficulties only to higher order terms (see Section 3.2 of Kovtun, https://arxiv.org/abs/1205.5040).

A causal and thermodynamically consistent formulation involving additional fields was given by Mueller and Ruggeri in their book Extended Thermodynamics 1993 and its 2nd edition, called Rational extended Thermodynamics 1998.

Note that the formula $\langle E\rangle = \frac32 k_B T$ is valid only under very special circumstances (nonrelativistic ideal monatomic gas in its rest frame), and does not generalize. In general there is no simple relationship between temperature and velocity.

One can say that your paradox arises because in the three scenarios, three different concepts of temperature are used. What temperature is and how it transforms is a matter of convention, and the dominant convention changed some time after 1968; after Balescu's paper mentioned above, which shows that until 1963 it was universally defined as being frame-dependent. Today both conventions are alive, the frame-independent one being dominant.

• Comments are not for extended discussion; this conversation has been moved to chat. – Manishearth Jun 24 '16 at 10:37

In the kinetic theory of gases, you only really define the temperature for molecules that are in constant, random, and rapid motion. So if you have a container with a gas at temperature $T$ you don't change the internal energy of the gas by uniformly moving the container. Uniformly moving the container gives all the molecules a non-zero average motion, but it doesn't affect the random motion of the molecules that make up the gas.

I think it is wrong to define the temperature by the average energy of the molecule in all frames of reference. The reason for that is clear: take all of your particles and send them at $100 m/s$ to the north. This won't make the gas hotter, just like the fan does not cool/heat the air (another great mystery!). The organized movement does not participate in the notion of temperature. Seemingly, it has to be defined in a rest frame of gas to make sense.

A simple calculation in mind shows a contradiction. If you define a temperature through the energy, you have to conclude that temperature transforms like an energy (which is a part of a momentum 4-vector). But if you try to expand it through velocity, you will immediately see that velocity squared transforms in a rather ugly way and won't simplify to the vector component transformation. First of all, I believe that the formula for energy you take is not correct in relativistic case.

Next, to the measurement! I think this answer is correct in distinguishing the observation of the temperature from its statistical definition. You can judge about the temperature of a body from the spectrum of black-body radiation it produces, but this is not the measurement of the temperature, but the measurement of the radiation which is subject to relativistic redshift.

• You commented the wrong answer – Andrii Magalich Jun 18 '16 at 16:30
• Sorry my bad, fixed now – user929304 Jun 18 '16 at 17:24

All kinds of weird things happen if you try to define temperature in a moving object. The paradox to me (not a generalized accepted answer) resolves by realizing that temperature should only be defined as measured when the object is stationary. Not only is not a scalar but it is not even well defined for areference frame in relative motion. Is temperature a Lorentz invariant in relativity?

Another problem to the ones you mentioned is , for a moving object that is at thermal equilibrium on its own reference frame, that as seen in the reference frame where the object is moving, the distribution of particles' velocities no longer follows the boltzmann distribution, not even relative to the center of mass: particles moving perpendicular to the motion of the object will not change its average velocity, but those moving parallel to the object will. In addition, because the compositions of velocities is non-linear, this deviation will neither be symmetrical between those moving in the same direction of the center of mass and those moving opposite to the center of mass.

I think you can avoid all these troubles if you define the temperature as proportional to the variance of velocity, i.e.

$$E[(v-\overline{v})\cdot(v-\overline{v})]=E(v\cdot v)-\overline{v}\cdot\overline{v}$$

Here $E$ means expected value, $v$ ranges over the velocities of the individual particles, and $\overline{v}=E(v)$.

Clearly this is frame-independent, because a change of frame adds some constant vector $v_0$ to each $v$ and to $\overline{v}$, leaving their difference unchanged.

I have a friend who likes to describe cold windy days by saying that on average, the air molecules have too much velocity and not enough speed. Clearly the measure of velocity --- that is $\overline{v}\cdot \overline{v}$ --- is frame-dependent: You don't feel the wind if you move along with it. But the measure of speed --- or, more accurately, of speed-minus-velocity --- that is, the displayed expression above --- is frame-independent, and I believe it's what is measured by your thermometer.

• what is measured by your thermometer is always measured in the rest frame of the thermometer. Thus it is meaningless to ask for a quantity that is frame independent, unless there are independent reasons for the latter. – Arnold Neumaier Jun 18 '16 at 14:45
• @ArnoldNeumaier : But if two thermometers are in uniform motion with respect to each other when they both encounter the same cloud of gas, it still makes sense to ask whether they'll show the same readings --- and it seems to me that if the answer is yes, we'd want to call that a frame-independent measurement, no? – WillO Jun 18 '16 at 14:58
• The question is meaningful but whether they would show the same reading is not clear. Wouldn't the high-speed flow across the thermometer heat it up, and the amount depends on the relative speed and on how long the measurement takes? So it is questionable whether the temperature is meaningfully defined through such a measurement. – Arnold Neumaier Jun 18 '16 at 15:21
• Thanks, I m not so sure about your last paragraph, i mean isn't speed just another name for the norm of the velocity vector? So when you speed minus velocity, i dont really understand. Would you a bit more clarity? – user929304 Jun 18 '16 at 17:25
• @user929304: The speed of a particle is the norm of its velocity vector. The average speed of a particle is the sum of the speeds of all the particles, divided by the number of particles. The average velocity of a particle is the sum of the velocities of all the particles, divided by the number of particles. When it's very windy, the average velocity is high. When it's very cold, the average speed is low. – WillO Jun 18 '16 at 17:54

The question may be about the covariance or otherwise of temperature, in which case have a look here. As well, have a look at the paper "Temperature in special relativity" by J. Lindhard, Physica Volume 38, Issue 4, 5 June 1968, Pages 635-640.