+ 8 like - 0 dislike
334 views

## 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.
This post imported from StackExchange Physics at 2016-06-18 20:38 (UTC), posted by SE-user user929304
retagged Jun 18, 2016
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.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Peter R
The temperature can be related to kinetic energy in the bulk flow rest frame, as I mention at http://physics.stackexchange.com/a/218643/59023. The issue of a relativistic distribution was raised at http://physics.stackexchange.com/q/216819/59023.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user honeste_vivere
The temperature can be related to kinetic energy in the bulk flow rest frame, as I mention at http://physics.stackexchange.com/a/218643/59023. The issue of a relativistic distribution was raised at http://physics.stackexchange.com/q/216819/59023.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user honeste_vivere
@PeterR I see what you mean. So if I understand correctly, the point here is that: looking at the object under study at a molecular level, one cannot meaningfully even define an inertial frame in which the molecules would all be at rest, not even their average motion. But where does that leave us? is this to say that temperature is not a relative quantity? sorry I'm still confused...

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user user929304
There is another issue. The conversion to temperature requires some equation of state, which is not trivial in relativistic fluids.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user honeste_vivere
@JohnRennie: Which other question?

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Arnold Neumaier
@ArnoldNeumaier: the same one that you linked above on June 18th

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user John Rennie

+ 9 like - 0 dislike

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.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Arnold Neumaier

answered Jun 17, 2016 by (13,979 points)
edited Jun 24, 2016

The standard form of the equilibrium distribution is $\beta u\cdot p$ where $\beta =1/T$ and $u_\mu$ is a relativistic 4-velocity, that means $u^2=1$. Of course you can define a new 4-velocity $\beta_\mu=\beta u_\mu$, so that $1/T$ is the zero'th component of a vector in the rest frame. This has no obvious advantage, but several disadvantages. It does not map onto standard relativistic hydro (which uses $u^2=1$), or relativistic kinetics, and is not used in any significant paper or book that I am aware of.

@Thomas: Standard relativistic hydrodynamics is a purely phenomenological theory, the simplest marriage of relativity and thermodynamics. It is valid only in the absence of dissipation, hence unrealistic. 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. There are various ways to patch the situation (Israel-Stewart is not fully consistent; Mueller-Ruggeri is better) but no solution from first principles is known.

The correct definition must of course come from quantum statistical mechanics, which is the underlying microscopic theory. Thus the question is the correct generalization of the ensemble that produces standard hydromechanics in the nonrelativistic approximation. This gives the term $\beta \cdot cp$ in place of $\beta H$. Simplifying this by assuming $\beta=\beta'u$ with a scalar $\beta'$ is already an extra assumption, hence most likely an approximation.

Regarding you first comment, I completely disagree and I and others (Schaefer, arxiv.org/abs/1403.0653, Schaefer & Teaney, arxiv.org/abs/0904.3107, Romatschke, arxiv.org/abs/0902.3663, Son and Starinets, arxiv.org/abs/0704.0240) have written extended reviews of the modern understanding of relativistic fluid mechanics.

@Thomas: Of the papers you cited in your comment, none derives from microscopic theory, and only Romatschke discusses a consistent dissipative formulation. It is no comprehensive review, it doesn't even mention the solution found by Mueller and Ruggeri much earlier. This shows that resolving the matter in a generally agreed way is still a topic of research.

They refer to the usual cases: i) weak coupling: QFT->kinetic theory-> fluid mechanics, ii) strong coupling holographic: QFT->AdS/CFT->fluid mechanics. Of course, in the general case there is no derivation, we cannot even solve QFT in equilibrium. More to the point, we view fluid dynamics as an EFT, so that what we have to provide is not a derivation, but a demonstration the fluid dynamics i) respects the underlying symmetries, ii) provides a systematic low energy expansion, iii) is consistent.

Regarding your first comment above: To the extent that all that Beccatini does is define $\beta_\mu =\beta u_\mu$ the answer is of course that both can be derived from weak coupling nonequilibrium QFT. All I was saying is that 99.976% of the literature uses $\beta u_\mu$, for good reasons (o.k., I'm exagerating, but other than hearing Francesco talk about it, I don't remember seeing it).

I reiterate my main comment. $\beta_\mu =\beta u_\mu$ is possible, but not what practioners do, for good reasons.

You misunderstand the work of Kovtun. This a feature, not a bug (i.e. it is a prediction of fluid dynamics, which people are actively trying to verify). It has nothing to do with relativity, the same effect exists in non-relativistic systems, where it has been verified, see for example here arxiv.org/abs/1209.1006.

You also mischaracterize the work of Romatschke, who showed, among other things, that higher order hydro follows not only from kinetics in the dilute limit, but from AdS/CFT in the dense limit.

@Thomas: The maximum entropy principle of statistical mechanics says that there is one multiplier for each relevant observable field. This gives the beta's. One needs a recipe for introducing $u$. normalizing beta is one possibility. Normalizing instead the heat current or the energy current is in general not equivalent. But some such identity is assumed in the derivations that start by generalizing the ideal fluid case.

@Thomas: A covariant scalar temperature as you want to insist on is also poorly adapted to general relativity, where accelerated observers interpret temperature differently (Unruh effect). Note that all your references are for flat space while van Weert, Becattini and Hayata et al. discuss a general relativistic context.

Standard Landau hydro follows from boosted black branes in AdS/CFT (as explained in all these reviews), there is a covariant definition of the Hawking temperature $T_H^2\sim \nabla_\mu\xi_\nu\nabla^\mu\xi^\nu$, numerical GR uses Landau hydro (as explained in Rezzolla's book), the Juttner distribution is a solution of the generallcovariant Boltzmann equation (as explained in Cercignani's book), second order conformal hydro is naturally embedded in curved space (as explained in Romatschke's review) ...

+ 7 like - 0 dislike

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.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Thomas
answered Jun 17, 2016 by (720 points)
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?

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user user929304

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Thomas
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.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Arnold Neumaier
Thanks for the addition, but is there a rest frame for a system of molecules? Given the difficulties pointed out by Peter R in the comments?

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user user929304
Kinetic theory always assumes a coarse graining procedure. For a given cell, I can always define a rest frame be demanding that the total three-momentum is zero. Indeed, relativistic kinetic theory is routinely applied to molecules, see the references given above. For systems not described by kinetic theory (strongly correlated fluids), we have relativistic fluid dynamics. Landau explained how to find the rest frame (the condition is $T_{0i}=0$).

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Thomas
Thus what temperature is and how it transforms is a matter of convention, and the conventions changed some time after 1968.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Arnold Neumaier
Arguably beginning with Juttner (1911), but, indeed, the modern era of relativistic kinetic theory begins in the early 1970s.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Thomas
+ 6 like - 0 dislike

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.

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) also introduces a relarivistic 4-velocity $u_\mu$ (with $u^2=1$), and a scalar temperature $T$, defined by thermodynamic identities, $dP=sdT+nd\mu$.

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 you need either a spectral line, or information on the absolute luminosity.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Thomas
answered Jun 17, 2016 by (720 points)
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?

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user user929304

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Thomas
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.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Arnold Neumaier
Thanks for the addition, but is there a rest frame for a system of molecules? Given the difficulties pointed out by Peter R in the comments?

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user user929304
Kinetic theory always assumes a coarse graining procedure. For a given cell, I can always define a rest frame be demanding that the total three-momentum is zero. Indeed, relativistic kinetic theory is routinely applied to molecules, see the references given above. For systems not described by kinetic theory (strongly correlated fluids), we have relativistic fluid dynamics. Landau explained how to find the rest frame (the condition is $T_{0i}=0$).

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Thomas
+ 3 like - 0 dislike

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.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user jim
answered Jun 18, 2016 by (40 points)
+ 2 like - 0 dislike

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.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Andrii Magalich
answered Jun 13, 2016 by (70 points)

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Andrii Magalich

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user user929304
+ 2 like - 0 dislike

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.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Andrii Magalich
answered Jun 13, 2016 by (70 points)
+ 1 like - 0 dislike

In relativistic thermodynamics, inverse temperature $\beta$ 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, where $\beta\cdot cp$ replaces the term $\beta H$ of the nonrelativistic canonical ensemble. See, e.g.,

F. Becattini, Covariant statistical mechanics and the stress-energy tensor, Phys. Rev. Lett 108 (2012), 244502.

In most models, the vector-valued inverse temperature is taken to be parallel to the comoving velocity field of the fluid, in which case it can be represented by this field dvided by $cT$ with a scalar temperature field $T$. The latter is 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 standard relativistic hydrodynamics (as described, e.g., by Misner et al.) is a purely phenomenological theory, the simplest marriage of relativity and thermodynamics. It is valid only in the absence of dissipation, hence unrealistic.

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. There are various ways to patch the situation, starting from a kinetic description (valid for dilute gases only): The first reasoanble formulation by Israel and Stewart turned out not to be fully consistent; an improved consistent formulation was given by Mueller and Ruggeri in their book Extended Thermodynamics 1993 and its 2nd edition, called Rational extended Thermodynamics 1998; see also the treatment in Romatschke, arxiv.org/abs/0902.3663

But no satisfying solution from first principles (i.e., quantum statistical mechanics) is known.

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.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Arnold Neumaier
answered Jun 17, 2016 by (13,979 points)
What are the other components of this field?

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Andrii Magalich

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Arnold Neumaier
Do you have any reference? I would like to understand the motivation

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Andrii Magalich

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Arnold Neumaier
Dear Neumaier, would you add a word on the paradox at hand?

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user user929304

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Arnold Neumaier
Thanks a lot, if I may ask, how does one in general define temperature correctly? If not in terms of the system's constituents' velocities.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user user929304
The standard form of the equilibrium distribution is $\beta u\cdot p$ where $\beta =1/T$ and $u_\mu$ is a relativistic 4-velocity, that means $u^2=1$. Of course you can define a new 4-velocity $\beta_\mu=\beta u_\mu$, so that $1/T$ is the zero'th component of a vector in the rest frame. This has no obvious advantage, but several disadvantages. It does not map onto standard relativistic hydro (which uses $u^2=1$), or relativistic kinetics, and is not used in any significant paper or book that I am aware of.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Thomas

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Thomas
@Thomas: Standard relativistic hydrodynamics is a purely phenomenological theory, the simplest marriage of relativity and thermodynamics. It is valid only in the absence of dissipation, hence unrealistic. 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. There are various ways to patch the situation (Israel-Stewart is not fully consistent; Mueller-Ruggeri is better) but no solution from first principles is known.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Arnold Neumaier
The correct definition must of course come from quantum statistical mechanics, which is the underlying microscopic theory. Thus the question is the correct generalization of the ensemble that produces standard hydromechanics in the nonrelativistic approximation. This gives the term $\beta \cdot cp$ in place of $\beta H$. Simplifying this by assuming $\beta=\beta'u$ with a scalar $\beta'$ is already an extra assumption, hence most likely an approximation.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Arnold Neumaier
Regarding you first comment, I completely disagree and I and others (Schaefer, arxiv.org/abs/1403.0653, Schaefer & Teaney, arxiv.org/abs/0904.3107, Romatschke, arxiv.org/abs/0902.3663, Son and Starinets, arxiv.org/abs/0704.0240) have written extended reviews of the modern understanding of relativistic fluid mechanics.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Thomas
@Thomas: Of the papers you cited in your comment, none derives from microscopic theory, and only Romatschke discusses a consistent dissipative formulation. It is no comprehensive review, it doesn't even mention the solution found by Mueller and Ruggeri much earlier. This shows that resolving the matter in a generally agreed way is still a topic of research.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Arnold Neumaier
They refer to the usual cases: i) weak coupling: QFT->kinetic theory-> fluid mechanics, ii) strong coupling holographic: QFT->AdS/CFT->fluid mechanics. Of course, in the general case there is no derivation, we cannot even solve QFT in equilibrium. More to the point, we view fluid dynamics as an EFT, so that what we have to provide is not a derivation, but a demonstration the fluid dynamics i) respects the underlying symmetries, ii) provides a systematic low energy expansion, iii) is consistent.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Thomas
Regarding your first comment above: To the extent that all that Beccatini does is define $\beta_\mu =\beta u_\mu$ the answer is of course that both can be derived from weak coupling nonequilibrium QFT. All I was saying is that 99.976% of the literature uses $\beta u_\mu$, for good reasons (o.k., I'm exagerating, but other than hearing Francesco talk about it, I don't remember seeing it).

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Thomas
+ 1 like - 0 dislike

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.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Wolphram jonny
answered Jun 17, 2016 by (30 points)
+ 1 like - 0 dislike

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.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user WillO
answered Jun 17, 2016 by (10 points)
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.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Arnold Neumaier
@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?

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user WillO
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.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Arnold Neumaier
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?

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user user929304
@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.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user WillO
No molecular model is needed, the equations for fluid mechanics date from 160 years ago, long before a molecular interpretation of temperature existed. Of course the results are invariant under Galilei boosts. The conceptual problems arise only in the relativistic regime, due to the frame dependence of the volume that enters the thermodynamics. For a nonrelativistic ideal monatomic gas, your formula is correct.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Arnold Neumaier
@ArnoldNeumaier: Thank you. I think the phrase I needed to hear was "due to the frame dependence of the volume that enters the thermodynamics".

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user WillO
+ 1 like - 0 dislike

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.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user jim
answered Jun 18, 2016 by (40 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.