A Temp of Moving Body: 110 Yrs of Debate | Current Tech Update?

[johnscitech]

TL;DR Summary

"Starting from Planck and Einstein, several authors have proposed their own reasoning, concluding that a moving body could appear cooler, hotter or at the same temperature as measured by a local observer" is there a solution to this problem ?

According to this scientific report ( https://www.nature.com/articles/s41598-017-17526-4 ) written in 2017 : " The construction of a relativistic thermodynamics theory is still controversial after more than 110 years. To the date there is no agreement on which set of relativistic transformations of thermodynamic quantities is the correct one, or if the problem even has a solution. Starting from Planck and Einstein, several authors have proposed their own reasoning, concluding that a moving body could appear cooler, hotter or at the same temperature as measured by a local observer."

So i would like to ask if anyone knows any update regarding the issue raised in this report. With current technologies, has there not been someone who has been able to make conclusive experiments about this question?

Thanks in advance for any answer.

 

[vanhees71]

As most apparent problems with relativity it's a question of conventions and definitions. In my field of research (relativistic heavy-ion collisions) it's commonly accepted that one should use van Kampen's manifestly covariant definitions (in the above cited great Nature article they quote Landsberg for this formalism, which indeed a bit earlier than van Kampen).

N. G. van Kampen, Relativistic thermodynamics of moving
systems, Phys. Rev. 173, 295 (1968),
https://doi.org/10.1103/PhysRev.173.295.

All intensive quantities like temperature and chemical potential(s) are defined in the local rest frame of the medium and thus automatically scalars. The same holds for the general (non-equilibrium as well as the equilibrium) phase-space distribution functions are described by vector fields (four-current densities like particle-number or general (conserved) charge four-currents) or tensor fields (e.g., energy-momentum-stress tensor field of 2nd rank and angular-momentum tensor of 3rd rank) in a manifestly covariant way.

E.g., to define the phase-space distribution $f(t,\vec{x},\vec{p})$ you need to define it as counting particles at a fixed time $t^*$ as measured in the rest frame of the particles which have momentum $\vec{p}$ in the "laboratory frame". In this frame the phase-space distribution function is defined as this number of particles $\mathrm{d} N$ (a scalar quantity) $\mathrm{d} N = f^* \mathrm{d}^3 x^* \mathrm{d}^3 p^*$. The phase-space distribution function is then defined in the lab frame by the same quantity. You only have to express the phase-space element in the rest-frame of the particles in terms of the lab-frame variables: The rest-frame volume element appears Lorentz contracted in the lab frame, i.e., $\mathrm{d}^3 x=\frac{1}{\gamma} \mathrm{d}^3 x^*$ or
$$\mathrm{d}^3 x^*=\gamma \mathrm{d}^3 x=\frac{p^0}{m c} \mathrm{d}^3 x,$$
where $p^0=\sqrt{m^2 c^2+\vec{p}^2}$ (the rest-frame components are of course $p^{*0}=mc$, $\vec{p}^*=0$). It's also clear that $\mathrm{d}^3 p^*/p^{*0}=\mathrm{d}^3 p/p^0$ and thus
$$\mathrm{d}^3 p^* = \frac{m c}{p^0} \mathrm{d}^3 p.$$
If interpreted in this way the phase-space volume element is an invariant
$$\mathrm{d}^3 x^* \mathrm{d}^3 p^*=\mathrm{d}^3 x \mathrm{d}^3 p$$
and thus (again just by conventional definition!) the phase-space distribution function $f(t,\vec{x},\vec{p})$ is a scalar field.

The particle number density is then covariantly given as the temporal component of a four-vector field, i.e., the current
$$j^{\mu}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 p \frac{p^{\mu}}{p^0} f(t,\vec{x},\vec{p}).$$
For a discussion of the Lorentz (Poincare) invariant formulation of the Boltzmann equation see

https://itp.uni-frankfurt.de/~hees/publ/kolkata.pdf

and the literature quoted therein.

I think a strong argument to use these manifestly covariant definitions of the thermodynamic quantities in SR is that this seems to be the only safe way to extend it the general relativity, where observables should be defined as scalars (or tensor fields) to have proper observational meaning.

Concerning experiments it's of course always important to clearly analyze which quantity is measured. E.g., to measure temperature as the quantity defined in the above described mechanism you have to fix the thermometer with the fluid, i.e., being at rest relative to the fluid element whose temperature you want to measure.

 

[Vanadium 50]

You want to know if a review paper written 112 years after relativity is somehow no longer valid 115 years after relativity? It sure would be odd if it were.

Here's what I wrote about it two months ago the last time this came up:

Anyway, one might say "but of course the phase space transformation formalism is the right way to look at it." But one might also say "but of course internal energy transforms as energy" or various other "of courses".

Fundamentally, the problem is how one determines if two objects are at the same temperature. You can touch them and see that there is no heat flow in either direction. Fine - but what do you do when you have well-separated objects? And moving at high relative velocity too? (That can't help) Non-relativistic thermodynamics was invented around a particular concept (two objects at the same T) that really doesn't exist in relativity.

 

[anuttarasammyak]

All intensive quantities like temperature and chemical potential(s) are defined in the local rest frame of the medium and thus automatically scalars. The same holds for the general (non-equilibrium as well as the equilibrium) phase-space distribution functions are described by vector fields (four-current densities like particle-number or general (conserved) charge four-currents) or tensor fields (e.g., energy-momentum-stress tensor field of 2nd rank and angular-momentum tensor of 3rd rank) in a manifestly covariant way.

That seems similar to mass of body whose value is measured in the rest frame of the body and shared with other FORs. Thermodynamic internal energy of a body is a part of its mass.

Say we observe speed of a rocket is almost c, temperature of tea that the pilot is tasting is almost
A. absolute zero
B. infinity
C. exact 70 degree Celsius as the pilot measures
Among these possible answers, C seems moderate and practical to me.

 

[Vanadium 50]

Among these possible answers, C seems moderate and practical to me.

Are we having a vote?

 

[vanhees71]

Well, if you don't like the outcome of the vote, you can just deny that the election was fair ;-).

 

[anuttarasammyak]

Say we observe speed of a rocket is almost c, temperature of tea that the pilot is tasting is almost
A. absolute zero
B. infinity
C. exact 70 degree Celsius as the pilot measures
Among these possible answers, C seems moderate and practical to me.

Say we prepare two bodies or heat reservoirs of same type whose temperatures are T at their rest IFRs. Let them touch and slide with relative speed of almost c. They can exchange heat through the sliding area but I do not think heat transfer from one to another takes place due to symmetry between the two. For any observers in any IFR if they keep the idea that heat transfer from high side to low side of temperatures even in such dynamic contact, I think they have no option but to say the two bodies have same temperature (of T).

 

[jbriggs444]

I do not think heat transfer from one to another takes place due to symmetry between the two.

As I understand the conundrum:

If they exchange energy and momentum [in the sliding side-by-side scenario] then they will each slow down and heat up. From each point of view, there is a transfer of heat from the other.

 

[anuttarasammyak]

Friction, if there is, would slow the relative speed and generate heat on the area. But I do not think friction is inevitable in the thought experiment. No friction but heat transfer on the areas.

 

[jbriggs444]

Friction, if there is, would slow the relative speed and generate heat on the area. But I do not think friction is inevitable in the thought experiment.

Well, consider if the two are radiating light at each other. If each radiates perpendicularly (in its frame) then this will transfer momentum. Relativistic aberration means that the arriving pulse will carry rearward momentum onto the target and that the light pulse will arrive blue-shifted(*). Each will slow down. Each will gain thermal energy as assessed in its new rest frame.

This may well not count as "friction", but it has much the same effect.

(*) I think that it is a blue shift, but I am not competent to actually compute it. It feels like it must be if energy is to be conserved.

 

[anuttarasammyak]

You say internal energy or heat energy measured in each IFR would increase and kinetic energy of system would decrease for energy conservation. This conversion of kinetic energy to heat energy might be inevitable even in the thought experiment. An irreversible process. Might be so. Thanks.

Do you have an idea how much percent of the generated heat would go into one and the rest to another to change their temperatures ? The ratio depends on IFR of the observers ?

 

[Vanadium 50]

Before going down the p;ath of "the answer is obvious, and any physicist who disagrees with my gut instict - including those who have studied this for years - is just wrong" we should consider that things are not so simple even without relativity.

Consider two identical volumes of gas at temperature T₁ moving towards each other at non-relativistic velocity v. They are in a remote, isolated and empty part of the universe so there is no opportunity for them to exchange heat with anything external. The gas meets and mixes, and what was kinetic energy is now thermal energy, so the system is now at temperature T₂.

Pre-collision, what is the temperature of the system? T₁? T₂? Some other temperature? No temperature? All of these have their problems.

Classical thermodynamics is built around the concept that temperature is defined by bringing objects into contact and watching which way the heat flows. It struggles with objects that are separated, In relativity, objects are usually well-separated.

 

[anuttarasammyak]

Consider two identical volumes of gas at temperature T1 moving towards each other at non-relativistic velocity v. They are in a remote, isolated and empty part of the universe so there is no opportunity for them to exchange heat with anything external. The gas meets and mixes, and what was kinetic energy is now thermal energy, so the system is now at temperature T2.

Pre-collision, what is the temperature of the system? T1? T2? Some other temperature? No temperature? All of these have their problems.

May I take you say that thermodynamics are not yet prepared to answer the questions in our daily life e.g. temperature of tea served in the train, steam temperature of running locomotive. Hot water pipes are designed and installed without physical principle. Our galaxy, which is in peculiar motion of 600km/s against CMB rest frame of reference, does not have well defined CMB temperature which is 2.725K in CMB frame of reference.
To answer these challenges I think a discipline called e.g. kinetic thermodynamics might be necessary.

 

[vanhees71]

It's far from trivial but can be, no surprise, handled best using the modern covariant formalism. See Sect. 11 in

N. G. van Kampen, Relativistic thermodynamics of moving
systems, Phys. Rev. 173, 295 (1968),
https://doi.org/10.1103/PhysRev.173.295.

In the meantime I've read over two papers on the very interesting history of Einstein's debate with von Laue about the subject in 1952/53. After first promoting Ott's formalism, contradicting himself (and Planck), he finally came to the conclusion that the modern covariant formalism is the best, based on the simple argument that entropy, as defined by the Boltzmann-Planck relation, $S=k \ln \Omega$, where $\Omega$ is the number of microstates compatible with the macrostate of the system (or in modern form $S=-k \langle f \ln f \rangle$), must be a scalar:

https://doi.org/10.1017/S0007087400028764
http://cscanada.org/index.php/ans/article/view/j.ans.1715787020120602.2121

 

[jbriggs444]

Do you have an idea how much percent of the generated heat would go into one and the rest to another to change their temperatures ? The ratio depends on IFR of the observers ?

To know how much heat is generated, one must have an idea of how much each object had to start with and how much they ended with. But then before one can find those answers, one must define the terms. Does the heat content of an object depend on the reference frame used to make the assessment?

Personally, I vote for judging the heat content of an object in its own rest frame. In which case, for a suitably symmetric setup, one would naturally expect a 50/50 split.

 

[vanhees71]

Just read van Kampen's paper. The key paragraph (directly after the un-numbered equation after Eq. (22))

When thermal energy and momentum are transferred, the heat lost by one system is not necessarily equal in amount to the heat gained by the other system. The reason is that the heat contents of the transmitting agency (e. g. electromagnetic waves) is not the same for all observers. For they use different frames when decomposing the thermal energy-momentum four-vector in energy and momentum.

(emphasis in the original).

 

[anuttarasammyak]

Mass is not additive. Thermodynamic internal energy, a part of mass, is neither. so sum of one's gain and other's loss of internal energy is not zero. I take it in this way.

As for the case of post #7, momentum and energy conservation before and after heat exchange are
\mathbf{p_1}+\mathbf{p_2}=\mathbf{p_1}'+\mathbf{p_2}'`
\sqrt{m_1^2+p_1^2}+\sqrt{m_2^2+p_2^2}=\sqrt{m_1'^2+p_1'^1}+\sqrt{m_2'^2+p_2'^2}
where $m_1=m_2=m$. The equation say sum of heat energy gain in one IFR and loss in another IFR
m_1'+m_2'-2m \neq 0 in general.

 

[Vanadium 50]

May I take you say that thermodynamics are not yet prepared to answer the questions in our daily life e.g. temperature of tea served in the train, steam temperature of running locomotive.

Actually, yes. How do you measure the temperature of the cup of tea in the passing locomotive? However, the scale of the problem in that case is v_train/v_thermal (squared, as it happens).

Essentially, this is an example of non-equilibrium thermodynamics, which is far from simple.

 

[Sagittarius A-Star]

(*) I think that it is a blue shift, but I am not competent to actually compute it. It feels like it must be if energy is to be conserved.

Yes, it is received blue-shifted.

Source and receiver are at their points of closest approach
...
Fig. 2b. It is much easier if, instead, we analyze the scenario from the frame of the source.
...
Since the receiver's clocks are time-dilated relative to the source, the light that the receiver receives is blue-shifted by a factor of gamma. In other words,
$f_r = \gamma f_s$

Source:
https://en.wikipedia.org/wiki/Relat...eiver_are_at_their_points_of_closest_approach

 

[anuttarasammyak]

Actually, yes. How do you measure the temperature of the cup of tea in the passing locomotive?

Thanks for teaching. I think now we should be satisfied with measuring temperature T only for homogeneous system in its rest frame and should share the value with all the IFR observers as I said in analogy of mass in relativity theory, at least until we may discover and define the way to measure the temperature of the cup of the tea in the passive locomotive.

 

[vanhees71]

You have to specify what you measure when measuring a temperature.

A nice example is the temperature of the cosmic microwave background radiation. We always quote a temperature $T \simeq 2.725 \; \text{K}$. So what is this temperature and how is it measured?

What's really measured by, e.g., the Planck satellite is the spectrum of the radiation. To understand what's measured it's sufficient to know that the cosmic background radiation (locally) has the spectrum of a black body and the satellite is moving, together with our entire solar system, relative to the restframe of this thermal radiation with a certain velocity $\vec{v}$. So which spectrum does the satellite measure?

In the local rest frame of the CMBR the spectrum, i.e., the number of photons per unit volume and per unit volume in momentum space (i.e., the phase-space distribution function) is given by
$$f^{*}(\vec{k})=\frac{2}{\exp(\beta |\vec{k}^*|)-1},$$
where the $2$ comes from the two polarization degrees of freedom of photons. Here and in the following I use natural units with $\hbar=c=k_{\text{B}}=1$ and $\beta=1/T$ with $T$ the temperature as defined in modern terms as a scalar quantity. A comoving CMBR satellite, i.e., a satellite at rest relative to the restframe of the CMBR would just measure the isotropic temperature $T$, which by definition is what we call temperature in the modern relativistic formulation of thermodynamics and statistical physics in terms of manifestly covariant quantities. In this formulation the phase-space distribution function of photons is a scalar field.

It's thus easy to find the phase-space distribution function for the real Planck satellite. We just have to write $f^*$ (the distribution function as seen from the point of view of an observer at rest relative to the restframe of the black-body radiation) in a manifestly covariant way. The restframe observer's four-velocity is simply $u^*=(1,0,0,0)$. Now the momentum dependence in $f^*$ is to be understood to be valid for the on-shell four-momentum of a photon obeying $k_{\mu} k^{\mu}=0$, i.e., $|\vec{k}|=k^0$ in any frame of reference. For the rest frame we thus have $|\vec{k}^*|=u^{*\mu} k_{\mu}^*$, and that's a manifestly covariant expression, i.e., we can immediately write the phase-space distribution function in any frame as
$$f(k)=\frac{2}{\exp(\beta u^{\mu} k_{\mu})-1}.$$
So what's measuring the Planck satellite? Since $u=\gamma (1,\vec{v})$ with $\gamma=1/\sqrt{1-\vec{v}^2}$ we have
$$u^{\mu} k_{\mu}=\frac{|\vec{k}|}{\sqrt{1-\vec{v}^2}}(1-|\vec{v}|\cos \vartheta),$$
where $\cos \vartheta=\vec{k} \cdot \vec{v}/(|\vec{k}||vec{v}|)$.
So for any fixed direction you observe a black-body spectrum with an effective temperature given by
$$\tilde{\beta}=\frac{1}{\tilde{T}}=\frac{\beta}{\sqrt{1-\vec{v}^2}}(1-|\vec{v}| \cos \vartheta)$$
or
$$\tilde{T}=T \frac{\sqrt{1-\vec{v}^2}}{1-|\vec{v}| \cos \vartheta}.$$
That's also easily interpreted. For $\vartheta=0$ what the satellite observes are photons coming from a source of black-body radiation coming towards the detector, i.e., you have a blue shift and thus you measure a higher temperature
$$\tilde{T}(\vartheta=0)=T \sqrt{\frac{1+|\vec{v}|}{1-|\vec{v}|}}.$$
For $\vartheta=\pi$ it's measuring photons from a source moving away from the detector with the corresponding red shift
$$\tilde{T}(\vartheta=\pi)=T \sqrt{\frac{1-|\vec{v}|}{1+|\vec{v}|}}.$$
For values of $\vartheta$ somewhere in between you get the corresponding blue and red shifted photons leading to a temperature given by the general formula. For $\vartheta=\pi/2$ the redshift is due to the transverse Doppler effect, where the observed frequency change from the frequency as measured in the restframe of the source is solely due to relativistic time dilation.

One should note that indeed the CMBR satellites measure this "dipole contribution" to the variation of the temperature with direction and what's shown as pictures of the CMBR temperature variations is corrected by this dipole contribution, such that you get a nearly isotropic temperature of $T=2.725 \; \text{K}$ quoted above. The slight variations in temperature in dependence of direction of $\Delta T/T \simeq 10^{-5}$, i.e., the deviations to a isotropic black-body spectrum are then the really interesting signal for cosmologists. From the dipole contribution to the derivation we can deduce that our solar system is moving with about $370 \; \text{km}/\text{s}$ towards the constellation Leo relative to the rest frame of the CMBR. For details see

https://en.wikipedia.org/wiki/Cosmic_microwave_background#Features

and the quoted reference from the PLANCK collaboration:

https://arxiv.org/abs/1807.06205

Another example, where the meausurement of temperatures of a moving medium becomes relevant is in my own field of research, ultrarelativistic heavy-ion collisions: If you collide heavy nuclei together (as done at the Relativistic Heavy Ion Collider at the Brookhaven National Lab and at the Large Hardon Colider at CERN) a very hot and dense medium of strongly interacting matter is formed which can be described with astonishing precision as a blob of collectively streaming almost ideal fluid, which implies that it is, most of the time, in a state close to local thermal equilibrium and thus a (local) temperature is, in principle, well defined. So how to measure "the temperature"? That's a not so simple story, and indeed you have to account for the flow of the medium, and you need different measurements to get a (rough) picture of the temperature evolution of this rapidly expanding and cooling fireball.

One meausurement is simply the abundance of all kinds of hadrons produced in the collision, and since the particle numbers are scalar quantities you can indeed infer a temperature from it by just fitting the measured abundancies of different hadrons to a model, where all hadrons are in thermal equilibrium at the time the "chemistry" of the fireball is fixed. That's called the chemical freezeout. The so deduced temperature at the highest collision energies turns out to be around $155 \; \text{MeV}$, close to the pseudocritical transition temperature for a zero net-baryon-number (baryo-chemical potential $mu_{\text{B}}=0$) medium of strongly interacting particles between a state where quarks and gluons are the relevant degrees of freedom ("quark gluon plasma") and a "hadron resonance gas", which is known from lattice-QCD calculations at finite temperature. That means that shortly after the fireball undergoes the (cross-over) transition from a QGP to a hadron-resonance-gas state the inelastic collisions cease and almost only elastic collisions stay relevant.

The other signal you can use to measure a temperature from hadronic observables are the transverse-momentum spectra of the hadrons. These hadrons have a momentum distribution as given after "thermal freeze-out", i.e., it's the spectrum at the time when also the elastic collisions don't play a role anymore in the fireball distribution and the particles are freely streaming towards the detector (the decay of unstable hadrons to stable ones has to be taken into account as well). This spectrum can be fitted well with a socalled "blast wave model", assuming that one has a collectively streaming medium at local thermal equilibrium. To get the true temperature you have to take into account the flow of the particles to get a true "scalar" temperature of about $100 \; \text{MeV}$.

Another way to measure an space-time weighted average temperature over the entire fireball evolution is to measure the spectrum of emitted photons. Since they don't strongly interact they come out of the fireball nearly undisturbed, i.e., they are emitted from a flowing source in thermal equilibrium, and again you have to take into account the blue shift from the radial motion of this source to get a true "scalar" average temperature.

Last but not least another related signal are "dileptons", i.e., the measurement of electron-positron or muon-antimuon pairs which come mostly from quark-antiquark annihilation in the QGP phase and from decays of the light vector mesons in the hot and dense hadron-resonance-gas phase of the medium. There you can, in addition to the transverse-momentum spectrum, also measure the invariant-mass spectrum of the dileptons. Since invariant mass is a scalar in principle you get directly the "scalar temperature" by measuring the slope of the invariant-mass spectrum, but there you also have the shape of the vector-meson resonance spectrum, and this shape is majorly changed in the hot and dense medium as compared to the vacuum. There's however a "mass window" between the $\phi$-meson mass and the $J/\psi$ mass where you have a pretty flat spectral function of the here relevant autocorrelation function of the electromagnetic current, such that you get a pretty much undistorted thermal spectrum from the part of the dileptons which is due to quark-antiquark annihilations in the QGP phase of the fireball expansion. In addition the mass range of about $1 \; \text{GeV}$ to $2 \; \text{GeV}$ which is quite a bit larger than the usual temperatures of the fireball (of around a few 100 MeV) the emission from the early hot phases of the fireball evolution is the dominant contribution. The only obstacle here is that in this mass region also another dilepton contribution is relevant, namely, the semileptonic decay of correlated charm-anticharm pairs (or D and anti-D mesons in the hadronic phase), which have to be somehow subtracted to get a clean spectrum from the thermal dilepton radiation from the QGP phase.