Thermodynamics and Relativity: observing gases, rel. effects

[giann_tee]

Few weeks ago a respected magazine brought some excitement with a story on missing bond between Thermodynamics and Relativity.

RELATIVISTIC THERMODYNAMICS. Einstein*s special theory of
relativity has formulas, called Lorentz transformations, that
convert time or distance intervals from a resting frame of reference
to a frame zooming by at nearly the speed of light. But how about
temperature? That is, if a speeding observer, carrying her
thermometer with her, tries to measure the temperature of a gas in a
stationary bottle, what temperature will she measure? A new look at
this contentious subject suggests that the temperature will be the
same as that measured in the rest frame. In other words, moving
bodies will not appear hotter or colder.
You*d think that such an issue would have been settled decades ago,
but this is not the case. Einstein and Planck thought, at one time,
that the speeding thermometer would measure a lower temperature,
while others thought the temperature would be higher. One problem
is how to define or measure a gas temperature in the first place.
James Clerk Maxwell in 1866 enunciated his famous formula predicting
that the distribution of gas particle velocities would look like a
Gaussian-shaped curve. But how would this curve appear to be for
someone flying past? What would the equivalent average gas
temperature be to this other observer? Jorn Dunkel and his
colleagues at the Universitat Augsburg (Germany) and the Universidad
de Sevilla (Spain) could not exactly make direct measurements (no
one has figured out how to maintain a contained gas at relativistic
speeds in a terrestrial lab), but they performed extensive
simulations of the matter. Dunkel
(joern.dunkel@physik.uni-augsburg.de ) says that some astrophysical
systems might eventually offer a chance to experimentally judge the
issue. In general the effort to marry thermodynamics with special
relativity is still at an early stage. It is not exactly known how
several thermodynamic parameters change at high speeds. Absolute
zero, Dunkel says, will always be absolute zero, even for
quickly-moving observers. But producing proper Lorentz
transformations for other quantities such as entropy will be
trickier to do. (Cubero et al., Physical Review Letters, 26 October
2007; text available to journalists at www.aip.org/physnews/select )

My question is what satisfies the definition of relativistic gas and how would that differ from single star systems with shifted spectra?

Take for example spinning galaxies that have different shifts in edges approaching or moving away from us. However we often hear the areas closer to the center to have whirls of gas - where does it count in?

Today in New Scientist there is a brief and mostly unrelated passage that says that cosmic rays are affected by the photons of CMB.
http://space.newscientist.com/article/dn12818

The problem arises because every cubic centimetre of space contains about 400 relic photons from the big bang fireball. They have little energy, but seen from the point of view of a speeding cosmic ray, they are enormously boosted in energy, becoming high-energy gamma rays. The interaction of cosmic rays with these gamma rays continually saps the very highest energy cosmic rays of energy. Therefore, we should detect none of them on Earth – but we do.

I guess that section is about collisions with CMB photons that seem to average the possible cosmic rays somehow.

Comments?

 

[OOO]

Interesting question. My feeling is that temperature measurements should somehow be boiled down to black body radiation because a relativistically moving mercury thermometer brings in some complications...

So what temperature does a relativistically moving black body appear to have ? I'd say it depends on the angle of observation.

Probably one has to take into account the following: in statistical mechanics every conserved quantity results in a intensive quantity/Lagrange parameter for the corresponding extensive quantity that is conserved. Energy->temperature, particle number->chemical potential, and well, Momentum->... Usually momentum isn't considered, but in this case, don't know. Probably the emitted radiation counts as a "momentum contact" ?

 

[pervect]

I have not investigated the subject in great depth. My favorite paper on relativistic thermodynamics is currently http://arxiv.org/abs/physics/0505004. It's short, simple, and so far I haven't found any problems with it. But I haven't read the entire literature on the topic, so my opinion is not fully formed. Here's a short quote (I've taken the liberty of including some of the bibliography text):

Theory of relativistic thermodynamics has a long and controversial history
(see, e.g., [1] Stuart, E. B., Brainard, A. J. and Benjamin, G-O. (ed), A Critical Review of
Thermodynamics (Mono Book Corp. Baltimore, 1970).and references therein). The controversy seems to have been settled more or less by the end of 1960s [2] C. K. Yuen, Amer. J. Phys., 38 (1970), 246. however, papers are still being published to this date (e.g., [[3] Ch. French and J. P. Vigier, Phys. Lett. A, 215 (1996), 247; H. Kolbenstvedt, Nuovo Cimento, 115B (2000), 257.).

I don't really understand why there was a controversy in the first place, nor do I have a really good feeling of the significance of the latest PRL paper. (I've only seen the abstract, I'd have to go to the library to get the complete paper).

I do know that I, personally, don't trust New Scientist to do a good job of reporting (of course that's a purely personal opinion). I would not be terribly surprised if there isn't as much controversy as some journalists try to make out there is.

The Nakamura paper that I like suggests that inverse temperature transforms as a 4-vector, and that the entropy of a system transforms as a scalar. This is exactly what one would expect for a quantity that is the log of the number of states.

Heat is a form of energy in non-relativistic thermodynamics, where the energy and momentum are distinct quantities. In relativity, however, the energy and momentum are
components of one physical entity, energy-momentum four vector namely, and
thus cannot be treated independently. Therefore we must treat the energy-
momentum exchange between the objects, not energy alone. Heat is a form of en-
ergy in non-relativistic thermodynamics, where the energy and momentum
are distinct quantities. In relativity, however, the energy and momentum are
components of one physical entity, energy-momentum four vector namely, and
thus cannot be treated independently. Therefore we must treat the energy-
momentum exchange between the objects, not energy alone. Consequently,
the inverse temperature must have four components, corresponding
to each component of energy-momentum four vector.

Here's the abstract of the recent PRL paper. It doesn't seem to be nearly as dramatic as the journalists try to make it appear:

http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=PRLTAO000099000017170601000001&idtype=cvips&gifs=Yes

There is an intense debate in the recent literature about the correct generalization of Maxwell's velocity distribution in special relativity. The most frequently discussed candidate distributions include the Jüttner function as well as modifications thereof. Here we report results from fully relativistic one-dimensional molecular dynamics simulations that resolve the ambiguity. The numerical evidence unequivocally favors the Jüttner distribution. Moreover, our simulations illustrate that the concept of “thermal equilibrium” extends naturally to special relativity only if a many-particle system is spatially confined. They make evident that “temperature” can be statistically defined and measured in an observer frame independent way.

 

[Chris Hillman]

Ditto pervect on the dubiousness of the alleged "controversy".

To the extent that valid issues exist, IMO they are too subtle for discussion except among highly knowledgeable experts; see our past discussion of the Ehrenfest "paradox" for another subject about which two seemingly contradictory but accurate characterizations hold true:

• the issues involved are elementary
• a dozen or more distinct common misconceptions are relevant

In general, there are several places in the literature where one can say that the presence of sufficiently many elementary issues in some topic, coupled with sufficiently many badly written papers by incautious authors who ignore (out of haste or ignorance) some of these issues, can lead to endless and ultimately pointless confusion. Unfortunately, such alleged "areas of controversy" can be almost impossible to tell from areas of genuine controversy, unless one has expert knowledge and has carefully studied the literature and thought about all the issues involved.

 

[OOO]

I have looked at pervect's favorite paper but I am still not sure if I have grasped these things.

Suppose I defined relativistic temperature by black body radiation. If I restrict on measuring temperature from an angle perpendicular to the direction of motion of the black body, the Doppler effect doesn't need not be considered.

What remains is the frequency shift due to time dilation. Transforming the frequency w0 that I have measured from my rest frame (k is zero in my frame since I observe perpendicular to the moving black body) gives

$$\omega = \omega_0 \cdot cosh \zeta > \omega_0.$$

So the temperature I got from measuring frequency w0 corresponds to a higher temperature (higher frequency w) in the frame of the moving black body and this effect is from time dilation (I see the black body's time slowed down).

Likewise the 0-component of the inverse temperature transformed to the frame moving with the black body yields

$$\beta = \beta_0 \cdot cosh \zeta > \beta_0.$$

So from the inverse temperature I have measured in my frame, I deduce a higher inverse temperature i.e. a lower temperature in the frame moving with the black body.

What's wrong with this argument ?

 

[giann_tee]

Sun is treated as a black body. I don't recall that other stars are any better.

WMAP image of the cosmic microwave background radiation anisotropy. It has the most perfect thermal emission spectrum known and corresponds to a temperature of 2.725 kelvin (K) with an emission peak at 160.2 GHz.

http://en.wikipedia.org/wiki/Black_body

Thinking that the black body spectrum is similar and connected to the Maxwell speed distribution then both must bend back somehow - can't say.

Is there such a hot gas on Earth?

 

[pervect]

I have looked at pervect's favorite paper but I am still not sure if I have grasped these things.

OK, suppose we have an object with a temperature of 300K in its rest frame. The paper says that the covariant way of representing the temperature is by representing 1/T as a 4-vector. In the rest frame of the object, the components of the 4-inverse-temperature will be

(1/300,0,0,0)

in non-relativistic units, to use relativistic units we'd have to multiply that by Gk/c^4.

In a moving frame, we just take a Lorentz boost of the inverse temperature to get the 4-vector that represents the inverse-temperature of the moving object.

Suppose the object absorbs a photon, which has an energy-momentum 4-vector P. To get the change in entropy, $\Delta\,S$, we just take the dot product of the energy-momentum 4-vector and the inverse temperature 4-vector (note: I've glossed over some sign issues - and I've also assumed that the system is large enough that the emission or absorption of a single photon will be reversible transformation).

So in the rest frame of the object, that works out to be $\Delta\,S = P_0/300$, where $P_0$ is the energy E of the photon, the other components of the 4-momentum of the absorbed photon other than energy are irrelevant, and we get the classic formula for entropy.

Because that the dot-product of two 4-vectors is invariant, this formulation is invariant, so an observer in any frame will work out the same result for $\Delta\,S$, the change in entropy, as the observer in the rest frame does.

If we look at what an observer in a different frame sees, they will represent the 4-vector for the photon with different numbers (i.e. it will have different components) which will be derived from the original 4-vector via a Lorentz boost. They will also represent the 4-vector for inverse temperature with a different components, which will also be a Lorentz boost of the original. But the dot product of the boosted vectors will be the same, so the change in entropy will be the same.

Most textbooks that I've read don't actually take this approach, though I think it's rather neat. The textbooks that I've happen to read mostly deal with the issue of temperature by considering it as defined only in the rest frame of the object (which is very often the rest frame of some fluid element of a relativistic fluid). But if you want to define the relativistic version of temperature for any observer and not just in the rest frame of the source, this seems to me to be the way to go.

 

[giann_tee]

I wonder if the center of Sun with it's 10 million degrees has a "thermal" spectrum of radiation. It seems according to all your writings that terms have gone out of the out of the box experience of thermodynamics so to speak.

 

[pervect]

The sun won't be a perfect blackbody. But as nearly as I can tell, the discrepancies from a perfect black body occur mainly because of absorption by the sun's atmosphere. The only condition for a black body is that it absorb all incoming light - because of its sheer size, I believe the sun does a good job of this (except for effects due to the solar atmosphere).

General relativistic effects occur because of gravitational time dilation, making the local thermodynamic temperature for a static observer close to the sun slightly higher than for one far away. But I do not believe that there are any systematic effects due to relativity which cause a deviation from Planck's law of black body radiation, i.e. all static observers will still see a Planck law.

The specifics of the relativistic behavior of the gas are basically irrelevant to the behavior of a black body, the only thing that is really important is that the gas have an absorption/emission coefficient that's close to unity. If this absorption condition is met, the gas will act like a black body. The black body law can be and usually is derived by considering the modes of a resonant cavity with no substance in it at all, just a vacuum - see for example the hyperphysics webpage

http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

Special relativity just isn't relevant to this computation, or to put it another way, Maxwell's equations are already fully compatible with special relativity.

It might be a good idea to move this to the astrophysics forum if you want more input, it appears that you're actually interested in the spectrum of the sun and what factors might make it depart from an ideal black body(?).

 

[bernhard.rothenstein]

temperature transformation

I have looked at pervect's favorite paper but I am still not sure if I have grasped these things.

Suppose I defined relativistic temperature by black body radiation. If I restrict on measuring temperature from an angle perpendicular to the direction of motion of the black body, the Doppler effect doesn't need not be considered.

What remains is the frequency shift due to time dilation. Transforming the frequency w0 that I have measured from my rest frame (k is zero in my frame since I observe perpendicular to the moving black body) gives

$$\omega = \omega_0 \cdot cosh \zeta > \omega_0.$$

So the temperature I got from measuring frequency w0 corresponds to a higher temperature (higher frequency w) in the frame of the moving black body and this effect is from time dilation (I see the black body's time slowed down).

Likewise the 0-component of the inverse temperature transformed to the frame moving with the black body yields

$$\beta = \beta_0 \cdot cosh \zeta > \beta_0.$$

So from the inverse temperature I have measured in my frame, I deduce a higher inverse temperature i.e. a lower temperature in the frame moving with the black body.

What's wrong with this argument ?

Do you think that homega/kT as an exponent in the distribution should be an invariant?

 

[OOO]

Do you think that homega/kT as an exponent in the distribution should be an invariant?

What do you mean ? From what I understand the distribution should contain $k_\mu \beta^\mu$ instead of $\omega/T$

 

[bernhard.rothenstein]

temperature and SR

What do you mean ? From what I understand the distribution should contain $k_\mu \beta^\mu$ instead of $\omega/T$

Bose Einstein
f(E)=1/(Aexp(E/kT)-1] ??

 

[OOO]

Bose Einstein
f(E)=1/(Aexp(E/kT)-1] ??

?
?x?
?

 

[giann_tee]

Black body (thermal, Planck) radiation is something deep! It's not just some chamber they used to simulate absorption. It is something fundamental.

Neon lamps and many other things don't have that spectrum.

Our Sun doesn't have exactly that spectrum but close. The curve FOR 5700 kelvins is where the Sun FITS.

I learned that models give that photon is travels million years to the surface of the Sun; and only the surface counts in as observational spectrum, Solar spectrum.

I understand that radiation can be shifted due to special relativistic Doppler shifting, or general relativistic gravitational.

The heat itself as proven experimentally produces a "standard distribution" which should rise and rise with the maximum speed present in that chamber with hot gas but not infinitely. Surely a gas at some temperature must obey to another distribution?

It is also possible to write down relativistic Boltzmann equations for systems in which a number of particle species can collide and produce different species. This is how the formation of the light elements in big bang nucleosynthesis is calculated. The Boltzmann equation is also often used in dynamics, especially galactic dynamics. A galaxy, under certain assumptions, may be approximated as a continuous fluid; its mass distribution is then represented by f; in galaxies, physical collisions between the stars are very rare, and the effect gravitational collisions can be neglected for times far longer than the age of the universe.
http://en.wikipedia.org/wiki/Boltzmann_equation

Also, Maxwell distribution, correct me if I'm wrong, is old stuff; Planck curve is better.

When they observe some radiation spectrum (relative intensities->energy) they can just divide it with Boltzmann k and get T, so defining T is not their immediate weakness.

On the other hand, pV=NkT
Something may happen with p, V, and energy density that is general relativity.

What remains is what about moving gas?

 

[bernhard.rothenstein]

temperature relativity

?
?x?
?

The Boltzmann constant is a relativistic invariant. The physical quantities that appear as exponents are dimensionless physical quantities and so relativistic invariants. In many distribution functions we find at exponents of e the physical quantity E/kT (E energy), resulting that T should transform as E does. In the case of distribution functions reffering to tardyons E=LE' and so T=LT' (L Lorentz factor). In the case when the distribution reffers to photons E=DE' and so T=DT' (D Doppler factor).
What I say are questions and not statements. Please tell where I am wrong. The transformation T=LT is derived by many Authors.

 

[bernhard.rothenstein]

temperature relativity

?
?x?
?

Looking in my collection of papers I have found
T. Greber and H. Blatter, "Aberration and Doppler shift; The cosmic backgrouind radiation and its rest frame," Am.J.Phys. 58 942 1990
in which it is shown that the temperature T in the Planck's radiation law transforms as
T'/T=1/L(1-bcostheta')
b=V/c, L Lorentz factor.

 

[giann_tee]

Well... the polarization of CMB is de-facto our own motion through space, visible as a change in spectrum/energy/kT.

 

[bernhard.rothenstein]

temperature relativity

Well... the polarization of CMB is de-facto our own motion through space, visible as a change in spectrum/energy/kT.

Please be more specific (CMB?) polarization?

 

[pervect]

Black body (thermal, Planck) radiation is something deep! It's not just some chamber they used to simulate absorption. It is something fundamental.

Neon lamps and many other things don't have that spectrum.

Our Sun doesn't have exactly that spectrum but close. The curve FOR 5700 kelvins is where the Sun FITS.

I have no disagreements with the above, but a few comments:

First observation: transparent bodies won't emit black body radiation by the definition of a black body.

Blackbody radiation" or "cavity radiation" refers to an object or system which absorbs all radiation incident upon it and re-radiates energy which is characteristic of this radiating system only, not dependent upon the type of radiation which is incident upon it.

Does the neon gas in a neon bulb meet this definition (of an object or system which absorbs all incident radiation)? I hope it's obvious that it does not.

Now, for the 64 dollar question. If you piled up a sun-size ball of neon gas, rather than having the small amount present in a bulb, would the resulting sun-sized ball of neon gas be transparent, or opaque? Would it be a black body, meeting the above definition?

You also ask:

Surely a gas at some temperature must obey to another distribution?

According to the PRL paper that was recently published, the velocity disbribution in a relativistic gas is different - it's the Juttner distribution rather than the Maxwell-Boltzman distribution.

But this won't matter as far as the observed spectrum goes...

 

[giann_tee]

Great. Here's what I found out. I haven't read anything special yet but in any case it was going to be off the textbook route.

There are relativistic effects and experimenting and calculating them would be a little complicated. One quick trick I encountered is that kT is actually very small! So to my amazement, temperature has to become insanely high.
http://militzer.gl.ciw.edu/diss/node5.html
Billions of kelvins are required...

"In our approach, we do not consider relativistic effects, which become important when the thermal energy or the Fermi energy become of the order of the rest mass energy of the electron, which corresponds to a temperatures of 4x10^9 K and a density of 1.5x10^3 gcm^(-3)"

As for 64$ I throw Sunyaev-Zeldovich Effect in. :-))

 

[giann_tee]

Reading from Satoshi Nozawaat Josai Junior College for Women... relativistic effects in sunyaev-zeldovich effect for galaxy clusters. So, if background radiation is passing through clusters it is changed a little bit.

"Relativistic corrections are very important for high-temperature CG Te> 10keV.
・ Relativistic corrections are significant for large X region (sub-millimeter region)For X>10, factor 4 effect!
・Fokker-Planck expansion approximation is OK for Te < 15keV"

 

[Dale]

OK, suppose we have an object with a temperature of 300K in its rest frame. The paper says that the covariant way of representing the temperature is by representing 1/T as a 4-vector. In the rest frame of the object, the components of the 4-inverse-temperature will be

(1/300,0,0,0)

in non-relativistic units, to use relativistic units we'd have to multiply that by Gk/c^4.

In a moving frame, we just take a Lorentz boost of the inverse temperature to get the 4-vector that represents the inverse-temperature of the moving object.

I have not been following this thread closely, nor do I have any previous experience with relativistic thermodynamics, but I find this post fascinating. What would the spatial components of the boosted 4-inverse-temperature represent?

 

[pervect]

I have not been following this thread closely, nor do I have any previous experience with relativistic thermodynamics, but I find this post fascinating. What would the spatial components of the boosted 4-inverse-temperature represent?

Mathematically, they describe how much changing the momentum of a moving system changed its entropy.

In non-relativistic thermodynamics, only energy changes the entropy of a system, changing the momentum of the system has no effect. In relativistic thermodynamics, energy and momentum "mix together" in a 4-vector, just as time and space do. So you need to know the change in momentum as well as the change in energy of a system to compute the change in entropy.

Suppose you have a speck of dust that emits a microwave photon. This changes the entropy of the speck of dust by (dE)/T, where T is the temperature of the speck. In some rest frame moving relativistically with respect to the speck of dust, the emitted microwave photon looks like a high energy x-ray. But the change in entropy of the system is still very small in spite of the fact that dE in that frame is very high. The 4-vector approach to temperature combines the energy and momentum of the emitted photon to properly compute the proper change in entropy which in this case is much smaller than dE/T.

There may be and perhaps is more physical significance than what I describe above, but I don't know what it would be offhand. Most of the textbooks I've read don't discuss temperature using this 4-vector approach, and the paper I mention is very short and doesn't go into more detail.

 

[Dale]

I read that paper you linked to. Very interesting. It looked to me like the key idea in the paper was to make a four-vector definition of "volume". So it seemed that his definition of volume would refer to the same collection of events in spacetime. Which is an odd definition since that collection of events would not be simultaneous in boosted frames. Am I misunderstanding the concept. Of course, later on the velocities dropped out and you were left with the "rest volume" which sounds like a perfectly reasonable relativistic quantity to me.

 

[pervect]

It's fairly standard to define a volume as a 3-vector, I believe, that's not something that's unique to this particular author. But your observation is right on about the fact that we mean a different collection of space-time points when we talk about a system at some time "t" in one frame and then talk about it as existing at some time "t'" in another frame - the specific collection of points we are talking about depends on whose defintion of simultaneity we use.

As far as representing velocity as a three-vector, using clifford algebra one can represent an oriented volume element by a three-form. In a 4-d space-time, a 3-form has a dual representation (Hodges dual) which is a 1-form, which is basically a vector.

Basically this means that you define a volume element by the space-time vector that's perpendicular to it, and you make the length of this space-time vector proportional to the proper volume you wish to represent.