# Thermodynamics and Relativity: observing gases, rel. effects

1. Oct 27, 2007

### giann_tee

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

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

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

2. Oct 27, 2007

### 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" ?

3. Oct 27, 2007

### pervect

Staff Emeritus
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):

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.

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/ser...00099000017170601000001&idtype=cvips&gifs=Yes

Last edited: Oct 27, 2007
4. Oct 28, 2007

### 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.

5. Oct 28, 2007

### 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 ?

Last edited: Oct 28, 2007
6. Oct 28, 2007

### giann_tee

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

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?

7. Oct 29, 2007

### pervect

Staff Emeritus
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.

Last edited: Oct 29, 2007
8. Oct 30, 2007

### 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.

9. Oct 31, 2007

### pervect

Staff Emeritus
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(?).

Last edited: Oct 31, 2007
10. Oct 31, 2007

### bernhard.rothenstein

temperature transformation

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

11. Nov 1, 2007

### OOO

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

12. Nov 1, 2007

### bernhard.rothenstein

temperature and SR

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

13. Nov 1, 2007

### OOO

???
?x?
???

14. Nov 1, 2007

### 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?

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?

15. Nov 1, 2007

### bernhard.rothenstein

temperature relativity

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.

16. Nov 1, 2007

### bernhard.rothenstein

temperature relativity

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.

17. Nov 1, 2007

### giann_tee

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

18. Nov 1, 2007

### bernhard.rothenstein

temperature relativity

Please be more specific (CMB?) polarization?

19. Nov 1, 2007

### pervect

Staff Emeritus
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.

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?

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...

Last edited: Nov 1, 2007
20. Nov 3, 2007

### 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. :-))