Re: Relativity and Thermodynamics
- From: Darwin123 <drosen0000@xxxxxxxxx>
- Date: Sun, 16 Aug 2009 15:28:47 -0700 (PDT)
On Aug 16, 5:43 pm, rabid_fan <r...@xxxxxxxxxxxxx> wrote:
The effects of relativity on thermodynamics seem littleNo. Temperature is not defined in terms of velocity. Temperature
discussed. How does temperature transform relativistically?
In a moving reference frame, a container of gas will appear
colder to a stationary observer because time is slower in
the moving frame. Slower time means a lower average particle
velocity and hence a lower temperature. Is this a correct
interpretation?
is defined in terms of the efficiency of a heat engine. A good
analysis of temperature has to avoid atomic-scale parameters. An
example of an atomic scale parameter would be velocity of an atom.
Temperature is a macroscopic quantity that represents a
statistical average of different atomic scale quantities. Not all
these quantities are velocity. The mass of the atoms also goes into
the temperature. Also the potential energy, which is related to the
distance between atoms.
As an example of where your reasoning fails, the average
kinetic energy per atom per mode is often said to be kT/2, where k is
Boltzmann's constant. However, the relativistic equation for kinetic
energy is not mv^2/2.
There is also that little problem about simultaneity. In special
relativity, the two systems are moving at a constant velocity. Thus,
there is a constant relative velocity. This means that two systems
exchanging heat can exchange heat only once during their trip. After
that they are moving farther apart. Thus, relativistic thermodynamics
intrinsically requires acceleration. There has to be a means for the
two reservoirs to come together more than once. According to some
definitions, this makes thermodynamics intrinsically a problem in
general relativity. To analyze a heat engine between two moving
reservoirs, one of the reservoirs has to make a round trip.
I forgot in which paper Einstein introduced relativistic
thermodynamics. I remember that he derived the transformation of
various macroscopic quantities. What he did was analyze the situation
where there were two heat reservoirs, one of which was on a high
velocity wheel that was rotating rapidly (near the speed of light). A
heat engine connected the two heat reservoirs only at one point in the
circle. Every time the moving heat reservoir was close to the
stationary heat reservoir, the heat engine would produce work by
moving a little bit of entropy from one reservoir to the other.
I forgot the details of the analysis. However, I do remember
this. The entropy is invariant to reference frame. The entropy, as
measured by all observers, is the same. There was some transformation
of the temperature that I forgot. Temperature had some variation with
the velocity of the observer.
For one thing, the black body formula is no longer valid. One can
But the temperature of a body can be measured by its
emission spectrum. How is this spectrum altered in
the relativistic transformation?
prove this easily in several ways.
Assume there is an observer moving near the speed of light in in
a stationary black body cavity. The black body formula says that the
spectrum of light inside the cavity is isotropic. The spectrum does
not vary with the direction that the observer is looking. However, the
relativistic Doppler shift contradicts this assumption. The spectrum
as measured by this moving observer is highly anisotropic. The
radiation propagating in the direction of motion is red shifted, and
the radiation propagating against the direction of motion is blue
shifted. So, if you use the black body formulas, the temperature for a
moving observer varies with direction. However, the temperature for a
stationary observer in the same cavity does not vary with direction.
This violates one of the postulates of relativity.
The contradiction is not a problem with relativity, but with the
thermodynamics. In order to use thermodynamics, the observer has to be
in a state that is nearly in equilibrium with its surroundings. A
stationary observer in a stationary black body cavity can be in a
state close to equilibrium with the radiation around it. However, an
observer moving close to the speed of light can not be in thermal
equilibrium inside a stationary cavity. The radiation from the
stationary cavity is going to slow the moving observer down. The
radiation will produce "friction" with the observer.
Another way to show the black body formula is invalid is this. The
black body radiation formula that we learned is not covariant. If the
observer is moving rapidly relative to the black body, the observer
will not see a black body spectrum. Therefore, there is no
thermodynamic temperature that can be determined from the system.
Strict thermodynamics can only analyze systems where all the
components are near equilibrium. The observer is part of the system.
Simple application of the Lorentz transformation to atomic-scale
quantities does not guarantee equilibrium.
.
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