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Re: [Phys-l] Photon Thermodynamics

On 06/17/2007 12:19 PM, Jeffrey Schnick wrote:
I have a question about Equation 45.17 in the Feynman Lectures on

Excellent question. You know folks are on the ball when
they find bugs in the Feynman lectures.

We shall see that the difficulty here is traceable to being
sloppy about what is held constant and what is not. This
seems to be a raging disease wherever thermodynamics is
discussed; not even Feynman is immune.

Feynman is applying thermodynamics to photons and has chosen
to treat T and V as independent variables and to treat U(T,V) and P(T,V)
as dependent variables.

Yes, that's what he /says/ he will do ... but no, that's
not what he really /does/ do. See below.

He takes the partial derivative of U=3PV (comes
from equation 39.17) holding T constant and obtains:
dU/dV (const T) = 3P
I was expecting:
dU/dV (const T) = 3P + 3V dP/dV (const T)

That expectation is correct.

Why is the second term missing? Its absence would suggest that P(T,V)
is actually only a function of T, P(T).

For photons under black-body conditions, P(T,V) is independent
of V. You might have surmised as much from a scaling argument:
We know P is intensive and T is intensive, so if there are no
other variables involved, you can't have an extensive variable
on one side of the equation and not on the other.

Remember: The idea that XXX is an intensive property is just
a particularly simple scaling property: It means that XXX
scales like the size of the system to the zeroth power.

Interestingly enough, for ordinary gas molecules under "conventional"
conditions, it is not true that P is independent of V in this

Feynman is pulling a fast one here, because there is actually
another variable running around, namely N, the number of photons.

When dealing with ordinary gas molecules, it is /conventional/
to assume that changes in T and/or V take place at constant N
unless otherwise stated. This is called a canonical ensemble.

When dealing with photons, it is hard to keep N constant over
long or even medium-long periods of time. It turns out that
Feynman is not saying
dU/dV (const T,N) = 3P
but rather
dU/dV (const T,mu) = 3P
where mu is the chemical potential. This is called a
grand canonical ensemble.

This is a trap for the unwary, due to conflicting conventions.

You can reconcile these two ideas by imagining that
a) you expand the photon gas at constant N for a very short time,
b) pause while little antennas in the walls radiate some
more photons into the box.
Step (a) is the same for molecules as for photons. Step (b)
is unconventional but not impossible for molecules; imagine
a reservoir of high-density liquid that evaporates so as to
make the gas density a function of (T) and not (T,V).

Actually there is another somewhat-related trap here. The
pressure is -- as always -- defined in terms of an isentropic
expansion, but here we are considering an isothermal expansion.
You can reconcile these ideas by imagining that the pressure
measurement is done sufficiently quickly that it is isentropic,
while the expansion is done slowly enough that it is isothermal.

Equation 39.18, PV^(4/3)=C
(where C is a constant), would lead one to believe that P is indeed a
function of V.

This can be considered another outbreak of the same disease:
being sloppy about what is held constant and what is not.
The law that P V^gamma = constant is not universally true;
it applies only to isentropic processes. To remind myself
of this, I prefer to write
P V^gamma = f(S)
where S is the entropy, and f(S) is "some function" of the

You know I like to visualize these relationships; in this case
the picture is that the contours of constant (P V^gamma) are
everywhere parallel to the contours of constant entropy.