Re: [Phys-l] Temp & Energy density

["David Bowman" <David_Bowman@georgetowncollege.edu>]

> Paul Lulai wrote:
>
> > Larger energy density implies increase in temp.
>
> Simply not so.  Others seem not to have directly addressed the 
> central misconception here so allow me.
>
> Example:  For a monatomic ideal gas at equilibrium
>
>     PV = Nkt = (2/3)U
>
> thus
>
>     energy density = U/V = (3/2)P
>
> That is, energy density is directly proportional to equilibrium 
> pressure and has no direct connection to temperature.
>
> John Mallinckrodt
 
I believe that John Decker's answer to Paul's original question was
complete and on target.  Paul's question (with the particles located
away from the piston) was clearly for a nonuniform particle
distribution, but a uniform one is prequisite for an equilibrium
situation for which the ideal gas equation of state would be
expected to hold.  So that equation of state can't be used to
describe the nonequilibrium situation asked about.
 
However, we can imagine an Ideal gas in a somewhat different more
equilibrated situation not originally asked about by Paul (i.e.
one that averages over a much larger collection of accessible
microstates).  Such a situation also requires that the internal
energy be U = (3/2)*N*k*T if the gas is monatomic.  Then the fact
that the particle number is constrained to be fixed (impermeable
container walls combined with a temperature far too cold for
particle-antiparticle pair creation) requires that, of necessity,
U/V = N*k*T/V.  Thus, in that (conserved particle) case, the energy
density is directly proportional to T/V.  So increasing the energy
density *does* imply an increase in temperature (as Paul, wrote
above).
 
The equation John M wrote above connecting U/V to P is a bit
different than a mere ideal gas situation.  An ideal gas is one that
in equilibrium obeys the equation of state: P*V = N*k*T.  In order
for that to happen all that is required is that the particle number
be conserved and that there be no interparticle positional
correlations in equilibrium.  I.e. all particles are uniformly
distributed throughout the available volume *and* their locations
are all mutually independent of each other.  Such a system has
no positional potential energy term in its Hamiltonian.  It doesn't
matter (much) what the kinetic energy term looks like  (other than
that there be no particle number changing processes and that the
kinetic energy as a monotone function of the particle momentum
magnitude |p| be bounded from below and unbounded as
|p| --> [infinity].  The kinetic energy can be a host of different
functional forms in terms of p.  For instance, the KE could be
proportional to p^4 instead of p^2 as for Newtonian particles.  In
that later case the gas would *still* be an ideal gas as long as
there were no spatial correlations (i.e. no positional PE & no
quantum correlations from interparticle exchange).  In fact, the
particles' KE term could use the full (special) relativistic
functional form: sqrt((m*c^2)^2 + (|p|*c)^2) - m*c^2 instead of the
Newtonian and the gas would still be ideal.  P*V is still N*k*T
(in equilibrium) even though it is *not* the case that p = (2/3)*U/V.
 
OTOH, suppose instead of requiring that there be no mutual
positional inter-particle correlations, we just require that the
Hamiltonian be just purely Newtonian translational kinetic energy.
I.e. each particle's energy is (1/2)p^2/m.  But now we don't care
whether or not there are any quantum inter-positional correlations
due to quantum exchange of identical particles.  Such positional
interparticle correlations are now allowed if desired.  In this case
the energy density is 2/3 of the pressure in equilibrium (i.e.
p = (2/3)*U/V) regardless of whether or not such interparticle
correlations are present.  If we have a low temperature Fermi or
Bose gas (with no potential energy terms) then the gas does *not*
obey the ideal gas equation of state even though it is *still* the
case that p = (2/3)*U/V.  In this case the 2 in the numerator of the
coefficient 2/3 is the exponent of the kinetic p^2 term in the
Hamiltonian, and the 3 in the denominator is the dimensionality of
the allowed region the particles are confined to.  (In general, if
the KE were proportional to |p|^r and the particles moved in d
Euclidean dimensions then the appropriate analog of the pressure p
in those dimensions would obey p = (r/d)*U/V where V is the
appropriate analog of the volume.)  Note that in this case we do not
have to require that the particle number be conserved.  Even a gas
of thermal photons obeys it once we change the ratio from 2/3 to 1/3
reflecting the fact that the photons are massless and have a kinetic
energy proportional to |p|^1.  Also a Bose-Einstein-condensed gas
obeys the formula.  A degenerate Fermi gas obeys it as well.
 
But if we have the *simultaneous* situation of the Hamiltonian being
of a Newtonian quadratic form in the momentum *and* the situation of
there being no inter-particle positional correlations *and* there
being no particle number changing processes, then we have *both*
that the Ideal gas equation of state is obeyed *and* the pressure is
(2/3) of the energy density in equilibrium.
 
David Bowman