Re: [Phys-l] Photon Thermodynamics

[carmelo@pacific.net.sg]

Quoting John Denker <jsd@av8n.com>:
> >  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.
> > 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)

Actually, Feynman was using
PV = (gamma – 1)U                        (39.11)
and
dU = (PdV + V dP) / (gamma – 1)          (39.12)
and deduce that
PV^gamma = C                             (39.14)

I don’t see any problem with Feynman’s calculations.

Feynman’s conclusion was on the compressibility of radiation. He  
related this to the contribution of radiation pressure in the stars.  
However, the compressibility of radiation may have significant impact  
on the attenuation of photons’ energy from distant stars or supernova!  
That is, the light may get “tired” because the vacuum in space is not  
perfectly empty; it is filled with background cosmic radiations.

One possible experiment is to illuminate a tunnel say, in a linear  
collider, by using lights with different wavelengths and intensities.  
By aiming a laser light through the tunnel with “large number of  
photons”, we may verify the attenuation and fine-tune Hubble’s Law. We  
may discover that the universe is not expanding so fast too. Whether  
Einstein’s profound mistake in the cosmological term, is really  
profound, it is still too early to tell.


Alphonsus