Thermodynamic equlibrium of radiating systems

[jumanicus]

Consider the following situation:

A sphere is surrounded by a shell of larger radius. Both are
concentric. Assume further that both are black bodies and made of the
same material and are of the same mass. Let both the sphere and the
shell be at the same temperature. Now according to Stefan-Boltzman's
law shell radiates more power inside (as well as outside) and the
sphere radiates less power (since its surface area is less than the
shell). So there will be a net absorption of heat by the sphere and net
emission of heat by the shell. Hence the temperature of the sphere will
increase and that of the shell will decrease. Does it mean that the
thermodynamic eqilibrium does not exist between two objects at the same
temperature if they are "connected" by radiation? I am not getting
this.

[Brian Elmegaard]

jumanicus <vishwesh.muzumdar@rediffmail.com> writes:

> temperature if they are "connected" by radiation? I am not getting
> this.
>
You need the view factor.

\dot{Q}_{sphere-shell}=\sigma A_sphere T^4
\dot{Q}_{shell-sphere}=\sigma A_shell F_{shell-sphere} T^4
F_{shell-sphere}=1\cdot A_sphere/A_shell
=>
\dot{Q}_{shell-sphere}=\sigma A_sphere T^4

and thus
\dot{Q}_{sphere-shell}=\dot{Q}_{shell-sphere}

--
Brian (remove the sport for mail)
http://www.et.web.mek.dtu.dk/Staff/be/be.html
http://www.rugbyklubben-speed.dk

[Jonathan Thornburg -- remove -animal to reply]

[[This looks like it might be a homework assignment, so I'm not
going to answer the question, but rather suggest further questions
to ponder...]]

jumanicus <vishwesh.muzumdar@rediffmail.com> wrote:
> A sphere is surrounded by a shell of larger radius. Both are
> concentric. Assume further that both are black bodies and made of the
> same material and are of the same mass. Let both the sphere and the
> shell be at the same temperature.
[[...]]

What's outside the shell? Or more precisely, what is the outer boundary
of the "system" which is to be in thermodynamic equilibrium, and if
this isn't "the entire universe", what is the outer boundary condition
there?


> Now according to Stefan-Boltzman's
> law shell radiates more power inside (as well as outside) and the
> sphere radiates less power (since its surface area is less than the
> shell).

What happens to the radiation emitted by the shell's inside: Does
*all* of it get absorbed by the sphere? If not, what happens to
the rest?

ciao,

-- "Jonathan Thornburg -- remove -animal to reply" <jthorn@aei.mpg-zebra.de>
Max-Planck-Institut fuer Gravitationsphysik (Albert-Einstein-Institut),
Golm, Germany, "Old Europe" http://www.aei.mpg.de/~jthorn/home.html
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam

[Squark]

jumanicus wrote:
> Consider the following situation:
>
> A sphere is surrounded by a shell of larger radius. Both are
> concentric. Assume further that both are black bodies and made of the
> same material and are of the same mass. Let both the sphere and the
> shell be at the same temperature. Now according to Stefan-Boltzman's
> law shell radiates more power inside (as well as outside) and the
> sphere radiates less power (since its surface area is less than the
> shell). So there will be a net absorption of heat by the sphere and net
> emission of heat by the shell. Hence the temperature of the sphere will
> increase and that of the shell will decrease.

No. Some of the radiation emmitted by the shell will be absorbed by the
shell (this is obvious even in the geometrical optics approximation: in
this
approximation one can imagine the equilibrium state as an ideal gas of
photons filling the space between the bodies, the emission / absorbtion
of photons at the boundaries having an effect equivalent to
reflection). In
equilibrium both bodies will be at the same temperature, and the space
between them will be filled with radiation at the same temperature.
Hence,
each will absorb precisely as much as radiate (since both proccesses
are
proprotional to area).

Best regards,
Squark

[Igor Khavkine]

jumanicus wrote:
> Consider the following situation:
>
> A sphere is surrounded by a shell of larger radius. Both are
> concentric. Assume further that both are black bodies and made of the
> same material and are of the same mass. Let both the sphere and the
> shell be at the same temperature. Now according to Stefan-Boltzman's
> law shell radiates more power inside (as well as outside) and the
> sphere radiates less power (since its surface area is less than the
> shell). So there will be a net absorption of heat by the sphere and net
> emission of heat by the shell. Hence the temperature of the sphere will
> increase and that of the shell will decrease. Does it mean that the
> thermodynamic eqilibrium does not exist between two objects at the same
> temperature if they are "connected" by radiation? I am not getting
> this.

No, thermal equilibrium will still be established in this situation.
The most general way to see this is to realize that there are three
bodies in thermal contact: the outer shell, the inner sphere, and the
cavity in between filled with EM radiation. Let A be the shell's inner
area and B be the sphere's outer area. The rate of emission from the
shell will be sigma*A*(T_sh^4-T_ca^4), whith T_sh and T_ca being the
shell and shell and cavity temperatures, respectively. While the rate
of emission from the sphere will be sigma*B*(T_sp^4-T_ca^4), with T_sp
being the sphere temperature. In both cases, sigma is the
Stefan-Boltzmann constant. So the net energy flux through the cavity
becomes zero once all three bodies are at the same temperature, T_sh =
T_ca = T_sp.

The above argument is quite general and works for any shape cavity
containing any shape object. However, the case you present is amenable
to a more direct geometric analysis. Here we must realize that a given
patch of the shell's inner surface will emit radiation in all
directions in proportion cos(th) dOmega per solid angle with respect
to the surface normal[1]. The integral of this quantity from th=0 to
th=pi/2,

/ /2pi /pi/2
| cos(th) dOmega = | dphi | cos(th) sin(th) dth = pi,
/ /0 /0

figures in the derivation of the Stefan-Boltzmann law[2]. However,
part of the radiation emitted by the shell will not even hit the
sphere, but be absorbed by the walls of the shell itself. In order to
caclulate the radiation actually absorbed by the sphere we need to find
the angle 2*chi subtended by the sphere at any point on the shell's
inner surface. The the radiation actually absorbed by the sphere will
be proportional to

/2pi /chi
| dphi | cos(th) sin(th) dth = pi*(1-cos(chi)^2).
/0 /0

It is an elementary exercise in geometry to see that
cos(chi)^2 = 1-(r/R)^2, wher r and R and the sphere and inner shell
radii, respectively. In other words, the fraction of the radiation
actually absorbed by the sphere is equal to (r/R)^2. But that's also
the ratio between the total radiation emitted by the sphere and the
shell. So, in thermal equilibrium geometry dictates that the shere
absorbs only as much radiation as it emits, which is as it should be in
equilibrium.

Like I mentioned above, in more complicated situations the geometric
analysis is harder to perform, but first argument above still applies.

Hope this helps.

Igor

[1] http://en.wikipedia.org/wiki/Lambert's_cosine_law
[2] http://en.wikipedia.org/wiki/Stefan-Boltzmann_law

[Blackbird]

jumanicus wrote:
> Consider the following situation:
>
> A sphere is surrounded by a shell of larger radius. Both are
> concentric. Assume further that both are black bodies and made of the
> same material and are of the same mass. Let both the sphere and the
> shell be at the same temperature. Now according to Stefan-Boltzman's
> law shell radiates more power inside (as well as outside) and the
> sphere radiates less power (since its surface area is less than the
> shell). So there will be a net absorption of heat by the sphere and
> net emission of heat by the shell.

This (wrong) assumption presupposes that the sphere absorbs all inside
radiation that is emitted by the shell, i.e., that direction of radiation is
perpendicular to the shells inner surface. But the radiation is in all
directions, and some radiation will miss the sphere, and be reabsorbed by
the shell, thus giving a net effect of zero. By symmetry arguments you can
establish that the amount of radiation actually hitting the sphere is
excatly the same as the amount emitted by the sphere.