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# Re: [Phys-l] Temperture profile in a graviational field

Hmmmm... if one defines thermodynamic equilibrium to imply that two bodies
in thermal contact adopt the same temperature - and one does indeed do that, then we can
kiss off that concept in considering a towering cylinder of air whose temperature
falls with height..

Brian W

On 1/18/2012 4:50 PM, John Denker wrote:
One more thing about chemical potential:

With great generality, we can say that at thermodynamic
equilibrium, the gradient of the entropy vanishes:

dS = 0 [1]

where "d" is the exterior derivative operator. We write
the gradient as dS rather than ∇S for technical reasons.
Keep in mind that dS is a vector in thermodynamic state-
space.

As usual, subject to mild conditions, we can expand dS
using the chain rule:

∂S ∂S
dS = -------- dN + -------- dE [2]
∂N | E ∂E | N

Conditions include: We assume constant volume throughout this
document. We also assume all the potentials are sufficiently
differentiable.

We recognize the partial derivative in front of dE as being
the inverse temperature:

∂S
β := -------- [3]
∂E | N

We can rewrite the other partial derivative in terms of the
celebrated cyclic triple partial derivative rule:

∂S ∂N ∂E
-------- -------- -------- = -1 [4]
∂N | E ∂E | S ∂S | N

Hence

∂S ∂E ∂S
-------- = -1 -------- -------- [5]
∂N | E ∂N | S ∂E | N

Note that if you weren't fastidious about keeping track of the
"constant E" "constant S" and "constant N" specifiers, it would
be very easy to get equation [4] and equation [5] wrong by a
factor of -1.

We recognize one of the factors on the RHS as the chemical potential:

∂E
μ := -------- [6]
∂N | S

Putting together all the ingredients we can write:

dS = 0
= - μ β dN + β dE [7]

Since we can choose dN and dE independently, both terms on the
RHS must vanish separately.

Also keep in mind that both N and E are conserved.

If we divide the system into two parcels, #1, and #2, then

dS1 = - dS2
dN1 = - dN2 [8]
dE1 = - dE2

and from that we conclude that, at thermal equilibrium:

β1 = β2 [9]
μ1 = μ2

assuming nonzero β i.e. non-infinite temperature.

In words:
-- In a system where the two parcels have reached equilibrium
by exchanging energy, they will have the same temperature.
-- In a system where the two parcels have reached equilibrium
by exchanging particles as well as energy, they will have the
same chemical potential.

On 01/17/2012 03:46 PM, Christopher M. Gould wrote:
The general principle is that any system in mass transport equilibrium
has a uniform chemical potential.
That is a second-rank somewhat-general principle.

The first-rank much-more-general principle is extremization of the
entropy. At equilibrium, the gradient of the entropy vanishes. See
equation [1].

The general principle is that any system in mass transport equilibrium
has a uniform chemical potential. If there are no external forces then
that means the temperature will be uniform. Here, with an imposed
gravitational potential, air at the top of a column will be colder than
at the bottom.
The requirement for uniform chemical potential does not supersede or
conflict with the requirement for uniform temperature. In the situation
we have been considering, *both* requirements apply.

You can visualize this in terms of the gradient vector dS. The fact
that dS=0 implies the projection of dS in *every* feasible direction
must vanish, including the dE direction and the dN direction among others.
Otherwise the system would be at non-equilibrium with respect to excursions
in the direction(s) of non-vanishing dS.
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