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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 equilibriumThat is a second-rank somewhat-general principle.
has a uniform chemical potential.
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 equilibriumThe requirement for uniform chemical potential does not supersede or
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.
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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