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*From*: brian whatcott <betwys1@sbcglobal.net>*Date*: Wed, 18 Jan 2012 17:24:57 -0600

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 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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**References**:**Re: [Phys-l] Temperture profile in a graviational field***From:*"Christopher M. Gould" <gould@usc.edu>

**Re: [Phys-l] Temperture profile in a graviational field***From:*John Denker <jsd@av8n.com>

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