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Copyright © 2010 jsd

Scenario: we have two regions R_{1} and R_{2} having energy E_{1}
and E_{2}, entropy S_{1} and S_{2}, et cetera. The total energy
is E_{12}, the total entropy is S_{12}, et cetera.

Scenario: The combined system R_{12} is thermally isolated,
i.e. there is never any energy or any entropy crossing the boundary of
R_{12}. Also we assume

| (1) |

Fundamental notion:

stable equilibrium ==> total entropy is maximized

For today, it is not our task to prove this. We take it as given.
Keep in mind that it refers to the total entropy S_{12}, not
to S_{1} or S_{2} separately.

Corollary:

(∂/∂x) S_{12} = 0
(2) |

for any reaction coordinate x. This is the usual way to quantify the notion of extremum.

Corollary:

(∂/∂x)(∂/∂x) S_{12} < 0
(3) |

which is the usual way to quantify the notion of maximum, as opposed to minimum.

Scenario: we shall be particularly interested in the case where x transfers energy to region 1 from region 2. In particular:

| (4) |

Also: we shall shortly be needing the following thermodynamic definitions and identities, so let’s get them written down:

| (5) |

... assuming we can express E as a differentiable function of S and V alone.

| (6) |

... assuming we can express E as a differentiable function of T
and V alone. This is a totally nontrivial assumption. For example,
C_{V} does not exist at a first-order phase transition, e.g. ice
water.

We assume that for any particular value of x, whatever is going on in region 1 is statistically independent of whatever is going on in region 2. Therefore the entropies are additive:

S_{12} = S_{1} + S_{2}
(7) |

Then equation 2 and equation 4 allow us to write:

| (8) |

Using equation 5 we can rewrite this as

| (9) |

... assuming T is not zero or infinity.

We can summarize this by saying that when we have two regions in equilibrium, they will have the same temperature.

We have proved this in a rather specific scenario. We remark that the same result – “equal temperature at equilibrium” – holds more generally.

Now we turn to the second derivative. From equation 3 we have

| (10) |

where we can take the second derivative of S_{2} either way,
(∂/∂E_{1}) or (∂/∂E_{2}), because
we will get two minus signs from equation 4 and they will
cancel. (By way of contrast, in equation 8 no such
cancellation occurred and an explicit minus sign appeared.)

Let’s consider separately the two terms on the LHS of equation 10. Using equation 5 and equation 6 we can rewrite the first term as:

| (11) |

We can plug this (and the corresponding expression for region 2) into equation 10 to find

| (12) |

... assuming the derivatives exist.

You might be tempted to argue (by symmetry or whatever) that both
terms on the LHS are separately positive, but alas that doesn’t work.
Suppose we have a sample of air (region 1) in equilibrium with a
sample of icewater (region 2). In that case,
(∂/∂E_{2})T_{2} is zero. This is a perfectly reasonable
equilibrium situation that happens every day, and equation 12
correctly describes it.

Now let’s suppose, temporarily, that we know that both terms are
nonzero, and both have the same sign. This might be a reasonable
supposition if the overall region R_{12} is homogeneous, and we
formed R_{1} and R_{2} by drawing an imaginary boundary. In such a
case, we can use equation 6 and multiply through by the
positive quantity (C_{V}_{1} · C_{V}_{2}) to conclude:

| (13) |

So we see that if we assume C_{V}_{1} and C_{V}_{2} both exist and both
have the same sign, then the sign must be positive.

However we should never be too eager to restrict attention to homogeneous systems. Thermodynamics is rightly celebrated for its power and generality. A result that is restricted to homogeneous systems is not nearly as useful as a result that applies more generally, so it is important to work out as much as we can without assuming homogeneity.Beware that C

_{V}_{2}does not exist for ice water. On the other hand, if we pretend that C_{V}_{2}is not infinite but merely a “large number” then we can pretend equation 13 is correct; it suggests that the heat capacity of the combined system is a “large number”.

If we rescind the assumption that R_{12} is homogeneous, then we need
to more carefully consider the case where we have a negative term on
the LHS of equation 12. This does not violate the inequality, so
long as the other term is sufficiently positive.

This situation is rather rare, but it can arise if one of the regions
is internally unstable. That is to say, we are given that R_{1} is in
stable equilibrium with R_{2}, i.e. stable with respect to exchanges
of energy between R_{1} and R_{2} ... but if we want to rule out possible
funny business we need to impose a stronger requirement, namely that
R_{1} must be internally stable, i.e. stable with respect to
rearrangement of energy *within* R_{1}. And similarly for R_{2}.

To say the same thing another way, the reaction coordinate x is not the only coordinate we need to consider. If we want to have true stability, the system must be stable against all perturbations, not just the first perturbation we think of.

Equation 12 says that if R_{1} is internally unstable and R_{2}
is also internally unstable, then the combined system R_{12}
violates the inequality, which means it is also unstable ... which is
a reasonable thing to say. It also says that if both regions are
internally stable, then the combined system is stable. (The tricky
case is where one region is internally stable and the other is
internally unstable.)

Ice water is not stable, i.e. it does not exhibit positive stability. In fact it is neutrally stable, i.e. it exhibits zero stability.

It is tempting to interpret equation 13 as saying that the overall heat capacity is positive ... but any such interpretation is on shaky grounds.

In particular, it would make no sense to say

| = C_{V}_{12}
(14) |

because E_{12} is constant by hypothesis. If you want to talk about
C_{V}_{12} you need to talk about some completely different apparatus.

Now you can appreciate why we carefully considered the two regions separately.

When stated this way, the silliness of equation 14 seems clear,
and you might think it would be hard to make such mistakes
... but actually it is easy to make such mistakes if you take a
shortcut by leaving off the subscripts. If you use E to mean E_{1}
sometimes and then use the same E to mean E_{1}2 sometimes, disaster
is likely.

Thermodynamics is intrinsically highly multi-dimensional. There are lots of variables, and you need to be fastidious and methodical about keeping track of them.

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Copyright © 2010 jsd