2
$\begingroup$

a concave-up entropy-energy graph

This is from Schroeder's An Introduction to Thermal Physics:

Let A and B be a system both with a concave-up entropy-energy graph. The systems will achieve thermal equilibrium when they are at the same temperature. Usually, however, the equilibrium will not be stable. Any small flow of energy from B to A will cause the temperature of B to increase while the temperature of A decreases. We then get a run-away effect, as more and more energy spontaneously flows from B to A. And if the initial fluctuation results in energy flowing from A to B, the run-away effect goes in the opposite direction.

If B is a large "reservoir" whose temperature doesn't change significantly when it absorbs or emits energy, then again any small transfer of energy from B to A will result in A becoming colder than B so we get a run-away effect. The only way for the equilibrium to be stable is if system B is "normal" and sufficiently small (more precisely, has a sufficiently small heat capacity) that a spontaneous transfer of energy from B to A causes B to cool off more than A does. Then A will become a bit hotter than B and the energy will spontaneously flow back.

I don't understand...Why would there be a run-away effect? When there's a flow of energy from B to A, the temperature of B increases, which by then means the internal energy of B increases as well. But how can B has its internal energy increases when it is losing some of its energy??

$\endgroup$
3
$\begingroup$

The curve shown in the picture has the property

$$\frac{\partial^2 S}{\partial U^2} > 0 $$

so that

$$\frac{\partial^2 S}{\partial U^2} = - \frac 1 {T^2} \frac{\partial T}{\partial U} > 0$$

Since $T^2>0$, this means that

$$\frac{\partial T}{\partial U} < 0$$

This means that, in this particular system, if we decrease the energy the temperature increases and if we increase the energy the temperature decreases.

Let's say that B and A are initially at the same temperature, and that a fluctuation makes a quantity $\delta Q$ of heat flow from B to A.

Since B has lost some energy, its temperature will rise. Since A has gained some energy, its temperature will decrease.

Now, since the rate of heat flow between two bodies is proportional to the temperature difference between them, this will mean that B will transfer some more energy to A, resulting in an even greater temperature difference.

You can see how this process creates a positive feedback that results in B becoming hotter and hotter (while losing energy) and A become colder and colder (while gaining energy), while the heat flux grows without bounds (1).

Since we don't observe such an unusual behavior, we must conclude that most of the thermodynamic systems we observe have the opposite property, i.e.

$$\frac{\partial^2 S}{\partial U^2} < 0 $$


(1) Eventually (and ideally), A will reach absolute zero, corresponding to its state of maximum energy, and the process will have to stop, because A cannot take any energy any more. But I am not sure about this last point, since funny things happen when we are close to absolute zero...

$\endgroup$
2
$\begingroup$

At constant volume, you have $dS=\frac{dU}{T}$, so $\frac{dS}{dU}=1/T$. Taking derivatives again we get

$\frac{d^2S}{dU^2}=-\frac{1}{T^2}\frac{dT}{dU}$

Thus, if the concavity of S vs U were positive then the temperature must diminish with an increase in internal energy. That is why, in general, the concavity is negative.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.