# Heat engines and “Angular momentum” engines?

We know that the theory of heat engines is that, if you accept the second law of thermodynamics, $\Delta S > 0$ then you can define temperature using $\frac{1}{T} = \frac{\partial S}{\partial E}$ And you would arrive at the conclusion that heat can only go from higher temperature to lower temperature reservoirs. And the best efficiency we can get is to use a reversible engine.

Now my question is "What's so special about energy?" If we replace energy with any conservative, for example, one component of angular momentum $L$. Then is it possible to define a "temperature" as $\frac{1}{T_L} = \frac{\partial S}{\partial L}$ And we would conclude that "in order to get useful angular momentum out of a reservoir, there is a maximum efficiency achieved by a reversible angular momentum engine" ?

• What's the point? Energy storage in rotating devices has already been looked at. Check out flywheels. – Torsten Hĕrculĕ Cärlemän Jan 5 '14 at 13:40

Yes. It turns out that your $T_L$ is equal to $-T/\omega$, where $\omega$ is the angular velocity and $T$ is the usual temperature. We normally work with the reciprocals of such quantities, and in the language of non-equilibrium thermodynamics we say that a gradient in $-\omega/T$ is the "thermodynamic force conjugate to" a flow of angular momentum.
We usually write the fundamental equation of thermodynamics with the energy on the left-hand-side, like this: $$dU = TdS - pdV + \sum_i \mu_i dN_i.$$ This equation can be extended with many other terms, including $\phi dQ$ (electric potential times change in charge) and $\omega dL$ (angular velocity times change in angular momentum). However, the "special" quantity here is the entropy, $S$, which is non-decreasing while all the other extensive quantities are conserved. We can rearrange this to put the special quantity on the left, and to get $$dS = \frac{1}{T} dU + \frac{p}{T}dV - \sum_i \frac{\mu_i}{T}dN_i + \dots - \frac{\phi}{T}dQ - \frac{\omega}{T} dL.$$ This observation is the basis of non-equilibrium thermodynamics. It follows immediately from this that $$\frac{\partial S}{\partial L} = -\frac{\omega}{T}.$$ It also follows that angular momentum cannot be spontaneously transferred from one body to another while keeping all other quantities constant unless the second body has greater $-{\omega}/{T}$.