On your devices, measure the following two derivatives and describe how they are related: [SWBQ] $\newcommand\myderiv[3]{\left(\frac{\partial #1}{\partial #2}\right)_{#3}}$ $$\myderiv{F_1}{x_2}{x_1} \qquad\qquad \myderiv{F_2}{x_1}{x_2}$$ Hiding one string and fixing one weight
We have spent considerable time talking about the potential energy $U$, which changes as $$dU = F_1dx_1 + F_2 dx_2$$ which we obtained by using energy conservation and tracking all the work done on our system. How would this have changed if we had a larger black box, such that one of our pulleys and weights was entirely hidden within the box? In this case, we would not have been able to measure $x_2$ or $F_2$, and the potential energy of the weight providing the force $F_2$would have been part of our system. This energy would be $$V \equiv U - x_1F_1$$ The second term added here is just the ”$mgh$” for the energy of those weights that we are now considering part of the system. This $V$ is a nice variable, as long as we keep $F_1$ fixed, i.e. if we don't reach inside and add some extra weight to the pulley.
As usual, when we have defined a variable, we want to know how it changes: write down the total differential of $V$. [SWBQ] $$dV=dU-F_1dx_1-x_1dF_1$$ $$=F_1dx_1+F_2dx_2-F_1dx_1-x_1dF_1$$ $$=F_2dx_2-x_1dF_1$$ This is interesting. It gives us a new definition for $F_2$: $$F_2 = \myderiv{V}{x_2}{F_1}$$ which makes sense. As long as we hold $F_1$ fixed, we can understand $V$ as the total energy of the system, in which case the work is simply related to the change in $V$. This also tells us that if we did a bunch of measurements of $F_2$ and $x_2$ with the same $F_1$ (which is an easy experiment), it would be easy to find $\Delta V$, but not so easy to find $\Delta U$.
Above we found a new definition for $F_2$ as a partial derivative of $V$. Interpret the total differential $dV$ to find a new definition for $x_1$. What does this mean? $$x_1 = -\myderiv{V}{F_1}{x_2}$$ It means xx is a partial derivative of $V$. It's not one that is easy to measure, though, so it's not particularly useful by itself. Yesterday we saw one Maxwell relation, which came out of mixed partial derivatives of $U$. We can now find a second Maxwell relation from $V$, by considering the two partial derivatives: $$\myderiv{\myderiv{V}{x_2}{F_1}}{F_1}{x_2} = \myderiv{\myderiv{V}{F_1}{x_2}}{x_2}{F_1}$$ Work out this Maxwell relation, and test it experimentally. [SWBQ] $$\myderiv{F_2}{F_1}{x_2} = -\myderiv{x_1}{x_2}{F_1}$$ The energy $V$ is a Legendre transform of the potential energy $U$. It changes the way we look at the energy, including the potential energy of one weight inside the system—provided we don't change that weight. There are two other Legendre transformations we can do to this system. The obvious next transformation to try would be $$W \equiv U - x_2F_2$$ which is very much like $V$, but treats the $x_2$ force specially rather than the $x_1$ force.
Find the total differential of $W$, and then find and test a Maxwell relation from $W$. [SWBQ]
Finally we have one last Legendre transformation for our system: $$Z \equiv U -x_1F_1 - x_2F_2$$ This looks a lot like we're putting all of our strings, pulleys and weights inside of our system. That seems pretty silly. There are a couple of uses for this transformation. One is to extract yet another Maxwell relation. The other reason to do this is if there happen to be even more strings coming out of our box. In that case, we would be able to keep $F_1$ and $F_2$ fixed, and yet still change our system. When we get to the thermodynamic equivalent of this energy (the Gibbs free energy), we will find that it is primarily useful when you have one or more additional degrees of freedom: e.g. in Chemistry, where you change the number of molecules.
Find the total differential of $Z$, and a Maxwell relation from it. Then test this relationship experimentally. [SWBQ]