Phase transition of a rubber band model

[mooshasta]

Homework Statement

In class we studied a phase transition in a model of a rubber band. For small applied tensions, $f < f^*$, the Helmholtz free energy of the band as a function of the length is $A_1(L) = (\kappa_1 /2)L^2$, while at large applied tensions, $f > f^*$, it is $A_2(L) = (\kappa_2 /2)L^2 + a$. The tension at which the two states coexist was given by $f^* = \left (2 a \Delta \kappa \right)^\frac{1}{2}, \Delta \kappa \equiv 1/\kappa_2 - 1/\kappa_1$. This analysis assumes that length and tension are controlled.

Now I have to consider an example where the net length of the system is $L_{total} = L + x$. The net free energy is given by
$$A_i(L) + (k/2)x^2, i = 1 or 2$$

where $i$ refers to the subsystem with variable $L$, called the primary subsystem. The secondary subsystem is that with variable $x$.

I wish to determine the net equilibrium free energy and equilibrium value of $L$ for a fixed value of $L_{total}$.

Homework Equations

The Attempt at a Solution

For the simple rubber band model, we determined the equilibrium free energy by defining a Legendre transform
$$\mathcal{G}_i(L_{total}; f) = A_i(L) + (k/2) x^2 - fL_{total}$$
and finding the minimum value of this by differentiating with respect to the length, setting the derivative to 0 and then evaluating the Legendre transform at this equilibrium length.

I attempted to do the same thing for this problem, and got as far as:
$$\left( \frac{\partial \mathcal{G}_i}{\partial L_{total}} \right)_f = \left[ \kappa_i L_{eq,i} - k \left( L_{total,eq,i} - L_{eq,i} \right) \right] \left( \frac{\partial L}{\partial L_{total}} \right)_f - f = 0$$

I am not sure what to do at this point. How could I go about using that equation to find the minimum of $\mathcal{G}_i$?

Thanks!

 

[mark726]

To find the equilibrium value of L, you can solve for L in the equation you have derived:
\kappa_i L_{eq,i} - k \left( L_{total,eq,i} - L_{eq,i} \right) = f

Then, you can substitute this value of L into the expression for the net free energy:
\mathcal{G}_i(L_{total}; f) = A_i(L) + (k/2) x^2 - fL_{total}

This will give you the net equilibrium free energy as a function of x. To find the minimum value of this function, you can take the derivative with respect to x and set it equal to 0, then solve for x. This will give you the equilibrium value of x, which you can then use to find the equilibrium value of L using the equation you derived earlier.

Alternatively, you can also use the method of Lagrange multipliers to find the minimum of \mathcal{G}_i. This involves introducing a Lagrange multiplier \lambda and solving the following set of equations:
\frac{\partial \mathcal{G}_i}{\partial L_{eq,i}} = \lambda \frac{\partial L}{\partial L_{total}}
\frac{\partial \mathcal{G}_i}{\partial x} = \lambda \frac{\partial L}{\partial L_{total}}
\frac{\partial \mathcal{G}_i}{\partial L_{total}} = 0

Solving these equations will give you the equilibrium values of L and x, as well as the value of the Lagrange multiplier \lambda, which can be used to find the minimum of \mathcal{G}_i.