tarea 6.2c
6.2c)
Calculate the internal energy, and the differential of work (for a change in η as well as in a).
For the first part I just put the equipartition funtion found in the last exercise into the formula for the internal energy.
 "U = -\frac{\partial}{\partial \beta}ln\Omega(T,\eta)")
so we have that
=&space;-\frac{\partial&space;}{\partial&space;\beta&space;}(-\frac{Nd}{2n}ln\beta&space;-\frac{Nd}{2}ln\beta)&space;=&space;\frac{Nd}{2n\beta}+\frac{Nd}{2\beta} "U = -\frac{\partial}{\partial \beta}ln\Omega(T,\eta)= -\frac{\partial }{\partial \beta }(-\frac{Nd}{2n}ln\beta -\frac{Nd}{2}ln\beta) = \frac{Nd}{2n\beta}+\frac{Nd}{2\beta}")
.
To find the differential of work for η we use
so we must find
=\frac{Nd}{2\beta&space;\eta&space;}&space;\frac{\partial&space;}{\partial\eta&space;}\left&space;(&space;-ln(\eta)&space;\right&space;) "\left \langle \frac{\partial H}{\partial \eta } \right \rangle = -\frac{1}{\beta } \frac{\partial }{\partial \eta } ln\Omega (T,\eta )=\frac{Nd}{2\beta \eta } \frac{\partial }{\partial\eta }\left ( -ln(\eta) \right )")
applying this to the equation for the differential of work and taking into consideration that β = 1/KT, we obtain.
Now we do the same for the parameter a, so
Calculate the internal energy, and the differential of work (for a change in η as well as in a).
For the first part I just put the equipartition funtion found in the last exercise into the formula for the internal energy.
so we have that
To find the differential of work for η we use
so we must find
applying this to the equation for the differential of work and taking into consideration that β = 1/KT, we obtain.
Now we do the same for the parameter a, so
In general your results are correct. But still you have a few errors in the derivation:
ResponderBorrar1. In some places (where you define the work differential) you use "\sum_{alpha=1}^n". This makes no sense, here. In our notes, we have "sum_{alpha =1}^K", where X_1, X_2, ..., X_K are the external parameters to be changed. Applying this to the present case, we would have K=2, X_1 = eta, and X_2 = a.
2. You have the wrogn sign in the definition of the work differential in terms of the partition function. You somehow forgot the sign when you were calculating DW for a change in a. But you got the wrong sign in DW for a change in eta.