Entropy changes in statistical mechanics : Introductory video
Before watching the video read the questions below. As you watch the video try to answer them
Questions
- Explain in your own words what the difference is between the extensive variables that I have labelled $\alpha^{(j)}$ and the variables I have labelled $B^{(k)}$ in the video.
- What derivative we trying to work out in this video? Use the space below to explain why and any details you think are relevant.
- Given that $\frac{S}{k_B} = \Psi + \sum_k \lambda_k \langle B^{(k)} \rangle$ write an expression for the derivative that we would like to calculate.
- Explain in your own words why: $\frac{\partial \Phi }{\partial \alpha^{(j)}} \textrm{d}\alpha^{(j)} = \textrm{d} \Phi$
- Give an expression that allows one to calculate $\Psi$.
- Hence, give an expression for the derivative $\frac{\partial \Psi}{\partial \alpha^{(j)}}$
- What is $\frac{e^{-\sum_k \lambda_k B_i^{(k)}}}{e^\Psi}$ equal to?
- Use this result to explain in your own words why: $\frac{\partial \Psi}{\partial \alpha^{(j)}} = - \sum_k \frac{\partial \lambda_k}{\partial \alpha^{(j)}} \langle B^{(k)} \rangle - \sum_k \lambda_k \left\langle \frac{\partial B^{(k)} }{\partial \alpha^{(j)}} \right\rangle$
- Give the final result for the change in infinitesimal change in entropy that accompanies an infinitesimal change in the fixed extensive thermodynamic variable $\alpha^{(j)}$
- Explain in your own words what the difference is between the extensive variables that I have labelled $\alpha^{(j)}$ and the variables I have labelled $B^{(k)}$ in the video.
- What derivative we trying to work out in this video? Use the space below to explain why and any details you think are relevant.
- Given that $\frac{S}{k_B} = \Psi + \sum_k \lambda_k \langle B^{(k)} \rangle$ write an expression for the derivative that we would like to calculate.
- Explain in your own words why: $\frac{\partial \Phi }{\partial \alpha^{(j)}} \textrm{d}\alpha^{(j)} = \textrm{d} \Phi$
- Give an expression that allows one to calculate $\Psi$.
- Hence, give an expression for the derivative $\frac{\partial \Psi}{\partial \alpha^{(j)}}$
- What is $\frac{e^{-\sum_k \lambda_k B_i^{(k)}}}{e^\Psi}$ equal to?
- Use this result to explain in your own words why: $\frac{\partial \Psi}{\partial \alpha^{(j)}} = - \sum_k \frac{\partial \lambda_k}{\partial \alpha^{(j)}} \langle B^{(k)} \rangle - \sum_k \lambda_k \left\langle \frac{\partial B^{(k)} }{\partial \alpha^{(j)}} \right\rangle$
- Give the final result for the change in infinitesimal change in entropy that accompanies an infinitesimal change in the fixed extensive thermodynamic variable $\alpha^{(j)}$