- #1

yucheng

- 232

- 57

- Homework Statement
- n/a

- Relevant Equations
- n/a

https://scholar.harvard.edu/files/schwartz/files/7-ensembles.pdf

https://mcgreevy.physics.ucsd.edu/s12/lecture-notes/chapter06.pdf

On page 3 of both the notes above, the author merely claims that $$P \propto \Omega_{\text{reservoir}}$$

But isn't $$P \propto \Omega_{\text{system}}\Omega_{\text{reservoir}}$$?

We have

$$\mathrm{ln} \ \Omega_{\text{res}} (E_{\text{tot}} - E_{\text{sys}})= \mathrm{ln} \ \Omega_{\text{res}} (E_{\text{tot}} ) - E_{\text{sys}} \frac{\partial \mathrm{ln} \ \Omega_{\text{res}}}{\partial E_{\text{sys}}} + \text{higher order terms}$$

Why are higher order terms negligible? We know that they are small compared to the total energy, and how large they are in numerical terms, depends on the unit of energy. Hence we can scale the unit such that ##E_{\text{sys}}<<1## then the higher order terms obviously disappear. Is reasonable?

$$\mathrm{ln} \ \Omega_{\text{sys}} (E_{\text{sys}})= \mathrm{ln} \ \Omega_{\text{sys}} (0) + E_{\text{sys}} \frac{\partial \mathrm{ln} \ \Omega_{\text{sys}}}{\partial E_{\text{sys}}} + \text{higher order terms} = E_{\text{sys}} \frac{\partial \mathrm{ln} \ \Omega_{\text{sys}}}{\partial E_{\text{sys}}} + \text{higher order terms}$$

so... in equilibrium, both the reservoir and system have the same temperature, but the partial derivatives are evaluated at different energies, but since they are in equilibrium,

Thus $$\frac{\partial \mathrm{ln} \ \Omega_{\text{sys}}}{\partial E_{\text{sys}}} \bigg \rvert_{0} = \beta_{sys}$$

$$\frac{\partial \mathrm{ln} \ \Omega_{\text{res}}}{\partial E_{\text{sys}}} \bigg \rvert_{E_{\text{tot}}} = \beta_{res}$$

But aren't $$\beta = 1/kT$$? So... aren't they equal?

Which just means that

$$P \propto \Omega_{\text{system}}\Omega_{\text{reservoir}} \propto e^{\beta({E_{\text{sys}} - E_{\text{sys}}})} \propto 1$$?

hence ##P \propto \Omega_{\text{system}}\Omega_{\text{reservoir}}## cannot be right!?

Thanks in advance!

https://mcgreevy.physics.ucsd.edu/s12/lecture-notes/chapter06.pdf

On page 3 of both the notes above, the author merely claims that $$P \propto \Omega_{\text{reservoir}}$$

But isn't $$P \propto \Omega_{\text{system}}\Omega_{\text{reservoir}}$$?

We have

$$\mathrm{ln} \ \Omega_{\text{res}} (E_{\text{tot}} - E_{\text{sys}})= \mathrm{ln} \ \Omega_{\text{res}} (E_{\text{tot}} ) - E_{\text{sys}} \frac{\partial \mathrm{ln} \ \Omega_{\text{res}}}{\partial E_{\text{sys}}} + \text{higher order terms}$$

Why are higher order terms negligible? We know that they are small compared to the total energy, and how large they are in numerical terms, depends on the unit of energy. Hence we can scale the unit such that ##E_{\text{sys}}<<1## then the higher order terms obviously disappear. Is reasonable?

$$\mathrm{ln} \ \Omega_{\text{sys}} (E_{\text{sys}})= \mathrm{ln} \ \Omega_{\text{sys}} (0) + E_{\text{sys}} \frac{\partial \mathrm{ln} \ \Omega_{\text{sys}}}{\partial E_{\text{sys}}} + \text{higher order terms} = E_{\text{sys}} \frac{\partial \mathrm{ln} \ \Omega_{\text{sys}}}{\partial E_{\text{sys}}} + \text{higher order terms}$$

so... in equilibrium, both the reservoir and system have the same temperature, but the partial derivatives are evaluated at different energies, but since they are in equilibrium,

Thus $$\frac{\partial \mathrm{ln} \ \Omega_{\text{sys}}}{\partial E_{\text{sys}}} \bigg \rvert_{0} = \beta_{sys}$$

$$\frac{\partial \mathrm{ln} \ \Omega_{\text{res}}}{\partial E_{\text{sys}}} \bigg \rvert_{E_{\text{tot}}} = \beta_{res}$$

But aren't $$\beta = 1/kT$$? So... aren't they equal?

Which just means that

$$P \propto \Omega_{\text{system}}\Omega_{\text{reservoir}} \propto e^{\beta({E_{\text{sys}} - E_{\text{sys}}})} \propto 1$$?

hence ##P \propto \Omega_{\text{system}}\Omega_{\text{reservoir}}## cannot be right!?

Thanks in advance!

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