If we define $\Omega$ as the following, where $E_r$ is the energy in state $r$, $$\Omega(E)=\int...\int d^{3N}p\ d^{3N}q\ \delta(E-E_r) \tag{1}$$ Then the laplace transform of $\Omega$ is the canonical partition function $Z(\beta)$, $$L\{\Omega(E)\}=\int_{0}^{\infty}\Omega(E)e^{-\beta E}dE=Z(\beta) \tag{2}$$ Motivated by this idea, I was wondering what the following would work out to, $$\beta \equiv \frac{1}{\Omega(E)}\frac{\partial \Omega}{\partial E} \tag{3}$$ $$\beta \Omega(E) = \frac{\partial \Omega}{\partial E} \tag{4}$$ $$L\{\beta \Omega(E)\}=L\left \{\frac{\partial \Omega}{\partial E}\right\} \tag {5}$$ $$\beta Z(\beta) = \beta Z(\beta) - \Omega(0) \tag{6}$$ Which implies that, $$\Omega(0) = 0 \tag{7}$$ This seems incorrect as the number of microstates associated with zero energy should be 1. Shouldn't it? Particularly for a microcanonical ensemble, $$P_r=\frac{1}{\Omega} \tag{8}$$ And taking (7) and (8) into account it seems it would be never possible to have a system with zero energy. It seems like I'm doing some rather loose mixing and matching of ideas from the canonical and microcanonical ensemble. I think that's where I'm getting myself in trouble.
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$\begingroup$ Where does $(3)$ come from ? In the Legendre transform $\beta$ is an independent variable. $\endgroup$– SolubleFishCommented Jun 5, 2021 at 18:26
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$\begingroup$ It's the definition of $\beta$ in the microcanonical ensemble, and by extension, in general. Alternatively written as, $\beta = \partial \ln \Omega / \partial E$. And yes, that's true, but by that argument so is the average energy, yet we can write an equation for it that makes it dependent on other quantities, such as $P_r$. $\endgroup$– michael bCommented Jun 5, 2021 at 19:28
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1$\begingroup$ Ok, but when you are computing the Laplace transform, you are switching to the canonical ensemble, where $\beta$ is a external parameter. You cannot use $(3)$ in the integral defining $Z(\beta)$ $\endgroup$– SolubleFishCommented Jun 5, 2021 at 19:59
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$\begingroup$ Is $\beta$ not an external parameter in the microcanonical ensemble? And yes, I do see that applying the Laplace transform inherently changes it to the canonical ensemble. But I would've thought that the definition in (3) holds in either case. Maybe this calculation is evidence that it does not. $\endgroup$– michael bCommented Jun 5, 2021 at 20:13
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$\begingroup$ Nevermind, I understand this now. In canonical ensembles ($NVT$), the temperature, and therefore $\beta$, is an external (independent) parameter, hence (2) is valid in that context. In the $NVE$ ensemble it isn't. That's a good enough answer for me. $\endgroup$– michael bCommented Jun 5, 2021 at 20:15
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1 Answer
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You are mixing equations from the microcanonical and canonical ensemble.
For clarity, let us write : $$\beta_*(E) = \frac{\partial \ln \Omega}{\partial E}(E)$$ for the temperature in the microcanonical ensemble and $E_*(\beta)$ for the average energy in the canonical ensemble.
Then equation $(4)$ is rewritten : $$\Omega(E) \beta(E) = \frac{\partial \ln \Omega}{\partial E}(E)$$ and because $\beta$ depends on $E$ you have : $$\beta Z(\beta)\neq \int_0^{+\infty}\beta_*(E)\Omega(E)e^{-\beta E} \text d E$$
