# Microcanonical Ensemble

• No Environment
• Idealized Walls (Only elastic collisions)
• Fixed Volume, Number of Particles, Energy

Q: Dice? Fixed number of dice, the sum of the dice on each row is fixed

When the energy is fixed, this shows that this formulation is useful for when we want to ignore the energy. Consider $$F=U-TS$$ and $$G=H-TS$$, then we can ignore $$U$$ and $$H$$ for $$|U|,|H|\ll TS$$.

Solids are when $$|U|\gg TS$$ and gasses for $$|U\ll TS$$. For solids, we can set $$T$$ to zero, for gasses we can set the energy to zero. Then the $$T$$ or $$U$$ are just the corrections to the model.

When $$|U|\approx TS$$ we get liquids.

$$N$$ atoms, classical. $$\mathbb{Q}=\{x_1,y_1,z_1,\cdots\} = \{q_1,\cdots,q_{3N}\}$$. $$\mathbb{P} = \{p_1,\cdots,p_{3N}\}$$.

$$H(\mathbb{P},\mathbb{Q}) = T+U = \frac{\mathbb{P}^2}{2m}+U(\mathbb{Q})$$.

Shell: $$\{E,E+\delta E\}$$ and limit $$\delta E\to 0$$.

[1] $$\Omega(E)\delta E = \int_{E We then get \(\Omega(E) = \int d\mathbb{P}d\mathbb{Q}\delta(E-H)$$ we get probability density $$\rho = \frac{\delta(E-H)}{\Omega(E)}$$.

\begin{align*} [1] \to \Omega(E) &= \lim_{\delta E\to 0}\frac{1}{\delta E}\int_{E $$\langle O\rangle_E = \frac{1}{\Omega(E)\delta E}\int_{E C.f. \(\langle O\rangle = \sum_\alpha p_\alpha O_\alpha$$.

## Ideal Gas

$$U(Q) = \mathbb{C}$$. We can set to zero. Then, $$H=\frac{\mathbb{P}^2}{2m}$$.

1. Configuration space: We want $$\int \rho dQ = 1$$. So, $$\int dQ = V^N = L^{3N}$$ (the last one is for a cube). So, $$\rho(\mathbb{Q}) = \frac{1}{V^N}$$. $$M=\frac{V}{V_0}$$, $$N\ll M$$. The difference with $$M$$ compared to an ideal gas is that an ideal gas is dilute but this gives particles a finite size of $$V_0$$. Note, $$\rho(\mathbb{P},\mathbb{Q}) = \rho(\mathbb{P})\rho(\mathbb{Q})$$ is separable since the momentum and position are independent.
2. Momentum space: $$E=\sum_{\alpha=1}^{3N}\frac{P_\alpha^2}{2m}$$. This gives a hypersphere of radius $$R=\sqrt{\mathbb{P}^2} = \sqrt{2mE}$$. The volume of the sphere is $$\mu(S_R^{\ell-1}) = \frac{\pi^{\frac{\ell}{2}R^\ell}}{\left(\frac{\ell}{2}\right)!}$$. For $$\ell=2$$, $$V_{R_2} = \pi R^2$$, $$\ell=3,\left(\frac{3}{2}\right)! = \frac{3\sqrt{\pi}}{4}\to V_{R_3}=\frac{4}{3}\pi R^3$$. $$\Omega(E) = \lim_{\delta E\to 0}\frac{\text{Shell Volume}}{\delta E} = \frac{1}{\delta E}\left[\mu(S_{\sqrt{2m(E+\delta E)}}^{3n-1}) - \mu(S_{\sqrt{2mE}}^{3N-1})\right] = \frac{d}{dE}\mu(S_{\sqrt{2mE}^{3N-1}}) = \frac{\pi^{\frac{3N}{2}}2m(2mE)^{\frac{3N}{2}-1}}{\left(\frac{3N}{2}\right)!} = \frac{3Nm\pi^{\frac{3N}{2}}R^{3N-2}}{\left(\frac{3N}{2}\right)!}$$

Example probability one of the particles has a particular momentum, on our hypersphere we get a thin strip around the sphere, $$\rho(p_1) = \rho(p_{1,x}) = ?$$ $$R'$$ is the radius of the strip, then $$p_1$$ is the height of the triangle with hypotenuse $$R$$ and leg $$R'$$. So, $$R'=\sqrt{2mE-p_1^2}$$. The annular area per thickness is given as $$\frac{\text{annular area}}{\delta E} = \frac{d}{dE}\mu(S_{\sqrt{2mE-p_1^2}}^{3N-2}) = \frac{(3N-1)m\pi^{\frac{3N-1}{2}}R'^{3N-3}}{\left(\frac{3N-1}{2}\right)!}$$. So, $$\rho(p_1) = \frac{\text{annular area}}{shell volume} = \frac{\cdots R'^{3N-3}}{\cdots R^{3N-2}}\propto \left(\frac{R'}{R}\right)^{3N}\frac{R^2}{R'^3}$$. Note, $$\left(\frac{R'}{R}\right)^2 = 1-\frac{p_1^2}{2mE}$$. Then, $$\left(\frac{R'}{R}\right)^{3N} = \left(1-\frac{p_1^2}{2mE}\right)^{\frac{3N}{2}}$$. Note, the rings are near the equator since most of the energy is in all of the particles, not isolated to just one or two. Expanding, $$\left(\frac{R'}{R}\right)^{3N} = \exp\left(-\frac{p_1^2}{2mE}\right)^{\frac{3N}{2}} = \exp\left(-\frac{3N p_1^2}{2m2E}\right) = \exp\left(-\frac{p_1^2}{2m kT}\right)$$. Note, $$\frac{1}{kT} = \frac{3N}{2E}$$.

Thus, $$\rho(p_1) = \frac{1}{2\pi m\left(\frac{2E}{3N}\right)}\exp\left(-\frac{p_1^2 3N}{2m 2E}\right) = \frac{1}{2\pi m\left(\frac{2E}{3N}\right)}\exp\left(-\frac{E_{1x}}{kT}\right)$$

## Results for Ideal Gas

• $$\rho(\mathbb{Q})^{-1} = V^N$$
• $$\rho(\mathbb{P})^{-1} = \frac{\text{Shell Volume}}{\delta E} = \frac{3Nm\pi^{3N/2}\sqrt{2mE}^{3N-2}}{\left(\frac{3N}{2}\right)!} = \frac{3Nm\pi^{3N/2}(2mE^{3N/2}(2mE)^{-1})}{\left(\frac{3N}{2}\right)!}$$.
• $$\Omega_{crude}(E) = \frac{V^N\pi^{3N/2}(2mE)^{3N/2}}{\left(\frac{3N}{2}\right)!}\left(\frac{3N}{2E}\right)$$
• $$2E=3Nk_B T$$
• $$\rho(p_1) = \frac{1}{\sqrt{2\pi(2m)\frac{kT}{2}}}\exp(-E_1/2(kT/2))$$, $$E_1=k_1=\frac{p^2}{2m}$$.

## Boltzmann Distribution

$$\langle E_1\rangle = \frac{kT}{2}$$ from the standard devition of the $$E_1$$ exponential distribution. Note that $$\langle p_1\rangle = \int \rho(p_1)\rho_1 dp_1 = 0$$.

## Microcanonical Ensemble

Closed, $$E,V,N$$. Divide the region into two parts, in the first part the state is $$S_{1j}$$ with energy $$E_1$$ and phase space volume $$\Omega_1(E_1)$$, The other part the state is $$S_{2i}$$ with energy $$E-E_1$$ and phase space volume $$\Omega(E-E_1)$$.

So, $$\rho(S_1)\propto \Omega_2(E-E_2)$$

$$\Omega(E) = \int dE_1\Omega_1(E_1)\Omega_2(E-E_1)$$.

$$\rho(E_1) = \frac{\Omega_1(E_1)\Omega_2(E-E_1)}{\Omega(E)}$$.

Assume that $$\rho(E_1)$$ is sharply peaked at $$E_1=E_1^*$$.

Extremum, $$\partial_{E_1}(\Omega_1(E_1)\Omega_2(E-E_1)) = \Omega_2(E_2)\left(\frac{\partial \Omega_1}{\partial E}\right)_{E_1^*} + \Omega_1(E_1)\left(\frac{\partial \Omega_2}{\partial E_2}\right)_{}\left(\frac{\partial E_2}{\partial E_1}\right)_{E_1^*} \Leftrightarrow \frac{1}{\Omega_1}\left(\frac{d\Omega_1}{d E_1}\right)_{E_1^*} = \frac{1}{\Omega_2}\left(\frac{d\Omega_2}{dE_2}\right)_{E-E_1^*}$$. $$\left(\frac{d\Omega_2}{dE_2}\right)_{E-E_1^*}=-1$$ $$\left(\frac{d\ln\Omega_1}{dE_1}\right)_{E_1^*} = \left(\frac{d\ln\Omega_2}{dE_2}\right)_{E_2^*}$$.

Define $$S_{equilibrium} = k_B\ln\Omega(E)$$

$$S = -k_B\sum_\alpha p_\alpha\ln p_\alpha$$.

### Is there a sharp peak and extensivity

$$\Omega_1(E_1)\Omega_2(E - E_1) = \exp\left(\frac{1}{k_B}(S_1(E_1) + S_2(E_2))\right)$$.

Taylor expanding: $$\exp\left(\frac{1}{k_B}(S_1(E_1^*)+\frac{1}{2}(E_1-E_1^*)^2\frac{\partial^2 S_1}{\partial E_1^2} + S_2(E-E_1^*) + \frac{1}{2}(E_1-E_1^*)\frac{\partial S_2}{\partial E_2})\right) = \Omega_1(E_1^*)\Omega_2(E-E_1^*)\exp\left(\frac{(E-E_1)^2}{2k_B}(\frac{\partial^2 S_1}{\partial E_1^2}+\frac{\partial^2 S_2}{\partial E_2^2})\right)$$. Note the linear terms vanish at the extrema.

$$\rho(E_1) = \frac{1}{\sqrt{2\pi\sigma_E^2}}\exp\left((E-E_1)^2/(2\sigma_E^2)\right),[\sigma_E]^{-1} = -\frac{1}{k_B}\left(\frac{\partial^2 S_1}{\partial E_1^2} + \frac{\partial^2 S_2}{\partial E_2^2}\right)$$. Note the minus only comes from the fact that it we know physically it must be positive.

Sign of $$\frac{\partial^2 S}{\partial E^2}$$?, $$\frac{\partial S}{\partial E} = \frac{1}{T}$$. $$\frac{\partial (1/T)}{\partial E}<0$$. This is the source of the minus sign.

$$\frac{\partial^2 S}{\partial E^2} \propto \frac{N}{N\cdot N}$$. Then, $$\sigma_E^2 \propto \frac{1}{N}$$ Hence $$\sigma_E^2/N\propto \frac{1}{\sqrt{N}}$$.

How do we know $$S$$ is extensive?

$$S_{tot}(E) = k_B\ln \Omega(E) = k_B\ln\Omega_1\Omega_2 \int\exp(-\frac{(E_1-E_1^*)^2}{2\sigma_E^2})dE_1 = S_1(E_1) + S_2(E_2) + k_B\ln(\sqrt{2\pi})\sigma_E$$. Note this last term is proportional to $$\sqrt{N}$$ but for most systems (large or weak long range forces), the linear term in N dominates.

Created: 2024-05-30 Thu 21:19

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