{From Phy 415: Thermal Physics Provided by Professor Alex Levchenko}
If the potential energy is a homogeneous function of coordinates and the motion takes place in a finite region of sapce, then there is a very simple relation between \bar{T} and \bar{U}
By (6.2) Euler's theorem on homoegeneous function (for derivation, refers to Marion & Thorton) : \Sigma \mathbf{v}_{a} \cdot \partial T / \partial \mathbf{v}_{a}=2 T
By definition in p 7, \mathbf{P} \equiv \sum_{a} \partial L / \partial \mathbf{v}_{a} = \sum_{a} \partial T / \partial \mathbf{v}_{a}
we get: 2 T=\sum_{a} \mathbf{p}_{a} \cdot \mathbf{v}_{a}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\sum_{a} \mathbf{p}_{a} \cdot \mathbf{r}_{u}\right)-\sum_{a} \mathbf{r}_{a} \cdot \dot{\mathbf{p}}_{a}
If we assume the system executes a motion in a finite region of space with finite velocities. Then \Sigma \mathbf{P_\alpha} \cdot \mathbf{r_\alpha} is bounded
then \bar{f} = \lim_{\tau \to \infty} \int_0^\infty d/dt(\Sigma \mathbf{P} \cdot \mathbf{r}) dt = \lim_{\tau \to \infty} \frac{\Sigma \mathbf{P} \cdot \mathbf{r}|^\tau_0}{\tau} = 0
Thus 2T = \Sigma P\cdot r = - \Sigma r\cdot \dot{P} = \Sigma r\cdot \frac{\partial U}{\partial r}
Thus 2 \bar{T}=\overline{\sum_{a} \mathbf{r}_{a} \cdot \partial U / \partial \mathbf{r}_{a}}
Refers to
for homoegeneous function definition and theorem 1:
We get Virial Theorem: If U\left(\alpha \mathbf{r}_{1}, \alpha \mathbf{r}_{2}, \ldots, \alpha \mathbf{r}_{n}\right)=\alpha^{k} U\left(\mathbf{r}_{1}, \mathbf{r}_{2}, \ldots, \mathbf{r}_{n}\right) is a homoegeneous function of degree k , then 2 \bar{T}=k \bar{U}
From statistical physics point of view, we can refers to Vol.5 LANDAU and LIFSHITZ, p 31, problem 2:
Derive the virial theorem for a macroscopic body for which the potential energy of interaction particles in a homogeneous function of degree n in their coordinates.
This derivation is clear to me, and it indeed gives me an idea on how to connect the material I learned so far.
Another problem is the hw assignment for this semester, elementary but not trivial:
Consider a classical particle of mass m, residing in the one-dimensional (1D) potential well, U(x) =a|x|^γ , with γ >0 , and kept in thermal equilibrium with its environment, at temperatureT. By using Maxwell-Boltzmann distribution prove the virial theorem, 〈p^2〉/2m=−〈x·F〉/2 , that relates thermal averages for the kinetic energy and force F acted upon particle from the potential.
\left\langle\frac{p^{2}}{2 m}\right\rangle=\frac{\int_{0}^{\infty}\left(\frac{p^{2}}{2 m}\right) e^{-\frac{p^{2}}{2 mkT}} d p}{\int_{0}^{\infty} e^{-\frac{p^{2}}{2 mkT}} d p}=T/2
2.
\int_0^\infty e^{-U(x)/T} dx = \frac{1}{\gamma}(\frac{T}{\alpha})^{1/\gamma}\Gamma(1/\gamma)
\int_{0}^{\infty} U(x) e^{-\frac{U(x)}{T}} d x=-\frac{\partial}{\partial \beta} \int_{0}^{\infty} e^{-\beta U(x)} d x=\frac{T^{2}}{\alpha \gamma^{2}}\left(\frac{T}{\alpha}\right)^{\frac{1}{\gamma}-1} \Gamma\left(\frac{1}{\gamma}\right)
3. \langle U \rangle = \frac{T}{\gamma}
Notice that x \frac{dU}{dx} = \gamma U(x) , then \langle x \frac{dU}{dx} \rangle =\langle \gamma U(x) \rangle
Thus \left\langle\frac{p^{2}}{2 m}\right\rangle=-\frac{1}{2}\langle x \cdot F(x)\rangle
