# Statistical mechanics

by arnesmeets
Tags: mechanics, statistical
 P: 18 This shouldn't be too hard but I'm struggling. Consider N identical particles in a box of volume V. The relation between their linear momentum and kinetic energy is given by $$E = c\left|\overrightarrow{p}\right|$$, where c is the speed of light. So, the Hamiltonian of the system is $$H_N\left(\overrightarrow{q_1},\,\cdots,\overrightarrow{q_N},\overrighta rrow{p_1},\,\cdots,\overrightarrow{p_N}\right) = c \sum_{i = 1}^N \left|p_i\right|$$ Now, I'm trying to find the average energy of the system. So, I need to get Z: $$Z = \int \exp\left(-\beta \cdot c \cdot \sum_i \left|p_i\right|\right) d\overrightarrow{q_1}\cdots d\overrightarrow{q_N} d\overrightarrow{p_1}\cdots d\overrightarrow{p_N} = V^{N} \cdot \left(\int \exp\left(-\beta \cdot c \cdot \left|\overrightarrow{p}\right|\right) d\overrightarrow{p}\right)^N$$ Now, I've got two questions: 1) Is there any mistake in the calculations for Z above? 2) How to evaluate the integral??
 P: 76 You can solve the integral quite easily by transforming to spherical coordinates (r,\theta, \phi). So the integral becomes: $$\int_0^{2\pi} \int_0^{\pi} \int_0^{\infty} e^{-\beta r} r^2\sin{\theta} dr d\theta d\phi = 4\pi \int_0^{\infty} r^2 e^{-\beta r}dr$$ I think you can now solve the rest on your own