Statistical mechanics

  1. May 22, 2006 #1
    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 [tex]E = c\left|\overrightarrow{p}\right|[/tex], where c is the speed of light. So, the Hamiltonian of the system is

    [tex]H_N\left(\overrightarrow{q_1},\,\cdots,\overrightarrow{q_N},\overrightarrow{p_1},\,\cdots,\overrightarrow{p_N}\right) = c \sum_{i = 1}^N \left|p_i\right|[/tex]
    Now, I'm trying to find the average energy of the system. So, I need to get Z:

    [tex]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[/tex]

    Now, I've got two questions:

    1) Is there any mistake in the calculations for Z above?
    2) How to evaluate the integral?? :blushing:
    Last edited: May 22, 2006
  2. jcsd
  3. May 25, 2006 #2
    Does nobody have a clue?:confused:
  4. May 25, 2006 #3
    You can solve the integral quite easily by transforming to spherical coordinates (r,\theta, \phi). So the integral becomes:

    [tex] \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[/tex]

    I think you can now solve the rest on your own :smile:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Statistical mechanics
  1. Statistical mechanics (Replies: 2)

  2. Statistical Mechanics (Replies: 1)

  3. Statistical Mechanics (Replies: 1)

  4. Statistical Mechanics (Replies: 0)

  5. Statistical mechanics (Replies: 0)