Register to reply 
Statistical mechanics 
Share this thread: 
#1
May2206, 05:38 PM

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 [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},\overrighta rrow{p_1},\,\cdots,\overrightarrow{p_N}\right) = c \sum_{i = 1}^N \leftp_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 \leftp_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?? 


#2
May2506, 07:18 AM

P: 18

Does nobody have a clue?



#3
May2506, 08:59 AM

P: 76

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 


Register to reply 
Related Discussions  
Statistical mechanics  Advanced Physics Homework  7  
Statistical Mechanics  Classical Physics  0  
Statistical mechanics  Classical Physics  1  
Statistical mechanics hw  Advanced Physics Homework  2  
Statistical mechanics  Classical Physics  0 