# Homework Help: Statistical physics - average potential energy and gravity field

1. Jun 15, 2010

### xz5x

The problem statement, all variables and given/known data

[PLAIN]http://img688.imageshack.us/img688/8140/86617607.jpg [Broken]

The attempt at a solution

[PLAIN]http://img63.imageshack.us/img63/9531/79093945.jpg [Broken]

This is my whole procedure:

[PLAIN]http://img685.imageshack.us/img685/5706/62291576.jpg [Broken]

[PLAIN]http://img706.imageshack.us/img706/7240/94560340.jpg [Broken]

This is the expected result:

[PLAIN]http://img143.imageshack.us/img143/1264/21782586.jpg [Broken]

Last edited by a moderator: May 4, 2017
2. Jun 16, 2010

### hikaru1221

I can see 2 problems:
1 - The second step in (b) is wrong.
2 - The expected result's dimension is wrong.

3. Jun 16, 2010

### zzzoak

Average energy may be rewriten in the form
$$E_p=\frac{1}{\beta}[1-\frac{x}{e^x-1}]$$
where
$$x={\beta}mgl$$
WolframAlpha for the series gives
$$\frac{1}{e^x-1}=\frac{1}{x}-\frac{1}{2}+\frac{x}{12}$$
and I get
$$E_p=\frac{x}{2\beta}(1-\frac{x}{6})$$

Last edited: Jun 16, 2010
4. Jun 16, 2010

### xz5x

@zzzoak

I didn't know that average energy may be rewriten in this form. Can you please show me the procedure of how can I get your average energy from mine?

I see that you got the right result. I'll try to solve the problem with this form of potential energy.
Thank you very much!

Last edited: Jun 16, 2010
5. Jun 16, 2010

### xz5x

@hikaru1221

Thank you for your response!
Yes, the second step in b) is not correct. This is my new result:
[PLAIN]http://img704.imageshack.us/img704/4834/85676997.jpg [Broken]

I canâ€™t get the expected result. Please tell me, is dimension of my new result OK?

Last edited by a moderator: May 4, 2017
6. Jun 16, 2010

### hikaru1221

It is the expect result's dimension that is wrong. I guess this should be the result: $$E_p=\frac{mgl}{2}(1-\frac{mgl}{6kT})$$.
So in your newest result, just omit the second power of (mgl/kT), you should get the expected result.

7. Jun 16, 2010

### zzzoak

$$E_p=\frac{1}{\beta}\frac{1-e^{-x}(1+x)}{1-e^{-x}}=$$
$$=\frac{1}{\beta}\frac{1-e^{-x}-xe^{-x}}{1-e^{-x}}=$$
$$=\frac{1}{\beta}[1-\frac{xe^{-x}}{1-e^{-x}}]=$$
$$=\frac{1}{\beta}[1-\frac{x}{e^{x}-1}]$$

Last edited: Jun 16, 2010
8. Jun 18, 2010

### xz5x

@zzzoak

@hikaru1221

Thank you very much!