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Statistical physics - average potential energy and gravity field

  1. Jun 15, 2010 #1
    Please help me solve this problem:

    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. jcsd
  3. Jun 16, 2010 #2
    I can see 2 problems:
    1 - The second step in (b) is wrong.
    2 - The expected result's dimension is wrong.
     
  4. Jun 16, 2010 #3
    Average energy may be rewriten in the form
    [tex]E_p=\frac{1}{\beta}[1-\frac{x}{e^x-1}][/tex]
    where
    [tex]x={\beta}mgl[/tex]
    WolframAlpha for the series gives
    [tex]\frac{1}{e^x-1}=\frac{1}{x}-\frac{1}{2}+\frac{x}{12}[/tex]
    and I get
    [tex]E_p=\frac{x}{2\beta}(1-\frac{x}{6})[/tex]
     
    Last edited: Jun 16, 2010
  5. Jun 16, 2010 #4
    @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
  6. Jun 16, 2010 #5
    @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 can’t get the expected result. Please tell me, is dimension of my new result OK?
     
    Last edited by a moderator: May 4, 2017
  7. Jun 16, 2010 #6
    It is the expect result's dimension that is wrong. I guess this should be the result: [tex]E_p=\frac{mgl}{2}(1-\frac{mgl}{6kT})[/tex].
    So in your newest result, just omit the second power of (mgl/kT), you should get the expected result.
     
  8. Jun 16, 2010 #7
    Your equation is
    [tex]E_p=\frac{1}{\beta}\frac{1-e^{-x}(1+x)}{1-e^{-x}}=[/tex]
    [tex]=\frac{1}{\beta}\frac{1-e^{-x}-xe^{-x}}{1-e^{-x}}=[/tex]
    [tex]=\frac{1}{\beta}[1-\frac{xe^{-x}}{1-e^{-x}}]=[/tex]
    [tex]=\frac{1}{\beta}[1-\frac{x}{e^{x}-1}][/tex]
     
    Last edited: Jun 16, 2010
  9. Jun 18, 2010 #8
    @zzzoak

    @hikaru1221


    Thank you very much!
     
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