Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

F(ε) = Σ e-nε/kT n = 0 to ∞

  1. Sep 22, 2008 #1
    What they said:

    f(ε) = Σ e-nε/kT n = 0 to ∞

    Write the following series in terms of f’(ε):

    g(ε) = Σ n ε e-nε/kT n = 0 to ∞

    Then use the geometric series results to show that g can be written in the form:

    ε + 2 ε2 + 3 ε3 + … = ε/(1- ε)2

    What I did:

    Let x = e-nε/kT

    1/(1- x) = 1 + x + x2 + ...

    1/(1- x)2 = 1 + 2x + 3x2 + 4x3 + ...

    x/(1- x)2 = x + 2x2 + 3x3 + 4x4 +...

    which leads me to:

    g(ε) = ε e-nε/kT / (1 - e-nε/kT)2 n = 0 to ∞

    How do they get, ε + 2 ε2 + 3 ε3 + … = ε/(1- ε)2

    and what does it mean?
     
  2. jcsd
  3. Sep 22, 2008 #2
    I'm not sure I understand your notation, please use proper parantheses.

    Do you mean:

    f(ε) = Σ e-nε/kT n = 0 to ∞

    [tex] f(\epsilon) = \sum_{n=0}^{\infty} e^{-\frac{n \epsilon}{kT} [/tex]

    OR

    [tex] f(\epsilon) = \sum_{n=0}^{\infty} \frac{e^{-n \epsilon}}{kT} [/tex]

    And what is your question?

    Is it why does this hold:

    [tex] \epsilon + 2 \epsilon^{2} + 3 \epsilon^{3} = \frac{\epsilon}{\left(1 - \epsilon \right)^2} [/tex]

    If so, that's just taking derivative of both sides. If you need more clarification, let me know
     
  4. Sep 23, 2008 #3

    Mute

    User Avatar
    Homework Helper


    The sum is the first one. The factor kT pegs this as a statistical mechanices problem.

    Basically, to the OP, what the problem amounts to is that you have an infinite sum,

    [tex]\sum_{n=0}^{\infty}\exp\left[-\frac{n\varepsilon}{k_BT}\right][/tex]

    which is really just a geometric series: if you let [itex]x = \exp
    \left[-\frac{\varepsilon}{k_BT}\right][/itex], you get

    [tex]\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}[/tex]

    (since the exponential is always less than one). Hence, if you take a derivative with respect to [itex]\beta = 1/(k_BT)[/itex] you get

    [tex]\sum_{n=0}^{\infty} n \varepsilon e^{-n \varepsilon \beta} = -\frac{\varepsilon e^{-\varepsilon \beta}}{(1-e^{-\varepsilon \beta})^2}[/tex]

    which allows you to easily find the average energy of the system.

    Or, even simpler, leaving it as x and differentiating with respect to that,

    [tex]\sum_{n=0}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}[/tex]

    so just multiply by [itex]\varepsilon e^{-\varepsilon \beta}[/itex] to get

    [tex]\sum_{n=0}^{\infty} n \varepsilon x^n = \frac{\varepsilon x}{(1-x)^2}[/tex]
     
    Last edited: Sep 23, 2008
  5. Sep 24, 2008 #4
    And what is your question?

    Is it why does this hold:

    ε + 2 ε2 + 3 ε3 + … = ε/(1- ε)2


    If so, that's just taking derivative of both sides. If you need more clarification, let me know

    How do they get, ε + 2 ε2 + 3 ε3 + … = ε/(1- ε)2

    and what does it mean?

    I understand how to proceed to get the expression

    g(ε) = ε e-nε/kT / (1 - e-nε/kT)2 n = 0 to ∞

    but epsilon is a constant, hυ, not e-nε/kT. Do they mean that ε = f(ε) = e-nε/kT? That makes sense then.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: F(ε) = Σ e-nε/kT n = 0 to ∞
Loading...