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Integrate Plancks Function to find Stefans Law

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data
    See uploaded file

    2. Relevant equations
    See Uploaded file


    3. The attempt at a solution
    Ok problem 1 I have with the solution why is (KT/HC)^4 on the outside of the integral and why is it to the 4th power. I assume it is to find (lamda)^5 but I would think if that was the case it would be to the 5th power. My second issue is when I apply the series formula 1/(1-epsilon) and make it 1/(1-e^-x) I end up x^3e^-nx but it appears to be missing another e^(-x). Then they do this derivative deal and get 1/a but I think it should be -1/a(e^-ax). Anyways they continue on this path until they get 6/a^4 which is the answer to the series but why do they stop there I am confused . Then how did the get (pi)^4/90 in there and how do I get the actual answer? Sorry I know its a ton of questions but I am hopelessly lost.
     

    Attached Files:

  2. jcsd
  3. Sep 4, 2012 #2
    They essentially do [itex]u = a\lambda[/itex], where [itex]a[/itex] is that mix of constants. Express [itex]\lambda = u/a[/itex] and [itex]d\lambda = du/a[/itex], and see what you get, then convert a to that constant mix.

    No, it's not. Observe that [itex]\frac 1 {1 - e^{-x}} = 1 + e^{-x} + e^{-2x} + ... [/itex]. That is, the first term of that is just 1. But when they integrate, the first term is [itex]e^{-x}[/itex], because they multiply the series by [itex]e^{-x}[/itex]
    This is the indefinite integral. Now plug in the integration limits - what do you get?
    They have found that all the integrals they need to compute are [itex]\frac 6 {n^4}[/itex].

    [itex]\sum_{n = 0}^{\infty}\frac 1 {n^4} = \frac {\pi^4} {90}[/itex]. This is known because on the left you haven the value of [itex]\zeta[/itex]-function at 4, which can be computed in a number of ways.
     
  4. Sep 4, 2012 #3
    Sweet thanks man you cleared up all my issues I get it now.
     
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