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Planck's Law

  1. Mar 11, 2004 #1
    Basically this problem is to derive Wien's displacement law from Planck's law.
    Specifically:
    a) Show that there is a general relationship between temperature and λmax stating that Tλmax = constant
    and
    b) Obtain a numerical value for this constant
    [Hint: Start with Planck's radiation law and note that the slope of u(λ,T) is zero when λ = λmax ]

    So I obediently try to differentiate u(λ, T) with respect to λ:

    I'll let C1 = 8Πhc and C2 = hc/(kBT)

    [tex] u(\lambda, T) = {C_1}\lambda^{-5} \left( e^\frac{C_2}{\lambda} - 1 \right)^{-1} [/tex]

    [tex] \frac{\partial u}{\partial \lambda} =
    C_1 \left((\lambda^{-5})(-1)({e^{\frac{C_2}{\lambda}} - 1)^{-2}({e^\frac{C_2}{\lambda}})(-\frac{C_2}{\lambda^2}) + (e^{\frac{C_2}{\lambda}} - 1)^{-1}(-5)\lambda^{-6}\right) = 0 [/tex]

    I divide through by C1 & get

    [tex] \frac{C_2e^\frac{C_2}{\lambda}}{\lambda^7({e^\frac{C_2}{\lambda}} -1)^2} - \frac{5}{\lambda^6({e^\frac{C_2}{\lambda}} -1)} = 0[/tex]

    [tex]\frac{C_2{e^\frac{C_2}{\lambda}} - 5\lambda({e^\frac{C_2}{\lambda}} - 1)}{\lambda^7({e^\frac{C_2}{\lambda}} -1)^2} = 0 [/tex]

    [tex]C_2{e^\frac{C_2}{\lambda}} - 5\lambda{e^\frac{C_2}{\lambda}} + 5\lambda = 0 [/tex]

    [tex] let x = \frac{C_2}{\lambda} = \frac{hc}{{k_B}T\lambda}[/tex]

    [tex]x{\lambda}e^x - 5{\lambda}e^x + 5{\lambda} = 0[/tex]

    [tex]x = 5 - \frac{5}{e^x} = 5(1 - e^{-x})[/tex]

    Now I substitute back for x = hc/(kTλ) and get:

    [tex]\frac{hc}{{k_B}T{\lambda}} = 5(1 - {e^\frac{-hc}{{k_B}T{\lambda}}}) [/tex]

    Oy!

    This, amazingly, is exactly the expression I'm supposed to end up with (for part a), according to the answer in the book.

    But how does this show that
    [tex]T\lambda = constant [/tex]

    and how do I solve for this constant?
     
  2. jcsd
  3. Mar 11, 2004 #2

    jamesrc

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    Well, this is kind of cheesy but:

    You've got x = 5(1-e-x)

    So plug that into a graphing calculator

    (y = x-5(1-e-x), find x where y = 0)

    Then you can write λT = hc/(kx), plugging in that value for x to find the value of the constant.

    I think there is a better way to find that answer, but I'm not a mathematician and I don't remember it. Do you think that's good enough?
     
  4. Mar 11, 2004 #3
    The equation is transcendental. You have no choice but to solve it numerically.

    As for how it's constant, the entire right hand of the equation is a constant and all but T and [itex]\lambda[/itex] are constants on the left.

    cookiemonster
     
  5. Mar 11, 2004 #4
    Yes, thanks, I did finally realize that everything but
    T and lambda are constant so T*lambda must be constant.

    But when I solve with the calculator
    x = 5(1-e^(-x))
    it gives me x=0
    which, come to think of it, seems perfectly correct.

    Except, it makes no sense, because how can it be that
    [tex]\frac{hc}{{k_B}T{\lambda}} = 0 [/tex]
    ?
     
  6. Mar 11, 2004 #5
    It can't! That's why you have to discard that solution. It's impossible to get [itex]\frac{hc}{{k_B}T{\lambda}} = 0 [/itex] because hc clearly is not zero. There is, however, another solution.

    Try plotting y = x and y = 5(1-e^(-x)) on the same screen and find where the two intersect.

    cookiemonster
     
  7. Mar 11, 2004 #6

    jamesrc

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    Oh, right. I guess I'm not as cheesy as I thought. What a shame, I really like cheese.
     
  8. Mar 11, 2004 #7
    That works -- after a bit of trial and error I get x = 4.965 as a pretty good approximation. That will let me solve for Tλ

    Is that the "usual" technique for solving equations of this form?

    How would you solve it if you didn't have a graphing device?
     
  9. Mar 11, 2004 #8
    I'd probably use bisection.

    I'd guess that it lies between 4 and 5, test if the function y = 5(1-e^(-x)) - x crosses the x-axis between these points. y(4) is positive and y(5) is negative. Try y(4.5). If it's positive (it is), then we have a new lower bound, since the function crosses between 4.5 and 5. So try 4.75. You get the idea.

    cookiemonster
     
  10. Mar 11, 2004 #9
    Thanks, but I like your graph trick better. :wink:

    Hope I remember it the next time a problem like this comes up.

    PS:
    James, if you graph
    (y = x-5(1-e^(-x)), find x where y = 0)
    on your calculator, you get (0,0).
    (At least, that's all I get on mine.)
     
    Last edited: Mar 11, 2004
  11. Mar 11, 2004 #10
    To solve this equation you could use an itterative method. You first assume the exponential term is small so that
    [itex]
    \frac{hc}{\lambda_mk_BT}=5
    [/itex]
    Then you put that in as the argument of your exponent to get the next itteration of
    [itex]
    \frac{hc}{\lambda_mk_BT}=5(1-e^{-5})
    [/itex]
    and continue this to get the root to desired accuracy.
     
  12. Mar 11, 2004 #11
    Super!

    It seems to take just a few iterations to get a good result, with very little thinking involved. What could be better?

    Thanks.
     
  13. Mar 12, 2004 #12

    jamesrc

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    Gold Member

    Really? I've got zero crossings at the origin and at x = 4.9651142317
    (please forgive the obscene # of decimal places; I've got a root finder on my calculator (it probably uses Newton-Raphson or some other iterative method to find the root)). The function looks kind of like a v tilted to the right.

    Anyway, none of that's important now since you solved it already. I hope I didn't add confusion.
     
  14. Mar 12, 2004 #13
    Not at all, James. Your suggestions are always appreciated. How were you to know that I have a crappy calculator?
     
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