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Wiens scaling law

  1. Aug 25, 2008 #1
    How does Wien's scaling law

    \frac{u(\lambda)}{T^5} = \frac{f(\lambda T)}{\lambda^5T^5}

    imply that if [tex]\frac{u(\lambda)}{T^5}[/tex] is plotted as a function of [tex]\lambda T[/tex], all experimental data will lie on a single curve?
  2. jcsd
  3. Aug 26, 2008 #2


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    I don't know anything about the physics here, but it seems obvious to me that if you plot [tex]y = u(\lambda)/T^5[/tex] versus [tex]x = \lambda T[/tex] then Wien's scaling law tells you that [tex]y = f(x) / x^5 [/tex]. So if you take a data point (x', y') and Wiens' law is true, then y' should be (within error) f(x')/(x')^5.
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