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Planck Spectrum and wavelength subsitution

  • Thread starter FortranMan
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  • #1
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All the below is presented in latex format. In Schroeder's intro to thermal, problem 7.38, he asks you to substitute

[EQ] \lambda = \frac{hc}{\epsilon} [/EQ]

into the following equation

[EQ] \frac{U}{V} = \int^{\infty}_{0} \frac{8 \pi \epsilon^{3}/(hc)^{3}}{e^{\epsilon / kT}-1} d \epsilon [/EQ]

from which you should get the Planck spectrum as a function of wavelength. So far I've been careful in replacing the differential of epsilon with

[EQ] d \epsilon = - \frac{hc}{\lambda^2} d \lambda [/EQ]

In preparation to plot and solve the integrand, he asks you to express the above equation in terms of hc/kT, which I'm guessing can be done by substituting the value x below.

[EQ] x = \frac{hc}{kT \lambda} [/EQ]

Again being careful to replace the differential of lambda,

[EQ] d \lambda = - \frac{hc}{kT x^2} dx [/EQ]

However as soon as arrange everything in terms of x, it looks exactly like this equation

[EQ] \frac{U}{V} = \frac{8 \pi (kT)^4}{(hc)^3} \int^{\infty}_{0} \frac{x^3}{e^{x}-1} dx [/EQ]

which, according to Schroeder, it shouldn't because when you plot the spectrum in terms of the wavelength the function should NOT peak at 2.82 and look exactly like the spectrum in terms of epsilon. So, what stupid differential mistake am I making?
 

Answers and Replies

  • #2
Dick
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I'm going to take your first expression, [EQ] \lambda = \frac{hc}{\epsilon} [/EQ] and substitute "[ tex ]" for [EQ] and "[ /tex ]" for [/EQ], and here's what I get [tex] \lambda = \frac{hc}{\epsilon} [/tex]. Eliminate the spaces in the tex /tex pair to get it to actually work.
 
Last edited:
  • #3
22
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All the below is presented in latex format. In Schroeder's intro to thermal, problem 7.38, he asks you to substitute

[tex] \lambda = \frac{hc}{\epsilon} [/tex]

into the following equation

[tex] \frac{U}{V} = \int^{\infty}_{0} \frac{8 \pi \epsilon^{3}/(hc)^{3}}{e^{\epsilon / kT}-1} d \epsilon [/tex]

from which you should get the Planck spectrum as a function of wavelength. So far I've been careful in replacing the differential of epsilon with

[tex] d \epsilon = - \frac{hc}{\lambda^2} d \lambda [/tex]

In preparation to plot and solve the integrand, he asks you to express the above equation in terms of hc/kT, which I'm guessing can be done by substituting the value x below.

[tex] x = \frac{hc}{kT \lambda} [/tex]

Again being careful to replace the differential of lambda,

[tex] d \lambda = - \frac{hc}{kT x^2} dx [/tex]

However as soon as arrange everything in terms of x, it looks exactly like this equation

[tex] \frac{U}{V} = \frac{8 \pi (kT)^4}{(hc)^3} \int^{\infty}_{0} \frac{x^3}{e^{x}-1} dx [/tex]

which, according to Schroeder, it shouldn't because when you plot the spectrum in terms of the wavelength the function should NOT peak at 2.82 and look exactly like the spectrum in terms of epsilon. So, what stupid differential mistake am I making?
rephrased. Now can anyone answer my question?
 
  • #4
10
0
Subtle difference

Your differentials are perfect. You just took them too far. Dont substitute with x. You want to solve for the energy distribution function in terms of lambda. That way you can take the derivative of the energy distribution function and set to zero to find maximum. Check out the wiki on Wien's Displacement Law and the derivation at the bottom.

en.wikipedia.org/wiki/Wien's_displacement_law
 

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