• Support PF! Buy your school textbooks, materials and every day products Here!

Tricky Derivation of Blackbody Equations.

  • Thread starter omegas
  • Start date
  • #1
9
0

Homework Statement



Starting with the Planck distribution:

[tex] R(\lambda,T) = \frac{c}{4} \frac{8 \pi}{\lambda^4} (\frac{hc}{\lambda})(\frac{1}{e^{hc/(\lambda kT)}-1})[/tex]

Derive the blackbody Stefan-Boltzmann law (ie total flux is proportional to T4) by integrating the above expression over all wavelengths. Thus show that

R(T) = [tex] \frac{2 \pi^5 k^4}{15h^3 c^2} T^4 [/tex]

and [tex]\int[/tex] [tex] \frac{x^3}{e^x -1} dx = \frac{\pi^4}{15}[/tex]




The Attempt at a Solution



I know I need to substitute [tex]x = \frac{hc}{kT} \frac{1}{\lambda}[/tex]. And somehow I think I can use the form KR([tex]\lambda[/tex],T) = A([tex]\lambda[/tex])B([tex]\lambda[/tex])
 

Answers and Replies

  • #2
258
1

Homework Statement



Starting with the Planck distribution:

[tex] R(\lambda,T) = \frac{c}{4} \frac{8 \pi}{\lambda^4} (\frac{hc}{\lambda})(\frac{1}{e^{hc/(\lambda kT)}-1})[/tex]

Derive the blackbody Stefan-Boltzmann law (ie total flux is proportional to T4) by integrating the above expression over all wavelengths. Thus show that

[tex] R(T) = \frac{2 \pi^5 k^4}{15h^3 c^2} T^4 [/tex]

and [tex]\int[/tex] [tex] \frac{x^3}{e^x -1} dx = \frac{\pi^4}{15}[/tex]




The Attempt at a Solution



I know I need to substitute [tex]x = \frac{hc}{kT} \frac{1}{\lambda}[/tex]. And somehow I think I can use the form KR([tex]\lambda[/tex],T) = A([tex]\lambda[/tex])B([tex]\lambda[/tex])
Just use that substitution and try to get it into the form of the integral provided.
 

Related Threads on Tricky Derivation of Blackbody Equations.

  • Last Post
Replies
4
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
783
Replies
6
Views
1K
Replies
7
Views
2K
Replies
1
Views
2K
Replies
3
Views
4K
Replies
18
Views
2K
Top