Tricky Derivation of Blackbody Equations.

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
omegas
Messages
9
Reaction score
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])
 
Physics news on Phys.org
omegas said:

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.