1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Planck's formula for Blackbody Radiation

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose that a blackbody spectrum is specified by Spectral Radiancy [tex]R_{T} (\nu) d\nu[/tex] and Energy Density [tex]\rho_{T} (\nu) d\nu[/tex] then show that
    [tex]R_{T} (\nu) d\nu = \frac {c}{4}\cdot \rho_{T} (\nu) d\nu[/tex]

    2. Relevant equations

    [tex]\rho_{T} (\nu) d\nu = \frac{8 \pi h \nu^{3} d\nu}{c^{3}(e^{\frac{h\nu}{kT}}-1)}[/tex]

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 7, 2008 #2


    User Avatar
    Homework Helper
    Gold Member

    What do you know about spectral radiancy for a blackbody? What have you tried?
  4. Oct 7, 2008 #3
    Spectral Radiancy [tex]R(\nu)[/tex] is defined such that [tex]R_{T}(\nu)d\nu [/tex] gives the energy absorbed per unit area per unit time when radiation has frequency between [tex]\nu[/tex] and [tex]\nu + d\nu[/tex]. It is essentially power
  5. Oct 7, 2008 #4
    Can someone please help me out with this one?
    Thanks in advance.
  6. Oct 7, 2008 #5


    User Avatar
    Homework Helper
    Gold Member

    If you integrate [itex]R_{T}(\nu)d\nu[/itex] over all possible frequencies you should get the total radiated power per unit area:

    [tex]\Rightarrow \int_0^{\infty} R_{T}(\nu)d\nu=\sigma T ^4[/tex]

    What quantity do you get when you integrate [itex]\rho_{T} (\nu) d\nu[/itex] over all possible frequencies?
  7. Oct 7, 2008 #6
    Since the latter quantity denotes energy density, the integral over all frequencies should give us total radiant energy contained in the cavity, which is [tex]kT[/tex] if we assume that [tex]T[/tex] is the absolute temperatre of the cavity.
    Am I right sir?
  8. Oct 7, 2008 #7


    User Avatar
    Homework Helper
    Gold Member

    No, the total radiated energy will not be kT; try carrying out the integration with the expression you provided for the energy density.
  9. Oct 7, 2008 #8
    hmm yes sir I did that, and got something like this:
    [tex] \frac{8(kT)^{4}\pi^{5}}{(hc)^{3}\cdot 15} [/tex]

    Ok so now in order that i prove my problem All i do is to divide Stefan's Law by the above right? keeping of course in mind stefans constant in terms of [tex] h, c , k[/tex].
    Is the relation true only between the integrals over [tex] 0 to \infty [\tex] or is it to even for spectral Radiancy?
    Last edited: Oct 7, 2008
  10. Oct 7, 2008 #9
    I fail to understand why that sentence of mine is appearing latexified :frown:
  11. Oct 7, 2008 #10
    backslash instead of forward slash?
  12. Oct 7, 2008 #11
    oh ok sir
  13. Oct 7, 2008 #12


    User Avatar
    Homework Helper
    Gold Member


    [tex]\sigma= \frac{2 \pi ^5 k^4}{15h^3c^2}[/tex]

    into the expression, and notice that you get:

    [tex]\int_0^{\infty} \rho_{T} (\nu) d\nu=\frac{4 \sigma T^4}{c}[/tex]

    [tex]\Rightarrow \sigma T^4=\int_0^{\infty} \frac{c}{4} \rho_{T} (\nu) d\nu[/tex]

    What does that mean [itex]R_{T}(\nu)d\nu[/itex] is?
  14. Oct 7, 2008 #13
    Hence it means that [tex]R_{T}(\nu}d\nu [/tex] is the power absorbed, per unit area by radiation with frequency lying in [tex]\nu[/tex] and [tex]\nu + d\nu[/tex] . But I dont quite get the physical interpretation of the result. Do you mind throwing some light on that sir?
  15. Oct 7, 2008 #14


    User Avatar
    Homework Helper
    Gold Member

    Actually. you already knew that [itex]R_{T}(\nu)d\nu[/itex] was the power radiated per unit area between [itex]\nu[/itex] and [itex]nu +dnu[/itex]. You used that to show that

    [tex]\int_0^{\infty} R_{T}(\nu)d\nu=\sigma T ^4[/tex]

    You then used your expression for the energy density to show

    [tex]\sigma T^4=\int_0^{\infty} \frac{c}{4} \rho_{T} (\nu) d\nu[/tex]

    Putting the two together you get:

    [tex]\Rightarrow \int_0^{\infty} R_{T}(\nu)d\nu=\int_0^{\infty} \frac{c}{4} \rho_{T} (\nu) d\nu[/tex]

    [tex]\Rightarrow R_{T}(\nu)d\nu=\frac{c}{4} \rho_{T} (\nu) d\nu[/tex]

    doesn't it?

    The physical interpretation of this result is that the radiated power is proportional to the energy density.
  16. Aug 7, 2010 #15
    An old thread, but hope this posting will give something to anyone confused on this topic and ended up here (just like me).

    I tried to derive this on my own but didn't get the factor 1/4. I got the proportionality value only c, which I thought physically sensible because the energy inside the cavity of a blackbody was carried away by EM waves whose speed is c. But my answer was not in agreement with the book I was reading, c/4.

    After searching for sometime, I finally found this:

    the factor of 1/4 was there because of two facts. Roughly, the two factors are:
    1. in thermal equilibrium, the radiated power is the same as the absorbed power. This account for 1/2 if we are going to calculate only the radiated power.
    2. the radiation going through the hole comes from every direction, and thus, it needs to be multiplied by cos^2 (comes from dot products of area and velocity of radiation A.v) which then averaged and account for another 1/2

    Hope this help anyone who ended-up here when searching the meaning for the factor 1/4.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook