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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

    gabbagabbahey

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    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

    gabbagabbahey

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    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

    gabbagabbahey

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    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

    gabbagabbahey

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    Substitute

    [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

    gabbagabbahey

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    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:
    http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html#c5

    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.
     
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