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Consider a black spheric shell of radius $R$ manteined at temperature $T$ in an empty space. By Stephan-Boltzmann's Law we know that the power radiated is given by $\dot{Q} = 4\pi R^2 \sigma T^4$ but now my question is what fraction of $\dot{Q}$ is radiated inward and outward?

So far I have found only instances in which it is claimed that the power radiated inward and outward is the same and it is equal to $\dot{Q}$ itself but I am not convinced of the result because the way we derive such Law requires taking an integral over the whole surface and considering both the heat radiated inward and outward at each element of the surface.

I would like to know if someone has a rigorous explanation for such a claim.

As always any comment or answer is welcome and let me know if I can explain myself clearer!

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Stefan-Boltzmann law gives the energy radiated per unit surface area of the black body as $$j=\sigma T^4.$$ The outer surface area of a spherical shell of a radius $R$ is $4\pi R^2$, which seems to be the area used for the formula given in the OP. If we were to consider emission from both outer and inner surface, we would have to double this area (provided that the shell is very thin).

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  • $\begingroup$ Now I see, so we always have to account for an infinitesimal width? $\endgroup$
    – Matteo Menghini
    Jan 9 at 15:30
  • $\begingroup$ @MatteoMenghini it is a shell. Take finite width, if it makes you less confused - in this case outer and inner surface would not be exactly the same, so you would have something like $4\pi(R-w/2)^2$ and $4\pi (R+w/2)^2$ for the inner and the outer surfaces. But for $R\gg w$ the difference is negligeable. $\endgroup$
    – Roger V.
    Jan 9 at 15:33
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    $\begingroup$ @RogerVadim and ofc $(R \pm w/2)^2 \approx R^2(1 \pm \frac w R)$ for $w \ll R$ $\endgroup$
    – JEB
    Jan 9 at 16:09

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