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Suppose I have a cavity with rough walls that is close to an ideal Black Body. For simplicity, assume that it is a square box. I cut a small hole in the box of size $dA.$

Energy would radiate away from the box, and the radiance is constant, so the irradiance per sterdian will follow Lambert’s cosine law.

What would be the vector field of the Poynting vector?

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  • $\begingroup$ AFAIK the Poynting vector is used in calculating the black body spectrum, so I suppose one could posit the black body curve and try to derive the poynting vector, . What for? $\endgroup$
    – anna v
    Commented Nov 2, 2022 at 5:37
  • $\begingroup$ @annav Mostly, I am interested in what the Poynting vector looks like at the hole of size $A,$ and inside the cavity. $\endgroup$
    – Jbag1212
    Commented Nov 2, 2022 at 13:08
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    $\begingroup$ The time averaged Poynting vector inside the cavity is zero. There is no net energy flux in a system in equilibrium. The time dependent Poynting vector fluctuates, so you would have to calculate a probability distribution or a series of correlation functions, I suppose. $\endgroup$ Commented Nov 2, 2022 at 13:39
  • $\begingroup$ If the time averaged poynting vector inside the cavity is zero, wouldn’t this imply a discontinuity at the exit hole? $\endgroup$
    – Jbag1212
    Commented Nov 2, 2022 at 14:11
  • $\begingroup$ Yes, you cannot have a hole of any finite size and still assume that the radiation inside the cavity is a blackbody. The cavity radiation approaches a blackbody as $dA \rightarrow 0$. $\endgroup$
    – ProfRob
    Commented Nov 2, 2022 at 14:17

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