Why does the Stefan-Boltzmann Law contain no information about the distance from the radiating body

electromagnetic-radiationelectromagnetismthermal-radiationthermodynamics

The Stefan–Boltzmann law states that the power radiated per unit area of a black body is equal to:

$$ \frac{dP}{dA} = \sigma \cdot T^4 \,\,\,\,\,\,\,\left[\frac{W}{m^{2}}\right] $$

This power per unit area is radiated through EM Waves because of the motion of the charge carriers due to the non-zero absolute temperature $T$.

Hence, my questions are:

1.) How can the value of $\frac{dP}{dA}$ be independent from the radial distance $r$ from the radiating body? I've seen people applying this law to find out the power per unit area at a generic distance from the radiating body.

2.) The quantity $\frac{dP}{dA}$ has the same unit of measurement of the Poynting Vector Magnitude. The last one describes the radiated power per unit area of an EM wave. Which is the relationship between these two quantities. I'm not sure they are exactly equal because:

  • Again, the Poynting Vector decays as $\frac{1}{r^2}$ in the far-field region, whilst $\frac{dP}{dA}$ does not appear to decay.

  • The Poynting Vector Magnitude maybe refers to a single wavelength, whilst $\frac{dP}{dA}$ to the whole spectrum of radiated wavelengths. But, can we say $\frac{dP}{dA}$ is the sum of all the Poynging vectors of each radiated wavelength?

3)Why is $\frac{dP}{dA}$ commonly referred as thermal power per unit area? It is just power per unit area. Then, if the receiving body is purely resistive, it will become heat. If it is reflective or inductive/capacitive, it won't become heat.

Best Answer

$\sigma T^4$ is the amount of energy the surface of a black body emits per square meter per second. The total energy emitted by a spherical black body with radius R per second is therefore $4\pi R^2\sigma T^4$. The energy flux at distance $r$ is thus $\frac{4\pi R^2\sigma T^4}{4\pi r^2} = \frac{R^2}{r^2}\sigma T^4$ so it does depend on r.

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