Stefan-Boltzmann Law

Understanding the radiation from a blackbody was a problem that many prominent physicists tacked in the late nineteenth and early twentieth century. Part of problem was to understand the total energy emitted by a blackbody, which is solely dependent on the temperature of the blackbody. It can be argued that there is no dependence on the direction of radiation i.e isotropy. Boltzmann finally came up with a physical basis to show that the total energy radiated by a blackbody is proportional to the fourth power of the temperature of the body.

Radiation can be described using the radiation density or the radiation pressure and take it from me, for now, that the radiation density is related to the radiation pressure as such

$$p = u/3$$

where $p$ is the radiation pressure and $u$ is the radiation density.

Now, let’s use

$$TdS = dU + PdV$$
$$TdS = d(u\times V) + PdV$$
where $u$ is radiation density per unit volume
$$TdS = Vdu + udV + PdV$$
$$TdS = Vdu + (u + P)dV$$
$$TdS = Vdu + (u + u/3)dV$$
$$TdS = Vdu + \frac{4u}{3}dV$$
$$dS = \frac{V}{T}du + \frac{4u}{3T}dV$$
$$dS = \frac{V}{T}\frac{du}{dT}dT + \frac{4u}{3T}dV$$
we can therefore say that
$$\frac{d}{dV}(\frac{V}{T}\frac{du}{dT}) = \frac{d}{dT}(\frac{4u}{3T})$$
$$\frac{1}{T}\frac{du}{dT} = -\frac{4u}{3T^2} + \frac{4}{3T}\frac{du}{dT}$$
$$\frac{4u}{3T^2} = \frac{1}{3T}\frac{du}{dT}$$
$$\frac{4u}{T} = \frac{du}{dT}$$
$$\frac{4dT}{T} = \frac{du}{u}$$
$$\ln u = 4\ln T + c$$
$$u = kT^4$$

and since $u = 4\pi I_b/c$
$$I_b = \frac{kc}{4\pi} T^4$$
$$I_b = \sigma T^4$$

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