The Stefan Boltzmann Law gives a relation between the total energy radiated per unit area and the temperature of a blackbody. Specifically it states that, $$ j= \sigma {T}^4$$ Now using the thermodynamic derivation of the energy radiated we can derive the above relation, which leads to $T^4$. But is there any intuitive reason for the $T^4$ term?
[Physics] Intuitive reason for the $T^4$ term in Stefan Boltzmann law
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Boltzmann built on two known properties of cavity radiation.
(1) The energy density, $u$, defined as $u = U/V$, depends only on temperature, $T$.
(2) The radiation pressure, $p$ is given by $p= u/3$. Radiation pressure was given a firm basis c1862 by Maxwell. The factor of 1/3 arises because of the three-dimensionality of the cavity, in which radiation is propagating in all possible directions. [It's easy for us now to derive this equation by considering the cavity as containing a photon gas.]
Boltzmann (1884) used a thought-experiment in which a cavity is fitted with a piston, and we take the radiation inside it through a Carnot cycle. On a $p–V$ diagram the isothermals are just horizontal lines, because $u$ is constant so $p$ is constant. The heat input along the top (temperature $T$) isothermal is $\Delta U + p \Delta V$. This works out to be $4 p \Delta V$. If the lower temperature isothermal is only slightly lower, at temperature ($T - d T$) then the cycle appears as a thin horizontal box, and the net work done during the cycle is simply $dp \Delta V$ in which $dp$ is the infinitesimal pressure difference between the two isothermals. We can then apply the definition of thermodynamic temperature in the form
$$\frac{dT}{T} = \frac{\text{work}}{\text{heat input}} = \frac{dp \Delta V}{4 p \Delta V} = \frac{dp}{4 p}~.$$
This integrates up easily to give the Stefan-Boltzmann law.
A modern thermodynamic derivation would probably start from the fundamental equation (embodying first and second laws of thermodynamics)… $$\text {d}U = T \text{d}S - P \text{d}V. $$ Therefore for an isothermal change $$\left(\frac{\partial U}{\partial V} \right)_T = T \left(\frac{\partial S}{\partial V} \right)_T - p .$$ Using one of the Maxwell relations, this becomes $$\left(\frac{\partial U}{\partial V} \right)_T = T \left(\frac{\partial p}{\partial T} \right)_V - p .$$ But for the radiation, $$\left(\frac{\partial U}{\partial V} \right)_T = \left(\frac{\partial (uV)}{\partial V} \right)_T = u = 3p \ \ \ \ \ \text{and} \ \ \ \ \ \left(\frac{\partial p}{\partial T} \right)_V = \frac {\text{d} p}{\text{d}T}. $$ Making these substitutions and tidying, $$ 4p = T \frac {\text{d} p}{\text{d}T}.$$ As before, separation of variables and integration gives the Stefan-Boltzmann law.
The Stefan-Boltzmann law is $$ \frac{P}{A}=\frac{2\pi^5 k^4}{15 h^3 c^2} T^4, \tag{1} $$ where $P/A$ denotes the emitted power per unit surface area of the blackbody. In equation (1), the coefficient of $T^4$ is the Stefan-Boltzmann constant. In contrast, the energy per unit volume ($E/V$) of the radiation inside the blackbody is $$ \frac{E}{V}=\frac{8\pi^5 k^4}{15 h^3 c^3} T^4. \tag{2} $$ In equation (2), the coefficient of $T^4$ is the radiation constant. Equation (2) is used, for example, to compute the energy density of the cosmic microwave background radiation, because in that case we are inside the blackbody (the universe). These two quantities are related to each other by $$ \frac{P}{A} = \frac{c}{4}\times \frac{E}{V}. $$ The question is, where does the factor of $4$ come from?
The essence of the answer is that inside the blackbody, the radiation at any point is coming and going equally in all directions; but outside the blackbody, that is no longer true. The distribution over directions is not uniform outside the body, and the difference ends up being a factor of $4$. Equation (1) describes the emitted power outside the blackbody, whereas equation (2) describes the energy density inside.
The derivation of both equations, (1) and (2), involves the quantity $B(\lambda)$ that was given in the OP. Consider the quantity $$ \frac{4\pi}{c}B(\lambda)\,d\lambda \tag{3} $$ where $d\lambda$ is an infinitesimal range of wavelengths. The quantity (3) is the energy density of the electromagnetic (EM) radiation that is contained inside the blackbody, within the given range of wavelengths. The factor of $4\pi$ accounts for the full sphere's worth of possible directions of the radiation at each point inside the blackbody. Integrating (3) over all wavelengths gives the total energy density (2).
To derive the Stefan-Boltzmann law (1), suppose for a moment that radiation is only able to escape through a small hole of area $A$ in the surface of the blackbody. How much energy leaks out per unit time? If we consider plane waves moving in a particular direction, then the amount that escapes through the hole depends on the direction in which the plane wave is moving compared to the orientation of the hole. So, to calculate this, we should start with the quantity $$ \frac{1}{c}B(\lambda)\,d\lambda, \tag{4} $$ which is the energy density per unit solid angle inside the blackbody. This is obtained from (3) by omitting the factor of $4\pi$. EM radiation travels at speed $c$, so the energy per unit time that escapes through the hole into a given narrow cone of solid angle $d\Omega$ is $$ \frac{1}{c}B(\lambda)\,d\lambda\times cA\cos\theta\,d\Omega \tag{4} $$ where $\theta$ is the angle of the radiation's direction of travel relative to the direction normal to the hole. The factor $A\cos\theta$ is the area of the hole projected orthogonally to the direction in which the radiation is traveling. This is the key. To calculate the rate at which energy escapes, we should integrate the quantity (4) over all directions with $\theta < \pi/2$. Directions with $\theta > \pi/2$ don't contribute, because that radiation is moving toward the inside of the body instead of toward the outside. Doing this integral (and cancelling the factors of $c$) gives $$ B(\lambda)\,d\lambda\times A \int_0^{2\pi}d\phi\int_0^{\pi/2} d\theta\,\sin\theta\,\cos\theta = B(\lambda)\,d\lambda\times A\pi, \tag{5} $$ so the quantity $$ \pi B(\lambda)\,d\lambda \tag{6} $$ is the power per unit surface area of the radiation that escapes from the hole. Even though this analysis considered radiation escaping through a "hole" in the surface, the same analysis applies to every piece of the surface of a blackbody that allows radiation to escape. The Stefan-Boltzmann law comes from integrating (6) over all wavelengths.
Best Answer
There's roughly $kT$ energy in each active mode. The active modes are characterized by momenta which live inside a sphere of radius proportional to $kT$, which has volume proportional to $T^3$. Multiplying these factors gives $T^4$, and the result clearly generalizes to $T^{d+1}$ in general dimension.