Question

Derive the Stefan-Boltzmann formula for the total energy density in blackbody radiation

$\frac{\mathrm{E}}{V}=\left(\frac{{\pi }^2{\mathrm{k}}_{\mathrm{B}}^4}{15{ħ}^3{C}^3}\right){\mathrm{T}}^4=\left(7.57\times {10}^{-16}{\mathrm{Jm}}^{-3}{\mathrm{K}}^{-4}\right){\mathrm{T}}^4$

Short answer

Deriving the Stefan-Boltzmann formula for the total energy density in the blackbody radiation is$$\begin{gathered} =\left[\frac{{\pi }^2{\left(1.3807\times {10}^{23}\mathrm{J}/\mathrm{K}\right)}^4}{15{\left(2.998\times {10}^8\mathrm{m}/\mathrm{S}\right)}^3\left(1.5046\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}\right]{\mathrm{T}}^4 \\ 7.566\times {10}^{-16}\frac{\mathrm{J}}{{\mathrm{m}}^3{\mathrm{K}}^4} \end{gathered}$$

Step-by-step solution

Definition of Stefan Boltzmann law

According to the Stefan-Boltzmann equation, the amount of radiation emitted by a dark substance per unit area is exactly proportional to the fourth power of the temperature.

Deriving the Stefan-Boltzmann formula for the total energy

From Equation 5.113:

$\frac{E}{V}=\underset{0}{\overset{\infty }{\int }}\rho \left(\omega \right)d\omega =\frac{h}{{\pi }^2{c}^3}\underset{0}{\overset{\infty }{\int }}\frac{{\omega }^3}{(e^{h\omega /k_{B}T}-1)}d\omega .$

Let $x=\frac{h\omega }{k_{B}T}.$then

$\begin{array}{rcl}\frac{E}{V}& =& \frac{h}{{\pi }^2{c}^3}{\left(\frac{k_{B}T}{h}\right)}^4\underset{0}{\overset{\infty }{\int }}\frac{x^3}{e^x-1}dx=\frac{{\left(k_{B}T\right)}^4}{{\pi }^2{c}^3{h}^3}\Gamma \left(4\right)\varsigma \left(4\right)\\ & =& \frac{{\left(k_{B}T\right)}^4}{{\pi }^2{c}^3{h}^3}.6.\frac{{\pi }^4}{90}\\ & =& \left(\frac{{\pi }^2{k_B}^4}{15c^3{h}^3}\right)T^4\\ & & \end{array}$

$$\begin{gathered} =\left[\frac{{\pi }^2{(1.3807\times {10}^{23}J/K)}^4}{15{(2.998\times {10}^{8}m/s)}^3{(1.5046\times {10}^{-34}J.s)}^3}\right]T^4\backslash \mathrm{hfill} \\ =\left[7.566\times {10}^{-16}\frac{J}{m^3{K}^4}\right]T^4 \end{gathered}$$