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# Blackbody radiation in 1, 2, or 3 dimensions

The total energy of a blackbody cavity is given by

$\int \frac{E_n}{e^{\beta E} - 1} \times g(n)\, dn.$

The fundamental wavelength of the photon is given by $$L$$ and we have $$\lambda_n = L/n.$$

The energy of the $$n$$th level is given by $$E_n = n\hbar\omega$$ and we have $$\omega = 2\pi c/L.$$

## 1d

In one dimension, there is just one mode per $$\omega_n$$ (ignoring polarization) and we have $$g(n) = 1.$$ Thus, \begin{align} E_\textrm{total} &= \int\frac{n\hbar \omega}{e^{\beta n\hbar\omega} - 1}\, dn\\ &= \hbar\omega \int\frac{n}{e^{\beta n\hbar\omega} - 1}\, dn. \end{align} Letting $$\gamma = \beta \hbar n\omega$$ we have \begin{align} E_\textrm{total} &= \frac{\hbar\omega}{\left(\beta\hbar\omega\right)^2}\int\frac{\gamma}{e^{\gamma} - 1}\, d\gamma \\ &=T^2 \times \textrm{const.} \end{align}

## 2d

In two dimensions, there are multiple modes for each $$n$$ since photons can oscillate in two dimensions and $$n = \sqrt{n_x^2 + n_y^2},$$ thus (ignoring polarization) $$g(n) = 2\pi n.$$

Thus our integral becomes \begin{align} E_\textrm{total} &= \hbar\omega \int\frac{n}{e^{\beta n\hbar\omega} - 1}g(n)\, dn \\ &= 2\pi\frac{\hbar\omega}{\left(\beta\hbar\omega\right)^3} \int\frac{\gamma^2}{e^{\gamma} - 1}\, d\gamma \\ &= T^3 \times \textrm{const.} \end{align}

## 3d

In three dimensions, $$n = \sqrt{n_x^2 +n_y^2 + n_z^2}$$ and $$g(n)$$ is generated by a spherical shell so that (ignoring polarization) $$g(n) = 4\pi n^2$$ and

\begin{align} E_\textrm{total} &= \hbar\omega \int\frac{n}{e^{\beta n\hbar\omega} - 1}g(n)\, dn \\ &= 4\pi\frac{\hbar\omega}{\left(\beta\hbar\omega\right)^4} \int\frac{\gamma^3}{e^{\gamma} - 1}\, d\gamma \\ &= \frac{4\pi^5}{15}k_b^4 T^4 \frac{L^3}{h^3c^3} \\ &= \frac{4\pi^5}{15} \frac{k_b^4 T^4}{h^3c^3} V \\ &= T^4 \times V\times\textrm{const.} \end{align}

Note by Josh Silverman
4 months, 1 week ago

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