Blackbody radiation in 1, 2, or 3 dimensions

The total energy of a blackbody cavity is given by

EneβE1×g(n)dn.\int \frac{E_n}{e^{\beta E} - 1} \times g(n)\, dn.

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

The energy of the nnth level is given by En=nωE_n = n\hbar\omega and we have ω=2πc/L.\omega = 2\pi c/L.


In one dimension, there is just one mode per ωn\omega_n (ignoring polarization) and we have g(n)=1.g(n) = 1. Thus, Etotal=nωeβnω1dn=ωneβnω1dn.\begin{aligned} 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{aligned} Letting γ=βnω\gamma = \beta \hbar n\omega we have Etotal=ω(βω)2γeγ1dγ=T2×const.\begin{aligned} 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{aligned}


In two dimensions, there are multiple modes for each nn since photons can oscillate in two dimensions and n=nx2+ny2,n = \sqrt{n_x^2 + n_y^2}, thus (ignoring polarization) g(n)=2πn.g(n) = 2\pi n.

Thus our integral becomes Etotal=ωneβnω1g(n)dn=2πω(βω)3γ2eγ1dγ=T3×const.\begin{aligned} 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{aligned}


In three dimensions, n=nx2+ny2+nz2n = \sqrt{n_x^2 +n_y^2 + n_z^2} and g(n)g(n) is generated by a spherical shell so that (ignoring polarization) g(n)=4πn2g(n) = 4\pi n^2 and

Etotal=ωneβnω1g(n)dn=4πω(βω)4γ3eγ1dγ=4π515kb4T4L3h3c3=4π515kb4T4h3c3V=T4×V×const.\begin{aligned} 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{aligned}

Note by Josh Silverman
3 years, 9 months ago

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