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I am reading the book "Laser Fundamentals" by William T. Silvast and he mentioned that the number of cavity modes can be calculated as an octant of a spherical volume, i.e.

$M=\frac{1}{8}\frac{4\pi}{3}(\frac{2L_x2L_y2L_z}{\lambda^3})$

where M is the number of modes, $L_x / L_y / L_z$ are the lengths of the rectangular cavity. I know the reason we have $\frac{1}{8}$ is because the numbers of modes are positive, but why do we use the spherical volume to calculate the number of modes at the very beginning? Thank you.

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I'm not quite sure about the context. However, if the goal is to calculate the number of modes having a wavelength $\lambda_k > \lambda$ then one proceeds as follows:

The wavevectors within a cavity must fulfill $$ k_i = \frac{\pi n_i}{L_i} $$ where $i \in \lbrace x,y,z\rbrace$. Valid wavevecors form a rectangular lattice within the $k$-space with the lattice distances of $\frac{\pi }{L_i}$ . This means that the volume in the $k$-space occupied by a single mode is $$ v_k = \frac{\pi^3}{L_xL_yL_z} $$ Now we know $|k| = \tfrac{2\pi}{\lambda_k}$, thus the volume in the $k$-space corresponding to all wavelengths greater $\lambda$ is $$V = \frac{4\pi}{3}\left(\frac{2\pi}{\lambda}\right)^3$$ Now $M$ follows: $$ M = \frac{V}{v_k} = \frac{4\pi}{3}\frac{8 L_x L_y L_z}{\lambda^3} $$ If one octant is considered, it will give us the factor $1/8$.

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  • $\begingroup$ Hi, thank you for your answer. Why is the number of mode defined to be $\frac{V}{v_k}$? In the book, the author tries to derive the formula of number of modes $M=\frac{4\pi}{3}\frac{L_xL_yL_z}{\lambda^3}$ from the boundary conditions that electromagnetic waves vanish at the boundaries of a rectangular cavity but he did not mention anything about the k-space volume. Could you please also elaborate on it a bit? Thanks! $\endgroup$ – ASAP but not simpler Apr 25 at 14:48
  • $\begingroup$ @ASAPbutnotsimpler The b.c. you mention establish the relation between $\vec{k}$ and $V$ i mentioned above: $k_i\propto n_i / L_i$. Now there are of course infinitely many modes possible within the cavity. However, if you assume an UV-cutoff at $\lambda$ it means that your wavevectors (w.v.) are limited by the condition i) $|\vec{k}|\leq\tfrac{2\pi}{\lambda}$. (Does the author mention such a cutoff?) So we are looking for the number of w.v.'s, which fufill the condition i). This number may be well approximated by how often the volume occupied by one mode fit into the volume defined by i). $\endgroup$ – denklo Apr 26 at 6:02
  • $\begingroup$ @ASAPbutnotsimpler btw. i think i found the missing factor 8 infron of the $L_i$. $\endgroup$ – denklo Apr 26 at 6:04

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