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  "body": "\n## Example\n\nLet $\\Psi$ be an even polynomial in $\\mu$, the characteristic function, with $\\Psi \\ge 0$ on $[0, 1]$ and $\\int_0^1 \\Psi(\\mu)\\, d\\mu \\le \\frac{1}{2}$.\n*Chandrasekhar's H-function* is the solution $H$ on $[0, 1]$ of\n$$H(\\mu) = 1 + \\mu\\, H(\\mu) \\int_0^1 \\frac{\\Psi(\\mu')\\, H(\\mu')}{\\mu + \\mu'}\\, d\\mu'.$$\nOne takes the solution that is analytic in $\\mu$ in the right half-plane and continuous on $[0, 1]$.\nThe simplest case is isotropic scattering, $\\Psi = \\frac{\\omega_0}{2}$ with albedo $0 \u003c \\omega_0 \\le 1$.\nThe unknown appears in a product with itself, so this is a [[nonlinear integral equation|BSCYBP6AS6]].\n\nCheck of a moment identity.\nLet $I = \\int_0^1 \\Psi(\\mu) H(\\mu)\\, d\\mu$.\nMultiply the equation by $\\Psi(\\mu)$ and integrate over $[0, 1]$.\nThe double integral $\\int_0^1 \\int_0^1 \\frac{\\mu}{\\mu + \\mu'}\\, \\Psi(\\mu) H(\\mu)\\, \\Psi(\\mu') H(\\mu')\\, d\\mu'\\, d\\mu$ does not change when $\\mu$ and $\\mu'$ swap, so it equals half the integral with $\\frac{\\mu + \\mu'}{\\mu + \\mu'} = 1$, that is $\\frac{1}{2} I^2$.\nSo $I = \\int_0^1 \\Psi\\, d\\mu + \\frac{1}{2} I^2$, and the physical root is\n$$\\int_0^1 \\Psi(\\mu) H(\\mu)\\, d\\mu = 1 - \\sqrt{1 - 2 \\int_0^1 \\Psi(\\mu)\\, d\\mu}.$$\nIn the conservative case $\\Psi = \\frac{1}{2}$ this gives $\\int_0^1 H(\\mu)\\, d\\mu = 2$.\nAlso $H(0) = 1$, directly from the equation.\n\n## Remarks\n\nThe function $H$ is tied to a Wiener–Hopf factorization.\nFor $z$ off the segment $[-1, 1]$, with the analytic continuation of $H$, one has\n$$\\frac{1}{H(z)\\, H(-z)} = 1 - 2 z^2 \\int_0^1 \\frac{\\Psi(\\mu)}{z^2 - \\mu^2}\\, d\\mu.$$\nThe right side is the dispersion function of the [[radiative transfer equation|EX9YYRMAJA]] in a half-space; $H(-z)$ is the factor holomorphic for $\\operatorname{Re} z \u003c 0$ and $H(z)$ the factor holomorphic for $\\operatorname{Re} z \u003e 0$.\nAs $z \\to \\infty$, both sides tend to $1 - 2\\int_0^1 \\Psi\\, d\\mu$, consistent with $H(\\infty) = 1/(1 - I)$ from the equation.\nIn the conservative case, $H$ gives the exact angular distribution of the light leaving a grey atmosphere, the companion of the [[Milne problem|KKT4Y0TBRZ]].\n\n## Explanations\n\n### L1 human\n\nLight that enters a thick cloud or a stellar atmosphere is scattered many times before some of it comes back out.\nChandrasekhar showed that the outgoing light is described by a single function $H$ of the cosine $\\mu$ of the exit angle.\nThis function solves an equation in which it appears twice, once outside and once inside an integral.\nSimple integral identities, such as $\\int_0^1 H\\, d\\mu = 2$ for pure scattering, follow by symmetry without solving the equation.\n",
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