Chandrasekhar's H-function

Chandrasekhar's H-function for different albedo

In atmospheric radiation, Chandrasekhar's H-function appears as the solutions of problems involving scattering, introduced by the Indian American astrophysicist Subrahmanyan Chandrasekhar. The Chandrasekhar's H-function $H(\mu )$ defined in the interval $0\le \mu \le 1$, satisfies the following nonlinear integral equation

$H(\mu )=1+\mu H(\mu )\underset{0}{\overset{1}{\int }}\frac{\mathrm{\Psi }({\mu }^{\prime })}{\mu +{\mu }^{\prime }}H({\mu }^{\prime })d{\mu }^{\prime }$

where the characteristic function $\mathrm{\Psi }(\mu )$ is an even polynomial in $\mu$ satisfying the following condition

$\underset{0}{\overset{1}{\int }}\mathrm{\Psi }(\mu )d\mu \le \frac{1}{2}$.

If the equality is satisfied in the above condition, it is called conservative case, otherwise non-conservative. Albedo is given by ${\omega }_o=2\mathrm{\Psi }(\mu )=\text{constant}$. An alternate form which would be more useful in calculating the H function numerically by iteration was derived by Chandrasekhar as,

$\frac{1}{H(\mu )}={[1-2\underset{0}{\overset{1}{\int }}\mathrm{\Psi }(\mu )d\mu ]}^{1/2}+\underset{0}{\overset{1}{\int }}\frac{{\mu }^{\prime }\mathrm{\Psi }({\mu }^{\prime })}{\mu +{\mu }^{\prime }}H({\mu }^{\prime })d{\mu }^{\prime }$.

In conservative case, the above equation reduces to

$\frac{1}{H(\mu )}=\underset{0}{\overset{1}{\int }}\frac{{\mu }^{\prime }\mathrm{\Psi }({\mu }^{\prime })}{\mu +{\mu }^{\prime }}H({\mu }^{\prime })d{\mu }^{\prime }$.

Approximation

The H function can be approximated up to an order $n$ as

$H(\mu )=\frac{1}{{\mu }_1\cdots {\mu }_n}\frac{\prod _{i=1}^n(\mu +{\mu }_i)}{\prod _{\alpha }(1+k_{\alpha }\mu )}$

where ${\mu }_i$ are the zeros of Legendre polynomials $P_{2n}$ and $k_{\alpha }$ are the positive, non vanishing roots of the associated characteristic equation

$1=2\sum _{j=1}^n\frac{a_j\mathrm{\Psi }({\mu }_j)}{1-k^2{\mu }_j^2}$

where $a_j$ are the quadrature weights given by

$a_j=\frac{1}{P_{2^{\prime }n}({\mu }_j)}\underset{-1}{\overset{1}{\int }}\frac{P_{2n}({\mu }_j)}{\mu -{\mu }_j}d{\mu }_j$

Explicit solution in the complex plane

In complex variable $z$ the H equation is

$H(z)=1-\underset{0}{\overset{1}{\int }}\frac{z}{z+\mu }H(\mu )\mathrm{\Psi }(\mu )d\mu ,\underset{0}{\overset{1}{\int }}|\mathrm{\Psi }(\mu )|d\mu \le \frac{1}{2},\underset{0}{\overset{\delta }{\int }}|\mathrm{\Psi }(\mu )|d\mu \to 0,\delta \to 0$

then for $\mathrm{\Re }(z)>0$, a unique solution is given by

$\ln H(z)=\frac{1}{2\pi i}\underset{-i\mathrm{\infty }}{\overset{+i\mathrm{\infty }}{\int }}\ln T(w)\frac{z}{w^2-z^2}dw$

where the imaginary part of the function $T(z)$ can vanish if $z^2$ is real i.e., $z^2=u+iv=u(v=0)$. Then we have

$T(z)=1-2\underset{0}{\overset{1}{\int }}\mathrm{\Psi }(\mu )d\mu -2\underset{0}{\overset{1}{\int }}\frac{{\mu }^2\mathrm{\Psi }(\mu )}{u-{\mu }^2}d\mu$

The above solution is unique and bounded in the interval $0\le z\le 1$ for conservative cases. In non-conservative cases, if the equation $T(z)=0$ admits the roots $\pm 1/k$, then there is a further solution given by

$H_1(z)=H(z)\frac{1+kz}{1-kz}$

Properties

• $\underset{0}{\overset{1}{\int }}H(\mu )\mathrm{\Psi }(\mu )d\mu =1-{[1-2\underset{0}{\overset{1}{\int }}\mathrm{\Psi }(\mu )d\mu ]}^{1/2}$. For conservative case, this reduces to $\underset{0}{\overset{1}{\int }}\mathrm{\Psi }(\mu )d\mu =\frac{1}{2}$.
• ${[1-2\underset{0}{\overset{1}{\int }}\mathrm{\Psi }(\mu )d\mu ]}^{1/2}\underset{0}{\overset{1}{\int }}H(\mu )\mathrm{\Psi }(\mu ){\mu }^{2}d\mu +\frac{1}{2}{[\underset{0}{\overset{1}{\int }}H(\mu )\mathrm{\Psi }(\mu )\mu d\mu ]}^2=\underset{0}{\overset{1}{\int }}\mathrm{\Psi }(\mu ){\mu }^{2}d\mu$. For conservative case, this reduces to $\underset{0}{\overset{1}{\int }}H(\mu )\mathrm{\Psi }(\mu )\mu d\mu ={[2\underset{0}{\overset{1}{\int }}\mathrm{\Psi }(\mu ){\mu }^{2}d\mu ]}^{1/2}$.
• If the characteristic function is $\mathrm{\Psi }(\mu )=a+b{\mu }^2$, where $a,b$ are two constants(have to satisfy $a+b/3\le 1/2$) and if ${\alpha }_n=\underset{0}{\overset{1}{\int }}H(\mu ){\mu }^{n}d\mu ,n\ge 1$ is the nth moment of the H function, then we have

${\alpha }_0=1+\frac{1}{2}(a{\alpha }_0^2+b{\alpha }_1^2)$

and

$(a+b{\mu }^2)\underset{0}{\overset{1}{\int }}\frac{H({\mu }^{\prime })}{\mu +{\mu }^{\prime }}d{\mu }^{\prime }=\frac{H(\mu )-1}{\mu H(\mu )}-b({\alpha }_1-\mu {\alpha }_0)$

See also

• Chandrasekhar's X- and Y-function

External links

• MATLAB function to calculate the H function https://www.mathworks.com/matlabcentral/fileexchange/29333-chandrasekhar-s-h-function