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The spatial solution is written as $$\Phi_k(r) = r^{1-\frac{d}{2}} \left(c_1 J_{1-\frac{d}{2}}(k r) + c_2 Y_{-1+\frac{d}{2}}(kr)\right).$$ In the case $d=3$, the solutions can be written as $$\Phi_k(r) = \hat{c}_1 j_0(k r) + \hat{c}_2 y_0(k r) = \hat{c}_1(k) \frac{\sin k r}{r} + \hat{c}_2(k) \frac{\cos k r}{r},$$ where $j_0$ and $y_0$ are the Spherical Bessel Functions.

My question is how $j_0$ and $y_0$ are co-related with $J_{1-\frac{d}{2}}$ and $J_{1+\frac{d}{2}}$. Canonical explanation with book reference will be helpful for me. Thanks in advance.

Reference post:Solution of a differentiation in integral form

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So let's look at what you have when $d=3$. When $d=3$,

$$\Phi_k(r)=r^{-\frac{1}{2}}\left(c_1J_{-\frac{1}{2}}(kr) + c_2Y_{\frac{1}{2}}(kr)\right).$$

In similar fashion to your previous post, we know that $J_{-\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}}\cos x $. You can prove this pretty easily using the exact same techniques as myself and the other user that answered. We also know that

$$Y_{\nu}(x) = \frac{J_{\nu}(x)\cos(\nu\pi)-J_{-\nu}(x)}{\sin(\nu\pi)}.$$

Letting $\nu = \frac{1}{2}$ we have that $Y_{\frac{1}{2}}(x) = -J_{-\frac{1}{2}}(x)$. So we see that in the case of $d=3$ that the only solution we need to concern ourselves with is $J_{-\frac{1}{2}}$.

Pursuing this, we have

$$\Phi_k(r) = Cr^{-\frac{1}{2}}J_{-\frac{1}{2}}(kr),$$

where $C = c_1-c_2$. This gives us that

$$\Phi_k(r) = Cr^{-\frac{1}{2}}\sqrt{\frac{2}{\pi kr}}\cos(kr) = -C'(k)\frac{\cos(kr)}{kr} = C'(k)y_0(kr).$$

Here I've defined $C'(k) = -\frac{C}{\sqrt{k}}$.

In the general case, $j_n(x) = \sqrt{\frac{2}{\pi x}}J_{n+\frac{1}{2}}(x)$ and $y_n(x) = \sqrt{\frac{2}{\pi x}}Y_{n+\frac{1}{2}}(x).$ From this it's clear that your solution can be written in terms of these since your Bessel functions are evaluated at half integers. I have to get to a seminar so if you have questions, I will address them later.

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  • $\begingroup$ Got it thanks. :-) $\endgroup$ – Complex Guy Sep 10 '13 at 19:15
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Form Arfken's Mathematical Methods for Physicist:

When the Helmholtz equation is separated in spherical coordinates, the radial equation has the form $$ R'' + \frac{2}{r} R' + \left(k^2 - \frac{n(n+1)}{r^2}\right)R = 0. \tag{1} $$ If we substitute $$ R(k r) = \frac{Z(k r)}{(k r)^{1/2}} $$ equation (1) becomes $$ Z'' + \frac{1}{r} Z' + \left(k^2 - \frac{(n + \frac{1}{2})^2}{r^2}\right)Z = 0, \tag{2} $$ which is Bessel equation. $Z$ is a Bessel function of order $n + \frac{1}{2}$ ($n$ an integer). Because of the importance of spherical coordinates, this combination, $$ \frac{Z_{n+1/2}(k r)}{(k r)^{1/2}} $$ occurs quite often.

Is convenient to label these functions as spherical Bessel functions with the following defining equations \begin{align} j_n(x) &= \sqrt{\frac{\pi}{2 x}} J_{n+1/2}(x), \\ \\ y_n(x) &= \sqrt{\frac{\pi}{2 x}} Y_{n+1/2}(x). \end{align}

Any book on Bessel functions or Special functions will have detailed relations and properties. Specifically, being $$ J_{n+1/2}(x) = \sum_{s=0}^\infty \frac{(-1)^s}{s!(s+n+\frac{1}{2})!}\left(\frac{x}{2}\right)^{2s + n + 1/2} $$ it's clear that $$ j_0(x) = \sqrt{\frac{\pi}{2 x}} \sum_{s=0}^\infty \frac{(-1)^s}{s!(s+\frac{1}{2})!}\left(\frac{x}{2}\right)^{2s + 1/2} = \frac{\sin x}{x}, $$ where we have used the Legendre duplication formula $$ s!(s+\tfrac{1}{2})! = \frac{\sqrt{\pi}}{2^{2 s + 1}}(2 s + 1)!. $$

Analogously, $$ y_0(x) = -\frac{\cos x}{x}. $$

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  • $\begingroup$ Thanks your addition makes me more comfortable to get. $\endgroup$ – Complex Guy Sep 10 '13 at 19:15

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