Expansion of Green's Function in Spherical Coordinates in Jackson

Question

So I was reading about the expansion of the Green function in Spherical coordinates from Classical Electromagnetism by J.D. Jackson and I'm really confused about a subtle step that he makes to go from equations $(3.117)$ and $(3.118)$ to equation $(3.119)$ (the image is attached at the bottom of the question). Summarising the discussion, since we can expand any function of $(r,\theta, \varphi)$ in terms of the Spherical Harmonics $Y_{lm}(\theta,\varphi)$ and the radial function $U_{lm}(r)$ as - $$F(r,\theta,\varphi)=\sum_{l=0}^\infty \sum_{m=-l}^{+l} (A_{lm}r^{l} +B_l r^{-(l+1)})Y_{lm}(\theta,\varphi) \tag{1}$$

then I can also write the Green function $G(\textbf{x},\textbf{x}')$ assuming that $\textbf{x}'$ is a fixed point and acts as some parameter. Then, I can expand the Green function too in terms of $Y_{lm}(\theta,\varphi)$ and $U_{lm}(r)$ as - $$G(\textbf{x},\textbf{x}')=\sum_{l=0}^\infty \sum_{m=-l}^{+l} A_{lm} (r|r',\theta', \varphi') Y_{lm}(\theta,\varphi) \tag{2} $$ where $A_{lm}$'s are the expansion coefficients with the $\textbf{x}'$ dependence wrapped into them ($r',\theta', \varphi'$ act like parameters in this expansion). Now, Jackson says that using the equation ${\nabla_{r}}^2 G(\textbf{x},\textbf{x}')=-\delta(\textbf{x}-\textbf{x}'$), and expanding both sides in terms of $Y_{lm}$'s and comparing the coefficients, we get that- $$A_{lm} (r|r',\theta', \varphi') =g_l (r,r'){Y_{lm}}^{*}(\theta ', \varphi ') \tag{3}$$ which I absolutely do not understand. There was no context at all about what $g_l$ is or why we could factorize $A_{lm}$ in this way. In fact, when I tried to do what Jackson says, i.e. expanding both sides of the Green Function equation and equating the coefficients, I instead get a differential equation as follows- $${\nabla_r}^2 A_{lm}(r|r',\theta', \varphi') - \frac{l(l+1)}{r^2}A_{lm}(r|r',\theta', \varphi')= -\frac{1}{r^2} \delta{(r-r')} {Y_{lm}}^{*} (\theta ', \varphi ') \tag{4}$$

and I have no clue how $(3)$, the separation Jackson quotes comes from the above differential equation $(4)$. I'm really stuck on this part for days and I've looked at various online resources, and the usual thing that is done in most lecture notes is that it is often assumed that you can separate the $A_{lm}$ but it logically does not make very much sense to me. Any sort of help would be really appreciated. Thanks!

Edit: By secavara's suggestion, I expanded the coefficient $A_{lm}(r|r',\theta', \varphi')$ in terms of the conjugate of spherical harmonics as (expanding with respect to $r',\theta',\phi'$ assuming r to be some parameter)- $$A_{lm}(r|r',\theta',\phi') = \sum_{\hat{l}=0}^\infty \sum_{\hat{m}=-\hat{l}}^\hat{l} H_{\hat{l}\hat{m}}(l,m,r,r') Y^*_{\hat{l}\hat{m}}(\theta',\phi') \tag{5}$$

Subbing this into the differential equation $(4)$ that I got, I get $$\implies \sum_{\hat{l}=0}^{\infty} \sum_{\hat{m}=-\hat{l}}^{+\hat{l}} \bigg( {\nabla_r}^2 H_{\hat{l}\hat{m}} -\frac{l(l+1)}{r^2} H_{\hat{l}\hat{m}} \bigg) {Y_{\hat{l} \hat{m}}}^{*} (\theta' ,\phi') = -\frac{\delta(r-r')}{r^2} {Y_{lm}}^{*} (\theta', \phi') \tag{6}$$

Now if we compare the coefficients of LHS with RHS, we see that-

$$\bigg( {\nabla_r}^2 H_{\hat{l}\hat{m}} -\frac{l(l+1)}{r^2} H_{\hat{l}\hat{m}} \bigg) =0 $$

for all $\hat{l} \neq l$ or $\hat{m} \neq m $. But this merely suggests that coefficients other than $H_{ lm}$ satisfy another differential equation and not that they are $\textbf 0$. Any further suggestions would be appreciated!

From Jackson Page 120

Answer 1

This is basically a compilation of the discussion I had with secavara and Anonjohn in the comment section so all credits go to them. Thanks, everyone!

So, starting from the differential equation $(4)$ which we get on substituting the expanded form of $G(\textbf{x},\textbf{x}')$ into the Green function equation ${\nabla_{r}}^2 G(\textbf{x},\textbf{x}')=-\delta(\textbf{x}-\textbf{x}')$ as
$${\nabla_r}^2 A_{lm}(r|r',\theta', \varphi') - \frac{l(l+1)}{r^2}A_{lm}(r|r',\theta', \varphi')= -\frac{1}{r^2} \delta{(r-r')} {Y_{lm}}^{*} (\theta ', \varphi ') \tag{1.1}$$

As suggested by secavara, just like we expanded $G(\textbf{x},\textbf{x}')$ in $(r,\theta,\phi)$ coordinates with the $\textbf{x}'$ dependence wrapped into the expansion coefficients $A_{lm}$'s , we can similarly expand $A_{lm}(r|r',\theta',\phi')$ by using the fact that as a function of $\theta′$ and $\phi′$, it can be expanded in terms of spherical harmonics (or their conjugates, for that matter) as follows- $$A_{lm}(r|r',\theta',\phi')=\sum_{\tilde{l}=0}^\infty \sum_{\tilde{m}=-\tilde{l}}^{\tilde{l}}H_{\tilde{l}\tilde{m}} (l,m,r,r') {Y_{\tilde{l}\tilde{m}}}^*(\theta',\phi') \tag{1.2}$$

Substituting the form of $A_{lm}$ from equation $(1.2)$ into the differential equation in $(1.1)$, the differential equation becomes-

$$\bigg({\nabla_r}^2-\frac{l(l+1)}{r^2}\bigg)\bigg(\sum_{\tilde{l}=0}^\infty \sum_{\tilde{m}=-\tilde{l}}^{\tilde{l}}H_{\tilde{l}\tilde{m}} (l,m,r,r') {Y_{\tilde{l}\tilde{m}}}^*(\theta',\phi') \bigg)=-\frac{\delta(r-r')}{r^2} {Y_{lm}}^{*} (\theta', \phi')$$

$$\implies \sum_{\tilde{l}=0}^{\infty} \sum_{\tilde{m}=-\tilde{l}}^{+\hat{l}} \bigg( {\nabla_r}^2 H_{\tilde{l}\tilde{m}} -\frac{l(l+1)}{r^2} H_{\tilde{l}\tilde{m}} \bigg) {Y_{\tilde{l} \tilde{m}}}^{*} (\theta' ,\phi')=-\frac{\delta(r-r')}{r^2} {Y_{lm}}^{*} (\theta', \phi') \tag{1.3}$$

By comparing the coefficients in LHS and RHS of equation $(1.3)$, we see that for all $\tilde{l}\neq l$ or $\tilde{m} \neq m$, we get the following condition-

$$\bigg({\nabla_r}^2 - \frac{l(l+1)}{r^2}\bigg) H_{\tilde{l} \tilde{m}}=0 \qquad \forall \tilde{l}\neq l, \tilde{m} \neq m \tag{1.4}$$

Now the critical physical input that is required in this problem, as was pointed out by Anonjohn was that the Dirichlet boundary conditions for the Green function require that it vanishes at infinity which in turn implies that $H_{\tilde{l} \tilde{m}} \rightarrow 0$ as $|r|\rightarrow \infty$. Since the differential equation still holds when we analyze its asymptotic behavior at $|r|\rightarrow \infty$, we get that - $${\nabla_r}^2 H_{\tilde{l} \tilde{m}}=0 \implies \frac{d^2}{dr^2}(H_{\tilde{l}\tilde{m}})=0 \implies H_{\tilde{l}\tilde{m}}=ar+b \qquad \text{(asymptotic behaviour)}\tag{1.5}$$

Thus, requiring that the green's function vanishes at infinity is tantamount to setting $a,b$ to $0$ which implies $H_{\tilde l \tilde m} =0$ for all $\tilde{l}\neq l$ or $\tilde{m} \neq m$. This trivially implies that the expansion of $A_{lm}$, as can be seen from equation $(1.2)$, would look like- $$\boxed{A_{lm}(r|r',\theta',\phi')= H_{{l}{m}} (r,r') {Y_{\tilde{l}\tilde{m}}}^*(\theta',\phi') = g_l(r,r'){Y_{\tilde{l}\tilde{m}}}^*(\theta',\phi')} \tag{1.6}$$

since all the other $H_{\tilde{l}\tilde{m}}$'s are zero. Hence, in $(1.6)$, we get our desired factorization if we call $H_{lm}(r,r')\equiv g_l(r,r')$ (since it ultimately depends only on $l$ and not $m$ as can be seen from the differential equation). The differential equation satisfied by $g_l$ can be obtained on substituting the form of $(1.6)$ into the differential equation of the Green function to get-

$$\boxed{\frac{1}{r^2} \frac{d^2}{dr^2}(rg_l(r,r'))-\frac{l(l+1)}{r^2}g_l(r,r')=-\frac{\delta(r-r')}{r^2}} \tag{1.7}$$

Edit: I discussed this with some of my friends and a flaw in this derivation is the assumption that the Green function does vanish at infinity. In the general case, there is no mention of the outer boundary being at infinity. The region of interest can have an inner boundary of radius $R_1$ and an outer boundary of radius $R_2$, and then the asymptotic analysis doesn't really make sense. So, I am still leaving the question as unanswered so that someone else could provide their suggestions on this too.

Answer 2

The factorization in equation $(3)$ is not justified in general.

In fact, there are boundary conditions for which the Green function cannot have the form given in Jackson. This expansion is valid in certain cases such as when the Green function is required to vanish on the entire boundary. These are the kinds of cases that Jackson discusses in this section, so he probably only had that in mind.

Suppose we have a boundary value problem where we require the potential to equal $V$ on the sphere of radius $a \gt 0$, where $V$ is given by $V(\theta, \varphi) = \sum\limits_{l=0}^\infty\sum\limits_{m=-l}^l 2^{-l}Y_{lm}(\theta, \varphi)$, and we are interested in the potential in the interior of the ball of radius $a$.

Using the Green function expansion assumed in Jackson and substituting $r = a$, we have

$$V(\theta, \varphi) = G(a, \theta, \varphi, r', \theta', \varphi') = \sum\limits_{l=0}^\infty \sum\limits_{m=-l}^l A_{lm}(\theta',\varphi')g_l(a, r')Y_{lm}(\theta, \varphi)$$

Then using orthogonality we have

$$A_{lm}(\theta', \varphi')g_l(a, r') = \int_0^{2\pi}\int_0^\pi V(\theta,\varphi)Y_{lm}^*(\theta, \varphi)\sin\theta\,d\theta\,d\varphi = 2^{-l}$$

This implies $g_l(a, r') \neq 0$ so dividing by it we get $$A_{lm}(\theta', \varphi') = \frac{1}{2^l g_l(a, r')}$$

The LHS is purely a function of $\theta'$ and $\varphi'$ while the RHS is purely a function of $r'$, so both sides must be constant.

Therefore, we see that $A_{lm}(\theta', \varphi') = c_l$ is a nonzero constant depending only on $l$.

Substituting back into the Green function expansion, we get

$$G(r, \theta, \varphi, r', \theta', \varphi') = \sum\limits_{l=0}^\infty c_l g_l(r, r')\sum\limits_{m=-l}^l Y_{lm}(\theta, \varphi)$$

But this is absurd since it does not even depend on $\theta', \varphi'$ and contradicts the requirement that $A_{lm}(\theta', \varphi') = Y_{lm}^*(\theta', \varphi')$ which was deduced from the delta function expansion. It will not be able to satisfy the equation $\nabla^2_{\mathbf{x}} G = -4\pi\delta(\mathbf{x}-\mathbf{x}')$.

So you are correct that this expansion is missing terms, and does not work for all such boundary value problems.

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But if we force $G(\mathbf{x}, \mathbf{x}') = 0$ on the entire boundary (which is what Jackson assumes in the sentence after equation (3.116)) then the expansion is valid. We further assume that our boundaries are always spherical, and in the general case consist of a sphere of radius $a$ and a sphere of radius $b$ with $0 \lt a \lt b$.

In that case, starting from your expansions (2) and (5), which are completely general, and substituting into the Poisson equation with the spherical harmonic expansion of $\delta(\mathbf{x} - \mathbf{x}')$, we get

$$A_{lm}(r | r', \theta', \varphi') = \sum\limits_{l'=0}^\infty\sum\limits_{m'=-l'}^{l'} H_{lml'm'}(r, r') Y_{l'm'}^*(\theta',\varphi')$$ $$\frac{1}{r}\frac{\partial^2}{\partial r^2}\left(r H_{lml'm'}(r, r')\right) - \frac{l(l+1)}{r^2}H_{lml'm'}(r, r') = 0\text{ if }(l', m') \neq (l, m)$$ $$\frac{1}{r}\frac{\partial^2}{\partial r^2}\left(r H_{lmlm}(r, r')\right) - \frac{l(l+1)}{r^2}H_{lmlm}(r, r') = -\frac{4\pi}{r^2}\delta(r - r')$$

The first equation has the general solution $$H_{lml'm'}(r, r') = B_{lml'm'}(r')r^l + C_{lml'm'}(r')r^{-(l+1)}\text{ if }(l', m') \neq (l, m)$$

Now we impose the boundary condition that $G$ vanishes at $r = a$ and $r = b$.

By orthogonality of the spherical harmonics $Y_{lm}(\theta, \varphi)$ we see that $A_{lm}(a|r',\theta',\varphi') = 0$ and then by orthogonality of the spherical harmonics $Y_{l'm'}(\theta',\varphi')$ we see that $H_{lml'm'}(a, r') = 0$ when $(l', m') \neq (l, m)$. Similarly, $H_{lml'm'}(b, r') = 0$ when $(l', m') \neq (l, m)$.

Therefore we have the system $$B_{lml'm'}(r')a^l + C_{lml'm'}(r')a^{-(l+1)} = 0$$ $$B_{lml'm'}(r')b^l + C_{lml'm'}(r')b^{-(l+1)} = 0$$ Its determinant is nonzero since $0 \lt a \lt b$ so it has the unique solution $$B_{lml'm'}(r') = 0$$ $$C_{lml'm'}(r') = 0$$ which holds when $(l', m') \neq (l, m)$.

Therefore, $H_{lml'm'}(r, r') = 0$ when $(l', m') \neq (l, m)$.

This means that our expansion for $A_{lm}(r|r',\theta',\varphi')$ simplifies to $$A_{lm}(r|r',\theta',\varphi') = H_{lm}(r, r')Y_{lm}^*(\theta',\varphi')$$ where $H_{lm}(r, r') = H_{lmlm}(r, r')$ for simplicity.

This is almost the factorization given in Jackson, except that we still have a dependence on $m$ in the radial factor. Showing that there is no dependence on $m$ takes longer, this is the next part.

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We now analyze the second radial equation: $$\frac{1}{r}\frac{\partial^2}{\partial r^2}\left(r H_{lm}(r, r')\right) - \frac{l(l+1)}{r^2}H_{lm}(r, r') = -\frac{4\pi}{r^2}\delta(r - r')$$

The general solution is $$H_{lm}(r, r') = \begin{cases} A_{lm}(r')r^l + B_{lm}(r')r^{-(l+1)} & r \lt r' \\ A_{lm}'(r')r^l + B_{lm}'(r')r^{-(l+1)} & r \gt r' \\ \end{cases} $$

We can write this more compactly with the Heaviside step function: $$H_{lm}(r, r') = [1 - H(r - r')]\left[A_{lm}(r')r^l + B_{lm}(r')r^{-(l+1)}\right] + H(r - r')\left[A_{lm}'(r')r^l + B_{lm}'(r')r^{-(l+1)}\right]$$

We need to substitute this in the radial equation to find a relationship between $A_{lm}(r')$ and $B_{lm}(r')$ that makes the equation also hold at $r = r'$ in the sense of distributions.

First we compute $$r H_{lm}(r, r') = [1 - H(r - r')]\left[A_{lm}(r')r^{l+1} + B_{lm}(r')r^{-l}\right] + H(r - r')\left[A_{lm}'(r')r^{l+1} + B_{lm}'(r')r^{-l}\right]$$

$$\frac{\partial}{\partial r}\left(r H_{lm}(r, r')\right) =$$ $$-\delta(r - r')\left[A_{lm}(r')r'^{l+1} + B_{lm}(r')r'^{-l}\right] + [1 - H(r - r')]\left[(l+1)A_{lm}(r')r^l - lB_{lm}(r')r^{-(l+1)}\right]$$ $$+\delta(r - r')\left[A_{lm}'(r')r'^{l+1} + B_{lm}'(r')r'^{-l}\right] + H(r - r')\left[(l+1)A_{lm}'(r')r^l - lB_{lm}'(r')r^{-(l+1)}\right]$$

$$\frac{\partial^2}{\partial r^2}\left(r H_{lm}(r, r')\right) =$$ $$-\delta'(r - r')\left[A_{lm}(r')r'^{l+1} + B_{lm}(r')r'^{-l}\right] - \delta(r-r')\left[(l+1)A_{lm}(r')r'^l - lB_{lm}(r')r'^{-(l+1)}\right]$$ $$+l(l+1)[1 - H(r - r')]\left[A_{lm}(r')r^{l-1} + B_{lm}(r')r^{-(l+2)}\right]$$ $$+\delta'(r - r')\left[A_{lm}'(r')r'^{l+1} + B_{lm}'(r')r'^{-l}\right] + \delta(r - r')\left[(l+1)A_{lm}'(r')r'^l - lB_{lm}'(r')r'^{-(l+1)}\right]$$ $$+l(l+1)H(r - r')\left[A_{lm}'(r')r^{l-1} + B_{lm}'(r')r^{-(l+2)}\right]$$ where I used the identity $\delta(r - r')f(r, r') = \delta(r - r')f(r', r')$.

Then we get $$r\frac{\partial^2}{\partial r^2}\left(r H_{lm}(r, r')\right) - l(l+1)H_{lm}(r, r') = $$ $$+\delta'(r - r')\left\{\left[A_{lm}'(r') - A_{lm}(r')\right]r'^{l+2} + \left[B_{lm}'(r') - B_{lm}(r')\right]r'^{-(l-1)}\right\} + \delta(r - r')\left\{(l+1)\left[A_{lm}'(r') - A_{lm}(r')\right]r'^{l+1} - l\left[B_{lm}'(r') - B_{lm}(r')\right]r'^{-l}\right\}$$

We want this to equal $-4\pi\delta(r - r')$ so we need

$$\left[A_{lm}'(r') - A_{lm}(r')\right]r'^{l+2} + \left[B_{lm}'(r') - B_{lm}(r')\right]r'^{-(l-1)} = 0$$ $$(l+1)\left[A_{lm}'(r') - A_{lm}(r')\right]r'^{l+1} - l\left[B_{lm}'(r') - B_{lm}(r')\right]r'^{-l} = -4\pi$$

Using Cramer's rule, this system has solution $$A_{lm}'(r') - A_{lm}(r') = -\frac{4\pi}{2l+1}r'^{-(l+1)}$$ $$B_{lm}'(r') - B_{lm}(r') = \frac{4\pi}{2l+1}r'^l$$

Therefore we can eliminate $A_{lm}'(r')$ and $B_{lm}'(r')$ to obtain $$H_{lm}(r, r') = A_{lm}(r')r^l + B_{lm}(r')r^{-(l+1)} - \frac{4\pi}{2l+1}H(r - r')\left(\frac{r^l}{r'^{l+1}} - \frac{r'^l}{r^{l+1}}\right)$$

Finally we substitute $r = a$ and $r = b$ and use the vanishing boundary condition to obtain for $a \lt r' \lt b$:

$$A_{lm}(r')a^l + B_{lm}(r')a^{-(l+1)} = 0$$ $$A_{lm}(r')b^l + B_{lm}(r')b^{-(l+1)} = \frac{4\pi}{2l+1}\left(\frac{b^l}{r'^{l+1}} - \frac{r'^l}{b^{l+1}}\right)$$

Using Cramer's rule again, we get $$A_{lm}(r') = \frac{4\pi}{2l+1}\frac{b^{-(l+1)}}{b^{2l+1} - a^{2l+1}}\left(\frac{b^l}{r'^{l+1}} - \frac{r'^l}{b^{l+1}}\right)$$

$$B_{lm}(r') = -\frac{4\pi}{2l+1}\frac{a^{2l+1}b^{l+1}}{b^{2l+1} - a^{2l+1}}\left(\frac{b^l}{r'^{l+1}} - \frac{r'^l}{b^{l+1}}\right)$$

We see that $A_{lm}(r')$ and $B_{lm}(r')$ do not depend on $m$. So we can write $A_{lm}(r') = A_l(r')$ and $B_{lm}(r') = B_l(r')$. Similarly we have $A_{lm}'(r') = A_l'(r')$ and $B_{lm}'(r') = B_l'(r')$.

We didn't need to solve the equations explicitly, we could have just noticed that both the boundary conditions and the constraint of the delta function at $r = r'$ involve some conditions on $A_{lm}(r'), A_{lm}'(r'), B_{lm}(r'), B_{lm}'(r')$ that do not involve $m$ explicitly.

Anyway, after all this work, we finally conclude that $H_{lm}(r, r') = H_l(r, r')$ is independent of $m$, so our factorization is $A_{lm}(r|r', \theta', \varphi') = H_l(r, r')Y_{lm}^*(\theta',\varphi')$ which is what is in Jackson.