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The general solution of the Laplace equation in spherical coordinates is (independant of $\phi$):

$$V(r,\theta ) = \sum ^{\infty} _{l=0} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l (\cos \theta ))$$

An example (Griffiths EM, 3rd edition, example 3.6):

The potential $V_0 (\theta )$ is specified on the surface of a hollow sphere of radius $R$. Find the potential inside the sphere.

Here $B_l$ must be zero because it would blow up at the origin, this is clear, but the next step loses me.

The Legendre polynomials constitute a complete set of functions on the interval $-1 \le x \le 1$ $(0\le \theta \le \pi)$

$$\int ^1 _{-1} P_l (x) P_{l'} (x) dx = \int ^\pi _0 P_l (\cos \theta ) P_{l'} (\cos \theta ) \sin \theta d\theta.$$

Because they are orthogonal functions if $l\ne l'$ the integral is zero, how ever if they are equal, we get an answer of $\frac{2}{2L+1}$, where did this answer come from? Griffiths skips this step.

Why was $\sin$ introduced into the second integral?

I think the Legendre polynomial is tripping me up. How are these dealt with under integrals?

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Feb 11 '14 at 23:09
  • $\begingroup$ I had thought about posting it there, although the question about boundary conditions would probably not be answered. $\endgroup$ – Astrum Feb 11 '14 at 23:22
  • $\begingroup$ The Legendre polynomials are standard orthogonal polynomials. They satisfy a lot of properties such as they solve a second-order differential equation, satisfy recursion relations and so on. Their normalization is decided by convention. For your computation, you need to take the normalization as given to you and impose your boundary conditions. The completeness is important and lets you work out the implications of the boundary conditions on the coefficients $A_l$ and $B_l$. $\endgroup$ – suresh Feb 12 '14 at 0:25
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To prove that Legendre polynomials are orthogonal, i.e. $$ \int_{-1}^1 P_n(x)P_m(x)\,\text{d}x = \frac{2\,\delta_{m,n}}{2n+1} $$ you can use Rodrigues' formula (eq 3.62 in Griffiths): $$ P_n(x) = \frac{1}{2^n n!}\left(\frac{\text{d}}{\text{d}x}\right)^n (x^2-1)^n. $$ We get $$ I_{mn} = \int_{-1}^1 P_n(x)P_m(x)\,\text{d}x = \int_{-1}^1 \frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^m \!X^m\,\left(\frac{\text{d}}{\text{d}x}\right)^n \!X^n\,\text{d}x, $$ where I used the short-hand notation $X = x^2-1$. Without loss of generality, we can assume that $n\geqslant m$. Let us integrate this by parts: $$ \begin{multline} I_{mn} = \left[\frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^m \!X^m \,\left(\frac{\text{d}}{\text{d}x}\right)^{n-1} \!X^n\right]_{-1}^1 \\- \int_{-1}^1 \frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^{m+1} \!X^m\,\left(\frac{\text{d}}{\text{d}x}\right)^{n-1} \!X^n\,\text{d}x. \end{multline} $$ However, since the derivative $$ \left(\frac{\text{d}}{\text{d}x}\right)^{n-1} (x^2-1)^n $$ will contain a factor $(x^2-1)$, the integrated part will vanish, so that only the integral remains: $$ I_{mn} = -\int_{-1}^1 \frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^{m+1} \!X^m\,\left(\frac{\text{d}}{\text{d}x}\right)^{n-1} \!X^n\,\text{d}x. $$ In this manner, we can integrate by parts $n$ times, until we obtain $$ I_{mn} = (-1)^n\int_{-1}^1 \frac{(x^2-1)^n}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^{m+n} \!(x^2-1)^m\,\text{d}x. $$ Now, $$ (x^2-1)^m = x^{2m} - mx^{2m-2} + \ldots $$ So that, if $n>m$, $$ \left(\frac{\text{d}}{\text{d}x}\right)^{m+n} \!(x^2-1)^m = 0, $$ and $I_{mn}=0$.

On the other hand, if $n=m$, we get $$ \left(\frac{\text{d}}{\text{d}x}\right)^{n+n} \!(x^2-1)^n = (2n)! $$ and $$ I_{nn} = (-1)^n\int_{-1}^1 \frac{(2n)!}{2^{2n} n!\,n!}(x^2-1)^n\,\text{d}x. $$ Substituting $x$ with $y = (x+1)/2$, we can rewrite this as $$ I_{nn} = 2\int_0^1 \frac{(2n)!}{n!\,n!}y^n(1-y)^n\,\text{d}y, $$ which is a beta function: $$ I_{nn} = 2\frac{(2n)!}{n!\,n!}B(n+1,n+1) = 2\frac{(2n)!}{n!\,n!}\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)} = 2\frac{(2n)!}{(2n+1)!} = \frac{2}{2n+1}, $$ and this concludes the proof.

With $x=\cos\theta$, you can write it as $$ \begin{multline} \int_{-1}^1 P_n(x)P_m(x)\,\text{d}x = \int_{-1}^1 P_n(\cos\theta)P_m(\cos\theta)\,\text{d}\cos\theta \\= \int_{0}^\pi P_n(\cos\theta)P_m(\cos\theta)\sin\theta\,\text{d}\theta. \end{multline} $$ Because of the orthogonality property, you can find the coefficients by integrating $V(r,\theta)$ with each Legendre polynomial: $$ \int_{0}^\pi V(r,\theta)P_l(\cos\theta)\sin\theta\,\text{d}\theta = \frac{2}{2l+1}\left(A_lr^l + \frac{B_l}{r^{l+1}}\right). $$

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For actually working out problems in Griffiths, it is sufficient to take the orthogonality relation as given, as suresh's comment suggests.

If you want to go a tiny bit deeper, you can take them Legendre polynomials as given in either Griffiths or Wikipedia, and plug them into the orthogonality integral, and actually work out the integral yourself to confirm the orthogonality for a few members of the sequence. Obviously, you can't do that for all the (infinite) members of the sequence, so that isn't really a formal proof of the relation. But it might help you see how the polynomials work.

The real reason the Legendre polynomials follow that normalization relation is because the Legendre equation is a form of the [Sturm-Lioville] equation. In Sturm-Liouville theory, you can write the orthogonality relation in terms of the $p(x), q(x)$, and $w(x)$ functions. You can then identify what those functions actually are for the Legendre equation, and directly plug them in to get the orthogonality relation that Griffiths cites. The actual Sturm-Liouville theory is typically left for math methods books like Arfken to explain.

The $\sin\theta$ factor comes from the substitution $x = \cos\theta, dx = \sin\theta\ d\theta$. This is just a matter of switching between the $x$ variable (easier to work with in the mathematical theory) and the $\theta$ variable that corresponds to a physical variable of interest.

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    $\begingroup$ There is a lot of beautiful mathematics underlying Legendre polynomials (and more generally orthogonal polynomials) as Colin points out. However, that might be lost on someone studying Electrodynamics at the level of Griffiths. So my muted comment was meant to remove the fear of the new and unknown. :-) $\endgroup$ – suresh Feb 12 '14 at 3:09
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Why was $\sin$ introduced into the second integral?

Looks like $x = \cos\theta$. Note, then, that $dx= -\sin\theta d\theta$. This also seems to match the new limits of integrations.

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