# What is the physical meaning/concept behind Legendre polynomials?

In mathematical physics and other textbooks we find the Legendre polynomials are solutions of Legendre's differential equations. But I didn't understand where we encounter Legendre's differential equations (physical example). What is the basic physical concept behind the Legendre polynomials? How important are they in physics? Please explain simply and give a physical example.

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en.wikipedia.org/wiki/… –  John Rennie Jan 24 '13 at 13:48
These polynomials are not really physics, they are simply a useful mathematical tool that appear in the solutions to many physical problems with spherical symmetries. I think the question is fine because they do come up a lot. –  dmckee Jan 24 '13 at 14:55
There is a very definite physical notion behind Legendre polynomials: a rank-$l$ Legendre polynomial corresponds to a spin-$l$ representation of the orthogonal group $SO(3)$. These are the usual traceless, symmetric tensors of rank $l$ we use in field theory all the time. If in QM you scatter two spinless particles, you measure the angular distribution and find that it is described by $P_2(\cos \theta)$ (for example), then you can be sure that the particles are exchanging a spin-2 resonance. –  Vibert Jan 25 '13 at 0:01
@Vibert That is a physical notion attached to the polynomials, but the math exists independently of the physics. The distinction here is that physicists must learn the math but mathematicians can know the polynomials while being ignorant of the physics. –  dmckee Jan 25 '13 at 5:02

The Legendre polynomials occur whenever you solve a differential equation containing the Laplace operator in spherical coordinates with a separation ansatz (there is extensive literature on all of those keywords on the internet).

Since the Laplace operator appears in many important equations (wave equation, Schrödinger equation, electrostatics, heat conductance), the Legendre polynomials are used all over physics.

There is no (inarguable) physical concept behind the Legendre polynomials, they are just mathematical objects which form a complete basis between -1 and 1 (as do the Chebyshev polynomials).

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I disagree with the last statement (see my comment above). Legendre polynomials correspond to $SO(3)$ (tensor) representations that are well-known, even to undergraduates. Just look up the partial wave expansion in a QM textbook. The Chebyshev polynomials play the same role, but in two dimensions. –  Vibert Jan 25 '13 at 0:03
If you do partial wave expansion, you do that because the Legendre polynomials are eigenfunctions of the $\vartheta$ part of the Laplace operator. The connection to the $SO(3)$ representation is interesting though. –  Rafael Reiter Jan 25 '13 at 0:06
Yes, that is exactly what I mean: Legendre polynomials are eigenfunctions of the Laplacian on $S^2$, and indeed they correspond to representations of $SO(3)$ - that's not an accident. You are free to think that this fact isn't important, but it's the 3D equivalent of classifying the representations of the Lorentz group - you do agree that it's meaningful to talk about scalars, spinors, currents, tensors etc. in particle physics, right? –  Vibert Jan 25 '13 at 0:12
I know too little about particle physics to give a qualified answer. But I think we disagree on the more fundamental question of how "physically" you interpret a mathematical object, which is more of a philosophical question. –  Rafael Reiter Jan 25 '13 at 0:19

Here's my 30 seconds hand waving argument for

"Why is it that we always encounter new special functions $f_n$ with orthogonality relations??"

$$\int f^*_n\cdot f_m=\delta_{mn}$$

Super broadly speaking, in physics we dealing with the dynamics of certain degrees of freedom. These often employ smooth symmetries, that is we're dealing with Lie groups, which are also manifold in themselfs. Take e.g. the Laplacian $\Delta=\nabla\cdot\nabla$ and the associated symmetries $R$ acting as $\nabla\to R\nabla$ in such a way that that $R\nabla\cdot R\nabla=\nabla\cdot\nabla$.

Now in case one is dealing with a "rotation" in the broadest sense of the word, one often has a compact manifold, where we can savagely define things like integration on the group, and these symmetry groups also permit pretty unitary matrix representations. That is there are necessarily matrices $U$ with

$$\sum_kU_{kn}^*U_{km}=\delta_{mn},$$

and well, the matrix coefficients $U_{kn}$ must be some complex functions.

To put it short, special functions are representation theory magic.

@zonk: Yes, it's the default theory. But of course, you only see the direct relation to special functions if you take the abstract Lie group theory and actually sit down and write down the matrices in some base. E.g. for the rotation group matrices $D$, you find

$$\begin{array}{lcl} D^j_{m'm}(\alpha,\beta,\gamma)&=& e^{-im'\alpha } [(j+m')!(j-m')!(j+m)!(j-m)!]^{1/2} \sum\limits_s \left[\frac{(-1)^{m'-m+s}}{(j+m-s)!s!(m'-m+s)!(j-m'-s)!} \right.\\ &&\left. \cdot \left(\cos\frac{\beta}{2}\right)^{2j+m-m'-2s}\left(\sin\frac{\beta}{2}\right)^{m'-m+2s} \right] e^{-i m\gamma} \end{array}.$$

Very sweet, right? Now here you have the Legendre Polynomials $P_\ell^m$

$$D^{\ell}_{m 0}(\alpha,\beta,0) = \sqrt{\frac{4\pi}{2\ell+1}} Y_{\ell}^{m*} (\beta, \alpha ) = \sqrt{\frac{(\ell-m)!}{(\ell+m)!}} \, P_\ell^m ( \cos{\beta} ) \, e^{-i m \alpha }$$

so that

$$\int_0^{2\pi} d\alpha \int_0^\pi \sin \beta d\beta \int_0^{2\pi} d\gamma \,\, D^{j'}_{m'k'}(\alpha,\beta,\gamma)^\ast D^j_{mk}(\alpha,\beta,\gamma) = \frac{8\pi^2}{2j+1} \delta_{m'm}\delta_{k'k}\delta_{j'j}.$$

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That is a very interesting view/line of argument. Do you find this in standard Lie groups literature? –  Rafael Reiter Jan 24 '13 at 15:47
...and how do the $f$'s relate to the $U$'s? –  Rafael Reiter Jan 24 '13 at 16:18