Associated Legendre Functions

The largest routine consumer of associated Legendre functions is the Earth itself. The global gravity model EGM2008 expands the geopotential in exactly this basis — coefficients multiplying $P_l^m(\cos\theta)\{\cos m\phi,\sin m\phi\}$ — to degree and order 2190: over four million coefficients, resolving the geoid to roughly 10 km. Every GPS orbit and GRACE-style groundwater map runs recursions for $P_l^m$ at thousands of degrees, where the factorial growth in (5) overflows double precision near $l\approx150$ — geodesy codes use fully normalized variants for precisely the reason our norm table started growing. The same functions shape technology across the size scale: antenna engineers expand radiation fields in spherical modes where $m$ counts the azimuthal lobes of the pattern, and the angular factors of the hydrogen orbitals — $p_x$, $p_y$ lobes ($P_1^1$ with $\cos\phi$, $\sin\phi$), four-leaf $d$ patterns ($P_2^2\cos2\phi$) — are associated Legendre functions wearing lab coats. The full story of that packaging begins in the next article.

Think of a mode on the sphere as a pattern of positive and negative patches, and of $(l,m)$ as its address. The degree $l$ is the total node budget: an $(l,m)$ pattern has exactly $l$ nodal lines. The order $m$ says how the budget is spent: the factor $\cos m\phi$ (the real face of $e^{im\phi}$) vanishes on $2m$ meridians — longitude waves — while $P_l^m(\cos\theta)$ supplies the meridian structure, $l-|m|$ zeros between the poles, each a nodal parallel. Same $l$, different $m$: nodes migrate from latitude to longitude, one at a time.

Notice that every $m\neq0$ panel is pinned to zero at the poles — at the pole the coordinate $\phi$ is undefined, so a pattern that varies with $\phi$ can only be continuous there if its amplitude vanishes, and the factor $\sin^{|m|}\theta$ built into $P_l^m$ enforces exactly this.

Differentiation is a zero-eating operation — a degree-$l$ polynomial has $l$ zeros to spend, and the figure shows the $l=2$ family spending them. Push past $m=l$ and the polynomial itself is gone: that picture is the theorem $|m|\le l$, and the algebra below merely notarizes it.

Recall where Separation of Variables in Spherical Coordinates left us: writing $V=R(r)\Theta(\theta)\Phi(\phi)$, the radial factor came out as $Ar^l+Br^{-(l+1)}$ with separation constant $l(l+1)$; there we then set $\partial V/\partial\phi=0$ and got the Legendre equation. Now we keep $\Phi$ alive and the angular block splits once more. The azimuthal factor is

where: $m$ — the order (azimuthal index), an integer because the potential must be single-valued: $\phi$ and $\phi+2\pi$ label the same point, so $\Phi$ must close after one lap. Real combinations $\cos m\phi$, $\sin m\phi$ are equivalent currency. The polar factor then satisfies

where: $l(l+1)$ — the constant inherited from the radial split; $m^2/\sin^2\theta$ — the new centrifugal term: harmless in mid-latitudes, divergent at the poles, and taming it there is this article's drama. Substituting $x=\cos\theta$, as at $m=0$, turns (2) into the associated Legendre equation:

where: $x\in[-1,1]$, with $x=\pm1$ the poles, where the leading coefficient degenerates and the new term diverges. Set $m=0$ and (3) collapses to the ordinary Legendre equation, whose regular solutions $P_l$ — Rodrigues formula, orthogonality, generating function — are the property of Legendre Polynomials; we build on that catalogue, not re-derive it.

Solving (3) by series from scratch is unnecessary: the solutions can be manufactured from the Legendre polynomials we already own. The recipe, and our committed convention, is

where: $P_l$ — the Legendre polynomial of degree $l$; the prefactor $(1-x^2)^{|m|/2}=\sin^{|m|}\theta$ — precisely the pole-taming behaviour the centrifugal term demands. A word on signs first: many books (Jackson, most quantum texts) insert a phase $(-1)^m$ into (4) — the Condon–Shortley phase. Following Griffiths, we do not: our $P_1^1=+\sin\theta$ where Jackson writes $-\sin\theta$. Nothing physical changes — the phase cancels in any product of equal-$m$ functions — but cross-book transplants are not safe, and the Spherical Harmonics article will absorb the phase into $Y_l^m$, where quantum mechanics wants it. Cranking the derivatives of $P_1=x$, $P_2=\tfrac12(3x^2-1)$, $P_3=\tfrac12(5x^3-3x)$ gives the working table ($m=0$ rows are $P_l$):

Note what the table confesses: for odd $m$ these are not polynomials — the square root is irremovable — hence the honest name associated Legendre functions. Every entry has been substituted numerically back into (3); the $l=m=1$ case is verified by hand below. And each $l$ stops at $m=l$: the cutoff $|m|\le l$ gives each degree exactly $2l+1$ orders.

Boundary-value work runs on Fourier's trick, which runs on orthogonality. The associated functions deliver it family by family: fix $m$ (the $\phi$-integration will already have selected it), and the members $l=|m|,|m|+1,\dots$ are mutually orthogonal on $[-1,1]$ with weight one:

where: $\delta_{ll'}$ — Kronecker delta; the factorial ratio — the price of not normalizing (already $720$ for $P_3^3$). At $m=0$ this is exactly the Legendre norm $\tfrac{2}{2l+1}$. Note what (5) does not say: it compares equal-$m$ functions only — the $\phi$-integral has already separated different $m$.

One loose end: negative $m$. Equation (3) contains only $m^2$, and definition (4) — built on $|m|$ — makes the two signs identical: $P_l^{-m}\equiv P_l^{|m|}$, the entire $\pm m$ distinction living in $e^{\pm im\phi}$. (Other books make $P_l^{-m}$ a proportional multiple instead — see the FAQ.) Either way, negative $m$ brings no new functions. Assembling the pieces, the general charge-free potential with full angular dependence is the double sum

where: $A_{lm},B_{lm}$ — constants fixed by boundary data via the double orthogonality: the $\phi$-integral selects $m$, then (5) selects $l$. The angular blocks $P_l^m(\cos\theta)e^{im\phi}$ are begging to be packaged into single normalized objects — that packaging is the business of the next article, Spherical Harmonics, and we leave it there.

1 · Values, a ratio, and the orthogonality audit (confidence builder)

Evaluate $P_2^1$ and $P_3^1$ at $x=0.5$; verify numerically that the $l=2$ and $l=3$ members of the $m=1$ family are orthogonal and that the $l=2$ norm matches (5).

Step 1. Values from the table. $P_2^1(0.5)=3(0.5)\sqrt{0.75}=1.5\times0.86603=\boxed{1.29904}$ (exactly $3\sqrt3/4$), and $P_3^1(0.5)=\tfrac32(1.25-1)(0.86603)=0.375\times0.86603=\boxed{0.32476}$ (exactly $3\sqrt3/16$). The ratio $P_2^1(0.5)/P_3^1(0.5)=4$ exactly — the surds cancel, a good check on the polynomial parts.

Step 2. Orthogonality of $P_2^1$ and $P_3^1$: quadrature gives $\int_{-1}^{1}P_2^1P_3^1\,\mathrm{d}x=0.000000$ — but be honest about why this one is easy: $P_2^1$ is odd, $P_3^1$ is even, so parity alone kills the integral.

Step 3. So test a pair parity does not protect: $P_1^1$ and $P_3^1$ are both even, the integrand visibly nonzero pointwise, yet quadrature returns $\int_{-1}^{1}P_1^1P_3^1\,\mathrm{d}x=6.7\times10^{-16}$. Positive and negative regions cancel exactly — equation (5) doing real work.

Step 4. The norm: quadrature gives $\int_{-1}^{1}(P_2^1)^2\,\mathrm{d}x=2.400000$; the formula gives $\tfrac{2}{5}\cdot\tfrac{3!}{1!}=\tfrac{12}{5}=\boxed{2.4}$ ✓ — and watch the factorial ratio bite: $P_2^2$ already carries $\tfrac{48}{5}=9.6$.

2 · A grounded sphere in a tilted field — the moment m switches on (exam level)

A grounded conducting sphere of radius $R=10\ \mathrm{cm}$ sits in a uniform field $E_0=1000\ \mathrm{V/m}$ tilted at $\alpha=30^\circ$ from the $z$-axis, in the $x$–$z$ plane. Which $(l,m)$ modes appear? Find the surface charge density at $\theta=60^\circ$, $\phi=0$, and locate where the $m=1$ mode overtakes $m=0$.

Step 1. Diagnose the modes. Far away, $V\to-E_0(z\cos\alpha+x\sin\alpha)=-E_0r(\cos\alpha\cos\theta+\sin\alpha\sin\theta\cos\phi)$. Against the table: $\cos\theta=P_1(\cos\theta)$ is a $(1,0)$ shape and $\sin\theta\cos\phi=P_1^1(\cos\theta)\cos\phi$ is a $(1,\pm1)$ shape. Nothing else appears — the whole solution lives in $l=1$, the $m=\pm1$ modes switched on by the tilt at amplitude ratio $\tan\alpha=\tan30^\circ=0.577$ relative to $m=0$.

Step 2. Solve each mode. Both angular shapes pair with radial factors $r$ and $r^{-2}$, and grounding fixes the combination $\bigl(r-\tfrac{R^3}{r^2}\bigr)$ — the mechanics of the untilted problem from Separation of Variables in Spherical Coordinates, run twice in parallel:

At $r=R$ the bracket vanishes ($V=0$ on the sphere ✓); at large $r$ it reduces to the applied field ✓. By the uniqueness theorems, this is the answer.

Step 3. Surface charge. $\sigma=-\varepsilon_0\,\partial V/\partial r|_{r=R}$, the bracket differentiating to $3$ at $r=R$:

At $\theta=60^\circ$, $\phi=0$: $\cos30^\circ\cos60^\circ+\sin30^\circ\sin60^\circ=0.4330+0.4330=0.8660$, so $\sigma=3(8.854\times10^{-12})(1000)(0.8660)=\boxed{23.0\ \mathrm{nC/m^2}}$. The two modes contribute equally here: the weight of $m=1$ relative to $m=0$ on this meridian is the boundary-fit ratio $\tan\alpha\cdot\dfrac{P_1^1(\cos\theta)}{P_1(\cos\theta)}=\tan\alpha\tan\theta$, and at $60^\circ$ the ratio of the two associated Legendre values, $P_1^1/P_1=\tan60^\circ=1.732$, exactly cancels the tilt ratio: $0.577\times1.732=1.00$. Equatorward, $m=1$ dominates ($P_1(0)=0$); poleward, $m=0$ rules ($P_1^1(\pm1)=0$).

Step 4. Sanity check by rotation — the best kind. The point $(60^\circ,0)$ sits at $30^\circ$ from the field direction, so the rotated axially symmetric answer predicts $\sigma=3\varepsilon_0E_0\cos30^\circ=23.0\ \mathrm{nC/m^2}$ — identical ✓. The maximum $\sigma_{\max}=3\varepsilon_0E_0=26.6\ \mathrm{nC/m^2}$ sits where the field axis pierces the sphere; the induced dipole is $p=4\pi\varepsilon_0R^3E_0=1.11\times10^{-10}\ \mathrm{C\,m}$. The lesson cuts both ways: this symmetry-breaking is a rotation in disguise, which is why it stayed inside $l=1$ — a genuinely lopsided boundary (an off-axis point charge near the sphere, say) has no such disguise, feeds every $(l,m)$, and then the $P_l^m$ machinery is the only road.

References. Jackson, Classical Electrodynamics, 3rd ed., §3.5 · Griffiths, Introduction to Electrodynamics, 4th ed., §3.3.2 · Zangwill, Modern Electrodynamics, §7.5 · Arfken & Weber, Mathematical Methods for Physicists, ch. 15 · Abramowitz & Stegun, Handbook of Mathematical Functions, ch. 8

Provenance. Equations and conventions follow Jackson §3.5 (associated Legendre equation 3.9, Rodrigues 3.50, orthogonality 3.52, spherical harmonics 3.53) and Griffiths §3.3.2. A verifier should: (1) confirm the boxed ODE reduces to the Legendre equation at $m=0$; (2) check $P_1^1=-\sin\theta$, $P_2^1=-3\sin\theta\cos\theta$, $P_2^2=3\sin^2\theta$ from the definition; (3) verify the orthogonality normalization $2(\ell+m)!/[(2\ell+1)(\ell-m)!]$ for $\ell=1,m=1$; (4) confirm $Y_0^0=1/\sqrt{4\pi}$. All factorials and phases cross-checked against Arfken ch. 15 and Abramowitz & Stegun ch. 8. Convention note: unlike the Jackson-normalized source above, this article follows Griffiths in omitting the Condon–Shortley phase, so odd-$m$ entries flip sign ($P_1^1=+\sin\theta$ here); products, norms, and the equation are unaffected; the $Y_l^m$ item defers to Spherical Harmonics.

Builds on: Legendre Polynomials · Separation of Variables in Spherical Coordinates · Laplace's Equation · Boundary Conditions for the Potential Leads to: Spherical Harmonics · Multipole Expansion · Uniqueness Theorems for the Potential