# How to use Legendre polynomials in order to determine the (an)isotropy of an on-lattice cluster aggregate?

I am currently testing various models of on-lattice (square lattice in two dimensions) cluster growth for anisotropy. I end up with a cluster, the boundary of which, in case of a truly isotropic model, should look like a circle. In anisotropic cases the cluster boundary will deviate from the circular shape.

I've been told that the Legendre polynomials can be used in order to determine the anisotropy of a cluster, but I'm not exactly sure of how I can use these Legendre polynomials in order to say something about the isotropy or otherwise of the clusters.

I gather that one can write the angular functional dependency of the cluster boundary in terms of a Legendre series due to the orthogonal nature of the Legendre polynomials, but once I've calculated the coefficients of the series I'm not sure how to interpret them in terms of anisotropy of the cluster or even if this is the correct thing to be doing.

Any hints, tips or further reading would be appreciated. I am a humble mathematician, so please go easy on the physics jargon as I may not be well versed in it.

Thanks for your time and help.

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If you do it in the standard way, which is known mostly as spherical harmonics in physics, then the coefficients in front of all Legendre polynomials except for the zeroth one $P_0$ - all spherical harmonics except for the constant $Y_{00}$ - represent anisotropies. The larger the coefficients are, the larger the anisotropy is. But there's nothing such as a "single measure of anisotropy": all the coefficients beyond the zeroth one contribute to it. –  Luboš Motl Mar 11 '13 at 17:52

It's not really the Legendre polynomials you should be calculating against, but more precisely the spherical harmonics $$Y_{lm}=\sqrt{\frac{(2l+1)(l-m)!}{4\pi(l+m)!}}P_l^m(\cos(\theta))e^{im\phi}.$$ Exactly what you integrate will depend on what exactly is the object you're studying, but the essentials will be the same. None of the coefficients by themselves will mean much, but in general bigger coefficients mean bigger anisotropies; the different varieties of spherical harmonics represent different modes of anisotropy.

For example, if the coefficients for $l=1$ are nonzero then it probably means you chose the wrong origin, and the cluster centre is in some particular direction. If the coefficients for $l=2$ are nonzero then you'll have some sort of ellipsoidal distribution; the moments then tell you if it's oblate or prolate, how much, and in which direction. (This is a common problem in nuclear physics, where the quadrupolar moments of nuclei are essential properties.) Nonzero $l=3$ moments could imply some sort of triangular or tetrahedral structure, and so on.

A good way to picture these anisotropies is to graph the surface $r=1+\epsilon Y_{lm}$:

You should also be careful because the different moments are not necessarily comparable, and it could be hard to tell which one is more important. An example of this is in electrostatics. Say you have some localized density of electrical charge $\rho(\mathbf{r})$ which you're trying to examine from far away. Then you can decompose the far field into its multipolar components as $$\phi(\mathbf{r})=\sum_{l=0}^\infty\frac{4\pi}{2l+1}\sum_{m=-l}^l Q_{lm}\frac{Y_{lm}(\theta,\phi)}{r^{l+1}},$$ where $$Q_{lm}=\int\mathrm{d}\mathbf{r} \rho(\mathbf{r})r^lY_{lm}^\ast(\theta,\phi)$$ are the multipole moments. These measure the different anisotropies of the distribution and imprint it on the different anisotropic components of the far field. However, their relative importance changes with the distance from the distribution, because the different anisotropic (multipolar) components decay at different rates. Thus the ratio between different coefficients essentially tells you the distance you need to be at for the more anisotropic component to be negligible.

In your case, I gather you want to study the cluster boundary as a surface $r=r(\theta,\phi)$. My first shot at that would be to attempt a decomposition of the form $$r(\theta,\phi)=\sum_{l=0}^\infty\sum_{m=-l}^l r_{lm}Y_{lm}(\theta,\phi),$$ where you can find the coefficients $r_{lm}$ using the orthogonality properties of the harmonics. Here the $r_{lm}$ are all distances and are thus comparable, so that if some coefficient is much smaller than another you can probably ignore it.

Each coefficient then tells you how much of some particular type of anisotropy (multipolarity) your surface has. You should be careful when judging them as absolutes because they are distances and therefore carry dimensional information; to remove that you could try dividing by $r_{00}$, which is up to constants the radius of the cluster. A useful measure of total anisotropy could then be $$\sum_{l\neq0}\left|\frac{r_{lm}}{r_{00}}\right|,$$ or squaring each term, or some such.

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Dear Emilio, thankyou very much for your detailed response. I'm sorry to have not replied before now. –  user21854 Apr 19 '13 at 9:06
That's OK! If this answers your question, consider accepting the answer ;). –  Emilio Pisanty Apr 19 '13 at 9:16
I think it partly answers my question, but my problem is that I am analysing cluster in two dimensions. As such do you think that the solutions to Laplace's equation in plane polar coordinates (i.e. the circular harmonics) would be more appropriate basis functions to use than the spherical harmonics? –  user21854 Apr 19 '13 at 10:43
In a word, yes. You still need to make sure that your anisotropy measures are comparable, and that you've chosen the right origin, but the essentials still apply, of course. –  Emilio Pisanty Apr 19 '13 at 12:28