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## Re: [Help-gsl] spherical harmonics for m<0 (m=-l)

 From: Drew Parsons Subject: Re: [Help-gsl] spherical harmonics for m<0 (m=-l) Date: Sat, 01 Oct 2005 00:31:24 +1000

On Wed, 2005-09-28 at 09:45 +1000, John Gehman wrote:
> There is a relationship -- look it up for sure, obviously, but I
> believe the negative values of m are simply the complex conjugate
> (relevant in the exponential) of the positive m values, and maybe the
> m = \pm 1 are opposite overall sign to each other.
>
> >
> > I think there is a relationship between +m and -m (Abramowitz &Stegun
> > 8.2.5)
> >
> >
> > I'm not sure why the original function is restricted to m>=0, maybe
> > there was a reason for that.
> >

Ah yes, I've found the relation I need now. I felt it was in some sense
a negative function like you said, John, but I couldn't prove it.

Abramowitz&Stegun 8.2.5 was not so helpful.  It gives the general form
for P_\nu^{-mu} (if I remember right mu could even be complex here.
Certainly it's at least real).  But the expression is given in terms of
the second order associated Legendre functions, Q_l^m, which is a real
bugger to work with.

I found the solution in Gradshtein and Ryzhik, 8.752.2, given
specifically for integer values:

P_\nu^{-m}(x) = (-1)^m \Gamma(\nu-m+1)/\Gamma(\nu+m+1) P_\nu^m(x)

or more simply for integer l,

P_l^{-m}(x) = (-1)^m (l-m)! / (l+m)! P_l^m(x)   ...(1)

So we can see the normalisation factor creeping in already.  Applying
this formula to the normalised associated Legendre functions, sfP_l^m,
it becomes very simply

sfP_l^(-1)(x) = (-1)^m sfP_l^m(x)               ...(2)

Then the relation between the spherical harmonics becomes apparent:

Y_l^{-m} = (-1)^m Y_l^m^*

taking the complex conjugate, just as John said.

Brian, since the GSL functions (gsl_sf_legendre_sphPlm) only work with
integer values of l and m, I suppose it would be trivial to include the
m<0 case using Eq. (2).  Only a test for m even or odd would need to be

I don't know if it would be worth generalising  gsl_sf_legendre_Plm as
well, depends how much trouble it is to evaluate (l-m)! / (l+m)!.

Anyway, thanks for the hints.

Drew