Lengendre Polynomials of cos(theta)

[pcalhoun]

Hey guys,

I've been working on a quantum related problem in my math physics class and I've run into a snag. When dealing with Legendre Polynomials ( specifically : $$P_{lm} (x)$$ ), I can find the general expression that can be used to derive the polynomial for any sets of l and m (wolfram math has that explicit equation on their website on this page, eq 65: http://mathworld.wolfram.com/LegendrePolynomial.html" .)

However, when trying to find the general expression for the Legendre polynomial of cos: $$P_{lm}(cos(\theta))$$, I find nothing. I tried to come up with the expression on my own by substituting for $$x = cos(\theta)$$ and $$dx = -sin(\theta)d\theta$$ , but I do not know how to handle the term with the derivative which goes like: $$\frac{d^{l+m}}{dx^{l+m}}$$.

My first guess was to try and replace the differential $$dx^{l+m}$$ with $$(-sin(\theta)d\theta)^{l+m}$$, but this didn't seem to give the correct final result.

If anybody knows of a website where the $$P_{lm}(cos(\theta))$$ formula is explicity given, or if someone knows how to actually derive the general form for these polynomials, let me know. I would really appreciate it. Thanks for your time.

PCalhoun

P.S. If more information is needed about the problem, I would be glad to elaborate.

 

[tiny-tim]

I tried to come up with the expression on my own by substituting for $$x = cos(\theta)$$ and $$dx = -sin(\theta)d\theta$$ , but I do not know how to handle the term with the derivative which goes like: $$\frac{d^{l+m}}{dx^{l+m}}$$.

My first guess was to try and replace the differential $$dx^{l+m}$$ with $$(-sin(\theta)d\theta)^{l+m}$$, but this didn't seem to give the correct final result.

Hey pcalhoun!

(have a theta: θ )

d/dx = d/dθ dθ/dx = (-1/sinθ)d/dθ

d²/dx² = (-1/sinθ)d/dθ [(-1/sinθ)d/dθ] = …

 

[Meir Achuz]

Just use (64) with cos theta=x. Then at the end go back to cos theta.