Integration of Legendre Polynomials with different arguments

[Alan Sammarone]

Hi everybody,
I'm trying to calculate this:

$$\sum_{l=0}^{\infty} \int_{\Omega} d\theta' d\phi' \cos{\theta'} \sin{\theta'} P_l (\cos{\gamma})$$

where $P_{l}$ are the Legendre polynomials, $\Omega$ is the surface of a sphere of radius $R$, and

$$ \cos{\gamma} = \cos{\theta'} \cos{\theta} + \sin{\theta'}\sin{\theta}\,\cos({\phi' -\phi}) $$

I am told that only the $l=1$ term survives due to orthogonality of Legendre polynomials (of course $\cos{\theta'} = P_{1}(\cos{\theta'})$), but I'm don't see why, since the Legendre polynomials have different arguments.

How can I show that this is true?

 

[Orodruin]

Your angle $\gamma$ is an angle from the direction defined by $\theta$ and $\phi$. The spherical harmonics of a fixed $\ell$ form an irrep of the rotation group and so your $P_\ell$ will be rotated into a linear combination of the spherical harmonics $Y_\ell^m$. The Legendre polynomial $P_1(\cos\theta') = \cos\theta'$ is directly proportional to $Y_1^0(\theta',\phi')$ and so your integral will project out this component of $P_\ell(\cos\gamma)$.

 

[math771]

Hi there! It looks like you are trying to calculate the sum of an infinite series and an integral involving Legendre polynomials. This can definitely be a tricky problem, so let me try to help you out.

First of all, it is important to understand the properties of Legendre polynomials. One of the most important properties is their orthogonality, which means that the integral of the product of two different Legendre polynomials over a range of values is equal to zero. In other words, if we have two different Legendre polynomials, say $P_{l}$ and $P_{m}$, and we integrate their product over a range of values, the result will be zero unless $l = m$. In that case, the result will be equal to some constant value.

Now, let's take a look at the expression you are trying to calculate. We have an infinite sum over different values of $l$, and an integral over the surface of a sphere. The integrand contains a product of Legendre polynomials, one with argument $\cos{\theta'}$ and the other with argument $\cos{\gamma}$. As you correctly pointed out, due to orthogonality, only the $l=1$ term will survive in this sum. This is because when we integrate the product of $P_{1}$ and $P_{l}$ over the range of values for $\theta'$ and $\phi'$, the result will be zero unless $l=1$.

To see this more clearly, let's use the expression for $\cos{\gamma}$ given in the problem. We can rewrite it as:

$$ \cos{\gamma} = \cos{\theta'} \cos{\theta} + \sin{\theta'}\sin{\theta}\,\cos({\phi' -\phi}) = P_{1}(\cos{\theta'}) \cos{\theta} + P_{0}(\cos{\theta'})\sin{\theta}\,\cos({\phi' -\phi}) $$

where $P_{0}$ is the Legendre polynomial with $l=0$. Now, when we integrate the product of $P_{1}$ and $P_{l}$ over the range of values for $\theta'$ and $\phi'$, we will get:

$$ \int_{\Omega} P_{1