Chapter 11: Q3P (page 404)

Prove Equation 11.33, starting with Equation 11.32. Hint: Exploit the orthogonality of the Legendre polynomials to show that the coefficients with different values of l must separately vanish.

Expert verified

It was proven (using orthogonality of Legendre polynomials), that for a boundary condition, $ψ (a, θ) = 0$ following must be valid:

$a_{l} = \frac{i j_{l} (k a)}{k h_{l}^{(1)} (k a)}$

Step by step solution

From boundary condition $ψ (a, θ) = 0$ , we have:

$\sum_{l = 0}^{\infty} i^{l} (2 l + 1) [j_{l} (k a) + i k a_{l} h_{l}^{(1)} (k a)] P_{l} (\cos θ) = 0$

We multiply the expression with $P_{l} (\cos θ) \sin θ d θ$ and integrate from 0 to $π$

$\sum_{l = 0}^{\infty} i^{l} (2 l + 1) [j_{l} (k a) + i k a_{l} h_{l}^{(1)} (k a)] \int_{0}^{π} P_{l} (\cos θ) P_{l} (\cos θ) \sin θ d θ = 0$

We can use expression for orthogonality of Legendre polynomials:

$P_{l} (\cos θ) P_{l} (\cos θ) \sin θ d θ = \frac{2}{2 l + 1} δ_{l, l}$

Where $δ_{l, l}$ is Kronecker-Delta symbol which vanishes for $l \neq l$ .

Only one term from the sum "survives", and that's for $l = l$

$2 i^{l} (2 l + 1) [j_{l} (k a) + i k a_{l} h_{l}^{(1)} (k a)] = 0$

In order for the last equation to be equal to zero, expression inside brackets must vanish:

$a_{l} = \frac{i j_{l} (k a)}{k h_{l}^{(1)} (k a)} .$

It was proven (using orthogonality of Legendre polynomials), that for a boundary condition $ψ (a, θ) = 0$ , following must be valid:

$a_{l} = \frac{i j_{l} (k a)}{k h_{l}^{(1)} (k a)} .$

Hence, it’s proved.

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