Finding Potential inside a hollow sphere

[guyvsdcsniper]

Homework Statement

The potential Vo(θ) is specified on the surface of a hollow sphere,
of radius R. Find the potential inside the sphere.

Relevant Equations

Vo(θ)

I am currently reading Griffiths book for electrodynamics and having trouble making a jump in one of the problems. I have attached the problem (3.6) in question.

In the part that is highlighted, I don't see how we go from (1-cosθ) to (P₀cosθ-P₁cosθ)?

I can see that from the Legendre Polynomials that P₀(cosθ) is equal to 1.
But how is cos₀ = to P₁cosθ ?

 

[kuruman]

You misunderstood. The Legendre polynomial of order 1, which is a function of $\cos\theta$, is just $\cos\theta$: $P_1(cos\theta)=\cos\theta$. The parentheses mean "function of" not multiplication.

 

[guyvsdcsniper]

You misunderstood. The Legendre polynomial of order 1, which is a function of $\cos\theta$, is just $\cos\theta$: $P_1(cos\theta)=\cos\theta$. The parentheses mean "function of" not multiplication.

Ok maybe I am confused.

I understood from the book the the Legendre Polynomial of P_n(x) could be defined by the Rodrigues Formula.

But if the Legendre polynomial is a function of cosθ, P_n(cosθ), this wouldn't apply?

I guess my thought process is if its a function of cosθ,
for n = 0, P₀(cosθ)=cos(0) = 1. Is this not the case?

 

[kuruman]

Ok maybe I am confused.

I understood from the book the the Legendre Polynomial of P_n(x) could be defined by the Rodrigues Formula.

But if the Legendre polynomial is a function of cosθ, P_n(cosθ), this wouldn't apply?

I guess my thought process is if its a function of cosθ,
for n = 0, P₀(cosθ)=cos(0) = 1. Is this not the case?

Not quite. See here for a list of the a list of the first 10 Legendre polynomials $P_n(x)$ where $x=\cos\!\theta$. When $\theta =0$, $\cos\!\theta =1$ and $P_n(1)=1.$ Specifically, $P_0$ is just a constant equal to 1.

 

[guyvsdcsniper]

Not quite. See here for a list of the a list of the first 10 Legendre polynomials $P_n(x)$ where $x=\cos\!\theta$. When $\theta =0$, $\cos\!\theta =1$ and $P_n(1)=1.$ Specifically, $P_0$ is just a constant equal to 1.

I was just going to respond to you.

I did find this article https://mathworld.wolfram.com/AssociatedLegendrePolynomial.html which essentially also list the first 10 Legendre polynomials P_n(x) where x=cosθ.

Im glad to accept the list and refer to it when needed, but do you know what formula is used to come to these solutions? I am searching and don't see much or maybe its going over my head? Or is this something beyond the scope of an Intro Electrostatics class?

 

[kuruman]

Most people look them up because once someone did the work to derive them, everybody else uses them without re-deriving them. They are tabulated for people to use. However, if you insist on deriving the Legendre polynomials yourself, you could use the Rodrigues formula $$P_n(x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}\left[(x^2-1)^n\right].$$Have fun.

 

[Orodruin]

In terms of how to arrive at the Legendre polynomials in the first place and their significance in expansions In spherical geometry: They are the (regular) solutions of the eigenvalue problem
$$
-\frac d{dx} (1-x^2) \frac d{dx} f = \lambda f
$$
on the interval [-1,1] with $\lambda$ turning out to be integers on the form $n(n+1)$. This in turn arises when looking for eigenfunctions of the angular part of the Laplace operator depending only on $\theta$ after making the substitution $x = \cos\theta$. It is therefore quite natural that $\cos\theta$ is used as the argument in the Legendre polynomials when solving the Laplace equation in spherical geometry.