# Finding Potential inside a hollow sphere

• guyvsdcsniper
In summary, the conversation discusses the confusion about the Legendre polynomials, specifically the functions of cosθ. The expert explains that the Legendre polynomial of order 1 is just cosθ and the parentheses mean "function of" not multiplication. They also point out that the first Legendre polynomial is just a constant equal to 1. The expert also mentions that most people look up the Legendre polynomials instead of deriving them and provides the Rodrigues formula for those who wish to derive them. They also mention the significance of the Legendre polynomials in solving the Laplace equation in spherical geometry.

#### 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 ?

#### Attachments

• Griffiths 3.6.pdf
79.8 KB · Views: 161
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.

kuruman said:
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 Pn(x) could be defined by the Rodrigues Formula.

But if the Legendre polynomial is a function of cosθ, Pn(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?

quittingthecult said:
Ok maybe I am confused.

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

But if the Legendre polynomial is a function of cosθ, Pn(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.

kuruman said:
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 Pn(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?

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.

guyvsdcsniper
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.

BvU