Spherical Boundary condition problem

[Mike Jonese]

Homework Statement

I'm trying to keep the post brief and will post more info if needed. But I am trying to understand how the value of two "A" constants were found. This is from Griffiths Electrodynamics.
In this part of the problem, I am given a boundary condition that is a function of theta. The problem requires legendre polynomials. Apparently you only need to first two terms when l=0 , l=1 in the sum to get a solution, not an infinite sum, I am not sure I understand why, or what that means.
Question 1) How did they get the value of A0, A1
Question 2) Why do we only need the terms when l=0 and l=1 and not all the other ones, how did we know we only need l=0,l=1

Homework Equations

1. is the boundary condition
2. is just the boundary condition rewritten using half angle top look like Pl(cos(theta))
3. is the solution sum that apparently you plug 2 into and get the value for constants A0, A1

The Attempt at a Solution

I think I a overlooking something obvious but can't figure out how they "read off" the values of those constants
Thank you very much for any insight - Mike

 

[Charles Link]

The boundary condition is the first equation, which can be rewritten with a trigonometric identity as the second equation. The boundary condition only contains the zero and first power of $\cos(\theta)$, so it only requires the $A_o$ and $A_1$ terms. $\\$ Note: $P_o(\cos(\theta))=1$, and $P_1(\cos(\theta))=\cos(\theta)$. $\\$ Note: In the summation, it is $A_i r^i P_i(\cos(\theta))$. (The copy you have there didn't seem to distinguish $i$ and $1$. It should also use a small $r$). $\\$ Additional note: The boundary condition is more properly written as $V(R, \theta, \phi)=k \sin^2(\theta/2)$, i.e. it gives $V(r,\theta, \phi)$ for $r=R$.

 

[Mike Jonese]

The boundary condition is the first equation, which can be rewritten with a trigonometric identity as the second equation. The boundary condition only contains the zero and first power of $\cos(\theta)$, so it only requires the $A_o$ and $A_1$ terms. $\\$ Note: $P_o(\cos(\theta))=1$, and $P_1(\cos(\theta))=\cos(\theta)$. $\\$ Note: In the summation, it is $A_i r^i P_i(\cos(\theta))$. (The copy you have there didn't seem to distinguish $i$ and $1$. It should also use a small $r$). $\\$ Additional note: The boundary condition is more properly written as $V(R, \theta, \phi)=k \sin^2(\theta/2)$, i.e. it gives $V(r,\theta, \phi)$ for $r=R$.

Oh ok! I used a lowercase l in mathmatica it probably would have been better to use i or n. Also (in the actual problem that this is coming from, that is that boundary condition on the surface of a sphere at radius R, so I r=R in this case. What you said makes sense but how do you actually find the values of A0 and A1? Thanks again

 

[Charles Link]

Oh ok! I used a lowercase l in mathmatica it probably would have been better to use i or n. Also (in the actual problem that this is coming from, that is that boundary condition on the surface of a sphere at radius R, so I r=R in this case. What you said makes sense but how do you actually find the values of A0 and A1? Thanks again

You simply set like coefficients of each power of $cos(\theta)$ equal on both sides of the equation where you write $V(R, \theta, \phi)=boundary \, equation=summation \, expression \, with \, r=R$. This one was simple. It can get more complicated for cases where powers of two and higher are involved, because the Legendre polynomials of higher order contain lower order terms of $cos(\theta)$ besides the nth power.