# Spherical Polar Coords with Laplacian

1. Jan 6, 2009

### Firepanda

http://img243.imageshack.us/img243/1816/laplaceds2.jpg [Broken]

Right I have a fair few questions on this, it's relating to question 7 only, although you need to refer back to the equation derived from question 6.

1) I used the equation from q6. as a fourier series substituting r=a. I end up with an An of:

An = (Vo * a^n * Pn)/pi

For n = odd (i.e 2n+1)

And An = 0 for n = even.

Now looking at the result I should get (at the end of q7.) my a^n should in fact be a^-n, it stays as a constant during my integration so I can't change it there, so where do I change it?

2) My fourier series was on the interal from 0 to pi/2, now from my definition of a fourier series the integral should go from -L to L with 1/L constant before the integration. What is my 1/L here? I used 1/pi, not sure if this is correct.

3) For the second fourier series in the integral of pi/2 to pi, I used 1/pi again for my 1/L (as defined before) but my An here doesn't get as far as being able to integrate it as I get 0 before I start, so I assume I leave this part out of the whole question?

Basically from my first integration I get close but I have no constants to my polynomials as I'm supposed to, which makes me think my question at 3) shouldn't be 0...

I see the orthoganality help at the bottom, how do I use this here?

ANY help is much appreciated!

Thankyou

Last edited by a moderator: May 3, 2017
2. Jan 7, 2009

### gabbagabbahey

Huh?! What function are you taking the Fourier series of and why?

This problem is about Legendre Polynomials not Fourier Series. Fourier Series are expansions of a function in terms of Cosines and Sines (or complex exponentials). Instead, the potential has been expanded
in terms of Legendre Polynomials:

$$V(r,\phi)=\sum_{n=0}^{\infty} A_n r^n P_n (\cos\phi)$$

Your task here is to find the coefficients $A_n$ of this expansion using the orthogonality properties of Legendre Polynomials, not any Fourier series equations you might have.

Hint: what happens when you integrate $$\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi$$ using first the Legendre expansion of your potential? And second, using your values of $V(a,\phi)$ given in Q7?

3. Jan 7, 2009

### Firepanda

Ahh, ok thanks, sorry the question I did previously had a summation very simalar which I used fourier for, and because it looked the same, and on the same question paper I thought I'd apply it here too.

I really haven't any experience with legendre polynomials, can i ask where the sin function came from in

$$\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi$$

Also, how would I integrate this? The limits dont match any in the hints so I don't see how I can use them.

4. Jan 7, 2009

### gabbagabbahey

Use the substitution $x=\cos\phi$...what does that make $dx$? How about the limits in terms of x? This is why I multiplied by $P_m(\cos\phi)sin\phi d\phi$ and integrated from zero to pi: it allows you to use the orthogonality conditon to extract your coefficients from the summation, allowing you to determine them.

5. Jan 7, 2009

### Firepanda

Ah ok, so my 1st case would be integral from 0 to pi of

Pn(cosx)Pm(cosx)sinx dx ?

Where x = phi

This is where I don't see how to integrate it, do I hold things constant or?

Thanks btw

6. Jan 7, 2009

### gabbagabbahey

No! Try again.

If $x=cos\phi$, then $P_n(\cos\phi)=P_n(x)$ and $P_m(\cos\phi)=P_m(x)$. What is $dx$? What is $x$ if $\phi=0$? What is $x$ if $\phi=\pi$?

7. Jan 7, 2009

### Firepanda

x = 1 and x = -1

So my integral should be the other way around? from pi to 0?

8. Jan 7, 2009

### gabbagabbahey

Yes.

Are you sure? What is $dx$ if $x=\cos\phi$? (would calculating $\frac{dx}{d\phi}$ make more sense to you?)

9. Jan 7, 2009

### Firepanda

sin(phi) d(phi) = dx...

lol, I'm sure you're trying to give me a subtle hint and im just not picking up on it

10. Jan 7, 2009

### gabbagabbahey

Don't you mean $-\sin\phi d\phi$?

That means that:

$$\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi=\int_{1}^{-1}V(a,\phi)P_m(x)(-dx)=-\int_{1}^{-1}V(a,\phi)P_m(x)dx=\int_{-1}^{1}V(a,\phi)P_m(x)dx$$ and so your integral is in the correct form to use the hints.

11. Jan 7, 2009

### Firepanda

Ok sorry just got back from the gym or I would have replied sooner, i get this now thanks.

So I put the summation of V(a,phi) into this integral, where An and r^n are constants, so my integral is:

$$\int_{0}^{\pi}P_n(\cos\phi)P_m(\cos\phi)\sin\phi d\phi$$

So i integrate this by parts?

Or perhaps I take

$$\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi$$

equal to 0?

So

$$V(a,\phi)(\sin\phi)=\sum_{n=0}^{\infty} A_n a^n P_n (\cos\phi) (\sin\phi) = P_n (\cos\phi)$$

means

$$V(a,\phi)=\sum_{n=0}^{\infty} A_n = \sum_{n=0}^{\infty}(\sin\phi)/(a^n)$$

Sorry lots of questions, I'm sure once I get my head around this I can do in the other integrals ranges.

Last edited: Jan 7, 2009
12. Jan 7, 2009

### gabbagabbahey

There's no need to actually carry out the integration. That's the whole point of the hint.

$$\int_{0}^{\pi}P_n(\cos\phi)P_m(\cos\phi)\sin\phi d\phi=\int_{-1}^{1}P_n(x)P_m(x)dx$$

with the substitution $x=\cos\phi[/tex] P.S. I'm assuming you already rearranged the integral when you substituted the series expression for V by moving the integral inside the summation and treating [itex]A_n$ as a constant?

13. Jan 7, 2009

### gabbagabbahey

No;

$$V(a,\phi)=\sum_{n=0}^{\infty} A_n a^n P_n (\cos\phi) \neq P_n (\cos\phi)(\sin\phi)$$

Why would you think that the two expressions were equal?

$$V(a,\phi)=\sum_{n=0}^{\infty} A_n a^n P_n (\cos\phi)$$

means that

$$\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi=\int_{-1}^{1}V(a,\phi)P_m(x)dx=\int_{-1}^{1}\left( \sum_{n=0}^{\infty} A_n a^n P_n (\cos\phi)\right)P_m(x)dx$$

$$=\int_{-1}^{1}\left( \sum_{n=0}^{\infty} A_n a^n P_n (x)\right)P_m(x)dx=\sum_{n=0}^{\infty} A_n a^n\left( \int_{-1}^{1} P_n (x)P_m(x)dx\right)$$

14. Jan 7, 2009

### Firepanda

Sorry ye I didn't mean those 2 were equal, I was in the middle of editing it but it wasn't saving it.

I was trying to use the sin somehow and end up with it in a taylor expansion, as the expansion has alternating signs just like the expression I'm trying to get to.

So I'm seeing now the summation has a zero in all the terms? And so for 0 < $$\phi$$ < pi, V(a, phi) = 0.

So now I just do the same for 0 < $$\phi$$ < pi/2 and pi/2 < $$\phi$$ < pi, and we'll see what happens?

15. Jan 7, 2009

### gabbagabbahey

Really? What about when m=n? Remember, you are summing over all values of n so there will be one value for which m=n is true.

Using the Legendre expansion of V you get:

$$\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi=\sum_{n=0}^{\infty} A_n a^n\left( \int_{-1}^{1} P_n (x)P_m(x)dx\right)=???$$

16. Jan 7, 2009

### Firepanda

$$\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi=\sum_{n=0}^{\infty} A_n a^n\left( \int_{-1}^{1} P_n (x)P_m(x)dx\right)=2/(2n+1)$$

$$\sum_{n=0}^{\infty} A_n=\sum_{n=0}^{\infty}2/((a^n)(2n+1))$$

17. Jan 7, 2009

### gabbagabbahey

Not quite;

$$\int_{-1}^{1} P_n (x)P_m(x)dx=\left\{ \begin{array}{lr} 0,& n\neq m \\ \frac{2}{2n+1}, & n=m \end{array}=\left\{ \begin{array}{lr} 0,& n\neq m \\ \frac{2}{2m+1}, & n=m \end{array}$$

So,
$$\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi=\sum_{n=0}^{\infty} A_n a^n\left( \int_{-1}^{1} P_n (x)P_m(x)dx\right)=\frac{2 A_m a^m}{2m+1}$$

Since all of the terms in the sum are zero except fr the n=m term.

Now Q7 tells you that:

$$V(a,\phi)=\left\{ \begin{array}{lr} V_0,& 0< \phi < \pi/2 \\ 0, & \pi/2 <\phi <\pi \end{array}$$

So you can also integrate $$\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi$$ using this expression and then compare your results.

18. Jan 7, 2009

### Firepanda

Ok great, I was working through the Vo one and I've hit a brick wall, the derivation is pretty much the same all the way to

$$\int_{0}^{1}V_0 P_m(x)dx$$

I can't see where to go from here, if I take Vo as V(a, phi) I can get to

$$\sum_{n=0}^{\infty} A_n a^n\left( \int_{0}^{1} P_n (x)P_m(x)dx\right)$$

How do I get this into $$P_2n$$ or $$P_2n+1$$?

19. Jan 7, 2009

### gabbagabbahey

$V_0$ is just a constant and can be pulled out of the integral:

$$\int_{0}^{1}V_0 P_m(x)dx=V_0\int_{0}^{1} P_m(x)dx$$

Now, you know that:

$$\int_{0}^{1} P_{2n}(x)dx=0$$

and

$$\int_{0}^{1} P_{2n+1}(x)dx=\frac{(-1)^n(2n)!}{2^{2n+1}n!(n+1)!}$$

So, when m is even (i.e. when m=2n) you have:

$$V_0\int_{0}^{1} P_m(x)dx=0$$

and when m is odd (i.e. when m=2n+1) you have:

$$V_0\int_{0}^{1} P_m(x)dx=\frac{(-1)^n(2n)!V_0}{2^{2n+1}n!(n+1)!}$$

Which tells you that :

$$\frac{2 A_m a^m}{2m+1}=\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi=\left\{ \begin{array}{lr} 0, & m=2n \\ \frac{(-1)^n(2n)!V_0}{2^{2n+1}n!(n+1)!}, & m=2n+1\end{array}$$

so $A_m=???$

20. Jan 7, 2009

### Firepanda

Greats thanks I <3 you, I won't ask for any more help, if I can't get the answer from that I'll prob just quit life altogether, great explanation!