Spherical Polar Coords with Laplacian

[Firepanda]

http://img243.imageshack.us/img243/1816/laplaceds2.jpg

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

 

[gabbagabbahey]

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...

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?

 

[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

<br /> \int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi<br />

please?

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

 

[gabbagabbahey]

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

<br /> \int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi<br />

please?

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

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.

 

[Firepanda]

Use the substitution x=\cos\phi...what does that make dx?

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

 

[gabbagabbahey]

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

Pn(cosx)Pm(cosx)sinx dx ?

Where x = phi

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?

 

[Firepanda]

x = 1 and x = -1

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

 

[gabbagabbahey]

x = 1 and x = -1

Yes.

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

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

 

[Firepanda]

Yes.



Are you sure? What is dx if x=\cos\phi?

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

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

 

[gabbagabbahey]

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

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

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.

 

[Firepanda]

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.

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:

<br /> \int_{0}^{\pi}P_n(\cos\phi)P_m(\cos\phi)\sin\phi d\phi<br />

So i integrate this by parts?

Or perhaps I take

<br /> \int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi

equal to 0?

So

<br /> V(a,\phi)(\sin\phi)=\sum_{n=0}^{\infty} A_n a^n P_n (\cos\phi) (\sin\phi) = P_n (\cos\phi)<br />

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.

 

[gabbagabbahey]

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:

<br /> \int_{0}^{\pi}P_n(\cos\phi)P_m(\cos\phi)\sin\phi d\phi<br />

So i integrate this by parts?

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]<br /> <br /> P.S. I&#039;m assuming you already rearranged the integral when you substituted the series expression for V by moving the integral inside the summation and treating A_n as a constant?

 

[gabbagabbahey]

So

<br /> V(a,\phi)=\sum_{n=0}^{\infty} A_n a^n P_n (\cos\phi) = P_n (\cos\phi)(\sin\phi)<br />

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.

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)

 

[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?

 

[gabbagabbahey]

So I'm seeing now the summation has a zero in all the terms?

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)=?

 

[Firepanda]

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)=?

\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))

 

[gabbagabbahey]

\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=2/((a^n)(2n+1))

Not quite;

\int_{-1}^{1} P_n (x)P_m(x)dx=\left\{ \begin{array}{lr} 0,&amp; n\neq m \\ \frac{2}{2n+1}, &amp; n=m \end{array}=\left\{ \begin{array}{lr} 0,&amp; n\neq m \\ \frac{2}{2m+1}, &amp; 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,&amp; 0&lt; \phi &lt; \pi/2 \\ 0, &amp; \pi/2 &lt;\phi &lt;\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.

 

[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?

 

[gabbagabbahey]

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?

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, &amp; m=2n \\ \frac{(-1)^n(2n)!V_0}{2^{2n+1}n!(n+1)!}, &amp; m=2n+1\end{array}

so A_m=?

 

[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!