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Spherical Polar Coords with Laplacian

  1. Jan 6, 2009 #1

    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!

  2. jcsd
  3. Jan 7, 2009 #2


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    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:

    [tex]V(r,\phi)=\sum_{n=0}^{\infty} A_n r^n P_n (\cos\phi)[/tex]

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

    Hint: what happens when you integrate [tex]\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi[/tex] using first the Legendre expansion of your potential? And second, using your values of [itex]V(a,\phi)[/itex] given in Q7?
  4. Jan 7, 2009 #3
    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.
  5. Jan 7, 2009 #4


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    Use the substitution [itex]x=\cos\phi[/itex]...what does that make [itex]dx[/itex]? How about the limits in terms of x?:wink: This is why I multiplied by [itex]P_m(\cos\phi)sin\phi d\phi[/itex] 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.
  6. Jan 7, 2009 #5
    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
  7. Jan 7, 2009 #6


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    No! Try again.:smile:

    If [itex]x=cos\phi[/itex], then [itex]P_n(\cos\phi)=P_n(x)[/itex] and [itex]P_m(\cos\phi)=P_m(x)[/itex]. What is [itex]dx[/itex]? What is [itex]x[/itex] if [itex]\phi=0[/itex]? What is [itex]x[/itex] if [itex]\phi=\pi[/itex]?
  8. Jan 7, 2009 #7
    x = 1 and x = -1

    So my integral should be the other way around? from pi to 0?
  9. Jan 7, 2009 #8


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    Yes. :smile:

    Are you sure?:wink: What is [itex]dx[/itex] if [itex]x=\cos\phi[/itex]? (would calculating [itex]\frac{dx}{d\phi}[/itex] make more sense to you?)
  10. Jan 7, 2009 #9
    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
  11. Jan 7, 2009 #10


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    Don't you mean [itex]-\sin\phi d\phi[/itex]? :wink:

    That means that:

    [tex]\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[/tex] and so your integral is in the correct form to use the hints.
  12. Jan 7, 2009 #11
    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[/tex]

    equal to 0?


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


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

    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
  13. Jan 7, 2009 #12


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

    [tex]\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[/tex]

    with the substitution [itex]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[/itex] as a constant?
  14. Jan 7, 2009 #13


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

    Why would you think that the two expressions were equal?

    [tex]V(a,\phi)=\sum_{n=0}^{\infty} A_n a^n P_n (\cos\phi)[/tex]

    means that

    [tex]\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[/tex]

    [tex]=\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)[/tex]
  15. Jan 7, 2009 #14
    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 < [tex]\phi[/tex] < pi, V(a, phi) = 0.

    So now I just do the same for 0 < [tex]\phi[/tex] < pi/2 and pi/2 < [tex]\phi[/tex] < pi, and we'll see what happens?
  16. Jan 7, 2009 #15


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    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:

    [tex]\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)=???[/tex]
  17. Jan 7, 2009 #16
    [tex]\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)[/tex]

    [tex]\sum_{n=0}^{\infty} A_n=\sum_{n=0}^{\infty}2/((a^n)(2n+1))[/tex]
  18. Jan 7, 2009 #17


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    Not quite;

    [tex]\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}[/tex]

    [tex]\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}[/tex]

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

    Now Q7 tells you that:

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

    So you can also integrate [tex]\int_{0}^{\pi}V(a,\phi)P_m(\cos\phi)\sin\phi d\phi[/tex] using this expression and then compare your results.
  19. Jan 7, 2009 #18
    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

    [tex]\int_{0}^{1}V_0 P_m(x)dx[/tex]

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

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

    How do I get this into [tex]P_2n[/tex] or [tex]P_2n+1[/tex]?
  20. Jan 7, 2009 #19


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    [itex]V_0[/itex] is just a constant and can be pulled out of the integral:

    [tex]\int_{0}^{1}V_0 P_m(x)dx=V_0\int_{0}^{1} P_m(x)dx[/tex]

    Now, you know that:

    [tex]\int_{0}^{1} P_{2n}(x)dx=0[/tex]


    [tex]\int_{0}^{1} P_{2n+1}(x)dx=\frac{(-1)^n(2n)!}{2^{2n+1}n!(n+1)!}[/tex]

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

    [tex]V_0\int_{0}^{1} P_m(x)dx=0[/tex]

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

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

    Which tells you that :

    [tex]\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}[/tex]

    so [itex]A_m=???[/itex]
  21. Jan 7, 2009 #20
    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!
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