Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Spherical Polar Coords with Laplacian

  1. Jan 6, 2009 #1
    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!

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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member


    [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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    [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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook