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Legendre polynomials, Jackson's book problem, potential

  1. May 4, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I'm stuck in evaluating an integral in a problem. The problem can be found in Jackson's book page 135 problem 3.1 in the third edition. As I'm not sure I didn't make a mistake either, I'm asking help here.
    Two concentric spheres have radii a,b (b>a) and each is divided into two hemispheres by the same horizontal plane. The upper hemisphere of the inner sphere and the lower hemisphere of the outer sphere are maintened at potential V. The other hemispheres are at zero potential.
    Determine the potential in the region [itex]a \leq r \leq b[/itex] as a series in Legendre polynomials. Include terms at least up to [itex]l=4[/itex]. Check your solution against known results in the limiting case [itex]b \to \infty[/itex] and [itex]a\to \infty[/itex].


    2. Relevant equations
    Below.


    3. The attempt at a solution
    I know I could solve this problem via the Green function for 2 spheres, I will try once I solve it via a less tricky way to see if I can get the same result.
    Attempt: I realize it's an azimuthal symmetric problem. Therefore the solution is of the form [itex]\Phi (r, \theta )= \sum _{l=0}^{\infty } [U_l^+ r^l + U_l^- r ^{-(l+1)}] P_l (\cos \theta )[/itex]. My goal is to determine these 2 "U" coefficients. In order to do this, I check out what the boundary conditions have to tell me.
    1)[itex]\Phi (a, \theta )= \sum _ {l=0}^{\infty } [U_l^+ a^l +U_l ^- a^{-(l+1)}]P_l (\cos \theta )[/itex]. I know that this expression can be worth either 0 or V depending upon the value of theta, I'll use this a step later. Now I use a trick, I multiply by [itex]P_{l'} (\cos \theta ' )[/itex] and I integrate as to use the orthogonality of the Legendre polynomials.
    [tex]\int _{-1}^1 \Phi (a,\theta )P_{l'} (\cos \theta ') d \cos \theta =\left ( \frac{2}{2l+1} \right ) [U_l^+ a^l +U_l ^- a^{-(l+1)}] [/tex]. But from cos theta worth -1 to 0, the integral vanishes because Phi is 0 there. Hence I'm left with [itex]\int _{-1}^0 \Phi (a,\theta )P_{l'} (\cos \theta ') d \cos \theta =V \int _{-1}^0 P_{l'} (\cos \theta ' ) d \cos \theta =-V \int _{\pi} ^{\pi/2} P_{l'} (\cos \theta ' ) \sin \theta d\theta [/itex]. I must solve this integral. I'm not sure it is "right", I probably have to consider [itex]-V \int _{\pi} ^{\pi/2} P_{l} (\cos \theta ) \sin \theta d\theta [/itex].
    I've been searching lots of relations/properties of the Legendre polynomials to see if one could help me. For example I can rewrite [itex]P_l (x)[/itex] as [itex]\frac{1}{(2l+1)} \left [ \frac{d}{dx} P_{l+1 }(x)-\frac{d}{dx} P_{l-1} (x) \right ][/itex]. But it doesn't seem to help me in any way.
    I'm out of ideas. If I had this integral calculated, I'd have one expression of the [itex]U_l ^+[/itex] and [itex]U_l ^-[/itex] in terms of "a".
    By looking at the other boundary conditions, 2)[itex]\Phi (b, \theta )[/itex], I guess I could do a similar job and find another expression of these U's in terms of b this time. This would give me 2 equations with 2 unknown, so I could solve it... and thus the problem.
    Any help is greatly appreciated.
     
  2. jcsd
  3. May 4, 2012 #2
    I'm not sure if it is possible to find a general expression for the coefficients, but in this problem you're only asked to find the coefficients up to [itex]l = 4[/itex]. So you could simply calculate the integral by hand for the first four Legendre polynomials, no?
     
  4. May 4, 2012 #3

    fluidistic

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    Hey, thanks for the help!
    Hmm I don't really understand even what the integral means. [itex]l'[/itex] is arbitrary? So that the integral could be for example [itex]\int _{\pi } ^{\pi/2 } P_{10239843} ( \cos \theta ) \sin \theta d \theta[/itex] or [itex]\int _{\pi } ^{\pi/2 } P_3 ( \cos \theta ) \sin \theta d \theta[/itex]. Am I mistaken?
     
  5. May 4, 2012 #4
    The point of the "trick" you've used is that it allows you to pick out any value of [itex]l'[/itex] that you want, and then the right hand side will be the coefficients for that choice. So yes, the idea is that you can use the equation you've derived to figure out any of the expansion coefficients that you wish -- but the catch, as you've found, is that you have to do them one at a time since there's no general answer for what the integral of an arbitrary Legendre polynomial is. That's why Jackson only asked you to compute the first four coefficients, instead of working out the entire series expansion.
     
  6. May 5, 2012 #5

    fluidistic

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    Ah I understand now, thanks a lot.
    I'm having a problem for the coefficients [itex]U_2[/itex]'s.
    From the first boundary conditions I've reached that [itex]V \int _{1}^0 P_l (x)dx=\frac{2}{2l+1} [U_l ^+a^l+U_l^-a^{-(l+1)}][/itex]. (Note that I got wrong the limits of this integral in my first post)
    While for the second boundary conditions, [itex]V \int _{0}^{-1} P_l (x)dx=\frac{2}{2l+1} [U_l ^+b^l+U_l^-b^{-(l+1)}][/itex].
    For [itex]l=0[/itex] this gives me [itex]U_0^+=\frac{V}{2}+\frac{b}{a-b}[/itex] while [itex]U_0^-=\frac{ab}{b-a}[/itex].
    For l=1, [itex]U_1^+=\frac{3V}{4b}+\frac{V}{2b^3} \left ( 1+\frac{a}{b} \right ) \left ( \frac{3a^2b^3}{2b^3-2a^3} \right )[/itex] and [itex]U_1^-=-\frac{V}{2}\left ( 1+ \frac{a}{b} \right ) \left ( \frac{3a^2b^3}{2b^3-2a^3} \right )[/itex].
    While for [itex]l=2[/itex], the system of equations to solve leads to an infinity of solutions.
    I reach in this case [itex]U_2^+=U_2^- \frac{\left ( \frac{1}{a^3} -\frac{1}{b^3} \right ) }{(b^2-a )}[/itex]. Thus any value of [itex]U_2^+[/itex] determine a value for [itex]U_2^-[/itex]. I know this cannot be right...
     
  7. May 5, 2012 #6
    You've still got the signs wrong; the original integral is from -1 to 1, so in either case you simply cancel out the part of the integral where the potential is zero, but you don't reverse the position of the bounds.

    As for the [itex]l = 2[/itex] case, it doesn't lead to infinite solutions, it actually leads to only one possible case, if I did the calculation correctly.
     
  8. May 6, 2012 #7

    fluidistic

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    Hmm.
    From theta equal 0 to pi/2, [itex]\Phi (a, \theta ) =V[/itex]. The integral goes from theta = 0 to pi. Or equivalently from [itex]\cos \theta[/itex] equal to 1 to -1... or I'm missing something? Why would it go from -1 to 1?
     
  9. May 6, 2012 #8
    Either way, this is not important, as long as you do the integral consistently. You must be making a calculational error if you think there are other solutions than the zero solution for [itex]l = 2[/itex].
     
  10. May 13, 2012 #9

    fluidistic

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    I've redone entirely half of the problem (only one boundary condition) but this time with taking extreme care of the limits of integration; in Jackson's book I've found a formula for [itex]\int _0 ^1 P_l (x)dx[/itex]. If I'm not wrong, it is worth [itex](-1) \left ( \frac{-1}{2}\right ) ^{\frac{l-1}{2}} \frac{(l-2)!!}{2 \left ( \frac{l+1}{2} \right ) }[/itex]. So that it seems I could get a general expression for the problem and from it I could get any term I would like.
    I'll try to finish the problem when I have some time for it.
     
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