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Checking my result in EM -finding a potential-

  1. May 13, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Consider 2 conductor hollow spheres that share the same center, of radii a and b (a>b). The hollow sphere of radius b is at zero potential while the hollow sphere of radius a has a potential of the form [itex]V(\theta, \phi ) =V_0 \sin \theta \cos \phi[/itex]. Where [itex]V_0[/itex] is a constant and theta and phi are the spherical coordinates. Find the potential in all the space.


    2. Relevant equations
    Tons.


    3. The attempt at a solution
    For the region [itex]0 \leq r \leq b, \Phi (r, \theta , \phi )=0[/itex] because the E field there is 0 (due to the conductivity of the inner sphere) and thus the potential is constant and worth the same as the surface of the inner sphere, namely 0.
    Now I attack the region between the 2 spheres.
    I notice that there isn't any azimuthal symmetry, hence the book of Griffith is of no big help, unless I missed out a chapter. I therefore use Jackson's book only to get some help.
    I propose a solution of the form [itex]\Phi (r, \theta , \phi ) = \sum _{l=0}^ \infty \sum _{m=-l}^l [A_{lm}r^l+B_{lm }r^{-(l+1 )}] Y_{lm} (\theta , \phi )[/itex] where [itex]Y _{lm} (\theta, \phi )[/itex] are the spherical harmonics.
    The first boundary condition is [itex]\Phi (a,\theta , \phi ) =0 \Rightarrow \sum _{l=0}^ \infty \sum _{m=-l}^l [A_{lm}b^l+B_{lm }b^{-(l+1 )}] Y_{lm}=0[/itex].
    Here I'm not 100% of what I've done. Namely, I say that since the spherical harmonics are linearly independent, the only way to get the infinite series as vanishing, is the condition that [itex]A_{lm}b^l+B_{lm }b^{-(l+1 )}=0[/itex]. Thus [itex]A_{lm}=-B_{lm}b^{-2l-1}[/itex].
    Now my goal is to determine [itex]B_{lm}[/itex], I'd be done with the problem if I succeed in doing so.
    I use the second boundary condition: [itex]V_0\sin \theta \cos \phi =\sum _{l=0}^ \infty \sum _{m=-l}^l B_{lm} [a^{-(l+1)}-b^{-2l-1 }a^l] Y_{lm } (\theta , \phi ) [/itex].
    I call [itex]C_{lm}=B_{lm}[a^{-(l+1)}-b^{-2l-1}a^l][/itex] for convenience.
    Seeking help in Jackson's book, [itex]P_l (\cos \gamma )=\sum _{m=-l}^l C_ m (\theta ', \phi ')Y_{lm} (\theta , \phi )=V_0 \sin \theta \cos \phi[/itex] and [itex]C_{ml}=C_m[/itex] doesn't depend on l. Gamma is the angle between 2 arbitrary vectors [itex]\vec x[/itex] and [itex]\vec x '[/itex] inside the region, namely [itex]\cos \gamma = \cos \theta \cos \theta ' + \sin \theta \sin \theta ' \cos (\phi - \phi ') [/itex].
    If I understood well Jackson's book, [itex]C_m (\theta ', \phi ' ) = \int Y^* _{lm} P_l (\cos \gamma ) d \Omega[/itex] (so that it's a double integral-surface integral- and the * denotes the complex conjugate).
    Therefore [itex]B_ {lm} = \frac{C_m (\theta ' , \theta ) }{a^{-(l+1 )}-b^{-2l-1}a^l}[/itex]. Replacing this into the original ansatz yields [tex]\Phi (r, \theta , \phi ) = \sum _{l=0}^ \infty \sum _{m=-l}^l \frac{\int Y ^ * _{lm}(\theta , \phi ) P_l ( \cos \gamma ) d \Omega }{a^{-(l+1 )}-b^{-2l-1}a^l} [r^{-(l+1)}-b^{-2l-1}r^l]Y_{lm} (\theta , \phi )[/tex] which would be my final answer.
    Since [itex]P_l (\cos \gamma ) =V_0 \sin \theta \cos \phi[/itex], at first glance when r=b, Phi is worth 0 as it should while when r=a, I think that [itex]\Phi[/itex] is worth [itex]V_0 \sin \theta \cos \phi[/itex] as it should too. I wonder if my answer is correct. That was a pretty tough problem to me and it's the first in the long list (~30) I should do.
    Thanks for any comment.
     
  2. jcsd
  3. May 13, 2012 #2
    For this type of problem, the usual route is to use the same trick as you used here: https://www.physicsforums.com/showthread.php?t=603238. From what I've read I don't think you've seen how to apply this trick, so bear with me if you understand it and are well ahead of me.

    For each choice of [itex]l[/itex] and [itex]m[/itex], you would use the orthogonality relationship between the spherical harmonics to determine the coefficients. That is, you integrate the potential over each of the two surfaces with respect to your given spherical harmonic, and use the fact that the right-hand side is zero by definition, except for when the two spherical harmonics match:
    [tex]\int_0^{2\pi} \int_0^{\pi} Y_{lm}^*(\theta,\phi) \Phi(r,\theta,\phi) d\theta d\phi = A_{lm} r^l + B_{lm} r^{-l-1}[/tex]
    Obviously this won't change much for the one surface where the potential is zero, but it will help for the other surface. Now, the way to get really tricky is to notice that the fixed potential on the outer surface ([itex]r=b[/itex]) can itself be expressed in terms of one particular spherical harmonic; and that means you only need to actually evaluate one integral :)
     
  4. May 13, 2012 #3

    fluidistic

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    I'm totally new in the subject, there's no way I have a clear understanding of what's going on (learned about the spherical harmonics yesterday).

    I think I did it explicitly thanks to Jackson's book but I'm not sure. Wouldn't that be my [itex]C_m (\theta ' , \phi ' )[/itex]?

    The outer surface in this problem (unlike the other problem I've tried to solve) is of radius a.
    Hmm I see, I guess I should look at a table of the first spherical harmonics.
     
  5. May 13, 2012 #4
    Well, the [itex]C_m[/itex] are not directly important in this problem insofar as you don't need to express any Legendre polynomials in terms of spherical harmonics when there's a much more direct route to solving this problem. If you do it your way, you should see that only one of those coefficients survives, instead of having to write the general series expression that you've written. But evaluating it for each possible value of [itex]m[/itex] is way too much work for this problem, when you can use the orthogonality relationship of the spherical harmonics. The way you've written it, it would not be obvious that only one term in that series actually survives.

    Yes, it's one of the first few spherical harmonics. Their orthogonality relationship is obviously a bit more complicated than for the Legendre polynomials, but the algorithm for solving the problem is basically the same.
     
  6. May 13, 2012 #5

    fluidistic

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    Hmm okay. Is my answer "right" though, but just "ugly" in the sense that it's not simplificated?
    You mean in my expression for Phi in the series with the index "m"?
    [tex]\Phi (r, \theta , \phi ) = \sum _{l=0}^ \infty \sum _{m=-l}^l \frac{\int Y ^ * _{lm}(\theta , \phi ) P_l ( \cos \gamma ) d \Omega }{a^{-(l+1 )}-b^{-2l-1}a^l} [r^{-(l+1)}-b^{-2l-1}r^l]Y_{lm} (\theta , \phi )[/tex]


    I think I understand their orthogonality and their normality. The inner product of 2 spherical harmonics is defined as an integral over a surface which is, if I'm not wrong, a sphere of radius 1.
    Also I don't find the particular spherical harmonics representing the potential of the outer sphere. I'm almost sure m must equal plus or minus 1.
     
  7. May 13, 2012 #6
    The steps you've taken are generally correct (although I don't think that last result is algebraically correct), but do not at all help you, because at the end you still don't have a form in which you can directly evaluate an integral relating to the potential at the outer surface. Let us write down the coefficients using the trick I described to figure them out:
    [tex]B_{lm} = \frac{1}{a^l + a^{-l-1}} \int \Phi(a,\theta,\phi) Y^*_{lm} d\Omega[/tex]
    [tex]A_{lm} = -b^{-2l-1} B_{lm}[/tex]
    Substituting this into the main solution:
    [tex]\Phi(r,\theta,\phi) = \sum_{l,m} \frac{\int \Phi(a,\theta,\phi) Y^*_{lm} d\Omega}{a^l + a^{-l-1}} \left[ -b^{-2l-1} r^l + r^{-l-1} \right] Y_{lm}[/tex]
    This is definitely where you want to go with solving this problem, instead of the excursion to rewriting the spherical harmonics in terms of the Legendre polynomials.

    Sorry for not being clearer about that -- look at [itex]Y^1_{\pm 1}[/itex]. The potential on the outer surface isn't quite either of those, but can you think of a linear combination of spherical harmonics that is?
     
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