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Computing Potential of a Spherical Shell w/o using Newton's theorems

  1. Feb 23, 2007 #1
    1. The problem statement, all variables and given/known data
    I'm well aware of how to compute the gravitational [electric] potential [tex]\Phi[/tex] due to a spherical mass [charge] distribution of radius [tex]R[/tex] by using Newton's theorems for spherical shells. However, how does one find an analytic expression for [tex]\Phi[/tex] without invoking these theorems?

    2. Relevant equations
    [tex]\rho(r, \theta, \phi) = \rho(r) = M \delta(r - R)[/tex]
    or equivalently:
    [tex]\rho(\vec{x}) = M \delta(x_1^2 + x_2^2 + x_3^2 - R^2)[/tex]

    [tex]\Phi(\vec{x}) = -\int_{\tau} \frac{G \rho(\vec{x}')}{|\vec{x}' - \vec{x}|} \; d^3x'[/tex]

    3. The attempt at a solution

    Now I can't simply transform the equation for potential into spherical coordinates because vector subtraction is involved. Furthermore, taking the line integral of gravitational force does not make things easier, since [tex]\vec{F} = \int_{\tau} \frac{G \rho(\vec{x}')(\vec{x}' - \vec{x})}{|\vec{x}' - \vec{x}|^3} \; d^3x'[/tex].

    Any help is welcome.
     
  2. jcsd
  3. Feb 23, 2007 #2
    why not integrate in polar coordinates? I'm sure you could get rid of [tex]\int d \theta[/tex] and [tex]\int d \phi[/tex] then.
     
  4. Feb 23, 2007 #3

    Meir Achuz

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    It is not clear ffrom your question whether the sphere is solid or just one shell.
    If the charge distribution is spherically symmetric (as you seem to imply), you can use Gauss's theorem, which poor Newton didn't kinow.
     
  5. Feb 23, 2007 #4
    My apologies. I had originally meant a spherical shell, from which the potential of an arbitrary spherically symmetric distribution can be computed. Gauss's theorem is certainly very useful, but the objective here is to compute the potential using the equation for [tex]\Phi[/tex] only, without applying any such simplifications.


    @ whatta: Converting to spherical/polar coordinates is the most efficient method, but how does one do so with factors of [tex]\vec{x}' - \vec{x}[/tex]?
     
  6. Feb 23, 2007 #5

    Jonathan Scott

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    If you want to compute the potential as seen at a given point, you can just use basic trigonometry to calculate the distance (vector magnitude) from that point to each infinitesimal volume of the sphere then integrate the corresponding expression for the potential over the shell.

    I remember that at school (something like 35 years ago) I was given this problem for a solid sphere as an integration exercise, but I noticed that if I integrated in some way over a thin segment (like a segment of an orange) where the axis of the segment pointed to the location where the potential was being measured, using some geometry with right angles and so on, so many things cancelled that I didn't have to actually do any integration. However, as far as I remember, it was probably easier to use a less ingenious model and do the integration.
     
  7. Feb 23, 2007 #6
    Thanks everyone, I finally figured it out. Converting to spherical coordinates and using the law of cosines to represent the norm of the vector difference is needed.
     
  8. Jan 4, 2009 #7
    Man, this topic is quite old but I'm very interested in knowing how this problem is solved. I hope the topic provider os still there. How did you figure this problem out if you don't mind that I ask? I'm in a rush actually and really want to know how this problem is solved. Thanks!

    PS: By the way, I'm the first time user. I think I made a mistake by sending the message via "report." Please ignore it the "report reply."
     
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