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Symmetry of mass distribution producing radial gravitational force field

  1. Dec 10, 2007 #1

    mma

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    I conjecture that only spherical symmetric mass distribution can produce radial gravitational force field (according Newton's law of gravitation).

    Can anybody prove or disprove this conjecture?
     
  2. jcsd
  3. Dec 10, 2007 #2

    quasar987

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    If [itex]\mathcal{G}(r)[/itex] is the gravitational field, that we assume depends only on r, then the field and the mass distribution are related by Gauss' law:

    [tex]\nabla\cdot\mathcal{G}(r)=\rho(r,\theta,\phi)[/tex]

    clearly this implies that rho depends only on r... i.e., the mass distribution is sperically symmetric.
     
  4. Dec 11, 2007 #3

    mma

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    I don't suppose that the gravitational field depends only on r. That's why I wrote "radial" and not "central". Radial means here that there is a scalar field s(p) and a point o that our force field is F(p) = s(p)(p-o) (p and o are space points, space is regarded to an 3-dimensional affine space). The conjecture is that the source of such gravitational field can only be a spherically symmetric mass distribution.
     
  5. Dec 11, 2007 #4

    quasar987

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    Oh, I understand. In this case, we can assume o is the origin. Then the hypothesis is that

    [tex]\mathcal{G}=s(r,\theta,\phi)r\hat{r}[/tex]

    and now the curlless nature of [itex]\mathcal{G}[/itex] (gravitostatic) gives that s is theta and phi independent. Then use Gauss's law as used as in post #2 gives that rho is theta and phi independent.
     
    Last edited: Dec 11, 2007
  6. Dec 12, 2007 #5

    mma

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    This proof seems very impressive and elegant!

    But, are you absolutely sure that [itex]\mathcal{G}[/itex] is curlless?

    I know that it must be curlless if it has a potential, and it must have a potential if it's divergence is zero. But it's divergence isn't zero at points where the mass density isn't zero.
     
  7. Dec 12, 2007 #6

    quasar987

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    Not 100% sure... for that I'd have to pick up my E&M book and confirm that all the steps taken to derive curl(E)=0 transfer to gravity, but I dont have my physics book anywhere near me. But I'm 95% sure.
     
  8. Dec 12, 2007 #7

    mma

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    I'm afraid that if I take an arbitrary smooth [itex]\mathcal{G}[/itex] vector field, then it's divergence is just a mass distribution which generates [itex]\mathcal{G}[/itex]. If [itex]\mathcal{G}[/itex] has a non-vanishing curl, then this example will be in your 5%.
     
  9. Dec 12, 2007 #8

    quasar987

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    Since we forgot the mathematical argument for curl(E)=0, look at it this way. It is a know fact from physics that gravitational fields are conservative. And we remember that this statement is equivalent to curl(G)=0. Voilà.
     
  10. Dec 13, 2007 #9

    mma

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    Yes, of course, a circulating force field would violate the conservation of energy. But how can we prove mathematically that it is impossible? What excludes such density distributions?
     
  11. Dec 13, 2007 #10
    Why does the violation of conservation not count as a mathematical proof?
     
  12. Dec 13, 2007 #11

    quasar987

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    Just grab Griffiths and flip through the first few pages. Or Simon's (Simmon?) mechanics.. and the chapter on gravitational fields.
     
  13. Dec 15, 2007 #12

    mma

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    I've got the correct answer:

    According Newton's law the gravitation al field of a given [tex]\rho[/tex] mass distribution is:

    [tex]\mathcal{G}(p)=\int_{R^3}{\rho(q)\frac{(p-q)}{|p-q|^{3}}{dq}[/tex]

    (the integral is a volume integral on the whole 3-dimensuional space)

    Taking an arbitrary smooth closed curve [tex]\mathcal{C}[/tex] , the circulation of [tex]\mathcal{G}[/tex] on it is:


    [tex] \int_{\mathcal{C}}{\int_{R^3}{\rho(q)\frac{(p-q)}{|p-q|^{3}}{dq}}{dp}

    = \int_{R^3}\rho(q){\int_{\mathcal{C}}{\frac{(p-q)}{|p-q|^{3}}{dp}}{dq}
    [/tex] (the integrals commute according Fubini's theorem)

    and the inner integral is zero for every q, because the [tex]\frac{(p-q)}{|p-q|^{3}}[/tex] function has a potential, hence it is circulationless. So, our [tex]\mathcal{G}[/tex] is also circulationless, i.e. curlless.


    With this small supplement, your proof is perfect, thank you, quasar987!
     
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