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Graviton Monopole

  1. Feb 9, 2006 #1

    What is the solution for a static graviton flux monopole through a closed surface? :rolleyes:

    Sphere surface area:
    [tex]dA = 4 \pi r^2[/tex]
    [tex]\Phi_G = \oint G_g \cdot dA = \oint G_g \cdot (4 \pi r^2)[/tex]
     
  2. jcsd
  3. Feb 9, 2006 #2

    dextercioby

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    What is Gg ...?

    Daniel.
     
  4. Feb 9, 2006 #3

    G_g is the gravitic field strength: (arbitrarily chosen symbol)
    [tex]G_g[/tex]
     
  5. Feb 10, 2006 #4
    Are you talking about arbitrary closed surface or about sphere. If you are talking about sphere then it is quite simple:
    First thing: Flux is [tex]\Phi=\oint_S \vec{G} \vec{dA}[/tex]. Second thing [tex]\vec{dA}[/tex] in spherical coordinates (which are most appropriate because problem has a spherical simmetry) is [tex]\vec{dA}=R^2\sin\theta d\phi d\theta\vec{e_r}[/tex]. Integration is trivial because [tex]\vec{G}=\frac{\gamma m}{R^2}\vec{e_r}[/tex].
    In case of arbitrary surface we have [tex]\oint_A \vec{G}\vec{dA'}=\int_V \nabla\vec{G}dV'=-\int_V \nabla^2{\phi_g}dV'=-\gamma m\int_V \nabla^2\frac{1}{|\vec{r}-\vec{r'}|} dV'=4\pi\gamma m\int_V \delta(\vec{r}-\vec{r'})dV'=4\pi\gamma m[/tex], where V is such that [tex]A=\partial V[/tex].
     
    Last edited: Feb 10, 2006
  6. Feb 12, 2006 #5

    [tex]G_g = \frac{\gamma m}{r^2}[/tex]

    Gauss' Law for gravitation:
    [tex]\Phi_g = - \left( \frac{\gamma m}{r^2} \right) (4 \pi r^2) = - 4 \pi \gamma m[/tex]
    [tex]\boxed{\Phi_g = - 4 \pi \gamma m}[/tex]

    [tex]\Phi_g = \oint_A \vec{G} \vec{dA} = - \oint_A \frac{\gamma m}{r^2} \vec{e_r} \cdot r^2 \sin \theta d\phi d\theta \vec{e_r} = - \gamma m \int_a^b \left[ \int_c^d \sin \theta d\phi \right] d\theta}[/tex]
    [tex]\boxed{\Phi_g = - \gamma m \int_a^b \left[ \int_c^d \sin \theta d\phi \right] d\theta}[/tex]
     
    Last edited: Feb 12, 2006
  7. Feb 12, 2006 #6
    First expression you wrote is meaningless. You should have some differential form under integration. Expression would be OK if you had integral sign removed. Third expression is OK but you didn't perform integration. You should integrate [tex]\theta[/tex] in boundaries from 0 to [tex]\pi[/tex] and [tex]\phi[/tex] in boundaries from 0 to [tex]2\pi[/tex].
     
  8. Feb 12, 2006 #7

    [tex]\Phi_g = \oint_S \vec{G} \vec{dA} = - \int_0^{\pi} \left[ \int_0^{2 \pi} \frac{\gamma m}{r^2} \vec{e_r} \cdot r^2 \sin \theta d\phi \vec{e_r} \right] d\theta[/tex]

    [tex]- \int_0^{\pi} \left[ \int_0^{2 \pi} \frac{\gamma m}{r^2} \vec{e_r} \cdot r^2 \sin \theta d\phi \vec{e_r} \right] d\theta = - \gamma m \int_0^{\pi} \int_0^{2 \pi} \sin \theta d\phi d\theta = -2 \pi \gamma m \int_0^{\pi} \sin \theta d\theta[/tex]

    [tex]-2 \pi \gamma m \int_0^{\pi} \sin \theta d\theta = - 4 \pi \gamma m[/tex]
    [tex]\boxed{\Phi_g = - 4 \pi \gamma m}[/tex]
     
    Last edited: Feb 12, 2006
  9. Feb 14, 2006 #8

    These are my equations for Yukawa Flux.

    [tex]g[/tex] - gauge coupling constant
    [tex]S_n[/tex] - Yukawa strong nuclear field strength

    Yukawa Strong Nuclear Flux:
    [tex]S_n = - \frac{g}{e^{\frac{r}{r_0}} r^2}[/tex]
    [tex]\Phi_n = \oint_S S_n dA = - \int_0^{\pi} \left[ \int_0^{2 \pi} \frac{g}{ e^{\frac{r}{r_0}} r^2} \cdot r^2 \sin \theta d\phi \right] d\theta = - g \int_0^{\pi} \left[ \int_0^{2 \pi} \frac{\sin \theta}{e^{\frac{r}{r_0}}} d\phi \right] d\theta[/tex]

    [tex]- g \int_0^{\pi} \left[ \int_0^{2 \pi} \frac{\sin \theta}{e^{\frac{r}{r_0}}} d\phi \right] d\theta = - 2 \pi g \int_0^{\pi} \frac{\sin \theta}{e^{\frac{r}{r_0}}} d\theta = - \frac{4 \pi g}{e^{\frac{r}{r_0}}}[/tex]

    [tex]\boxed{\Phi_n = - \frac{4 \pi g}{e^{\frac{r}{r_0}}}}[/tex]

    However, every other flux through a closed surface, such as Maxwell's equations and Gauss' Law for gravitation is only dependent on gauge. This flux equation is dependent on both gauge and radius. Are these equations correct?
     
    Last edited: Feb 14, 2006
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