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Is there something intrinsic about the structure of space that gravity is proportional to 1/r^2 instead of, for example, 1/r^2.143 ? What makes the exponent turn out to be a nice even number?

Qmechanic
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Shookster
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1 Answers1

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It's just geometry. At any given distance $r$ from a massive particle, the force from that particle is spread out over the surface of a sphere of radius $r$. The surface of a sphere scales as the square of the radius. If you assume gravitational flux[1] is conserved, then flux density, and thus force at a given point, must decrease proportional to the increase in surface area over which it distributed. Thus, inverse-square.

As for why it's a nice even number, it's because we live in a universe with a nice integral number of dimensions. If we lived in a fractal space with some weird non-integral dimensionality, then maybe the exponent would indeed be something weird.

[1] Not a particularly commonly encountered concept, but one which shows up in the Gauss's Law formulation of gravity as the surface integral of gravitational field strength, analogous to magnetic flux.

  • But why is it that gravitational flux is perfectly conserved in our universe? I.e. why should laplace's equation hold for gravity? – Razor Jan 27 '24 at 23:34
  • @Razor That depends on what the fundamental theory of gravity actually is. If we assume gravitons exist, then it's because changes in the gravitational field propagate at c, so gravitons are massless, which means virtual gravitons can have zero energy and thus do not have an exponential decay term limiting their range. – Logan R. Kearsley Jan 28 '24 at 05:59
  • Hmmm, I was also thinking something along these lines. Can we invert this and assert that an exact integer exponent for $r$ is a good argument for existence of gravitons? Do you know if someone predicted existence of gravitational particles before the advent of GR and QFT? – Razor Jan 28 '24 at 14:59
  • Apparantly Laplace did – Razor Jan 28 '24 at 17:09