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I was seeing the generalisation of Newton's shell theorem to GR, including the answer given here: Is spacetime flat inside a spherical shell?, but I don't understand why proving that the metric inside the hollow shell is the flat spacetime metric tells us that the external shells have no gravitational effect on the mass inside them.

Could someone elaborate on it?

Qmechanic
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MicrosoftBruh
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1 Answers1

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We calculate the motion of a freely falling object using the geodesic equation:

$$ {d^2 x^\mu \over d\tau^2} = - \Gamma^\mu{}_{\alpha\beta} u^\alpha u^\beta $$

where the symbols $\Gamma^\mu{}_{\alpha\beta}$ are the Christoffel symbols, which are calculated from the metric. If the metric is the Minkowski metric and assuming we are using the usual Cartesian coordinates $(t, x, y, z)$ the Christoffel symbols are all zero and the geodesic equation becomes simply:

$$ {d^2 x^\mu \over d\tau^2} = 0 $$

And this is the equation of motion of a free particle.

So if the geometry inside the shell is described by the Minkowski metric this implies there are no gravitational forces inside the shell.

John Rennie
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    "if the geometry inside the shell is described by the Minkowski metric this implies there are no gravitational forces inside the shell". But this has nothing to do with shells. If the geometry is described by the Minkowski metric this implies there are no gravitational forces. – Charles Francis Dec 10 '20 at 19:20
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    But since we obtain the metric in spherical coordinates the Christoffel symbols would not be zero. How do we handle that part? – MicrosoftBruh Dec 12 '20 at 12:16
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    @MicrosoftBruh you can work in the rest frame of the particle. Since the geometry is Minkowski it is unaffected by a Lorentz transformation. In this frame $\mathbf U = (c,0,0,0)$ so the only non-zero terms are the ones involving $\Gamma^0{}_{\alpha\beta}$, and these are all zero even in polar coordinates. – John Rennie Dec 12 '20 at 12:39