Transition of spherically symmetric gravitation

  • #1
Imya
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TL;DR Summary
Contraction of a spherically symmetric body. An observer outside looks at the sphere and the metric around it. An observer at the center looks at distant stars.
Outside a static spherically symmetric body in vacuum lies the Schwarzschild metric. However, when observing a contractive sphere from a point near the sphere, it no longer appears spherical. Its opposite end appears stretched for an observer stationary relative to the center of the sphere. In this case, should gravity on the surface of the contractive sphere instantly transition to the Schwarzschild metric? Or is there a transitional metric above the surface of the contractive sphere with a thickness approximately ##2 \cdot r \cdot \frac{{dr}}{{dt}} \cdot \frac{1}{c}~?##
The initial radius of the sphere is considered a lot more than the gravitational Schwarzschild radius. The internal structure of the sphere is not limited in the above question.

And the related question. At the center of a spherically symmetric thin solid static shell lies a point observer. For this observer, distant stars appear violet shifted slightly more ##\frac{{G \cdot M}}{{{c^2} \cdot r}}## (##\frac{{G \cdot M}}{{{c^2} \cdot r}}## is for an observer on the surface of the sphere). The initial radius of the shell is much larger than the Schwarzschild gravitational radius (##r \gg {r_{\rm{g}}} = 2\frac{{G \cdot M}}{{{c^2}}}##). The shell instantly (simultaneously across the entire sphere from the center's perspective) changes its structure from solid to non-interacting cold dust (without emitting radiation to simplify the problem). The dust begins to accelerate and fall towards the sphere center.
For the observer at the sphere center, will the second derivative of the violet shift with respect to time appear at the same moment when he sees the transition of the sphere from solid to dust? Or slightly earlier by approximately ##<\frac{r}{c}##?
How will the light from distant stars further change for the observer at the center when the circumference of the sphere reaches approximately ##2 \cdot \pi \cdot \frac{{2 \cdot G \cdot M}}{{{c^2}}}##? And how will light change further?
Distant stars are assumed to be stationary.
 
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  • #2
Imya said:
In this case, should gravity on the surface of the contractive sphere instantly transition to the Schwarzschild metric?
Yes - see Oppenheimer-Snyder collapse for a fleshed-out example.
Imya said:
For the observer at the sphere center, will the second derivative of the violet shift with respect to time appear at the same moment when he sees the transition of the sphere from solid to dust?
It'll start as soon as he sees something start to move, which would be the inner surface of the sphere.

Note that I'm not sure you can have something instantly transition from solid to dust. You'd have to have some of the components of the stress-energy tensor be discontinuous in time, which I would expect to violate ##\nabla_aT^{ab}=0##. You could certainly consider a spherical shell of dust that was instantaneously stationary at the start of the model.
 
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  • #3
Imya said:
TL;DR Summary: Contraction of a spherically symmetric body. An observer outside looks at the sphere and the metric around it. An observer at the center looks at distant stars.

when observing a contractive sphere from a point near the sphere, it no longer appears spherical.
Symmetries in spacetime are given by Killing vector fields. These are standard GR vectors, so they are covariant. They reflect the underlying geometry and are independent of any coordinate system or observer.

The basic premise here is flawed. If the spacetime is spherically symmetric then it has the corresponding Killing vector fields in all coordinates for all observers.
 
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  • #4
See also the discussion about spacetime symmetries starting from this thread.
 

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