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From http://jila.colorado.edu/~ajsh/bh/schwp.html The amount by which radial distance is "squished" is:

$ \dfrac{1}{\sqrt{ 1 - r_s/r}}$

where the Schwarzschild radial distance (or "circumferential radius") r is the point at which measured circumference is $2 \pi r$ and the Schwarzschild radius is

$r_s=2 G M/c^2$

is the Scharzchild radius at which we get a black hole for that amount of mass ($M$). The amount by which time is dilated is the inverse of this (so pretty similar to the case for moving objects in flat space). Usually, this equation is integrated over dr to calculate the total distance between 2 points near a dense gravitational body.

Lets assume we have a thin spherical shell that is dense enough to make 50% time dilation immediately above the surface, which I believe comes out to

$r=(4/3)r_s$

The radial distance will be squished by 50% as well (though some people refer to this as "expansion" since you can fit more into the same space relative to the reference frame outside this gravitational well) and the circumference of the outside of the sphere will be

$2\pi(4/3)r_s$

According to Birchoff's theorem, the apparent gravitational field should be zero inside (just as it is for Newtonian gravity).

According to an answer to the question at Does a massive spherical shell expand the time inside itself?, "gravitational time dilation depends on the gravitational potential" so should be the same inside (which makes since if you think of photon red-shift as an indication of time difference).

It seems like you would still have the same length contraction inside as well (if it is just as much an effect of gravitational potential as time dilation) but it would be the same in all directions, which implies that the measured inner circumference of the sphere would be twice the outer circumference (so $2 (2\pi(4/3)r_s)$ ) and the inner (measured) radius would be $2 ((4/3)r_s)$.

Is this right?

Dustin Soodak

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    $\begingroup$ Welcome to physics.SE. Please mark up your question using mathjax. That means basically putting dollar signs around it and then using LaTeX math inside. There are tutorials you can find online for how to do math in LaTeX. $\endgroup$ – Ben Crowell Nov 28 '18 at 1:30
  • $\begingroup$ Got the LaTeX in...much easier than I thought it would be. $\endgroup$ – Dustin Soodak Nov 28 '18 at 18:24
  • $\begingroup$ Please note that the "Schwarzschild Metric" section in the first link is incorrect. $\endgroup$ – safesphere Nov 29 '18 at 14:05
  • $\begingroup$ what about it in particular? A typo in the equation for Schwarzschild Metric itself? $\endgroup$ – Dustin Soodak Dec 3 '18 at 22:41
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As stated in On a common misunderstanding of the Birkhoff theorem:

In Newtonian gravity (NG) it is known that the gravitational field anywhere inside a spherically symmetric distribution of mass is determined only by the enclosed mass. This is also widely believed to be true in general relativity (GR), and the Birkhoff theorem is often invoked to support this analogy between NG and GR. Here we show that such an understanding of the Birkhoff theorem is incorrect [...] The correct metric, matching continuously to the location of an external observer, is determined both by the enclosed mass and mass distribution outside. The effect of the outside mass is to make the interior clock run slower [...]

Further in the section 2.0 the author shows that "the time term of the metric is always maintained continuous, but the space term is not". There is no length contraction or expansion inside an empty shell, only time dilation.

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  • $\begingroup$ Just to make sure I understand this, if the circumference outside the sphere is $2\pi(4/3)r_s$, the circumference inside will be the same and the inner measured radius will be $(4/3)r_s$ but the time dilation will still be at 50%. This is the same as immediately outside where there is also 50% radial length contraction. Correct? $\endgroup$ – Dustin Soodak Nov 29 '18 at 7:28
  • $\begingroup$ Yes, this is correct, except it is a radial length expansion instead of contraction outside. The closer you get to the event horizon, the further you "see" it moving away from you. If you get really close, it would appear light years away. Also please note that 50% is valid only for a static shell. The concept remains the same for a collapsing shell, but the % would be different. See the answer of A.V.S. to: physics.stackexchange.com/questions/414695/… $\endgroup$ – safesphere Nov 29 '18 at 14:04
  • $\begingroup$ Thanks for the link and explanation. This has been bugging me for a while and the only answer I found turned out to be incorrect. I think of it as "contraction" because it is contracted relative to the outside reference frame so I'm glad you put in an example to clarify. Your response, however, brings up another question: how does something ever make it into a black hole according to their own reference frame if space keeps expanding exponentially in front of it as they get closer to the event horizon? Also, is the % difference for a collapsing sphere due to frame dragging? $\endgroup$ – Dustin Soodak Nov 30 '18 at 23:46
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    $\begingroup$ @DustinSoodak I haven't done the math (although it's not complicated), but intuitively, you "see" the event horizon so far away only if you hover over it using rocket engines to hold you steady. However, if you are in a free fall, then your speed approaches the speed of light and the length contraction due to the relativistic speed cancels out the gravitational length expansion. The % difference ($f(t)$ in the link, 50% for $\dot{R}=0$) for a collapsing sphere also is due to the relativistic speed of the collapse ($\dot{R}$). Frame dragging refers to rotation, but there is none in this case. $\endgroup$ – safesphere Dec 1 '18 at 1:05
  • $\begingroup$ So according to to an outside observer you might never reach it but from from your frame of reference, relativistic length contraction in your direction of motion will surpass the expansion. I didn't think to add in SR effects. $\endgroup$ – Dustin Soodak Dec 1 '18 at 5:11

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