In general relativity, $\sqrt{1-v_{e}^2}$ (with $v_e$ the escape velocity as a fraction of the speed of light) is the factor by which time is dilated in a gravitational field. Is it also the factor by which objects get contracted?
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2$\begingroup$ Gravitational length contraction is not a thing. See John Rennie's answer here for why such a thing is not defined. $\endgroup$– J. MurrayCommented Feb 27, 2018 at 1:21
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$\begingroup$ @JMurray: My question does not involve black holes or event horizons. Your reference does mention radial lengths expansions. I'm asking: what is the factor as a function of $v_e$. $\endgroup$– Pierre BerriganCommented Feb 27, 2018 at 2:50
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1$\begingroup$ The presence of black holes is irrelevant. The point is that the idea of gravitational length contraction does not make sense. $\endgroup$– J. MurrayCommented Feb 27, 2018 at 3:34
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$\begingroup$ I fail to understand why you would insist on the question not making sense. John Rennie's answer that you refer me to, says (I quote:) "...there is a sense in which radial distances are stretched not contracted..." and even gives the Schwarzschild metric deriviate to measure it. $\endgroup$– Pierre BerriganCommented Feb 27, 2018 at 13:17
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As mentioned in the answer linked in the comments, the notion of gravitational length contraction doesn't make sense.
Special Relativity
Imagine the two of us standing next to one another, holding identical, synchronized clocks. You then start sprinting around in circles while I sit still. When you are finished, you come back and compare your clock to mine, and find that they no longer agree. This is time dilation.
Now imagine the two of us standing next to one another, holding identical meter sticks. You start sprinting around again. This time, as you pass me, we each hold out our meter sticks, and when we line them up next to each other, we each observe the other's meter stick to be a bit shorter. This is length contraction.
General Relativity
Imagine the two of us standing next to each other, holding identical, synchronized clocks. Now you take your clock to a location of different gravitational potential - maybe to the top of a mountain - and sit there for a while. When you come back down and we compare our clocks, we see that the clocks no longer agree. This is gravitational time dilation.
Now, I'm sure you knew all of that. What I'm trying to make clear is that there is no analogous way to talk about gravitational length contraction. We might start off as before, but then it breaks down:
Imagine the two of us standing next to one another, holding identical meter sticks. You then travel to a region of different gravitational potential. When you come back ... the gravitational potential is now the same as it was before, so nothing has happened, and your stick is the same length as mine.
The trouble is that distances need to be compared locally - but if two meter sticks are in the same region of specetime, then they have the same gravitational potential, so there is no way for any gravitational length contraction effect to cause a discrepancy between them.
In the linked question, John Rennie makes the statement
[...] there is a sense in which radial distances are stretched not contracted [...]
but this is not the effect you are talking about. This refers to the fact that if you draw a circular loop around the Earth, then you could measure its circumference $C$ and divide by $2\pi$ to get the distance to the Earth's core.
However, the proper distance to the Earth's core is not the same the number you just calculated. It would be equal if the space enclosed by the loop were flat, but because of the curvature, the proper distance to the core is a bit larger than $C/2\pi$.
This does not refer to length contraction in any meaningful way - rather, it expresses the fact geometry in curved space is not the same as it is in flat space.
answered Feb 27, 2018 at 16:39
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$\begingroup$ "This does not refer to length contraction in any meaningful way" The formula for the proper distance at $r$ is $ds = \gamma dr$ if the escape velocity at $r$ is used as $v$. This implies to me that you can replace the notion of curved spacetime with that of space flowing into the planet—an accelerating river (see "the river model") that flows through and accelerates everything in it and causes the very time dilation and length contraction that moving through space causes in SR. So isn't it possible that the same (or related) effects are at play and that it's just a matter of interpretation? $\endgroup$ Commented Apr 4, 2022 at 19:02
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$\begingroup$ @GumbyTheGreen The expression $\mathrm ds = \sqrt{1-2GM/r} \mathrm dr$ arises because the so-called "radial coordinate" $r$ of a point $p$ is not the distance between $p$ and the coordinate origin. Instead, we observe that $p$ lies on a sphere which is centered at the origin; $r$ is then the surface area of that sphere divided by $4\pi$. On a flat spacelike slice of spacetime, this is also equal to the radial distance to the origin, but in general it is not; in this case, the slice obtained by setting $t=$constant is not flat. The larger point which I try to express in my answer is [...] $\endgroup$ Commented Apr 4, 2022 at 19:36
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$\begingroup$ [...] that the entire concept of gravitational length contraction is ill-defined. If you start with two identical rulers and transport one into a gravity well, then any local measurements of their lengths will return the same values. $\endgroup$ Commented Apr 4, 2022 at 19:44
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$\begingroup$ I understand all that and I think I've accounted for it. I'm just saying that each slice of curved space corresponds to a slice of flat space and is longer than it to a degree that arguably matches the length contraction formula of SR. E.g., if $ds=2dr$ at some $r$, then an external observer would see lengths contract to half their normal value at that point. I'm not referring to local measurements. I know you've said that "distances need to be compared locally" but I don't see why that should be the case when the metric allows us to convert coordinate distances to and from [...] $\endgroup$ Commented Apr 5, 2022 at 2:57
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$\begingroup$ [...] proper distances as shown in this answer and this one. For reference, the equality of formulas I'm referring to can be seen here. Btw, minor correction: I assume you meant to say that "$r$ is then the square root of the surface area of that sphere divided by $4\pi$." $\endgroup$ Commented Apr 5, 2022 at 4:10