As space-time is distorted in a gravitational field, relativistic effects such as time dilation and length contraction take effect.

Time dilation is explained simply enough: closer to the source of gravity, slower the time passage.

However, space contraction gives no such clear answer. for example, a single thread on the matter contains the following:

  • gravitational field would produce a transverse expansion of distances
  • gravitational fields produce increased distances in the direction perpendicular to the field
  • [gravitational fields] produce decreased distances in the radial direction

So, to pose the question as bluntly and directly as I can:

Ignoring all other effects (relativistic or otherwise), which is shorter: a measuring stick closer to the source of gravity or a farther one?

If the orientation of the stick plays a part, try to answer for both the case where it's pointing towards the source of gravity, as well as perpendicular.

Similarly, if the position of the observer relative to source of gravity plays a part try to take that into account (but it doesn't seem to in the case of time dilation so I don't expect it here either)

  • $\begingroup$ for example, a single thread on the matter contains [...] Please give a link so we can see what you're talking about. $\endgroup$ – user4552 Nov 7 '14 at 2:10
  • $\begingroup$ possible duplicate of Gravitational Length Contraction $\endgroup$ – user4552 Nov 7 '14 at 2:10
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    $\begingroup$ @BenCrowell - the thread I quoted was physicsforums.com/threads/… but i fear that it does not clarify my question $\endgroup$ – barney.tearspell Nov 7 '14 at 2:15
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    $\begingroup$ @BenCrowell - should not be a duplicate since that question specifically asks about the difference between static and free-falling length dilation (or at least from the answers point of view), even though i think the person who asked the question tried to find out the same thing i do $\endgroup$ – barney.tearspell Nov 7 '14 at 2:22
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    $\begingroup$ i was under the impression that the curvature of time-space gives rise to effects analogous to movement at relativistic speed.. and that that is the source of the space-time dilation-contraction effects. Furthermore, I'm pretty certain that the time dilation due to gravity is a real effect: as-in clocks at different altitudes will tick at different speeds (requiring adjustments for GPS satellite clocks). Benrg's answer seems to disagree with that. $\endgroup$ – barney.tearspell Nov 7 '14 at 3:17

Your question isn't well defined, because there is no way to compare the two measuring sticks.

I can take two clocks, put them at different heights, then bring them back together and I will find that they now show different times due to time dilation. However if I take two rulers, put them different distances from a black hole then bring them back together I will find that they are still the same length. The reasons for this are covered in the answer Ben Crowell linked above.

However there is a sense in which radial distances are stretched not contracted (as rockandAir8747 says in their answer). But to understand what this means you need to understand what we mean by the Schwarzschild radial distance. The Schwarzschild coordinates use a radial coordinate $r$ that we naively interpret as the distance from the centre of the black hole. However we cannot measure the distance to the centre of the black hole because there's an event horizon in the way. So what we do is lay out our measuring tape in a circle centred on the black hole and measure the circumference of this circle. Because we know that in flat space the circumference is $2\pi r$ we can take our measured circumference and divide by $2\pi$ to get the Schwarzschild radial distance $r$.

But this $r$ coordinate is calculated assuming space is flat, so it shouldn't be any surprise that it is different from the distance you'd measure if you let down a measuring tape towards the event horizon. There's a nice image showing this difference in the question Difference between coordinate and proper distance in Schwarzschild geometry:

Radial distance

The $dr$ in the bottom diagram is the distance in Schwarzschild coordinates while the $dr$ in the top diagram is the distance you'd measure with a ruler, and it's obviously bigger than the Schwarzschild $dr$.

I suppose in this sense you could claim that the ruler is length contracted near to the black hole, because it takes more rulers than you expect to cover the distance $dr$.

If you want the gory details, including an equation for how to calculate lengths measured towards the event horizon then see the questions:

The measured distance, s, between the two Schwarzschild radii $r_1$ and $r_2$ is given by:

$$ s = \left[ r\sqrt{1-\frac{r_s}{r}} + \frac{r_s}{2} log \left( 2r \left( \sqrt{1-\frac{r_s}{r}} + 1 \right) - r_s \right) \right]_{r_1}^{r_2} $$

  • $\begingroup$ Thank you for this. At first glance, the answer seems to be (or contain) what i'm looking for, though it's a lot to digest on first read for a layman. Can you help me refine the question? -- in movement at relativistic speed you can decidedly say that length is contracted in direction of motion, even though a unit is still a unit, and a sphere will 'look' like a sphere (even though it becomes a disc) from all reference frames unless you discount all effects except relativity. I was hoping such a perspective can be applied in this case as well. $\endgroup$ – barney.tearspell Nov 7 '14 at 9:33
  • $\begingroup$ I apologize in advance for crack-sci-talk, but it's hard to pose the question correctly without knowing and understanding what form the answer should be. However, another way to pose the same question might be: if curvature of space is viewed as space density (does that make any sense?) is it getting denser or rarer in presence of a gravitational field? $\endgroup$ – barney.tearspell Nov 7 '14 at 9:37
  • $\begingroup$ @JohnRennie: "However if I take two rulers, put them different distances from a black hole then bring them back together I will find that they are still the same length.". Why SR does not seem to have this problem? $\endgroup$ – bright magus Nov 7 '14 at 9:53
  • $\begingroup$ @brightmagus: because I can measure the length of a moving ruler as it passes me i.e. I can do a local measurement to determine its length. $\endgroup$ – John Rennie Nov 7 '14 at 10:02
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    $\begingroup$ trying to propose a thought experiment: if a distant observer measures two sticks that are equidistant to the observer, which would cover more of an arc: one closer to a source of gravity or a farther one. (assuming again that all other effects - e.g. gravity lensing - are discounted) $\endgroup$ – barney.tearspell Nov 7 '14 at 10:08

length is stretched radially relative to the longitudinal and latitudinal directions. It is stretched, not contracted. Its best to think of what is happening as do to a curvature of spacetime. Every place in curved space looks locally like Minkowski spacetime. I've not come across of accounts that i remember as describing general relativity effects as do to time dilation and such. The norm is by far to use curved space time to describe events. notice near a black hole that there are two direction that have axial symmetry. The radial and time directions do not have that symmetry.

  • $\begingroup$ so if I'm reading this correctly, a measuring stick parallel to the surface of earth, at sea level would be marginally longer than the same stick in orbit? (discounting all other effects except gravity) however they would be equal if they were perpendicular to the surface? $\endgroup$ – barney.tearspell Nov 7 '14 at 2:59
  • $\begingroup$ This answer throws together a bunch of ideas, but I don't see any logical thread relating them. The first sentence is the only one that appears to relate to the question, and it's incorrect for the reasons given in benrg's correct answer to the question that this one duplicates. $\endgroup$ – user4552 Nov 7 '14 at 3:00
  • $\begingroup$ @barney.tearspell: a measuring stick parallel to the surface of earth, at sea level would be marginally longer than the same stick in orbit? This isn't a meaningful statement, for the reasons discussed in the other question. $\endgroup$ – user4552 Nov 7 '14 at 3:00
  • $\begingroup$ well, i wouldn't use stretching or contracting to describe it. This is being made to complicated. What you have is something like space being distorted into a funnel shape near a black hole or the earth. Length is kind of a relative thing. Its not easy to think of how exactly distant lengths relate to each other, The two directions perpendicular to the radial direction are more immediate things that can be used. What you have here is not so much as a contraction or stretching, but a tilting of a surface embedded in higher dimensions. $\endgroup$ – rockandAir8747 Nov 7 '14 at 3:22
  • $\begingroup$ @rockandAir8747, "Its not easy to think of how exactly distant lengths relate to each other," and yet somehow SR does not seem to perceive this as a problem. $\endgroup$ – bright magus Nov 7 '14 at 7:47

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