# Gravitational Length Contraction

General Relativity predicts that a clock at rest in a gravitational field will run slower than a clock in free fall. Similarly, will a vertical ruler on the earth's surface be shorter than a ruler in free fall? Why or why not and by how much?

• possible duplicate of How to calculate spatial distance in space-time? – John Rennie Aug 9 '14 at 9:39
• Yes. I've linked a previous question that explains how you calculate the length of the ruler in a gravitational field. – John Rennie Aug 9 '14 at 9:40
• @JohnRennie : It is not so simple. Your calculus has been made at fixed Schwarzschild radial coordinates, and this is not the case for a free fall ruler. However, the question seems ill-posed. If it is the free fall observer which is observing the free fall ruler, there is no gravitationnal effect. If it is an observer at fixed radial coordinates which is observing the free fall ruler, we have to take in account (admitting that the free fall ruler has zero speed for the synchronization at one extremity), speed acquired by the ruler when the other extremity synchronizes with the fixed observer. – Trimok Aug 9 '14 at 10:34
• @Trimok: Ah yes, I didn't read the question carefully enough. My calculation compares the lengths of rulers fixed at different radial distances, not a fixed and freely falling ruler. I'll remove the VTC. – John Rennie Aug 9 '14 at 10:36

The reason it makes sense to talk about gravitational time dilation is that the gravitational field solution (Schwarzschild geometry) has a time-translation symmetry. If you're hovering at a certain altitude (all your Schwarzschild coordinates are constant except for $t$) and emit two light pulses at times separated by $\delta t$, it follows immediately from symmetry that any hovering receiver that detects those signals will detect them at a separation of $\delta t$. But that's coordinate time in both cases, which is related to proper time by the local metric. So actually the receiver will see a redshift or blueshift given by the ratio of (the square root of) the appropriate component of the metric at each location. Thus you can consistently think of this metric component as a "time dilation factor" and get the right answer.

There's nothing analogous for length contraction. This metric doesn't have a spatial translation symmetry. Even if it did, the idea of two light beams being emitted from different coordinate positions, and received with the same coordinate separation but a different proper separation, doesn't seem as deserving of the name "length contraction" as the former case seems to deserve the name "time dilation". FLRW cosmology is symmetric under spatial translations (and not time translations), and you can make exactly the above argument with $\delta r$ instead of $\delta t$. In fact this is the easiest (and usual) way of deriving the cosmological redshift formula. But no one calls it "length contraction". Maybe they would if the universe were contracting instead of expanding.

(edit: Since the Schwarzschild metric does have rotational symmetry, you can make this argument with an angular displacement $\delta\theta$, showing that the emitted and received physical distances are related by the ratio $r_\mathrm{emitter}/r_\mathrm{receiver}$. But it would be silly to call that length contraction—for starters, it's true in Newtonian physics also. It's kind of silly to call the $\delta t$ case time dilation too, since it's geometrically similar to this case, and as far as the universe is concerned, that intrinsic geometry is all that matters. The laws of physics, being local, can't even see the global time-translation symmetry that leads us to introduce the Schwarzschild coordinates that make the time-dilation picture seem sensible.)

Yes. We can figure out

$dr = ds \sqrt{1-\frac{2GM}{rc^2}}$.

Here is a straightforward derivation: http://www.physicsforums.com/showthread.php?t=404153

• You can calculate that quantity, but it doesn't mean anything in isolation. It's the ratio between the physical length of the ruler and the unphysical coordinate length. If you picked different coordinates you'd get a different, equally meaningless number. The only way this particular formula might make sense is if you were considering, say, a rigid hoop encircling a gravitating body (like Ringworld). In that case $r$ has physical meaning as the reduced circumference (= physical circumference divided by $2\pi$). – benrg Aug 9 '14 at 17:39
• I don't see how this is meaningless; you could compare the ends of the rulers the moment the free fall one passes the stationary one. – jeffythedragonslayer Aug 9 '14 at 23:35
• @dacodemonkey: The experiment you propose is a local experiment. Locally, GR is the same as SR. Therefore the results of your experiment would be the same as in Minkowski space. – Ben Crowell Nov 7 '14 at 3:02