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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?

nahano
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    possible duplicate of How to calculate spatial distance in space-time? – John Rennie Aug 09 '14 at 09:39
  • Yes. I've linked a previous question that explains how you calculate the length of the ruler in a gravitational field. – John Rennie Aug 09 '14 at 09:40
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    @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 09 '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 09 '14 at 10:36

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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.)

benrg
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  • How about an observer high in the gravitational field measuring the length of a long rod low in the field by determining the angle which it subtends when viewed from his position? – R.W. Bird Sep 30 '20 at 14:03
  • Sorry, but your symmetry argument is absurd. The gravitational length contraction directly follows from the Schwarzschild metric: $$ d\tau^2 = (\text{time dilation}),dt^2-(\text{length contraction}),dr^2 $$ and is a representation of the fact that a zero Ricci tensor preserves an infinitesimal spacetime hyper-volume in vacuum: $$(\text{time dilation})(\text{length contraction})=1$$ – safesphere Jun 23 '23 at 15:41
  • @safesphere The effect of Weyl (non-Ricci) curvature near a black hole is called spaghettification, and acts opposite the way you may be thinking: it stretches in the radial direction (and compresses in the two angular directions). It's not length contraction, since it doesn't affect objects with internal forces strong enough to maintain their shape, while length contraction is universal. The product of the $dt^2$ and $dr^2$ coefficients can be changed by rescaling $r$, so it isn't physically meaningful. – benrg Jun 23 '23 at 19:31
  • I never said the length contraction and time dilation were the effects of the Ricci curvature. I said the Ricci tensor was zero with no Ricci curvature. So yes, they are the effects of the Weyl curvature, but this is not the point. The point is that their relation expressed in my second formula above is indeed defined by the fact that the Ricci tensor is zero. This relation will change in the presence of matter where this tensor is not zero. The relativistic (velocity in SR or gravity in GR) length contraction does not physically affect objects like the tidal forces do in your example. – safesphere Jun 23 '23 at 20:39
  • I am puzzled why you object the obvious. Consider you are 6 feet tall and hover at, say, one tenth of the Schwarzschild radius from the horizon of a supermassive black hole with negligible tidal forces. If I observe you from afar in approximately Schwarzschild coordinates, what is your height in my coordinates? Do the math, it is a fraction of your actual height. And if you fall close enough to the horizon, you become Planck length thin in just one second. You see no change yourself and are not physically affected. This is the gravitational length contraction whether or not you like this term. – safesphere Jun 23 '23 at 20:48
  • In our coordinates, all matter (stars, etc.) that has ever fallen to an supermassive black hole remains as a layer above the horizon. What is the thickness of this layer in our coordinates? It is less than the Planck length while the stars are not physically compressed. Also, rescaling the radial coordinate means changing to different coordinates, which is a meaningless claim in this case. In SR we have a velocity length contraction that is canceled by changing to the rest frame, so what? When you apply some math procedure, you should always think of how it translates to physical measurements. – safesphere Jun 23 '23 at 20:56
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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

Qianyi Guo
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    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 09 '14 at 17:39
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    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. – nahano Aug 09 '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. –  Nov 07 '14 at 03:02
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Here's a non GR explanation. Those who prefer to stick to majority views may wish to ignore it.

Mass can be thought of as a collection of waves, each with the potential to do work. The total potential of all the waves to do work is the total potential of the mass to do work.

If we describe this in terms of the distance over which it could potentially apply a force, we get the well known mass-energy equivalence equation:

$E=mc^2$

If we describe it in terms of the time over which it could potentially apply a force, we get the less familiar equation:

$\rho=mc$

Whichever way we choose to describe it, we are describing the same potential to do work.

When a mass falls freely through a gravitational field, it neither does work nor has work done on it. It's potential to do work therefore remains constant as it falls. However, we know that it gains momentum [aka KE] and loses energy [aka GPE] on the way down (what we are seeing is a result of the changing ratio of total energy to total momentum). From this, and the above equations it can be logically deduced that the unit of time is dilating while the unit of length is contracting.

During free fall $ |dE/ds| = |d\rho/dt| = mg$

Which leads to the conclusion that the rate of length contraction is the same as the rate of time dilation, but we don't know how to measure it.

Alan Gee
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  • “the rate of length contraction is the same as the rate of time dilation - This is exactly correct in vacuum where the gravitational equations simplify to state that the Ricci tensor is zero. This tensor describes a change in an infinitesimal spacetime hyper-volume. Zero means this hyper-volume is conserved: if you stretch it in one dimension (time dilation), it would equally shrink in another dimension (length contraction). – safesphere Jun 23 '23 at 15:19