# Gravitational Length Contraction and Dilation?

1. May 18, 2010

### D.S.Beyer

My interest surrounds the qualities and affects of gravitational length contraction (GLC).

Like most other General Relativity circumstances GLC is difficult to measure because of frame of reference issues (such that a meter in space is longer than a meter on earth but in both frames of reference a meter measures a meter).

But what really confounds me are the affects of GLC as it relates to a broader view of spacetime. Below is a question regarding distance:

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A, B and C are points in 'free space'. B views the distance between A and C to be 1 parsec. A and C both agree that they appear 1 parsec from each other. The 'fence posts' are simply a graphical way of saying that every distance is the same relative to each other.

A stray neutron star has moved between A and C. The neutron star has contracted the space around it.

My questions are:
"Does the distance between A and C increase?"

"If the distance does not increase, then, since the lengths around the neutron star are contracted are other distances dilated (stretched) to maintain a constant distance?"

"If the distance does not increase and other space is not dilated, do A and C 'pull' closer to each other, even if they are well outside the neutron stars gravitational influence? And does B view this?"

(Thanks for putting up with my photoshop skills. Slowly I am beginning to understand the bizarre construct of spacetime, in great part due to the physics forums. Thanks everyone!).

2. May 19, 2010

### bcrowell

Staff Emeritus
You've clearly made an effort to formulate this clearly so that the question has an unambiguous answer, but I don't think you've quite succeeded.

Let's consider this simply in the case of Newtonian mechanics. If your scenario is going to be well posed, then it should provide a way of verifying, in the Newtonian context, that there is no length contraction or expansion. But A, B, C, and the neutron star are all physical objects that move in certain ways. You can't just use them as signposts that mark certain locations in space, because they would be violating the laws of physics if they were standing still.

In GR, there is a general expectation that gravitational fields produce increased distances in the direction perpendicular to the field. This is not a rigorously provable statement, or even one that has a rigorously well defined meaning, but there are reasonable plausibility arguments for thinking about it that way. For a good presentation of this argument, see Max Born, "Einstein's Theory of Relativity," Dover, 1962, pp. 318-320. Briefly, the argument is that if you use rulers to measure the cicumference of a rotating disk, the rulers are length contracted, so the circumference comes out to be greater than $2\pi r$. By the equivalence principle, the acceleration felt by someone on the disk is equivalent to a radial gravitational field. Therefore we would expect that any gravitational field would produce a transverse expansion of distances. But none of this can be made really rigourous, because concepts like gravitational fields don't behave in GR the way you'd think they would.

3. May 19, 2010

### yuiop

There are a number of things to take into consideration here. First, the light paths taken between A, B and C are curved when the neutron star is present. This would make the triangle look bloated, like the rotor of a wankel engine. The gravitational lensing effect of the curved paths makes the visual distance between A and C look greater from the point of view of an observer at B using subtended angle. A photon travelling from A to C takes longer than would be crudely expected even after taking the curved path into account. This is know as the "Shapiro time delay" and is worth looking up, as it relates closely to the questions you ask. Next, you have to consider the difference between distance and infinitesimal rod lengths. Vertical gravitational length contraction of measuring rods makes distances between widely separated locations appear longer in the vertical direction, because it takes more length contracted rods to span the distance when placed end to end. So with the neutron star present, the distance between A and C as measured by short rulers and the time it takes a photon to travel that distance are both longer, than when the neutron star is not present. They also "look" further apart from the point of view of B. So hopefully you understand now that when you ask if "lengths" close to A are "stretched" you have to make it clear whether you mean measuring rods are stretched in coordinate terms (they are not) or if you mean lengths (as in extended distances) are stretched as measured by short rulers or radar measurements (in which case they are). Basically, in just about every way you can make a practical measurement, the distance between A and C is greater when the neutron star is present.

Last edited: May 19, 2010
4. May 19, 2010

### yuiop

Valid points, Ben. However, we can for example consider something like the Solar system where orbiting bodies have velocities that are significantly lower than the speed of light, that we can ignore the relativistic effects of their orbital velocities and focus on the gravitational relativistic effects.

You are right that an observer on the rim of a rotating disk would measure the circumference to be longer than the measurement made by a non rotating observer, if the rotating observer measures the circumference using short rulers placed end to end and it would be greater than $2\pi r$. However, in the gravitational context, an observer on a ring of Schwarzschild coordinate gravitational radius r, would measure the circumference of the ring to be be $2\pi r$ if he measures the distance using short measuring rods and this would agree with the circumference calculated by the Schwarzschild observer at infinity. If the local observer calculates the circumference by timing the circular orbital period of a satelite and assumes Kepler's third law holds, then he also gets the same answer. If vertical radial distances are measured using local rods or by radar measurements then the value of radius of the ring is not the same as thr r assumed by the Schwarzschild observer and different results are obtained. It is not possible to apply either of these last two radial measurement methods in the case of a black hole.

5. May 19, 2010

### bcrowell

Staff Emeritus
What you're saying is technically correct, but misleading IMO. The r in the Schwarzschild coordinates is simply *defined* as the circumference divided by 2pi, so there is no sense in which you can compare circumference and radius to see if their ratio is 2pi. This is completely different from the case of the rotating disk, where you really can measure the length of a radial line and the length of the circumference, and verify that the ratio is greater than 2pi.

What makes this kind of discussion ultimately a dead end, IMO, is that it revolves around trying to connect changes in lengths to gravitational fields, and yet we know that by the equivalence principle gravitational fields are not really meaningful things to talk about in GR. By the equivalence principle, we can pick any point in spacetime and say that it has any gravitational field we like -- including zero, in the case of a free-falling observer.

6. May 21, 2010

### D.S.Beyer

This is all great news. Thanks!

I still have some questions about the observed distance from the view point of B.

Lets say B is all of us on Earth. And A and C are stars on either side of a spiral galaxy. What can we say about the distance between them? Are they further apart from each other than the appear? Due to the length contraction of the galactic center? Or are they closer to each other than they appear, due to the galactic center spacetime geodesics?

We can assume that both the geodesics and the length contraction are at play in distorting the 'reality' of what we on Earth are seeing. The geodesics make objects close to the mass appear further away, and the length contraction makes everything appear closer than it measurably is.

So at some distance from the galactic center the geodesics 'balance out' the length contraction and there after the objects (stars) are further from the center, and other side, than they appear. ...????

I feel way off base with this but I had to ask.

7. May 21, 2010

### yuiop

As I said earlier, the light coming from A and C follows a curved path and give the appearance of A and C being further apart when viewed from B, due to gravitational lensing.
Imagine a straight line from A to C and two more stars (D and E) at the intersection with the line with the blue circle (and make the circle a bit smaller). A mirror is placed on the near the surfaces of A,C D and E and a light signal is sent from B. The light signal takes longer to return from D and E than it takes to return from A and C. It will also take more standard rulers to span the distance from B to D than from B to A, if the galaxy is sufficiently massive and D and E are close enough to the galaxy centre.

8. May 21, 2010

### yuiop

I concede that there are many ways to measure and interpret r. If we have an ordinary massive body like the Earth, we could in principle drill a hole to the centre and measure the distance with rulers and in that case the circumference of a ring around the Earth will be less than 2Pi*r where r is the measured radius from the centre of the Earth similar to the case of a rotating solid disc. In the case of solid concentric rings around a black hole, the distance between consecutive rings as measured by rulers or radar is also greater than the Euclidean calculation based on 2Pi*r.

Last edited: May 21, 2010
9. May 24, 2010

### starthaus

Here is an easy way to get the expressions for length contraction and time dilation in GR. Start with the Schwarzschild metric (for the weak field approximation):

$$-ds^2=\left(1+\frac{2\Phi}{c^2 }\right)^{-1}dr^2+r^2(d \theta^2 +\sin^2 \theta d \phi^2)-c^2 \left(1+\frac{2\Phi}{c^2}\right)dt^2$$

If you want to calculate time dilation, make $$dr=d\theta=d\phi=0$$ . The reason is that you are trying to determine the relationship between two co-located events.

$$\frac{ds}{cdt}=\sqrt{1+\frac{2\Phi}{c^2}}$$

Length contraction for the case of a non-rotating object is obtained by making $$d\theta=d\phi=0$$ (absence of rotation) and $$dt=0$$ (object endpoints are marked simultaneously bu the distant observer).

$$\frac{ds}{dr}=\frac{1}{\sqrt{1+\frac{2\Phi}{c^2}}}$$

Particularization:

For the radial gravitational field :

$$\Phi=-\frac{GM}{r}$$

so:

$$cdt=\frac{cd\tau}{\sqrt{1-\frac{2GM}{rc^2}}}$$
i.e. coordinate time appears dilated wrt proper time

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

i.e. coordinate length appears contracted wrt proper length.

Last edited: May 24, 2010
10. Jun 4, 2010

### Passionflower

My copy of this book only has 293 pages but anyway I wonder why you say "produce increased distances in the direction perpendicular to the field"?

It seems more logical to me to say produce decreased distances in the radial direction.

By analogy, in the case where matter is present, would you say that: "volume is increased away from the center of the mass"?

I agree with that, r has not physical relationship whatsoever, it is simply a measure of curvature.

11. Jun 4, 2010

### yuiop

It depends upon who is doing the measuring and what method they are using. A local observer using a ruler to measure the radial distance (x) between two concentric rings around the gravitational source, will measure the distance to be greater than that calculated by assuming Euclidean geometry i.e x > (c2-c1)/2Pi where c2 and c1 are the circumferences of the rings.

An observer on the larger ring using a radar device will also reach the same conclusion.

An observer on the smaller ring using a radar device will reach the opposite conclusion.

Last edited: Jun 4, 2010
12. Jun 4, 2010

### Passionflower

Right, but based on that measurement he would not be able to make a definitive conclusion, he could postulate that the radius has a (non Euclidean) deficit or that the circumference expanded in relation to the radius.

Idem ditto for those cases.