# Gravity distortion at the center of a massive object?

1. Nov 14, 2013

### darkhorror

I am not totally sure what I am trying to ask, but wondering about gravity at the center of a massive object such as the sun, earth, highly massive star. If you look at the time dilation at the center vs the surface or far in space time at the center time is running slowest. Now obviously for person at the center he will still see his clock ticking normally, and the other clocks running faster.

Now it seems there also has to be some sort of "length contraction" though it doesn't seem the same as length contraction in SR. More that there is more space crammed into a small space. Where the person far away in space would see a meter stick in the center of the massive object looking smaller than their meter stick. The person in the center would see an opposite effect.

Are these correct or where am I going wrong?

2. Nov 14, 2013

### WannabeNewton

Are you asking if the gravitational field makes objects look bigger than they actually are to observers at infinity? If so then yes it's true. There's also the fact that proper distances get affected by the gravitational field.

Consider for simplicity the Schwarzschild geometry external to a static spherically symmetric star and an observer hovering in place outside the star at some Schwarzschild radial coordinate $r$; let $p = (t,r,\theta,\phi)$ be some event on the observer's worldline. The event $q = (t,r + dr,\theta,\phi)$ lies in the local simultaneity slice at $p$ relative to this observer; this is because the separation vector between $p$ and $q$ is given by $\xi = (0,dr,0,0)$ which is orthogonal to the 4-velocity $u = ((1-2M/r)^{-1/2},0,0,0)$ of the observer. So now imagine that this observer is using an ideal ruler to measure the spatial distance between these simultaneous events $p$ and $q$. The ruler measurement will of course just be the length of the separation vector $\xi = (0,dr,0,0)$ connecting these two events and this comes out to $ds = (g_{\mu\nu}\xi^{\mu}\xi^{\nu})^{1/2} = (g_{rr})^{1/2}dr = (1 - 2M/r)^{-1/2}dr$
(you could have arrived at this straight from the line element but I feel the above is more intuitive). In other words, the ruler measurement is affected by the curved geometry due to the gravitating central mass $M$.

As for the deflection of light making objects look bigger to an observer at infinity, consider a light ray that just grazes the surface of a star $M$ of Schwarzchild coordinate radius $R$. Let $k = (\dot{t},\dot{r},0,\dot{\phi})$ be the tangent vector to the null geodesic describing this light ray (we are restricting ourselves to trajectories in the equatorial plane). At the very instant that it grazes the surface of the star, we have $\dot{r} = 0$. Since $g(k,k) = 0$ we get that $(1 - 2M/R)\dot{t}^{2}= R^{2}\dot{\phi}^{2}$ at this instant. We know from conservation of energy and angular momentum that $\dot{t} = (1 - 2M/R)^{-1}e$ and $\dot{\phi} = \frac{l}{R^{2}}$ so $b:=\frac{l}{e}=\frac{R}{(1 - 2M/R)^{1/2}}$ where $b$ is the impact parameter for light ray trajectories. Note that $b > R$ so what this means is that an observer at infinity will see the star as having a larger diameter than it actually has because of gravity (more directly, because of the deflection of light that results from the presence of a gravitational field).

Last edited: Nov 14, 2013
3. Nov 15, 2013

### A.T.

Yes.

This one is more subtle. Around a mass there is a spatial distortion, which can be understood in terms of real physical radial distance (ds) vs. the radial coordinate computed by dividing a real physical circumference by 2PI (dr). This is a picture for the space outside of a black hole explains how ds & dr are related:
http://ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/Black_Holes.htm

This includes the interior:
"commons.wikimedia.org/wiki/File:Schwarzschild_interior.jpg" [Broken]

So if you put a very long ruler through the star it will appear to be shortened. This is a cumulative effect of the local distortion (ds/dr) integrated along the radial line. But you asked about spatial distortion locally at the center, and there you have ds/dr = 1. So locally at the center the spatial distortion vanishes.

And if you create a spherical cavity at the center, space will be flat throughout the entire cavity, and clocks will run at the same rate throughout the entire cavity, but still slower than at the surface. And there is no "pull" towards the center throughout the entire cavity.

Last edited by a moderator: May 6, 2017
4. Nov 15, 2013

### Bill_K

What about the transverse direction? Does it contract in the transverse direction?

5. Nov 15, 2013

### pervect

Staff Emeritus
I'd have to disagree that it's a good idea to call what goes on with the metric inside a black hole as "length contraction".

As long as you use the same simultaneity convention, a 1 meter ruler / line segment has a length of 1 meter, it doesn't matter whether you are in a high gravitational potential or a low gravitational potential.

Length is observer dependent, but the observer dependence comes from simultaneity issues, if you use the simultaneity convention of a static observer, a proper length of 1 meter will always be a proper length of 1 meter. There isn't any sort of length "correction" to proper length due to gravitational potential.

There is a metric correction when you convert coordinates (which are not lengths) into lengths, but I think it's very confusing to call this length contration. Amongst the other issues, this depends on the details of the coordinates one uses.

You'll see something very lie the usual SR effects if you use the simultaneity convention appropriate for a moving observer (say, an infalling one) rather than a static one. But this is different from saying that there is a length contraction effect due to gravity, it's more akin to the "Lorentz contraction".

You can also say that the radial distance between two concentric spherical shells, one of circumference C and one of circumference C + 2*pi*delta is greater than delta due to distorted geometry of a black hole (unlike the situation in flat space-time). But I woldn't call this "length contraction" either.

Last edited: Nov 15, 2013
6. Nov 15, 2013

### PAllen

Hmm, I interpret the OP as asking a completely different question. They should come back and clarify. To simplify what I think they are asking, imagine a transparent planet with a small hollow cavity at its center. I think they want to know about clock and ruler comparisons for someone in the hollow versus at infinity.

Then, it is clear that clocks will be running slower in the hollow than at the surface (and this comparison will be mutual), and slow compared to distant clocks by amount greater than clocks at the surface. The amount of difference will, of course, depend on your assumptions about the planet, but the weak field approximation of surface to center would be straightforward to compute.

As for distance, inside the hollow you have an exact Minkowski metric for the local experience. However, to write a metric for a global coordinate patch covering all of space time, you would have to scale the flat metric to mesh with metric at the cavity boundary. Then, what you would really need to ask is the angle subtended by 1 meter of proper length in the cavity viewed orthogonally on approach to infinity, compared to Euclidean expectation. This would be a lot more work to compute, even approximately.

[edit: I see AT understood the question the same way I did.]

7. Nov 15, 2013

### darkhorror

You are correct this is more of what I was asking, same with AT. My questions aren't so much about getting a specific numbers but understanding what explains how gravity slow's down time and how it works with the consistency of light in these situations.

8. Nov 15, 2013

### A.T.

According to local clocks and rulers light still propagates at c. But when you do measurements over large distances based on distant clocks, you get different average speeds than c.

9. Nov 15, 2013

### Bill_K

Ok, if everybody understands it, and agrees on how they understand it, maybe somebody can answer the question I raised. A.T. described the contraction in the radial direction. What about the transverse direction?

10. Nov 15, 2013

### yuiop

Using this interior solution for the Schwarzschild metric and setting dr=dt=0 I would say no. The result is the same as for the exterior Schwarzchild solution when dr=dt=0, so no transverse length contraction. For an extended transverse ruler, the ruler would have to curved such that all points along the ruler are at constant r from the centre.

Last edited: Nov 15, 2013
11. Nov 15, 2013

### PAllen

I don't know the answer, but I described what I would consider to be observable length contraction. For me, the indicated calculation would take far more time than I am willing to spend, and I have no intuition of how the answer would come out because I've never calculated (or followed someone else's) for an equivalent scenario.

12. Nov 15, 2013

### Staff: Mentor

But that scaling, counterintuitively, does *not* change the space coordinates at all!

The general metric for a static, spherically symmetric spacetime can be written in this form:

$$ds^2 = - J(r) dt^2 +\frac{1}{1 - 2m(r) / r} dr^2 + r^2 d \Omega^2$$

where $r$ is the standard Schwarzschild radial coordinate and $m(r)$ is the mass inside radius $r$. (This result is derived, for example, in MTW; I can look up the reference if needed.) At the outer surface of the object, $J(r) < 1$ and $m(r) = M$, where $M$ is the total mass of the object; so there is both "time dilation" due to being in the gravity well, and "space distortion" due to the presence of mass.

However, at the *inner* surface of the object (i.e., the outer edge of the hollow cavity), $J(r)$ has decreased, but $m(r) = 0$--there is no mass left inside that radius. So there is still "time dilation" compared to infinity (more time dilation than at the outer surface), but there is *no* "space distortion" any more, even in this global metric; the coefficient of $dr^2$ is $1$! To put this another way, to transform from this global metric to the local metric at any point inside the hollow shell, the *only* thing you need to do is rescale the time by the factor $\sqrt{J}$; you don't need to "rescale the space" at all.

Even if you computed it and it showed that 1 meter of proper length subtended less angle (which is my guess as to what it would show), that wouldn't show that the hollow cavity was "length contracted"; see above. All it would show is that the paths of light rays are distorted when the pass through the massive object to get from the hollow cavity to the region outside.

13. Nov 15, 2013

### yuiop

As mentioned in #10, that you might of missed, when dr=dt=0, $r d\Omega$ is a measure of transverse length and there is no effective transverse length contraction, just as in the regular Schwarzschild metric.

P.S. What is is J(r) in the equation you quote? I have seen a similar equation stated as:

$$ds^2 = - (1-2m(r)/r) dt^2 +\frac{1}{1 - 2m(r) / r} dr^2 + r^2 d \Omega^2$$

but I cannot recall where I saw it, so I am not sure of that equation's authenticity.

Wait ... It might have been this document by Sean Carroll (See eq 7.32).

Last edited: Nov 15, 2013
14. Nov 15, 2013

### PAllen

J(r) is dependent on matter. What Peter gives covers any form of matter with possibly radially varying density and pressure as long as it exactly spherically symmetric and static.

Last edited: Nov 15, 2013
15. Nov 15, 2013

### WannabeNewton

I don't see how this relates to your original question at all. The physical motivation behind gravitational time dilation is a completely different question and one that is covered in every GR text.

There is also no clash with the fact that $c = 1$ in local Lorentz frames. If we have an observer with 4-velocity $u^{\mu}$, a light wave with tangent vector $k^{\mu}$, and an event at which the light wave passes by the observer, then $E = -k_{\mu}u^{\mu}$ is the energy of the wave relative to this observer at that event and $p^{\mu} = g^{\mu}{}{}_{\nu}k^{\nu} + u^{\mu}u_{\nu}k^{\nu}$ its 3-momentum relative to this observer at said event.
Therefore $\left \| p \right \| = [(g^{\mu}{}{}_{\nu}k^{\nu} + u^{\mu}u_{\nu}k^{\nu})(g_{\mu\alpha}k^{\alpha} + u_{\mu}u_{\alpha}k^{\alpha})]^{1/2} = [k^{\nu}k_{\nu} + u_{\alpha}u_{\nu}k^{\nu}k^{\alpha} ]^{1/2} = E$
hence $c = 1$, as we would expect.

Gravitational time dilation is a global effect and is just a manifestation of gravitational redshift.

16. Nov 15, 2013

### PAllen

I think the only meaningful definition of length contraction/expansion is that there exists a measurement you would accept as a valid length measurement such that the locally measured length of an object (comoving if required - thus local rest length) differs from some other observer's measurement. In the case of SR, there are many ways (in principle) to realize the length measurement of a moving object so as to show the contraction.

Thus, to my mind, there are two defensible statements about length change observed at a distance:

- either it is undefined, because there is no procedure you would accept as valid
- or there is at least one procedure for remote length measurement that you would accept.

What I don't accept is that grr=1 in the fitted flat metric has any meaning at all for length contraction without an accepted way for a distant observer to measure length. If there is no procedure you would accept, then we should agree the term has no possible meaning.

17. Nov 15, 2013

### Naty1

How about tidal effects, 'spaghettification' ??

18. Nov 15, 2013

### PAllen

The discussion concerns the interior of an ordinary massive body, not the near singular region of a BH.

19. Nov 15, 2013

### PAllen

Actually, though your equation does come from Carroll, it looks fishy to me. Since it is normally assumed that m(r)/r->0 as r->0, it suggests clocks speed up as you go deeper into a massive body, with gravitational time dilation vanishing at the center. This seems absurd to me. This behavior is also completely different from that given by the SC interior matter solution (which, though physically implausible, would expect to get qualitative features right).

Last edited: Nov 15, 2013
20. Nov 15, 2013

### Naty1

On rereading the OP, I see while he mentioned 'surface' early in his post that was not part of his final questions. It seemed to me that everywhere except the center, some tidal effects even if minor, are at play...