# Time sped up in G-field

1. Dec 20, 2009

### quZz

Hello everyone!

Does time always slow down in a gravitational field?
If yes, is there a way to prove it according to GR? If not, are there any known configurations of g-field with time sped up?

2. Dec 20, 2009

### espen180

Yes, the presence of a gravitational field is a cause for time dilation. In GR we don't think of gravity as a force, and a gravitational field therefore does not exist. Instead, an expression relating time dilation to the distance from a point mass has been found from the Schwartzschild solution to the Einstein Field Equations.

The scale factor for gravitational time dilation is $$t=\tau \sqrt{1-\frac{2GM}{c^2r}}$$ where $$t$$ is coordinate time and $$\tau$$ is proper time.

Meaning that as an observer infinitely far away from the point mass measures a process to take a time $$\tau$$, an observer in the gravitational field will measure the process to take a time $$t$$.

Thus, an observer close to point mass will measure all processes farther away from the point mass to speed up.

3. Dec 20, 2009

### meopemuk

Yes, time always slows down in a gravitational field, because gravity is always attractive.

Rather than saying that "time slows down", it is more correct to say that all physical processes become slower in the gravitational field. For any quantum system placed in the field the separation between its energy levels is reduced (due to the attractive nature of the field). This means that the speed of all time-dependent processes in the system is reduced too.

Eugene.

4. Dec 20, 2009

### espen180

When you say that the energy levels converge, do you mean for an outside observer or in the system's rest frame, or both?

So close the Schwartzschild radius, where time dilation approaches infinity, all energy levels of quantum systems converge, then? Meaning that all atoms become ionized close to the event horizon. But it is also possible that electrons are embedded into the nuclei of the atoms. Nuclear reactions between electrons and protons should create neutrons and photons (radiation). A quick calculation gives me a wavelength of $$1.585\, pm=1.585\cdot10^{-12}m$$ for the wavelength, which is within the gamma part of the spectrum. If such a process occurs, by how much would the radiation probably be redshifted when it gets far away from the horizon?

5. Dec 20, 2009

### meopemuk

I think it is more correct to take the point of view of the outside observer. Observer in the gravity field has all its measuring devices (clocks, rulers, calorimeters, etc.) affected by gravity. So, he may not even notice the time dilation and red shift.

That's right.

Not sure about that. It is true that strong gravitational potential effectively leads to the weakening of binding forces in atoms, nuclei, etc. However, at the same time, all effects that can break the binding become weaker as well. For example, all radiation becomes red-shifted (because it is emitted in transitions between "compressed" energy levels). So, it might happen that atoms are not ionized spontaneously.

How did you reach this conclusion?

6. Dec 20, 2009

### espen180

It was probably the fault of incorrect assumptions. As the binding energy of an atom decreases (becomes more negative) the electrons close in on the nucleus, and when the energy levels converge, the (unstable) levels where the electrons are very close to the nucleus are spontaneously reached. Though, this argument might be invalid if the convergance is not present in the system's reast frame.

7. Dec 21, 2009

### Ich

That's definitely less correct, in terms of GR.

8. Dec 21, 2009

### meopemuk

The advantage of my point of view is that there is no need to assume curved space-time and other GR stuff. The red shift and gravitational time dilation follow directly and exactly from Newtonian gravity plus quantum mechanics.

Eugene.

9. Dec 21, 2009

### yogi

The G field of a spherically symmetrical clump of matter is an isotropic acceleration field - mass affects time in a gravitational situation for the same reason clocks placed at opposite ends of an acceleraing rocket will log different times ....because the two rocket clocks are at different potentials a la the direction of the acceleration vector

10. Dec 21, 2009

### Ich

You don't need curved spacetime to produce "gravitational" time dilatation.
No, they don't.
This point of view works as a first order approximation, and it breaks down exactly when Newtonian gravity breaks down. It is sterile, you can't go further.
In the same approximation, you can derive both time dilatation and redshift from the principle of relativity, constant speed of light and the equivalence principle - and Newtonian gravity (see e.g. Einstein 1907). If you continue with this approach, you can replace Newtonian gravity by a much more accurate theory, GR. You can then derive why time dilatation and energy have so much in common.

Changing energy levels, derived from Newtonian gravity, as a cause for time dilatation resemble very much an ether, IMHO.

11. Dec 21, 2009

### meopemuk

I agree that curved 4D space-time manifold and Einstein's "geometric" approach to gravity is a neat idea. Though it has is one serious problem. It is very difficult (I would say impossible) to unify with quantum mechanics.

I can also agree that the Hamiltonian approach in Newtonian gravity does not look general enough. One can add extra velocity-dependent terms to the Newton's potential to describe relativistic effects such as the Mercury's perihelion shift or frame dragging (see, for example, the Einstein-Infeld-Hoffmann Hamiltonian). However, these additions seem to be ad hoc, without any fundamental guiding principle. Nevertheless, there is one important advantage of the relativistic Hamiltonian theory: it is naturally consistent with quantum mechanics. You can simply replace all dynamical variables in the Hamiltonian by corresponding operators and you immediately obtain a relativistic quantum theory of gravity.

Eugene.

12. Dec 22, 2009

### Ich

Great, a theory that predicts nothing but is consistent with quantum mechanics.

Can we agree that, as long as there is no TOE, GR is the gold standard in Gravity, that this subforum is devoted to the discussion of relativity, and that we therefore should try to get across the underlying (heuristically extremely successful) philosophy rather than disputing it?

13. Dec 22, 2009

### Chronos

Time is constant for any given observer, so you must define time in terms of the coordinate system of the person holding the watch. Anything less is . . . irrelevant.

14. Dec 22, 2009

### quZz

Is there a general proof that time slows down?

15. Dec 22, 2009

### Naty1

Leonard Susskind in THE BLACK HOLE WAR,2008,

This is the frame of the distant observer.

Seems like the above conflicts with the classical view that a free falling atom just passes through the event horizon...and it does in its own frame...but in order to resolve the free falling versus distant frame conflict an observational photon would require so much energy as to vaporize the atom under scrutiny(!!) and scatter it all over the stretched horizon....so there seem to be two descriptions even BEFORE the atom crosses the event horizon!!!! (Susskind, pg 257)

And more: BLACK HOLES AND TIME WARPS, KIP THORNE, 1994 (pgs 491-492)
I do not know if this remains a modern, current, view.

16. Dec 22, 2009

### bcrowell

Staff Emeritus
Time dilation relates to the gravitational potential, not the gravitational field. Differences in gravitational potential relate to differences in the rate of flow of time.

One way to get the effect you want is simply to obtain some exotic matter that has negative mass ( http://en.wikipedia.org/wiki/Energy_condition ). But we don't know if this type of exotic matter exists in our universe.

The real issue is "sped up" relative to what? The structure of the universe is clumpy. Therefore there are some regions that have a higher gravitational potential and others that have a lower gravitational potential. The regions of higher gravitational potential are sped up relative to the regions of lower gravitational potential. The regions of lower gravitational potential are slowed down relative to the regions of higher gravitational potential.

Consider configuration A, in which you have a region of space that is empty except for a clump of negative mass in subregion R.

Now consider configuration B, in which you have a region of space containing a uniform density of hydrogen has, except for an empty subregion R.

Configurations A and B are identical in terms of the rate of flow of time. In both cases, time flows more quickly in subregion R than in the surrounding areas.

17. Dec 22, 2009

### Phrak

Some of the answers you're getting are a bit tangential to explaining your problem. You are asking about elasped time in the context of relativity. In relativity, elapsed time is Relative to the observer.

Given the way you ask, there are contradictory answers; each equally true. Context is very important in relativity.

1) Time does does not slow down on the watch of the observer holding the clock in a gravity well.

2) The observer at a distance from a gravity well will see, via a speed-of-light, time delayed signal from a telescope that the clock in the gravity well is running slower.

3) With two clocks initially showing the same time and at the same place the clock in the gravity well will display less elapsed time (usually) when they are compared again, and compared at the same place.

4) Should one observer dash off half way to the nearest star and back at really high speed, his elasped time will be less than the one in the gravity well when they meet again.

Last edited: Dec 22, 2009
18. Dec 23, 2009

### quZz

I don't really know the best way to state the problem... =)
In the simplest case... assume that all masses are located in the finite volume of space V and that g-field is stationary.
By choosing a proper reference frame and coordinates it is possible to have
$$g_{0\alpha} = 0$$
and
$$g_{00}(|\vec{x}| \to \infty) \to 1$$.
Now the question comes... is it possible to prove that for any energy-momentum tensor $$T_{ik}$$ satisfying energy conditions (btw which of them?) we have $$g_{00} \le 1$$?

19. Dec 23, 2009

### bcrowell

Staff Emeritus
One thing to point out is that the actual universe doesn't obey these assumptions (see #16). But that doesn't mean that these assumptions can't be interesting, or that they can't be useful approximations for discussing some part of the universe.

I'm not so sure that it makes sense to treat $$g_{00}$$ as a measure of the rate of flow of time, without any further assumptions. E.g., in a solution with closed timelike curves, such as the Godel metric, I don't think it's valid. I would guess that you can also have coordinate singularities where $$g_{00}$$ goes nuts, becomes negative, etc., without constituting a fast-time region of the kind you have in mind.

If you add the assumption of weak fields, and maybe some kind of assumption that the topology is well behaved, then I think it's probably true that there are no fast-time regions. After adding these assumptions, you can essentially treat everything according to Newtonian gravity, and the only question is whether you can ever get a gravitational potential that's greater than the potential at infinity. Now suppose you did have such a gravitational potential. Since we're assuming weak fields, the potential is a continuous, well-behaved function everywhere, and therefore it attains its maximum somewhere. (This is the point at which I think some kind of topological assumption may become necessary as well. You need some kind of generalization of the extreme value theorem to calculus on a manifold, and I think that requires topological assumptions.) At this maximum, the divergence of the gravitational field has to be positive, so by Gauss's law you must have some negative mass there. This would presumably violate whatever appropriate energy condition we'd put on our list of assumptions.

Last edited: Dec 23, 2009
20. Dec 23, 2009

### Ich

You don't need negative energy, e.g. a negative trace of T is sufficient.
Look at http://en.wikipedia.org/wiki/De_Sitter_space" [Broken], where the "potential" has its maximum at the origin.
If you have a finite Dark Energy region, you also have g00>1. But I don't know if and how this is possible.

Last edited by a moderator: May 4, 2017