# B About gravitational time acceleration

1. Sep 17, 2016

### Kai91

Hello.

I have a theoretical doubt about the gravitational time dilation in the particular case of the acceleration in regard to the scenario of a planet in wich the pass of the time is faster than in the Earth (e.g., a planet X where a year is equivalent to a second in Earth) .

The fact of the time going much slower than the Earth in a planet in the vicinity of a massive black hole it's known, but I want to know if the opposite case it's possible, i.e., if Earth can have a pass of time much slower than other planet if it satisfies any type of conditions, and which are these conditions.

Thank you.

2. Sep 17, 2016

### Staff: Mentor

There is no such planet. The fastest relative to Earth would be a massless planet whose time would be a factor of 1.0000000007 faster than Earth.

You could go much more massive than Earth to get time arbitrarily slower than Earth, but not the other way.

3. Sep 17, 2016

### Ibix

In short, time dilation depends on the gravitational potential. The further down you go the slower your clocks tick compared to ones higher up. The problem is that the Earth doesn't have very strong gravity - so however far up you go you can't increase your gravitational potential very much compared to the surface of the Earth. So you can't get your clocks going very much faster than ones on Earth's surface.

On the other hand you can find much stronger sources of gravity and decrease your gravitational potential as much as you like. Therefore you can make the clocks run as slow, compared to ones on the surface of the Earth, as you like.

4. Sep 17, 2016

### Kai91

Perfect responses, my question is solved

Thank you so much

5. Sep 17, 2016

### A.T.

Does this apply to the observable universe that we know? Can we be sure that there is no place beyond that, with a much higher gravitational potential than what we observed so far?

6. Sep 17, 2016

### Staff: Mentor

The concept of "gravitational potential" doesn't apply to the universe as a whole, because it isn't static. "Gravitational potential" is only meaningful in a static region of spacetime (more precisely a stationary region). So strictly speaking, the answer to your question is "mu": there is no way to compare the "gravitational potential" of someone in intergalactic space outside the Milky Way (which is the highest gravitational potential in our region of spacetime), and at rest relative to that galaxy, with someone in the intergalactic space outside a galaxy a billion light-years away, so we can't even ask the question meaningfully.

Yes, in the sense that the question has no meaning. See above.

7. Sep 17, 2016

### Ibix

Beyond the observable universe, there be dragons. But we can never see them, so whether their watches are ticking fast or slow is a matter of speculation (assuming that the question can even be posed formally at all).
Is there even a way to compare their clock rates unambiguously?

8. Sep 17, 2016

### A.T.

That makes sense, thanks.

9. Sep 18, 2016

### Staff: Mentor

Strictly speaking, no, because they are moving relative to each other. But if they were both "comoving" observers, i.e., if they both saw the universe as homogeneous and isotropic, then you could argue that their clock rates were the same based on the properties of the congruence of "comoving" observers in FRW spacetime. That would still not be a fully satisfactory (to some) way to compare clock rates, but at least it would be based on some aspects that are not coordinate-dependent.

10. Sep 18, 2016

### stevendaryl

Staff Emeritus
Well, you can ask the question in terms of "twin paradox" type scenarios. The typical twin paradox has a traveler blasting off from the Earth, going on a long voyage, and returning to Earth only 10 years older than when he left, while his twin who has remained at home has aged 20 years. The question is whether it is possible that there is a reverse twin paradox: Is it possible for a traveler to return much older than the stay-at-home twin?

Special Relativity gives no way; the traveling twin is always younger than the one who remains unaccelerated. General Relativity gives a way to have a tiny such effect: If the traveler gets well away from the Earth and the Sun, and spends a decade or so floating in space far from any stars or planets, then when he returns to Earth, he might be (very slightly) older than a twin who remained on Earth the whole time. So the question can be phrased in this way: is there an upper bound to the ratio:

$\frac{\tau_{traveler}}{\tau_{stay-at-home}}$

where $\tau_{traveler}$ is the elapsed time on the traveler's clock, and $\tau_{stay-at-home}$ is the elapsed time on the stay-at-home twin's clock?

Getting "higher" in the Earth's gravitational well increases the ratio slightly, and then getting higher in the sun's gravitational well increases the ratio a tiny bit more. Maybe getting outside of our galaxy would increase it more. Maybe our galaxy is inside of some cluster of galaxies, and getting outside of that cluster might increase it even more. So is there an upper bound to the ratio? (The assumption is that nothing dramatic happens to the stay-at-home twin--he doesn't accelerate much, and he doesn't fall into a black hole, or anything.)

11. Sep 19, 2016

### Staff: Mentor

Yes, that's true; if the two twins start and end together, then whether spacetime is static or not doesn't matter.

Yes. Here's why.

We'll assume that the stay at home twin is at rest relative to a comoving observer, but that he is inside a gravity well so his rate of time flow is dilated by some factor $\sqrt{1 - 2M / r}$ as compared with an observer that is comoving and not inside any gravity well. So we can rephrase your question as follows: is it possible for any observer to have more elapsed proper time between two given events than a comoving observer that passes through both? If the answer to that is no, then there is obviously an upper bound to the ratio, namely the factor $1 / \sqrt{1 - 2M / r}$ by which the comoving observer's clock runs faster than the clock in the gravity well.

And the answer to the question as I've rephrased it above is in fact no, because a comoving observer's worldline is a geodesic, and for any portion of that worldline after the Big Bang, there are no conjugate points along it (i.e., there are no other geodesics that cross it at two separate events). And between any two events along a timelike geodesic with no conjugate points, that geodesic has maximal proper time--IIRC there is a theorem to this effect in Hawking & Ellis. (The "no conjugate points" condition is the crucial one; without it the theorem no longer holds.)

Dropping the assumption that the stay-at-home twin is at rest relative to a comoving observer doesn't change anything, as long as we assume that his motion is still geodesic (or more precisely that the motion of the center of mass of the gravity well he is in is geodesic). That's because the "no conjugate points" condition above applies to any geodesic in FRW spacetime, not just comoving ones.

12. Sep 19, 2016

### stevendaryl

Staff Emeritus
Okay, but that's for the case of a single symmetrical gravitational source of mass $M$. What I was wondering was the possibility of having gravitational effects at many different scales. Near the Earth, you can raise your "gravitational potential" by getting farther from the center of the Earth. You can raise it further by getting farther from the sun. You can raise it further by getting outside of our galaxy. If our galaxy is inside a gravitational cluster, you can raise it further by getting outside of that cluster. Etc. So is it mathematically possible (consistent with GR) to have infinitely many different scales, of clusters of clusters of clusters, etc.?

I'm not sure I understand the connection with geodesics. The stay-at-home twin is not following a geodesic, since he's accelerating away from the center of the Earth, relative to freefall.

13. Sep 19, 2016

### Staff: Mentor

Yes, but this doesn't go on indefinitely. Once you get outside superclusters (roughly, one more level above the cluster level), you are a comoving observer experiencing comoving proper time; there are no further gravity wells to get out of.

If the universe as a whole is consistent with FRW spacetime, to a good approximation, I don't think so, because in FRW spacetime, the universe is modeled as a homogeneous mass/energy distribution, and a homogeneous mass/energy distribution, from a time dilation perspective, is the equivalent of no mass/energy at all; comoving observers have the maximal rate of time flow, and that rate is bounded. To have infinitely many levels of structure, you would have to have observers whose rates of time flow were unbounded.

Yes, but there will be some comoving observer, moving on a geodesic, who has the same motion relative to the universe as a whole, at least to a good enough approximation. If you're not comfortable with imagining a comoving observer who is spatially co-located with Earth but magically is unaffected by the gravity wells of Earth, Sun, Milky Way, our cluster, our supercluster, etc., then imagine a comoving observer who is just outside all those gravity wells. He is moving along a geodesic, and his rate of time flow has a finite, bounded ratio to the stay at home twin's rate of time flow, because there are only a finite number of gravity wells in between them (roughly the ones I just named). And the comoving observer's rate of time flow is maximal, as I've already shown.