# Maximum possible gravitational time speed-up in the universe?

1. Jun 15, 2014

### Arbitrageur

This question needs to be framed:

We track time at the Earth's surface using atomic clocks.
We know that objects in orbit have clocks that run slower because of velocity and faster because of reduced gravity, and which effect is dominant depends on altitude as this illustration shows:

https://en.wikipedia.org/wiki/Error_analysis_for_the_Global_Positioning_System
http://upload.wikimedia.org/wikipedia/commons/3/36/Orbit_times.svg

Eyeballing the figures on the chart, at GPS orbit, it looks like time speeds up 520 picoseconds per second due to lower gravity, and time slows down 100 picoseconds per second due to velocity, for a net increase of perhaps 420 picoseconds per second.

I would like to disregard the velocity figure for the purpose of my question and focus on gravity because I don't want to talk about orbits around Earth, but rather about a hypothetical observer who is stationary relative to the cosmic microwave background.

My question is,

Q: What is the maximum value in picoseconds per second the gravitational speed-up plot on this chart can reach anywhere in the universe?

To find this value I imagine the hypothetical observer would be not only outside the solar system, but also outside the Milky Way galaxy, and also outside the local group. For starters we could put them at the lowest gravitational point between the local group and a neighboring galaxy cluster. Finding lower gravitational areas elsewhere in the universe may not differ much from such a remote area, though I'm not really sure about that.

I assure you this isn't a homework question, just something I'm curious about. While I might be able to do some math for the effects of the sun and moon since I know their distance and mass, and I can find some of that type of information online, I really have no idea what the time dilation effect of the Milky Way's gravitational field is, nor that of the local group (local galaxy cluster) and I didn't have much luck finding it.

Usually I can find answers I'm searching for like this online within 5 minutes, but I searched about 90 minutes and found some answers but they were wrong. I know they were wrong because the question asked the same thing I did about the gravitational time dilation, and the answers I found talked about the velocity of the sun going around the Milky Way and the corresponding time dilation, which is not relevant to the gravitational time dilation question, at least not directly.

My guess would be that the maximum value might reach something like an order of magnitude higher between galaxy clusters, than the 520 picoseconds per second at GPS orbit, but that guess could be way off, though I'd be a little surprised if it reached two orders of magnitude higher. Has anyone ever done the math on this or at least made a real estimate better than my guess?

2. Jun 16, 2014

### Staff: Mentor

This question can't really be answered, because the underlying concept of "gravitational time dilation" doesn't have meaning with respect to the universe as a whole. Gravitational time dilation only makes sense for stationary, isolated system; in other words, a system consisting of a central gravitating body (possibly with other much smaller bodies orbiting it) surrounded by empty space all the way out to "infinity", that is not changing with time (not that nothing moves at all, but that things like orbital motions are periodic, with no average change). We can (at least as a fairly good approximation) model the Earth, or the solar system, or even our galaxy or a galaxy cluster, this way, but the universe as a whole can't be modeled this way, because it's expanding and because it's not isolated--there is no empty space out to "infinity" (according to our current "best fit" model, it's spatially infinite, but even if it were closed and therefore had a finite spatial volume, there would still be no empty space out to "infinity").

For the case of a stationary, isolated system, your question can be answered easily: the maximum value of "rate of time flow" is just its value at infinity. For example, in the case of the Earth, someone "hovering" at the altitude of the Earth's surface (I say "hovering" to eliminate any effects of orbital motion) has a time dilation factor, relative to infinity, of

$$\phi = \sqrt{1 - \frac{2 G M}{c^2 R}}$$

where $R$ is the radius of the Earth and $M$ is its mass. If you plug in numbers, you get $\phi \approx 1 - 6.9 \times 10^{-10}$. The answer to your question is just the inverse of this, i.e., the maximum "rate of time flow", relative to the rate for someone "hovering" at the Earth's surface, is about 7 parts in $10^{-10}$ larger, i.e., about 700 picoseconds per second larger.

However, this answer doesn't mean what you might think it means. To see why, try evaluating $\phi$ for someone "hovering" at the Earth's distance from the Sun, i.e., use the mass of the Sun for $M$ and the radius of the Earth's orbit for $R$. You will get $\phi \approx 1 - 9.8 \times 10^{-9}$. That means that, with respect to the solar system, the answer to your question is that the "rate of time flow" at infinity is about 9800 picoseconds per second larger than it is on Earth (again, we're ignoring all effects due to orbital motions, the Earth's rotation, etc.). And actually, since for values this small you can just add $\phi$ values linearly, the combined effect of the Earth's and the Sun's gravitational fields is the sum of 9800 and 700, i.e., the maximum "rate of time flow" at infinity due to both effects is about 10,500 picoseconds per second larger.

But even that's not quite right, because the solar system is in the Milky Way's gravity well. Plugging in the numbers for someone "hovering" at the Sun's distance from the center of the Milky Way gives $\phi \approx 1.6 \times 10^{-6}$, i.e., a maximum rate of time flow at infinity of about 1,600,000 picoseconds per second larger. (The additional effect due to the Earth and Sun, from above, is too small to matter here, given that we're only making a very approximate calculation.) We could go on and try to calculate $\phi$ relative to the Local Group of galaxies, or the Virgo cluster, or the supercluster of which it is a part.

You can see where this is going, of course. In fact, I've basically just re-stated your question in different terms. But the common feature of all the above calculations is, as I noted before, that we are modeling all the systems as stationary and isolated. That shows up in the formula for $\phi$ as the assumption that there is a well-defined value of $M$ and a well-defined value of $R$ that we can use. Relative to the universe as a whole, this doesn't work; there are no well-defined values of $M$ and $R$ to plug into the formula. The universe has no "center of mass" that we can measure or estimate Earth's distance $R$ from, and it has no "total mass" $M$ because $M$ is what you measure from the outside, by putting test bodies in orbit about the system and measuring their orbital parameters, and there's no way to put a test body in orbit about the universe. (And in any case, according to our current best-fit model of the universe, it's spatially infinite, so $M$ would be infinite even if we could measure it "from the inside", and $R$ would be changing even if we could define it, because the universe is expanding.)

3. Jun 16, 2014

### bcrowell

Staff Emeritus
This isn't quite right. You seem to be saying that a necessary and sufficient condition for a gravitational potential is that the spacetime is stationary and asymptotically flat. Actually the necessary and sufficient condition is that it's static.

An asymptotically flat spacetime containing orbiting bodies actually does not have a well-defined potential, but it may *approximately* have a well-defined potential.

There are cosmological models that include rotation and are stationary but not static. They don't have a potential.

The spacetime surrounding a rotating body like the earth can be modeled as a Kerr metric, which is stationary and asymptotically flat, but not static. It doesn't have a potential, but in the weak-field limit you can approximate it as having a potential.

Asymptotic flatness is neither necessary nor sufficient.

4. Jun 16, 2014

### WannabeNewton

Actually Peter is correct. A stationary space-time is all that's needed. The potential is defined by the time-like Killing field as $\phi = \frac{1}{2}\ln (-\xi_{\mu}\xi^{\mu})$. There is no need for $\xi_{[\gamma}\partial_{\mu}\xi_{\nu]} = 0$. In fact $\phi$ is analogous to the gravitoelectric potential so it is completely independent of $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\partial_{\alpha}\xi_{\beta}$ which through $\xi_{i} = g_{i t}$ in adapted coordinates is actually related to the gravitomagnetic potential.

Last edited: Jun 16, 2014
5. Jun 16, 2014

### Arbitrageur

Thanks for the very informative answer! I accept this, and I suspected as much, which is why I tried to frame a more answerable scenario than the thread title immediately following the question, namely:

So to clarify, I was considering reference frame A to be an observer on Earth's surface which is the baseline of the plot I referenced, and reference frame B to be a low gravitational area far outside the Milky Way, which is stationary relative to the cosmic microwave background.

To better illustrate this concept, allow me to mark these two reference frames A and B on an image of the Virgo supercluster:

http://www.brighthub.com/science/space/articles/92389.aspx
http://freepicninja.com/img/thumb/gravitational_time_dilation.jpg

Fair enough, then scratch the universe as a whole idea and let's focus on whether it's possible to make some guesstimate of the scenario I mentioned immediately following my question, illustrated above, and a guesstimate is good enough, I don't need an exact calculation.

Based on the Milky Way figure you cited of 1,600,000 picoseconds per second and the idea that other galaxies are contributing to a lesser amount, would you care to make a guesstimate on what the gravitational time speed up might be at point B relative to the Earth's surface at point A? Obviously something greater than 1,600,000 picoseconds per second, maybe 10,000,000 picoseconds per second, or 10 microseconds per second?

6. Jun 16, 2014

### Staff: Mentor

The "far outside the Milky Way" part is OK, if you add the qualifier "and at rest with respect to the Milky Way's center of mass". The answer for that is the 1,600,000 picoseconds per second that I gave (more or less--my calculation was very rough). If you really mean "far outside the Local Group of galaxies", or even "far outside the local supercluster of galaxies", the answer would be different--probably larger, but I would hesitate to guess how much larger. I would have to look up some more data to calculate it.

But for any of the cases I mentioned above, the "stationary relative to the cosmic microwave background" part is a problem, because such an observer will *not* be at rest relative to the center of mass of the Milky Way, or the Local Group, or the local supercluster. (At least, they won't according to our best current understanding.) An observer that sees the CMBR as isotropic in the Milky Way's vicinity will be moving at, IIRC, several hundred to a few thousand kilometers per second relative to the Milky Way's center of mass. The relative velocity for the cases of the Local Group and local supercluster is probably about the same, possibly even somewhat larger.

Also, on the length scale of the local supercluster and larger, the expansion of the universe is significant, because it means that observers who see the CMBR as isotropic, but are spatially separated on these scales (say, a hundred million light years, roughly the size of the local supercluster), will be in relative motion; that is, they will be able to observe a significant redshift in each other's light signals. And this, as I said before, invalidates the whole concept of "gravitational time dilation". So the answer to even your revised question is not really meaningful on scales much larger than the Milky Way, or possibly the Local Group (though even on that scale the relative motion due to the universe's expansion starts to be significant).

7. Jun 16, 2014

### Staff: Mentor

Are you thinking of the Godel universe? That's the only one I can think of off the top of my head that includes rotation. AFAIK the Godel universe is not asymptotically flat.

I agree with WBN's comment here; the norm of the timelike KVF in Kerr spacetime should meet all the requirements of a "potential". Even if we take into account that this KVF is not timelike everywhere, we can simply restrict the use of its norm as a "potential" to the region where it is timelike, similar to what we have to do in the static case (since the corresponding KVF in Schwarzschild spacetime is only timelike outside the horizon).

I agree it's not sufficient, because it's certainly possible to have an asymptotically flat spacetime that does not have a timelike KVF.

However, I'm curious about the necessary part. Can you give an example of a spacetime that *does* have a well-defined potential but is not asymptotically flat? (I know there can be stationary, or even static, spacetimes that are not asymptotically flat, but how would you interpret the norm of their timelike KVF, physically, if there is no "value at infinity" to compare it to?)

8. Jun 16, 2014

### bcrowell

Staff Emeritus
It sounds like we're talking about different definitions. There's the question of whether, as in Newtonian gravity, the entire field can be recovered from knowledge of a scalar potential. In a general stationary spacetime, you need more than one scalar potential, e.g., any stationary vacuum spacetime can be written in terms of two potentials: http://en.wikipedia.org/wiki/Stationary_spacetime (In a general spacetime, the thing that plays the role of a potential is the whole metric.)

Well, first off there's the less exotic issue that in your examples involving periodic orbits, the spacetime simply isn't stationary.

Anyway, for cosmological models including rotation, the Godel universe is the best known example, but it has some weird features like CTCs. There are more realistic models that include rotation:

Bunn et al., "How Anisotropic is our Universe?," http://arxiv.org/abs/astro-ph/9605123
Su and Chu, "Is the universe rotating?," 2009, http://arxiv.org/abs/0902.4575

Empirically, we can only place upper bounds on the universe's rate of rotation. More info here: http://physics.stackexchange.com/a/12781/4552

Einstein's static universe (with the cosmological constant fine-tuned to prevent collapse) is not asymptotically flat, and it's homogeneous, so in that sense you could describe it using a constant gravitational potential. The gravitational potential, like any potential, is arbitrary up to an additive constant, so it's irrelevant whether you have a value at infinity to compare to. In this example, the constant potential is a potential that allows you to make statements about time dilation (there is none), but it's not a potential from which you can recover complete knowledge of the metric. (It's not a vacuum metric.)

So I think the general picture is: (1) If you want a scalar potential from which you can recover an entire vacuum metric, you need a static spacetime; stationarity is insufficient, and asymptotic flatness is neither necessary nor sufficient. (2) If you want a scalar potential that just gives a measure of relative time dilation, then stationarity suffices, and asymptotic flatness is neither necessary nor sufficient. Examples like bodies in periodic orbits are not stationary.

But keep in mind that in case #2, time dilation isn't the whole story; if you have rotation, then clock synchronization isn't transitive, so time dilation isn't the only nonnewtonian effect.

Last edited: Jun 16, 2014
9. Jun 16, 2014

### Staff: Mentor

Yes, that's definitely a different definition than I was using. I was only using "potential" to refer to the quantity that goes into the "time dilation factor"; I agree that in a spacetime that is stationary but not static, you can't recover the entire metric just from knowledge of the time dilation factor, i.e., from knowledge of a single scalar.

Yes, I should have clarified that this is always an approximation; for example, in the case of the solar system, I was assuming that every object except the Sun was a "test body" that doesn't affect the overall metric, which is obviously not exactly true but is a good approximation for many purposes.

Thanks for the links! I'll check them out.

Hmm, yes, this is a good example. So really I was implicitly assuming that the "potential", at least for the purposes of answering the OP's question about gravitational time dilation, has to vary. See further comments below.

The gravitational time dilation factor is usually thought of as a ratio, with the value at infinity being 1; but I see now that a case like the Einstein static universe could always be normalized so that the constant potential was 1 (just rescale the metric), so that the potential is still arbitrary in that sense. I'll have to think more about the non-asymptotically-flat stationary but not static case after I've read the links you gave on rotating universe models.

Agreed, except that I don't think the "vacuum" qualifier is necessary, is it? The Einstein static universe is not vacuum (it contains dust as well as a cosmological constant).

I agree with this now, since the Einstein static universe is not asymptotically flat.

Agreed.

10. Jun 16, 2014

### bcrowell

Staff Emeritus
In the case of the Einstein static universe, you can't recover knowledge of the whole metric from a scalar potential (which would just be constant). The Riemann tensor is nonzero. Actually, I'm not even sure that requiring a vacuum spacetime suffices. See the final paragraph of this WP article, presumably written by CH: http://en.wikipedia.org/wiki/Stationary_spacetime You may also need separate information about the spatial metric.

Last edited: Jun 16, 2014
11. Jun 16, 2014

### Staff: Mentor

Ah, ok. On thinking it over, I think another example would be the metric of the interior of a static gravitating body like a non-rotating star; I don't think knowledge of a single scalar potential would suffice there either.

I think you mean the Ricci tensor is nonzero, correct? The Riemann tensor is nonzero even in vacuum cases where the metric *can* be recovered from a scalar potential.

12. Jun 16, 2014

### bcrowell

Staff Emeritus
Just saying it's not flat, and you can't reconstruct any curvature information from a constant potential.

Maybe you're right, but it's not obvious to me why it wouldn't suffice in that particular case...?

13. Jun 16, 2014

### pervect

Staff Emeritus
I would expect that one could find some sort of vector potential for a stationary space-time, similar to the way that there is a vector potential for electromagnetism. I don't have any proof at this point, I haven't dug up anything to confirm this hunch.

I also know that vector potentials have been added to the Newtonian scalar potential U in the IAU 2000 recommendations for the Barycentric Celestial Reference System aka BCRS (not sure if this will turn out to be relevant).

14. Jun 16, 2014

### Staff: Mentor

Consider the general line element for a static, spherically symmetric spacetime, not necessarily vacuum; it can be written in this form (MTW gives the derivation, not sure which chapter but somewhere in the mid-20's I think; my notation is a bit different from theirs):

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

The two scalars $J(r)$ and $m(r)$ are, in general, independent; neither one determines the other. In vacuum, $m(r)$ becomes a constant and there is only one scalar left, $J(r) = 1 - 2M / r = 1 - 2 \phi (r)$.

15. Jun 16, 2014

### WannabeNewton

Yes see Ben's wiki link above on stationary spacetimes. The vector potential so to speak is actually a 1-form and it is given by $g_{it}$ whose twist (generalization of curl) leads to gravitomangetic effects. Also see:

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