# Gravitational effects on time

1. Aug 20, 2014

### pforeman

If, when the universe was young, and you had all of today's universe worth of matter compacted into a space only a few thousand/ million light years in diameter the gravitational effect on time would be much stronger than it is today. Does this affect our calculations on how old the universe is?

Is time dilation factored in when cosmologist predict down to the 1 X 10-30 sec. what happened when in the first milliseconds after the big bang.

When we look at a spinning galaxy, and say why aren't the arms spinning slower than the center, would the time dilation enter into the calculation?

When we look at super massive stars, Is there a significant time dilation to affect our calculation on it's age.

Thanks,
Paul

2. Aug 21, 2014

### Staff: Mentor

No, because the "gravitational effect on time" that you are talking about doesn't apply to the early universe; for that matter, it doesn't apply to the universe as a whole today. It only applies to a specific type of scenario, a stationary, isolated gravitating system surrounded by empty space.

Relativity is certainly factored in; but since we're still dealing with the universe as a whole, not a stationary, isolated system surrounded by empty space, "time dilation" is not a good description of the effects of relativity.

If you're in one of the solar systems that are rotating about the galaxy's center, like we are in our galaxy, then in principle, yes, since the galaxy can be viewed as a stationary, isolated gravitating system surrounded by empty space. But the effect is too small to matter.

If we're looking at a distant galaxy, then no, because we're well outside the entire system.

No; supermassive stars are not very compact, and you need a compact object--one whose actual radius is not much larger than the Schwarzschild radius associated with its mass--for gravitational time dilation to be significant when comparing the "rate of time flow" within the object to the rate of time flow far away. Other than black holes, which aren't really "objects" in the usual sense, the only objects we know of that are compact enough for gravitational time dilation effects to be at all significant are neutron stars.

3. Aug 21, 2014

### Chalnoth

Nope. The full equations of General Relativity are used to calculate the age of our universe, so any time dilation is included automatically.

4. Aug 21, 2014

### Tanelorn

How big would a supermassive star need to be to prevent its own light from leaving it? ie. An active star which is also a black hole.

I read, here I think, that if our solar system was made entirely of water it would be a black hole.. Perhaps it needs to collapse first to be dense enough?

Last edited: Aug 21, 2014
5. Aug 21, 2014

### Staff: Mentor

No such object is possible. Any star which gets compact enough to prevent its own light from escaping will collapse, thereby becoming a black hole; when this happens, it will no longer be a star.

6. Aug 21, 2014

### Tanelorn

http://www.astrosociety.org/edu/publications/tnl/24/blackhole4.html

MYTH: Black holes are very dense.
TRUTH: Small and medium black holes are very dense, but a supermassive black hole with a 100 million solar masses, for example would have a density the same as water. [You can work this out from the mass of the black hole and the radius of its event horizon; this assumes that all of the matter is distributed within the entire event horizon, not just in the singularity.]

I could hardly believe it either!

Which is why I wondered how close the mass of the largest stars get to this situation?
http://en.wikipedia.org/wiki/List_of_largest_known_stars

What is the effect of the mass of these stars on the light leaving them? Does the hydrogten line get shifted lower in frequency for example? If so, would such an effect also mean time dilation?

Last edited: Aug 21, 2014
7. Aug 21, 2014

### Staff: Mentor

It also assumes that the volume enclosed by the event horizon is given by the usual Euclidean formula, which is not really a valid assumption, although a lot of pop science presentations make it.

If you're just interested in density, the average density of a supermassive star is much less than that of water. The average density of the Sun is about 1.4 times that of water.

If by "this situation" you mean "having a significant time dilation effect", no star, no matter how massive, is going to be anywhere close. See below.

The light emitted by stars is redshifted as it climbs out of their gravity well, yes; and the term "time dilation" is often used to refer to this effect. But the effect is very small for any star; for the Sun, it's about 3 parts per million.

For an object to have a significant time dilation effect on light leaving it, it has to be very compact: that is, its actual radius has to be as close as possible to the Schwarzschild radius corresponding to its mass. However, there's a limit to how close a static object can get; no static object can have a radius smaller than 9/8 of the Schwarzschild radius associated with its mass. Any smaller and the object collapses and becomes a black hole. The only objects we know of that come close to this limit are neutron stars.

8. Aug 21, 2014

### Tanelorn

9. Aug 21, 2014

### Staff: Mentor

It isn't. Time dilation is given by $\sqrt{1 - 2GM / c^2R}$, where $M$ is the object's mass, $R$ is its radius, $G$ is Newton's gravitational constant, and $c$ is the speed of light. (Relativists often use units where $G = c = 1$, which is why you will often see this formula written without them appearing.) $GM / c^2$ is the Schwarzschild radius associated with the mass $M$, so, as I said before, time dilation is based on how compact the object is--how close its radius $R$ is to the Schwarzschild radius associated with its mass.

Last edited: Aug 21, 2014
10. Aug 21, 2014

### Tanelorn

[You can work this out from the mass of the black hole and the radius of its event horizon; this assumes that all of the matter is distributed within the entire event horizon, not just in the singularity.]

"It also assumes that the volume enclosed by the event horizon is given by the usual Euclidean formula, which is not really a valid assumption, although a lot of pop science presentations make it."

Peter, What would be a more valid assumption?

11. Aug 21, 2014

### Staff: Mentor

That the concept of "volume enclosed by a black hole event horizon" doesn't really make sense; at any rate, there is no unique value that can be assigned to it. There are various ways to interpret the spacetime structure inside the horizon; there is one way that leads to assigning a spatial volume inside the horizon that's equal to the Euclidean formula; but there's another way that leads to assigning an *infinite* spatial volume inside the horizon! Spacetime inside the horizon simply does not work the way we are used to.

12. Aug 21, 2014

### marcus

Peter could I make one or two cosmetic changes? Tell me if I'm wrong. I think that the R here is the distance of the observer or the CLOCK from the center of mass. And there should be a 2 in front, in the formula for the Schw. radius, no? You probably wrote the original in haste and didn't have time to fix the typos. Here's the original post.
If it's OK I'll fix what I think are typographical errors.
So if the object is a black hole and the observer is right down at the horizon then his distance R equals the Schw. radius and the ratio is UNITY and square root of (1 - 1) is zero so the speed of the observer's clock appears to distant observers to be zero.

13. Aug 21, 2014

### Staff: Mentor

No, it isn't. It's the radial coordinate of the observer, but the radial coordinate does not translate directly into proper distance in the radial direction. To get the proper radial distance from the center of mass (at $r = 0$), you have to compute the integral

$$D = \int_0^R \sqrt{g_{rr}} dr$$

where $g_{rr}$ is the radial metric coefficient, and is not in general equal to 1, so the value of the integral, and hence the distance, will not be $R$. For an object like the Earth or the Sun, the difference is very small (about one part in 10^8 for the Earth, and about one part in 10^5 for the Sun); but for an object like a neutron star, it's significant.

Yes, you're right, there should.

No; in fact, for a black hole, the concept of "distance from the center of mass" doesn't really make sense; the "center of mass", if it's anywhere, is at the singularity, and the singularity is in your future if you're just passing the horizon of the hole, so the proper notion of "distance" to it is really the time it will take you to fall to it, which is a timelike "distance", not a spacelike distance.

Not really; distant observers can't see a clock that is crossing the horizon. Also, an object at the horizon can't be static (it must be falling into the hole), and the formula for time dilation in terms of $R$ only really applies to static objects. A better way to say it would be that the time dilation factor of a static object at $R$ approaches zero in the limit as $R$ approaches the horizon.

14. Aug 21, 2014

### Tanelorn

"but there's another way that leads to assigning an *infinite* spatial volume inside the horizon! Spacetime inside the horizon simply does not work the way we are used to."

Hence the idea that a black hole could contain another Universe, and our Universe could be a black hole in another Universe. This is just one of many possible solutions though correct?

15. Aug 21, 2014

### marcus

Maybe this is a silly example but the sun is not very compact. I think its density is comparable to that of water. Its radius is around 700,000 km and its Schw. radius is around 3 km.

Suppose we have a timepiece, a heat resistant clock of some type, dangling somehow 900,000 km from the center of the sun. So it is above the surface

Then the ratio would be 3/900000 = 1/300000

1- that ratio = 299999/300000 and the square root would be 0.99999833…

So that clock would be running slower by the factor 0.99999833… compared with a a clock out much farther so that the dilation effect was negligible.

So it would lose roughly 2 seconds every 1000000 seconds---or about 2 seconds every 12 days.

Anyone interested could make a similar calculation for a the clock of an observer hovering a few hundred km above the surface of the earth. All one needs to know is the radius of the earth and the earth's Schw. radius. (the latter is about 9 millimeter so one only needs to look up the actual radius).

16. Aug 21, 2014

### Staff: Mentor

Yes, but that's pure speculation. Also, such a universe would not look anything like ours does; see below.

Not really. The reason you can get an infinite spatial volume is that you can pick out a spacelike dimension inside the horizon that is infinitely long; but it's only one dimension. The other two spatial dimensions are finite, and the full spacelike hypersurface is basically a hypercylinder. This is a large asymmetry between the spacelike dimensions, and we don't observe anything like it in our universe.

No, at least not if you mean one of many possible solutions for a black hole. What I'm describing is just the region inside the event horizon in the standard Schwarzschild solution for a black hole; that's the only solution for a non-spinning, uncharged hole.