If place A experiences time at twice the rate in place B, then shouldn't place A be twice as old as place B? So doesn't that mean that the Universe could contain places older than itself? Or at least older than the 15 billion year date that we have gathered?
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$\begingroup$ Near the horizon of a black hole $\endgroup$– Bob BeeCommented May 26, 2017 at 23:31
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$\begingroup$ Related, and possible a duplicate: Effect of space time relativity on the age of the universe? $\endgroup$– John RennieCommented May 27, 2017 at 4:21
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$\begingroup$ Also: Does a spaceship travelling at near lightspeed see the universe aging slow or fast? $\endgroup$– John RennieCommented May 27, 2017 at 4:21
2 Answers
We should be careful when we make statements such as "twice as old as....". A better phrase could be "clocks at location B tick at half the rate as clocks at location A".
Let us first consider a location that is infinitely far away (or at least very, very far away) from any source of stress-energy (e.g. matter), where the metric line element is $ds^2=dt^2-a(t)^2 dx^2$. In this location, there is no (significant) gravitational time delay, and clocks tick at a certain rate (e.g. 1 tick/second).
Now I may compare that rate at which clocks tick at this location (infinitely far away from any matter) to the rate at which clocks tick at a location close to a massive object. Close to a massive object, we will assume that there is a non-negligible gravitational potential $\phi$ (for this discussion, this is the same gravitational potential as $\phi=G\frac{M}{r}$ for a spherical body from intro physics). In a region with a large gravitational potential, clocks will tick at a rate that is different compared to clocks infinitely far away from any source of stress-energy (where $\phi\approx 0$).
The difference in readings ($\Delta t$) between these two clocks after a time $t$ has passed is $\frac{\Delta t}{t}\sim \phi$. As it turns out, $\phi$ is very small. At the surface of the Earth, $\phi\sim 10^{-6}$ ($\leftarrow$ I think....too lazy to check right now).
As to your question about locations where the age of the universe is different, the only location where $\frac{\Delta t}{t}$ would be significant would be in regions of a strong gravitational field: near a neutron star or black hole. (Then again, we would no longer be working in the weak field regime of gravity, which is not very well understood.) The universe in these regimes would not be twice as old, but rather, a person, or particle will have aged half as much as it otherwise would have if it were not located in a strong gravitational field.
The universe age is defined with respect to some preferred reference frame, with time in that frame being comoving time, or cosmic time, or equivalently how time passes for a test particle moving with the average expansion, also called the cosmic flow. And the age of the universe in that time reference frame is about 13.8 billion years. It's a very useful reference frame to use that would be observed to be roughly the same anywhere in the universe.
But in Relativity time observed, or time measured by a (say perfect) clock depends on what reference frame it is in, and at rest. So different observers could measure times passing at different rates.
An observer moving with the cosmic flow will measure cosmic time. The gravitational field for that observer is the average field for the universe. If that observer measures the geometry of the universe, on the average, it will say it is the geometry we consider the cosmological solution. It's the FLRW metric, in the coordinate system that is comoving with the flow. If that observer measures invariant curvatures of that universe it will get the same curvatures we measure.
In fact, we have to adjust what we measure because we have a peculiar velocity of a few hundred Kms/sec with respect to the cosmic flow.
And the rate of time passing can be different for an observer who is in a gravitational field which is different than the average that determines the cosmic flow.
For instance, an observer hovering over the horizon of a black hole will have his/her time pass much slower than the universe aging. He/she might only age 50 years while the universe ages 100 years (all depending on how far above the horizon he/she is). Or about 6.9 billion years (if alive, or conceptually) while the universe ages 13.8 billion years. There have been questions as to what that observer sees, does he/she see the universe ending, or other stars exploding, while he/she ages less? The answer is a bit tricky, because the observer is just not able to see that much in this case, see eg the LAST answer for the PSE question at Falling into a black hole
And it is different if the observer is falling in or just hovering.
And yes, time can pass differently and things age differently depending on the gravitational field you are in. The times measured by our GPS satellites passes faster compared to the time on earth, and those have to be adjusted, although it is a very small difference
Similarly an observer can very very far from any nearby objects, say in what is called the great void, and time will pass faster. But we can only measure cosmological times approximately, and so far those have not made any difference, except as we note the redshifts or blueshifts associated with those. The universe we measure as about 13.8 billion years plus or minus about 35 million years 1 sigma (so 3 sigma is about 100 million years). Time on earth or for the solar system would just not be noticeably different for astrophysical aging measurements (like the age of rocks).
If you were to somehow have had an asteroid orbiting a black hole near it, would it's rocks age slower? Yes, but before that would make much difference it would have radiated enough gravitational energy and decayed into the black hole.
Not clear you could find a case where you could measure the difference. But if you did its explainable by all those arguments above. The universe age is discussed at https://en.m.wikipedia.org/wiki/Age_of_the_universe