I understand that time is relative for all but as I understand it, time flows at a slower rate for objects that are either moving faster or objects that are near larger masses than for those that are slower or further from mass.

So, the illustrative example I always see is that if I were to leave earth and fly around at near light speed for a while or go into orbit around a black hole the time I experience would be substantially shorter than for those I left behind at home on earth and I'd come back to find that I've only aged however long I felt I was gone by my own clock but that on Earth substantially more time would have elapsed.

Following through on this model, the stars in orbit around the black hole at the center of the Milky Way are aging much much slower (relative to us), right? So does it not follow that the center of the galaxy is some appreciable (I have no idea how to go about putting this in an equation so will avoid guessing at the difference) amount "younger" than the stuff further away from the center?

If this is not true, could someone please explain why not, and if it is true, can someone please point me to where I can calculate the age of the center of the galaxy :-)

And to be clear... what I'm asking is... if there was an atomic clock that appeared at the center of the galaxy when the center was first formed, and we brought it through a worm hole to earth today - how much time would have elapsed on that clock vs. the age we recon the galaxy currently is? (13.2 billion years)

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    $\begingroup$ A more interesting question would be, looking at the recent movie Interstellar as a guide, would any planets/stars extremely close to Sagittarius A* be younger, as this question suggests, even if the 2 current answers are correct and most of the ones only somewhat close to it are not much younger? Sorry if this comment spoiled anything in the movie. $\endgroup$
    – trysis
    Jan 25, 2015 at 21:55
  • $\begingroup$ @trysis, could you elaborate? I'm unable to parse from your comment what your "more interesting" question actually is - I didn't see the movie (no worries about the spoiler :-). $\endgroup$ Jan 26, 2015 at 0:16
  • $\begingroup$ Well, in the movie, there is a planet that the protagonists go to that is extremely close to a black hole. You can be close enough to the planet to orbit it and still experience time "normally", but when you get really close, and, in particular, when you are on the planet, you experience 1 hour for every 7 years that passes away from the planet. I don't know how realistic this is, but I have heard that the creators tried to make it realistic. I suppose by "more interesting" I meant is it possible :). $\endgroup$
    – trysis
    Jan 26, 2015 at 4:45
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    $\begingroup$ @trysis: the star nearest to Sagittarius A$^*$ is S2. I make its time dilation (including the effect of orbital speed) about 0.999 i.e. time runs about 0.1% slower. $\endgroup$ Jan 26, 2015 at 6:59
  • $\begingroup$ @John Rennie: S2 gets to about 1% of the speed of light in perihelion. Adding the time dilation from Sagittarius A*'s gravity field, wouldn't the total be more than 0.1%? $\endgroup$
    – Blackscale
    Mar 31, 2015 at 11:47

5 Answers 5


The gravitational potential of the disk of the Milky Way can be approximated as:

$$ \Phi = -\frac{GM}{\sqrt{r^2 + (a + \sqrt{b^2 + z^2})^2}} \tag{1} $$

where $r$ is the radial distance and $z$ is the height above the disk. I got this equation from this paper, and they give $a$ = 6.5 kpc and $b$ = 0.26 kpc.

In the weak field approximation the time dilation is related to the gravitational potential by:

$$ \frac{\Delta t_r}{\Delta t_\infty} = \sqrt{1 - \frac{2\Delta\Phi}{c^2}} \tag{2} $$

At the centre of the galaxy $r = z = 0$ and equation (1) simplified to:

$$ \Phi = -\frac{GM}{a + b} \tag{3} $$

No-one really knows the mass of the Milky Way because we don't know how much dark matter it contains, but lets guesstimate it at $10^{12}$ Solar masses. With this value for $M$ and using $a$ + $b$ = 6.76 kpc equation (3) gives us:

$$ \Phi = 6.4 \times 10^{11} \text{J/kg} $$

Feeding this into equation (2) gives:

$$ \frac{\Delta t_r}{\Delta t_\infty} = 0.999993 $$

So over the 13.7 billion year age of the universe the centre of the Milky Way will have aged about 100,000 years less than the outskirts.

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    $\begingroup$ Probably a stupid question, but… we have a tangential velocity that the centre of the galaxy doesn't have, haven't we? Does this fact reduce the ratio you've calculated, or is the impact negligible? $\endgroup$
    – Blackhole
    Jan 25, 2015 at 23:15
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    $\begingroup$ @Blackhole: no, that's a good question and as it happens I answered it recently. The time dilation due to our orbital velocity is about an order of magnitude less than the time dilation due to the gravitational potential. So if you include this the centre will have only aged about 90,000 years less than the outskirts. However take this figure with a pinch of salt as we don't know the dark matter distribution. $\endgroup$ Jan 26, 2015 at 6:27
  • $\begingroup$ @Blackhole while the "center" doesn't have a tangential velocity, the stars (and everything else) near the center still orbit, and I think they'd have velocities higher than ours, so the velocity difference would actually add to the gravitational one rather than subtract. $\endgroup$
    – Rick
    Jan 26, 2015 at 15:44
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    $\begingroup$ It might be worth taking that comment (re 90,000) and making it an endnote to the question itself. $\endgroup$ Jan 26, 2015 at 15:52

The centre of the galaxy will indeed appear to pass through time more slowly than the edges, but the effect will not be great.

Because the Einstein field equations are very difficult to solve, it is not possible to calculate the exact magnitude of the time dilation, but we can make an approximation. By assuming that the black hole at the centre of the galaxy is electrically neutral and non-rotating, and ignoring the effects of all other mass/energy, we can calculate the time dilation at a distance $r$ from the galactic centre, as seen by an observer at infinity.

The formula for this time dilation is $\Delta t_0 = \Delta t_\infty \sqrt{1 - \frac{r_S}{r}}$, where $t_0$ is the proper time at a distance of $r$ from the galactic centre; $t_\infty$ is the proper time measured at infinity, and $r_S$ is the Schwarzschild radius of the black hole that lives at the centre of the galaxy. Because $r_S$ is many times smaller than $r$ (except for any unlucky stars finding themselves being eaten by the black hole), we would not see any appreciable difference in the rate at which time passes between stars close to the centre and those far away.

All of this analysis assumes that Sagittarius A* is exactly at the centre of the Milky Way, which is not exactly true. The distance between the two will cause the actual centre to be slowed by the gravity of the black hole, just like anything else. This will be highly dependant on the proper distance between the centre and the hole, but could be calculated - with some approximation - by the above formula.

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    $\begingroup$ The mass of the Milky Way is about $10^{12}$ Solar masses and the mass of Sagittarius A$^∗$ is around $10^6$ to $10^7$ Solar masses. Ignoring the mass of the Milky Way seems a poor approximation to me ... $\endgroup$ Jan 25, 2015 at 20:21

Time Dilation Effects for Stars on the Outer Edge of The Milky Way First, we tackle gravitational time dilation: whereby stars closer to the center will age more slowly because they are in stronger gravity. Ignoring galactic rotation to simplify, general relativity tells us that t’ = t (1-2GM/rc^2)^1/2 where t’ is the dilated (slowed) rate of time passage and t would be the rate of time passage if the mass M (of the galaxy in this case) was not present – or at an “infinite distance from it” – and r is the distance from the center of the mass of the galaxy. We will assume the point in question lies at the outer edge – so that the entire mass of the galaxy lies inside it. We will ignore any hypothetical dark matter - which in any case is thought to lie farther out in "a halo". Using binomial expansion, where (1+x)^n is approximately equal to 1+nx, to simplify we get t’ = t (1 - GM/rc^2) = t – tGM/rc^2 Letting the change in time between the two situations equal ∆t, then ∆t = t’ – t = t – t - tGM/rc^2 = tGM/rc^2 Therefore ∆t/t = GM/rc^2 Going with orders of magnitude to get a rough sense of it, let M = the mass of the Milky way = approximately 10^12 solar masses, which equals about 10^42 kilograms. And, let r = 60,000 light years (a broadly accepted radius for the Milky Way - but subject to debate depending on where we define the "edge") which equals 5.6 x 10^20 meters. Therefore ∆t/t = (6.673 x 10^-11 m3/kg-s2 x 10^42kg)/5.6 x 10^20 meters x 9 x 10^16 m2/s2 = 1.32 x 10^-6

And, if we accept the age of the Milky way at about 13.20 billion years, then 1.32 x 10^-6 x 13.2 x 10^9 years = 17.42 x 10^3 years or approximately 17,000 years In other words, the stars near the center would be about 17,000 years younger than those at the outer edge due to gravitational time dilation.

Now to address velocity time dilation: whereby stars moving faster with respect to the center will age more slowly. The speed of the outer stars of the Milky Way is about 210 km per second wrt to the center of the galaxy. From special relativity we get t’ = t (1 – v^2/c^2)^-1/2 Using binomial expansion to simplify we get: t’ = t (1 + ½ v^2/c^2) and rearranging for ∆t/t, as we did before, ∆t/t = ½ v^2/c^2 = ½ (2.1 x 10^5/3 x 10^8)^2 = 2.45 x 10^-7

And 2.45 x 10^-7 x 13.2 x 10^9 years = 3234 years less time passed because of the star's motion. So, the net time dilation for a star at the edge of the galaxy would be roughly equal to the years gained by gravitational time dilation minus years lost due to velocity time dilation which, rounding off to the nearest thousand, is about 17,000 – 3000 = 14,000 years

  • $\begingroup$ The self-promotion seems pretty gratuitous, to be honest. It's not at all clear how it's related to the question at hand. $\endgroup$ Apr 27, 2022 at 15:30
  • $\begingroup$ "stars closer to the center will age more slowly because they are in stronger gravity" this statement is conceptually incorrect. $\endgroup$
    – ProfRob
    Apr 27, 2022 at 17:55
  • $\begingroup$ I appreciate the answer and a little self-promotion is never a bad thing! $\endgroup$ Jun 7, 2022 at 1:40

A simple answer would be that since the universe expanded from a singularity, everywhere is the center of the universe.

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Sep 6, 2022 at 6:32
  • $\begingroup$ I didn't ask about the center of the universe - I asked about the center of the galaxy. You should delete this answer before it gets a ton of downvotes. $\endgroup$ Oct 18, 2022 at 3:09

Time on earth move 1 second per week faster than time in orbit because of the lack of mass in orbit so that said the closer you go to the center of the galaxy the more mass is there so time should move faster but also the closer you go to the center the closer you are to the black hole so you would naturally orbit faster to remain in orbit and not fall and the faster you go the slower time moves so I wonder if the orbiting speed will overcome the mass and neutralize the flow of time to test that someone would need to test the flow of time on the moon to see if it's the same as Earth's time or better yet test the speed of time on the Jupiter's moon if you find it is moving faster than Earth's time then that would be proof that time moves faster in the center of the Galaxy because there is more mass around Jupiter's moon without the mass directly on Jupiter's moon


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