Just curious to know what will be the time difference between someone living on the outer layer of our galaxy and us, considering the known facts that

  • Time is relative to gravity
  • Our sun goes around at 483,000 miles/hour(792,000 km/hr) to complete its galactic year
  • Star systems at outer layer travel faster than ours due to Dark Matter and Dark Energy - couldn't find speed of those star systems though!

We can assume that both (Solar system and outer layer of star system) are on the same side of the galaxy.

EDIT: In order to re-phrase this question more clearly, as per Joan.bdm, it is related to time dilation - what is the aging rate difference if we left one of two twins on the Earth and the other one on a planet at outskirts of the Milky Way. Conversely, what is aging rate difference between closer to center of our galaxy and us?

  • $\begingroup$ What do you mean by time difference? Do you mean something like: "if we shoot a photon towards that system, how long does it take to arrive?"? $\endgroup$ – Py-ser Aug 29 '14 at 5:37
  • $\begingroup$ I think he refers to time dilation. Some kind of extreme twin paradox: what if we left one of two twins on the Earth and the other one on a planet at outskirts of the Milky Way. I guess he is asking something like which would be the aging rate difference. $\endgroup$ – Joan.bdm Aug 29 '14 at 6:35
  • $\begingroup$ Fredy, assuming @Joan.bdm is right (and I think that is the case), can you re-phrase the question to make it clearer? $\endgroup$ – HDE 226868 Aug 29 '14 at 17:45
  • $\begingroup$ Sorry-couldn't follow up quickly enough - thought I will get an email update for responses. Yes - Joan.bdm correctly interpreted my quest. I will re-phrase my question accordingly. $\endgroup$ – Fredy Sep 1 '14 at 2:54

Your third question has a very well known answer here:

enter image description here

What means, that although a big time difference had to exist, there is a much smaller one, and in the opposite direction. This is considered the effect of the dark matter.

Our Sun is around 30000 ly from the galactic center, the edge of the galaxy is around 50000 ly, so we can read on the diagram, that the actual speed difference is around 15 km/s, which is 1/20000 of the speed of light. This causes a (not really big) time dilation due to the special relativity.

There is another source of the time dilation, which is caused by the gravity of the galaxy, which differs in the case of us and in the case of somebody living on the outer edge. On such weak gravitation (i.e. for from any black hole, neutron star, etc) it practically depends on the newtonian gravitational acceleration.

But this acceleration is very small - it takes a half galactic year (some hundred millions of year) to revert the orbital speed of the Sun around the galactic core! So, we can consider that actually negligible.

Thus only the special relativistic effect remains. 15 km/s is around 1/20000 of the speed of the light. In cases of speed which are much smaller as the light speed, we can use the speed dilation formula 1/(2*(v/c)^2). Substituting 1/20000 into this formula, we get 1:800 000 000 .

For laymans, we can say, it takes around 25 years for the clocks of this people living on the outer edge of our galaxy to dilate 1 second.

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    $\begingroup$ This plot is not the rotation curve of the Milky Way, which is less accurately known (mainly because we are observing it from within, making these type of measurements difficult). $\endgroup$ – Walter Aug 29 '14 at 16:09
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    $\begingroup$ This is incorrect. First, gravitational time dilation depends on the gravitational potential, not the acceleration. Second, its contribution is actually an order of magnitude more significant than that of velocity. Thus, one cannot consider gravitational time dilation negligible in any situation in which the special-relativistic contribution is not counted as negligible. $\endgroup$ – Stan Liou Aug 31 '14 at 1:47
  • $\begingroup$ Um, doesn't that diagram just show the tangential component of a star's orbit around the center of the galaxy? If so, then it has only an indirect effect on (velocity based special relativity) time dilation, which is dependent only on the radial component. (I had to join just to ask this.) $\endgroup$ – Spencer Aug 12 '17 at 18:09
  • $\begingroup$ @Spencer Time dilation is defined in an inertial frame, i.e. you have to fix a point from which you see that. The question - indirectly - defines it to a non-moving distant observer. I am not sure, but the definition of the proper time, and thus, the time dilation, looks to me independent from the motion direction in the SR. $\endgroup$ – peterh Aug 12 '17 at 19:49

It's not really clear what you're aksing. There is no "relative time difference" between anywhere and somewhere else in the universe. There are, of course, time differences between different events (space-time points).

The motion of stars in the Milky Way (or any other) galaxy are sub-relativistic, implying that GR effects (such as time dilatation) are negligible. Of course, the Galaxy is rather big and it takes much longer than a human life time for light, let alone space craft, to travel accross it.

  • $\begingroup$ time dilatation is not negligible, even our own GPS system corrects for its time dilatation, but it is probably small. $\endgroup$ – Kári Gunnarsson Apr 19 at 5:06

In the slowly-changing, weak-field regime, the effects of general relativity can be approximated by the following metric (in units of $c=1$): $$\mathrm{d}s^2 = -\left(1+2\Phi\right)\mathrm{d}t^2 + \left(1-2\Phi\right)\mathrm{d}S^2\text{,}$$ where $\mathrm{d}S^2$ is the metric is Euclidean $3$-space and $\Phi$ is the Newtonian gravitational potential. Thus, the rate of proper time vs coordinate time is $$\frac{\mathrm{d}\tau}{\mathrm{d}t} = \sqrt{1+2\Phi-\left(1-2\Phi\right)\frac{\mathrm{d}S^2}{\mathrm{d}t^2}} = 1+\Phi - \frac{1}{2}v^2 + \mathcal{O}(v^4,\Phi^2,v^2\Phi)\text{.}$$ At low speeds, we can drop the higher-order terms to infer a relative time dilation between two stars to be $$\frac{\mathrm{d}\tau_1}{\mathrm{d}\tau_0} \approx \frac{1+\Phi_1-v_1^2/2}{1+\Phi_0-v_0^2/2} \approx 1+(\Phi_1 - \Phi_0) - \frac{1}{2}(v_1^2-v_0^2)\text{.}$$ The gravitational potential of the Milky Way can be modeled as the sum of an spherically symmetric bulge, an axisymmetric disk, and a spherically symmetric dark matter halo. Many galactic models exist, but for simplicity I will be using ones from Irrgang et al. [arXiv:1211.4353] for the Sun:

  1. Model I (after Allen & Santillan (1991)): $v_0 = 8.072\times 10^{-4}$, $\Phi_0 = -2.113\times 10^{-6}$.
  2. Model II (after Wilkinson & Evans (1999) and Sakamoto et al. (2003)): $v_0 = 8.019\times 10^{-4}$, $\Phi_0 = -1.845\times 10^{-6}$.
  3. Model III (after Navarro et al. (1997)): $v_0 = 7.996\times 10^{-4}$, $\Phi_0 = -3.664\times 10^{-6}$.

Because the velocity-dependent contribution is squared while the potential contribution is not, the gravitational time dilation could be more significant. Note that this does not include the contribution from the Sun itself, the vicinity to which is going to be more significant. However, if we're comparing time dilation for observers near similar stars at similar orbits from their stars, that difference will cancel out.

Unfortunately, the gravitational potential is rather model-dependent due to the many uncertainties involved in the observations, but you can play with some of the values given in that paper if you wish.


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