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Main Question: If you were to average out the rate of the passage of time in the observable universe relative to earth, what would it be?

Alternative Precise Question: What is the rate of passage of time at the halfway point between Andromeda and The Milky Way relative to earth? I imagine it must be very fast.

Assume 1 time unit = 1 earth time unit.

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I think it would be interesting to find out that earth experiences time vastly different from the rest of the universe.

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    $\begingroup$ I just figure that we don't have enough mass here on earth to make a significant difference $\endgroup$
    – Krumuvecis
    May 10, 2021 at 18:36
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    $\begingroup$ It sounds like you are being asked to work out the gravitational time dilation on earth relative to that in flat space. $\endgroup$
    – m4r35n357
    May 10, 2021 at 19:40
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    $\begingroup$ I imagine it must be very fast. Why do you think that? $\endgroup$
    – G. Smith
    May 10, 2021 at 19:49
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    $\begingroup$ Time passes fine in the absence of gravity. $\endgroup$
    – G. Smith
    May 10, 2021 at 22:05
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    $\begingroup$ I think it would be interesting to find out that earth experiences time vastly different from the rest of the universe. It doesn’t, because Earth’s gravity is extremely weak: $\varphi/c^2\ll 1$, where $\varphi$ is the Newtonian gravitational potential. $\endgroup$
    – G. Smith
    May 10, 2021 at 22:14

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Gravitational time dilation in a gravity well is equal to the relativistic time dilation due to the speed required to escape that gravity well (see this Wikipedia article for more information). Escape velocity from the Earth is about 11.2 km/s. Solar escape velocity from Earth's orbit is about 42.1 km/s. Escape velocity from the Milky Way is about 550 km/s. So total escape velocity from Earth to a point halfway between Andromeda and the Milky Way is somewhere in the vicinity of 600 km/s, which produces a time dilation factor of about 1.00000200277612; that is, for each second that passes on Earth around 1.00000200277612 seconds pass at your hypothetical distant point.

(This calculation assumes the observers are at rest relative to one another. If the distant observer is at rest relative to the cosmic microwave background, then there's another factor of about 600 km/s to account for due to the Earth's movement relative to the CMB.)

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    $\begingroup$ You don't add the velocities you add their squares and then square root (for velocities much less than light). You also include the motion of the earth around the sun (30 km/s) and the sun around the galactic center (250 km/s). The "total" still works out to about 600 km/s for an observer co-moving with the milky way but far from it. $\endgroup$ May 11, 2021 at 3:44
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To calculate time dilation with respect to a point midway between the Milky Way and Andromeda you need to account for motion of the Earth relative to that point and for the difference in gravitational potential between the surface of the Earth and that point.

Because both these effects are small they can be treated as additive (in the same way that sometimes people talk about a SR correction and a GR correction to the clocks on the GPS satellites).

Roughly speaking, Andromeda is approaching the Milky Way at $\sim 110$ km/s. The tangential component of the relative velocity is much smaller than that. In addition, the Sun orbits the galactic centre at something like 220 km/s and these two velocities combine to give a speed of about 250 km/s or a Lorentz factor of $\gamma -1 = 3.5\times 10^{-7}$, assuming that the midpoint is co-moving with Andromeda.

If we assume that the point midway between the Milky Way and Andromeda is at zero potential, then if the Milky Way has a mass of $\sim 10^{11} M_{\odot}$ interior to the Sun's galactic radius, then the potential is of order $\Phi \sim -5\times 10^{10}$ J/kg and an associated dilation factor of $$1 - \Phi/c^2 = 1 + 5\times 10^{-7}\ .$$ Note that we don't need to include the gravitational potential due to the Earth itself or due to the Sun because these potentials are about 2 orders of magnitude smaller.

i.e. The two effects are small and of similar size. The total time dilation is $\sim 10^{-6}$. i.e. $1 + 10^{-6}$ seconds pass on Earth for every 1 second on a clock midway between the Milky Way and Andromeda.

To do the calculation between the Earth and an average bit of the universe then I think you would have to go with working out a time relative to the co-moving rest frame. The Sun moves at 370 km/s with respect to the Cosmic Microwave Background (which defines the comoving rest frame). The motion of the Earth around the Sun is much smaller and can be neglected. This yields a time-dilation factor of $\gamma -1 \simeq 7.6\times 10^{-7}$.

For the potential we need to think about not just the Milky Way but the Local Group of galaxies. However, it turns out that the mass of the local group is only a few times that of the Milky Way and is of course spread over a much larger volume. So the additional gravitational potential due to all the other local galaxies is rather small compared to that contributed to mass interior to the Sun's galactic radius. Therefore the time dilation number for the gravitational potential will only be marginally bigger than that to the midpoint between Andromeda and the Milky Way.

Thus my answer for the time dilation factor with respect to the co-moving frame is about $1.3\times 10^{-6}$.

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Time at the midpoint between Andromeda and the Milky Way (assuming that no gravity is present) passes barely faster than time on Earth, contrary to what you imagine. I'll give a technical derivation.

The spacetime surrounding the Earth is the Schwarzschild metric:

$$ ds^2=\left(1-\frac{r_s}{r}\right)dt^2-\left(1-\frac{r_s}{r}\right)^{-1}dr^2-r^2d\Omega^2, $$

Here $r$ is the radial distance from the center of the Earth and $r_s$ the Schwarzschild of the Earth (which we can set to be $8,7(mm)$). The only part of the metric we have to take into consideration is the $dt^2$ part, as $r$ and $\Omega$ (representing the angles in the radial coordinates) stay constant. From this, we can see that passage of time is $r$-dependent. Now, let's say that:

-time passage at a point between Andromeda and the Milky Way: 1
-time passage at the surface of the Earth (where $r=6000 000$): $\sqrt{\left(1-\frac{r_s}{r}\right)}=\sqrt{\left(1-\frac{0,0087}{6000000}\right)}$

As you can see clearly, the rate at which time on Earth flows is practically the same as the rate at which it flows at the midpoint between Andromeda and the Milky Way (so time doesn't flow much faster at the midpoint). If a black hole were present at the midpoint, then time on Earth would flow much faster compared to the time passage around the hole. If the Eart revolved around a black hole, then time on Earth could indeed flow considerably slower than at the midpoint (so time at the midpoint would indeed flow much faster than on Earth, though it of course depends what you call much).

What about the flow of time on Earth when compared to the average flow of the whole universe? Considering the fact that space is flat almost everywhere, we can safely say that the average rate of time of the whole universe will be no different from the rate of time at the midpoint between Andromeda and the Milky Way. So time on Earth flows at the same rate wrt to the flow of the whole universe as it does wrt to the midpoint. That is, barely slower. Note that I don't take the gravitational influence of the sun and the Milky Way into consideration. But these contributions don't make time on Earth go substantially slower, so even when you take them into consideration this would not change the conclusion.

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  • $\begingroup$ This doesn't take into account several important effects. $\endgroup$
    – ProfRob
    May 11, 2021 at 7:19
  • $\begingroup$ @ProfRob Arent'these the only effects? $\endgroup$ May 11, 2021 at 13:09
  • $\begingroup$ Yes, that a clock on the Earth experiences dilation due to the Earth's motion and due to the total difference in potential compared to a point midway between the Milky Way and Andromeda. Your second answer fails to account for (i) the galactic potential and (ii) the motion of the Milky Way relative to the co-moving reference frame. $\endgroup$
    – ProfRob
    May 11, 2021 at 13:18
  • $\begingroup$ @ProfRob But I mentioned in the answer that I didn't take into account the reduction in the "velocity" of time due to the sun and the Milky Way. i stated that they are of the same order and thus don't contribute significantly to the difference between time flow on Earth and the midpoint. $\endgroup$ May 11, 2021 at 13:29
  • $\begingroup$ The gravitational potential due to the Earth is negligible (2 orders of magnitiude smaller) than the galactic potential. $\endgroup$
    – ProfRob
    May 11, 2021 at 13:56
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Here is an answer that comes from this paper: https://www.sugarsync.com/pf/D7718709_68878570_6399503 (which I wrote)

To show how this works in a gravitational field, we want to compare the permittivity of free space versus the permittivity of space on the surface of the gravitational body:

• In Free Space: c=√(1/µ0ɛ0), where µ0=1.25663706 x 10-6 and ɛ0=8.85418782 x 10-12

• And using the Schwarzschild metric to determine time dilation relative to free space: t!=t/√(1-2Gm/rC2)

On the Earth: Mass=5.9722x1024kg Radius=6.371x106m We find a time dilation factor of tE=1.000,000,000,699,68 Thus we find a permittivity factor of ɛE = ɛ0 x tE=8.85418783x10-12 This works out to 21 centimeters per second in the relative speed of light

On the Sun: Mass=1.989x1030kg Radius=6.9634x108m We find a time dilation factor of tS =1.000,002,121,041,69 Thus we find a permittivity factor of ɛS = ɛ0 x tS=8.85420660x10-12 This works out to 635 metres per second in the relative speed of light

On star R136a1: Mass=6.263x1032kg Radius=2.089x1010m (The largest known star) We find a time dilation factor of tS =1.000,002,226,755,19 Thus we find a permittivity factor of ɛR = ɛ0 x tR=8.85438498x10-12 This works out to 6,675 metres per second in the relative speed of light

Thus we can see the extremely small distortion of spacetime surrounding Earth when compared to the larger distortion around the Sun and much large distortion surrounding star R136a1.

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    $\begingroup$ You seem to think that the “relative speed of light” takes various values. Relative to what? $\endgroup$
    – G. Smith
    May 10, 2021 at 19:50
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    $\begingroup$ You are expected to use MathJax on this site. 10-6 is an unacceptable way to write a power of ten. $\endgroup$
    – G. Smith
    May 10, 2021 at 19:52
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    $\begingroup$ If you think that a gravitational field affects the permittivity of free space, then this is a non-mainstream personal theory. $\endgroup$
    – G. Smith
    May 10, 2021 at 19:54
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    $\begingroup$ Their values have absolutely nothing to do with nature. They depend 100% on humans’ choice of units. In Gaussian units these supposed “constants of nature” don’t even exist because Coulomb’s constant and the corresponding magnetic constant are 1. P.S. I have no idea who Arvin Ash is, but he apparently doesn’t understand EM or SR. $\endgroup$
    – G. Smith
    May 10, 2021 at 19:59
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    $\begingroup$ The question is about time dilation. You use the Schwarzschild metric to calculate some examples - fine - and then for some reason add confusing extra calculations which seem to imply that the permittivity and permeability of vacuum is different in gravitational fields. There is no support for this and it is not relevant to the question. $\endgroup$
    – geshel
    May 11, 2021 at 2:12

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