<p>This is similar to previous questions you've asked, especially <a href="/q/38799/30366">this one</a>. I think you're still clinging to the idea that there's a quasi-Newtonian master clock that defines the <em>real</em> passage of time, and specific clocks run slower than the master clock by a ratio that can be calculated.</p>
<p>In reality, there's no master clock, and every physical clock ticks at one second per second. It's like the fact that every meterstick is one meter long. There's little difference between spatial intervals and time intervals in a relativistic world.</p>
<p>If you have two metersticks touching at one end, but with their other ends at different locations, which one is longer? The simple answer is that they're both one meter long. But if you felt like making the problem pointlessly complicated, you could construct two planes perpendicular to meterstick <span class="math-container">$S$</span> and containing the ends of meterstick <span class="math-container">$S'$</span>, and define the distance between those planes to be the length of <span class="math-container">$S'$</span> &quot;as measured by&quot; <span class="math-container">$S$</span>. To make it a little more confusing, instead of computing that length in terms of the angle <span class="math-container">$θ$</span> between the metersticks (which would be <span class="math-container">$1\text{m}\cdot\cos θ$</span>) you could instead use a slope, <span class="math-container">$v=\tan θ$</span>, in terms of which the length would be <span class="math-container">$1\text{m}\cdot γ$</span> where <span class="math-container">$γ = 1/\sqrt{1+v^2}$</span>.</p>
<p>That's exactly what's going on in special relativity with gamma factors and <span class="math-container">$v = dx/dt$</span> velocities. It's just geometry. It looks much more different from Euclidean geometry than it really is because of the different conventions, particularly the use of slopes (3-velocity) instead of angles (<a href="https://en.wikipedia.org/wiki/Rapidity" rel="nofollow noreferrer">rapidity</a>).</p>
<p>The metersticks sharing a common endpoint are like special-relativistic clocks that move inertially away from a common point, and the length of <span class="math-container">$S'$</span> &quot;as measured by&quot; <span class="math-container">$S$</span> is like time dilation.</p>
<p>The Hubble recession is the same, except that at large scales there is significant spacetime curvature. There are many people, including professional astronomers, who think there's some fundamental difference between the two, but there isn't.</p>
<p>As I said in my other answer, the Hubble expansion is like lines of constant longitude on the earth's surface. At the poles, they radiate out from a common point, like the metersticks. Far from the poles but also far from the equator, they're still at a nonzero angle to each other. At the equator, they're parallel (that's the maximum size of a recollapsing universe). If you consider a small enough portion of the earth, it's roughly flat and you can use the <span class="math-container">$\cos θ$</span> or <span class="math-container">$1/\sqrt{1+v^2}$</span> formula to get the length of one line &quot;as measured by&quot; another. It's geometrically the same situation as before, aside from the curvature of the earth which is negligible at small scales.</p>
<p>At larger scales where curvature is significant, you have to decide how to define &quot;as measured by&quot; length, before you can calculate it. There's no definition that seems to make sense. But then, even the original flat-space definition made little sense. That's not a problem, because none of the laws of physics depend on a notion of &quot;as measured by&quot; length, even at small scales.</p>

CC BY-SA 4.0
2020-09-17T20:03:52.793Z
No, I really just want to know that the math of special relativity time dilation due to speed difference, as when having to accelerate  to relativistic speeds within the galaxy is the same as the math of special relativity time dilation due to speed difference, caused by Hubble expansion. The earlier answer than yours was a confusing Yes/No "Yes, but it has little to do with the Special Relativistic time dilation of a moving body".
ParityViolator
CC BY-SA 4.0
2020-09-17T20:21:43.597Z
The next question would then be if a being there were to leave now - from close to the Hubble Horizon and travel at near light speed to here, (arrive in some 14 billion years). How old would he say the universe was when he left? It would be 27.8 billion years here where he arrived but would he say it was 13.8 billion years old where he was when he left? -  I know this is not a well formed question. And what now means when I say if he left now would probably cause a debate.
ParityViolator
CC BY-SA 4.0
2020-09-16T20:55:22.740Z
2020-09-16T21:08:15.340Z
2020-09-16T21:08:15.340Z
benrg
<p>Yes, but it has little to do with the Special Relativistic time dilation of a moving body - in the sense that recession velocity is due to the expansion of space between galaxies rather than their relative motions.</p>
<p>The cosmological expansion leads to a redshift <span class="math-container">$z$</span> and also to time dilation by a factor <span class="math-container">$(1+z)$</span>. i.e. Those distant clocks <em>appear to run slower to us</em> by a factor <span class="math-container">$(1+z)$</span>. And vice versa - our clocks would appear to run slow from the perspective of an observer in a distant galaxy.</p>
<p>The time dilation effect has been observed in the light curves of high redshift supernovae (<a href="https://arxiv.org/abs/0804.3595" rel="nofollow noreferrer">Blondin et al. 2008</a>) and the durations of gamma ray bursts (<a href="https://arxiv.org/abs/1309.5612" rel="nofollow noreferrer">Zhang et al. 2013</a>) and is consistent with this prediction. This is an important observation since it is very difficult to account for this time dilation effect using a &quot;tired light&quot;  explanation.</p>

CC BY-SA 4.0
2020-09-16T16:19:35.610Z
This isn't really answering the intent of my question. Let me put it this way.... In that distant galaxy, that has a relativistic speed difference from here, if there were alien beings, there would they be experiencing time flowing more slowly from us? that is, just as we feel a traveler in a spacecraft that is at high velocity, relative to us, would experience slower time. (And likewise we would seem slower to them)
ParityViolator
CC BY-SA 4.0
2020-09-19T00:08:54.873Z
Nobody experiences "time flowing more slowly". I believe I have answered the question as asked. @ParityViolator
ProfRob
CC BY-SA 4.0
2020-09-20T08:21:01.547Z
I mean if we on Earth are the reference, we would see such people in a spacecraft at high velocity move more slowly, age more slowly etc...likewise  alien beings in a distant galaxy that is moving at relativistic speeds we would see such people as moving more slowly etc -- is the question. Internally sure they feel the passage of time the same.
ParityViolator
CC BY-SA 4.0
2020-09-20T10:24:12.460Z
And the answer is yes. I can't see how it can be made any plainer. @ParityViolator
ProfRob
CC BY-SA 4.0
2020-09-16T16:00:09.593Z
2020-09-19T17:20:12.787Z
2020-09-19T17:20:12.787Z
ProfRob
<p>If time passes more slowly, relative to Earth, for a traveler at relativistic speeds, say at .8c, traveling between stars inside this galaxy, does time also pass more slowly for a distant galaxy that is moving away at relativistic speeds, where the speed difference is due to the Hubble expansion of space?</p>

<general-relativity><expansion><time-dilation><special-relativity>
Does time also pass more slowly for a galaxy that is traveling at relativistic speeds, where the speed difference is due to Hubble expansion?
ParityViolator
# Does time also pass more slowly for a galaxy that is traveling at relativistic speeds, where the speed difference is due to Hubble expansion?

(Asked by: username_0 on 2020-09-16)

If time passes more slowly, relative to Earth, for a traveler at relativistic speeds, say at .8c, traveling between stars inside this galaxy, does time also pass more slowly for a distant galaxy that is moving away at relativistic speeds, where the speed difference is due to the Hubble expansion of space?

**Tags:** general-relativity, expansion, time-dilation, special-relativity

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## 2 answers

### Answer by username_1 on 2020-09-16. Score: 4

This is similar to previous questions you've asked, especially [this one](/q/38799/30366). I think you're still clinging to the idea that there's a quasi-Newtonian master clock that defines the *real* passage of time, and specific clocks run slower than the master clock by a ratio that can be calculated.

In reality, there's no master clock, and every physical clock ticks at one second per second. It's like the fact that every meterstick is one meter long. There's little difference between spatial intervals and time intervals in a relativistic world.

If you have two metersticks touching at one end, but with their other ends at different locations, which one is longer? The simple answer is that they're both one meter long. But if you felt like making the problem pointlessly complicated, you could construct two planes perpendicular to meterstick $S$ and containing the ends of meterstick $S'$, and define the distance between those planes to be the length of $S'$ "as measured by" $S$. To make it a little more confusing, instead of computing that length in terms of the angle $θ$ between the metersticks (which would be $1\text{m}\cdot\cos θ$) you could instead use a slope, $v=\tan θ$, in terms of which the length would be $1\text{m}\cdot γ$ where $γ = 1/\sqrt{1+v^2}$.

That's exactly what's going on in special relativity with gamma factors and $v = dx/dt$ velocities. It's just geometry. It looks much more different from Euclidean geometry than it really is because of the different conventions, particularly the use of slopes (3-velocity) instead of angles ([rapidity](https://en.wikipedia.org/wiki/Rapidity)).

The metersticks sharing a common endpoint are like special-relativistic clocks that move inertially away from a common point, and the length of $S'$ "as measured by" $S$ is like time dilation.

The Hubble recession is the same, except that at large scales there is significant spacetime curvature. There are many people, including professional astronomers, who think there's some fundamental difference between the two, but there isn't.

As I said in my other answer, the Hubble expansion is like lines of constant longitude on the earth's surface. At the poles, they radiate out from a common point, like the metersticks. Far from the poles but also far from the equator, they're still at a nonzero angle to each other. At the equator, they're parallel (that's the maximum size of a recollapsing universe). If you consider a small enough portion of the earth, it's roughly flat and you can use the $\cos θ$ or $1/\sqrt{1+v^2}$ formula to get the length of one line "as measured by" another. It's geometrically the same situation as before, aside from the curvature of the earth which is negligible at small scales.

At larger scales where curvature is significant, you have to decide how to define "as measured by" length, before you can calculate it. There's no definition that seems to make sense. But then, even the original flat-space definition made little sense. That's not a problem, because none of the laws of physics depend on a notion of "as measured by" length, even at small scales.

> **Comments:**
> **username_0**: No, I really just want to know that the math of special relativity time dilation due to speed difference, as when having to accelerate  to relativistic speeds within the galaxy is the same as the math of special relativity time dilation due to speed difference, caused by Hubble expansion. The earlier answer than yours was a confusing Yes/No "Yes, but it has little to do with the Special Relativistic time dilation of a moving body".  
> **username_0**: The next question would then be if a being there were to leave now - from close to the Hubble Horizon and travel at near light speed to here, (arrive in some 14 billion years). How old would he say the universe was when he left? It would be 27.8 billion years here where he arrived but would he say it was 13.8 billion years old where he was when he left? -  I know this is not a well formed question. And what now means when I say if he left now would probably cause a debate.  

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### Answer by username_2 on 2020-09-16. Score: 3

Yes, but it has little to do with the Special Relativistic time dilation of a moving body - in the sense that recession velocity is due to the expansion of space between galaxies rather than their relative motions.

The cosmological expansion leads to a redshift $z$ and also to time dilation by a factor $(1+z)$. i.e. Those distant clocks *appear to run slower to us* by a factor $(1+z)$. And vice versa - our clocks would appear to run slow from the perspective of an observer in a distant galaxy.

The time dilation effect has been observed in the light curves of high redshift supernovae ([Blondin et al. 2008](https://arxiv.org/abs/0804.3595)) and the durations of gamma ray bursts ([Zhang et al. 2013](https://arxiv.org/abs/1309.5612)) and is consistent with this prediction. This is an important observation since it is very difficult to account for this time dilation effect using a "tired light" explanation.

> **Comments:**
> **username_0**: This isn't really answering the intent of my question. Let me put it this way.... In that distant galaxy, that has a relativistic speed difference from here, if there were alien beings, there would they be experiencing time flowing more slowly from us? that is, just as we feel a traveler in a spacecraft that is at high velocity, relative to us, would experience slower time. (And likewise we would seem slower to them)  
> **username_2**: Nobody experiences "time flowing more slowly". I believe I have answered the question as asked. @ParityViolator  
> **username_0**: I mean if we on Earth are the reference, we would see such people in a spacecraft at high velocity move more slowly, age more slowly etc...likewise  alien beings in a distant galaxy that is moving at relativistic speeds we would see such people as moving more slowly etc -- is the question. Internally sure they feel the passage of time the same.  
> **username_2**: And the answer is yes. I can't see how it can be made any plainer. @ParityViolator  

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