<p>There is no Aristotelian/frictional force that drags things along with expanding space. The motion of objects is governed by</p>
<ul>
<li>inertia</li>
<li>ordinary gravitation, which is an inward acceleration (note: acceleration, not velocity)</li>
<li>the cosmological constant, which is an outward acceleration (if <span class="math-container">$Λ&gt;0$</span>)</li>
<li>nongravitational forces</li>
</ul>
<p>and nothing else. The reason matter has expanded for most of the universe's history is that it began with outward inertia (for unknown reasons – perhaps related to inflation), and the attractive force of gravity was too small to slow it to a stop. In the present era, attractive gravity has become weak enough that the repulsive <span class="math-container">$Λ$</span> force dominates.</p>
<p>If you plop a very long rod (of negligible mass) into this system, in the present era, there will be a net outward force on the ends, so it will be under tension. The result will be the same as if you were tugging on both ends: if the rod is too weak to survive the tension, then it will break; otherwise, it will be slightly longer than it would have been if it weren't under tension, but it won't grow over time (except see below).</p>
<p>If it's an earlier era, before <span class="math-container">$Λ$</span> dominance, then there will be a net inward force, and the rod will be under pressure. This is basically the same but with everything negated: either it will break or it will be slightly shorter than if it weren't under pressure.</p>
<p>If you leave the rod for a very long time, the matter responsible for the attractive force will continue to expand and thin, so the attractive force will decrease over time, and the rod will get slightly longer. Eventually, the matter density goes to zero, while the repulsive force remains constant, and that's the longest the rod will ever get (unless it breaks or deforms plastically). The repulsive force over a distance of 1 billion light years, given the measured <span class="math-container">$Λ$</span>, is an acceleration of under <span class="math-container">$10^{-10}\text{ m/s}^2$</span>.</p>

CC BY-SA 4.0
2022-09-09T10:53:25.150Z
I think we should also take into account the fact that masses in space tend to take up a certain shape for a reason. For example, the planets ended up as spheres, and galaxies ended up as spirals etc. I'm pretty sure there would be more forces at play on the ruler.
Julian
CC BY-SA 4.0
2022-09-09T11:19:06.653Z
Good thing you added "of negligeable mass" there, since, if I read your answer correctly, a sufficiently massive ruler would be in the regime of "ordinary gravity applies," in which case we might have to invoke the "mole of moles" rule (what-if.xkcd) and watch the ruler collapse into a spherical object, probably quickly into a black hole
Carl Witthoft
CC BY-SA 4.0
2022-09-10T00:00:44.880Z
@CarlWitthoft (and Julia) The whole thing is ridiculous, I'll grant you, but OP asked for a billion-light-year long stick and I was trying to honor that in some way.
benrg
CC BY-SA 4.0
2022-09-10T23:04:57.257Z
This is what I always thought, but there are a lot of people who write narratives that are "in a zillion years, galaxies don't hold together, then stars don't hold together, then planets don't hold together, then atoms hold together, and eventually everything is lonely subatomic particles". I take it the real version is clumps of matter with ever larger and emptier expanses of nothing between them.
hobbs
CC BY-SA 4.0
2022-09-11T00:58:01.243Z
@hobbs There are two ways they might be correct. First, if they're talking about a [Big Rip](https://en.wikipedia.org/wiki/Big_Rip), which is not standard cosmology. Second, even in standard cosmology, everything will eventually fall apart because of thermal and quantum fluctuations, but it takes an extremely long time. At time scales of tens of billions of years from now, it's clumps of matter separated by huge voids as you said, and some clumps are very large (superclusters).
benrg
CC BY-SA 4.0
2024-09-11T11:52:37.253Z
If we drop the stick at, say, the time of recombination, where only matter dominated, then as I understand you answer there is no "drag" (and in fact the inward pressure would make the rod collapse). Yet, the expansion velocity at this time at the distance of 1 Glyr is ~1600 times the speed of light. So if my end is stationary wrt. me, it seems to me that your answer imples that an alien 1 Glyr away would see the end move by them at 1600 c? What am I missing?
pela
CC BY-SA 4.0
2024-09-11T20:14:20.363Z
@pela You can't really have an object during recombination that is much larger than $c$ times the duration of recombination. Cosmological time and comoving distance are like latitude and longitude respectively. The recombination epoch is an annulus encircling the pole. You can't fit a straight (geodesic) line segment much longer than the width of the annulus into it. Farther from the pole, you can fit longer geodesics, and a constant-latitude curve (of length $a$ times the comoving distance) better approximates a geodesic.
benrg
CC BY-SA 4.0
2024-09-11T20:23:42.133Z
@pela This is an interesting metric so I worked it out: in the approximation that $\dot a$ is constant over the time interval, the longest geodesic has length $c\sqrt{8\Delta t/H}$ where $H = H((t_i+t_f)/2)$. For the recombination epoch this is $\sim 10^6\text{ ly}$. For significantly shorter lengths you can probably come up with a reasonable heuristic definition of length remaining constant over time. For significantly longer lengths there's such a difference between constant-latitude and geodesic distance, and space along the geodesic is so nonuniform, that it's probably meaningless.
benrg
CC BY-SA 4.0
2024-09-12T14:37:32.840Z
I suspect the problem is that you really *cannot* drop a long rod out of nowhere, but even if the rod is only 1 Myr at z ~ 1100 (where H ~ 1.6e6 km/s/Mpc), this would still mean the far end of the rod would be moving at v = 1.6c in local coordinates.
pela
CC BY-SA 4.0
2024-09-13T06:35:04.007Z
@pela I got distracted. What I should have said is that in the constant-$\dot a$ approximation that I used, the annulus (actually a 1+1 slice of it) is flat, so you can think of it as a region of 1+1 Minkowski space (that's how I calculated the maximum geodesic length), so any rod small enough to fit can manifestly stay the same length without exceeding $c$. It's just a worldtube with parallel edges. The comoving-length-times-$a$ of the rod changes with time because it's measured along a hyperbolic arc of time-varying radius, but the Minkowskian proper length is constant.
benrg
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2024-09-13T11:42:14.573Z
Wait, isn't "comoving-length-times-" the physical length? What is "Minkowskian proper length"? I think you're describing a 2+1 spacetime diagram with conformal time as a function of comoving distance, is that correct? Although a rod's worldtube wouldn't have parallel edges there. Sorry, I'm confused; the rod doesn't change its physical length, right?
pela
CC BY-SA 4.0
2024-09-14T01:51:41.567Z
@pela "Comoving length times $a$" is like distance measured along one of the curved parallels of a map like [this one](https://commons.wikimedia.org/wiki/File:1935_base_rev2.png). "Minkowskian proper length" is like the straight-line distance between two straight and parallel lines on that map. That map is a Lambert conformal conic projection, and I had in mind the analogous projection in 1+1 dimensions with hyperbolic parallels, though I was considering only a region of it where the conformal factor is constant so it's isometric in Minkowski space (which happens when $\ddot a=0$).
benrg
CC BY-SA 4.0
2024-09-17T13:19:59.620Z
I'm not quite sure I understand this, but it seems this is a way of saying that dropping a rod so long that its end would be traveling a $v>c$ isn't possible, not because $v>c$ isn't possible, and not because dropping a long rod isn't possible, but because you can't send the "information of dropping" far enough to have a long enough rod. That would sort of make sense, although it feels a little unsatisfactory…
pela
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2024-09-18T00:31:43.927Z
@pela I was talking about the difficulty of defining "maintain a constant length", not implementing it (you could use preprogrammed rockets all along the rod's length). I can draw a constant-width strip on a plane/sphere (Minkowski/de Sitter), but what does it mean to draw one on a more complicated shape? That problem vanishes in a small, roughly flat/spherical region, but the second issue is that on a sphere, constant $RΔλ\cos\phi$ won't produce a constant-width strip in the natural geometric sense. The FLRW equivalent is wrong enough to suggest you need $v>c$ when you really don't.
benrg
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2024-09-22T14:27:14.977Z
Okay, thanks. Did you by any chance write any papers on this where I can dive more into it?
pela
CC BY-SA 4.0
2022-09-09T00:06:02.570Z
2022-09-09T00:06:02.570Z
benrg
<p>I'll give it a try, and I'm perfectly willing to take my answer back if it's wrong beyond repair.</p>
<p>The first thing I take issue with is your wording &quot;the space surrounding the ruler expands&quot;. I would rather say &quot;the space the ruler is embedded in expands&quot;. The ruler doesn't &quot;know&quot; anything about it.</p>
<p>Does the expanding space change the material properties of the ruler, e.g. by weakening or stressing the electromagnetic forces holding its atoms together? I don't think so. I have not heard that star physics or chemistry have changed during the past phase of &quot;normal&quot; (past the inflationary period) expansion. (Of course it's possible that I just missed it, I'm open for additions and corrections.)</p>
<p>That the space in and around clusters of galaxies doesn't expand is owed to the fact that <em>gravity works on space proper,</em> &quot;holding it together&quot;, if you want. Electromagnetism doesn't (apart from the comparably tiny amount of energy which electromagnetic fields contain).</p>
<p>Lastly you establish a nice paradox: How do we measure the expansion of the space a ruler is embedded in? It's a paradox because in the typical <em>gedankenexperiment</em> the ruler is <em>used</em> to measure distances, but now we want to gauge our gauge.</p>
<p>The only way to measure the expansion of space, I would think, is to measure the time needed to traverse it with a given fraction of the speed of light. If it takes longer than before, space has expanded; one could say &quot;new space&quot; has been mixed in.</p>
<p>In this sense the ruler would lengthen, yes. But it is not an elastic stretching (&quot;against&quot; the space it is embedded in); instead, the space itself has lengthened.</p>

CC BY-SA 4.0
2022-09-09T17:11:15.077Z
+1 just for "That the space in and around clusters of galaxies doesn't expand is owed to the fact that gravity works on space proper, "holding it together", if you want. Electromagnetism doesn't", a key concept I hadn't understood before (I thought space expanded but the galaxies clung together in it) simply explained.  But I'm still a bit unclear on your conclusion, "the ruler would lengthen ... [because] the space itself has lengthened."  If I've understood, you mean, the ruler (like everything) contains space "between" its fundamental particles, and since the ruler's gravity over that ...
lessthanideal
CC BY-SA 4.0
2022-09-09T17:14:48.740Z
... length is too small to hold its interior space together, that space will expand and so the ruler's constituent particles will move further apart?  [I'm assuming fundamental particles don't get bigger in diameter as space expands since the other forces prevail, and so we can say the ruler got longer measured against, e.g., width of one of its atoms.  Using the model of a particle can be said to have a defined diameter for simplicity]
lessthanideal
CC BY-SA 4.0
2022-09-09T10:18:57.667Z
2022-09-09T10:18:57.667Z
Peter - Reinstate Monica
<p>Let's see..  suppose the ruler masses  0.001 g per meter.  1E9 ly * 0.001 g <span class="math-container">$\approxeq$</span> 9.5E18 kg , roughly equal to 324 Bamberga.<br />
Leaving aside the obvious questions about rigidity in the presence of galactic gravity fields, if we take Benrig's value <span class="math-container">$\Lambda_{accel} = 10^{-10} \frac{m}{s^2}$</span> , and then compare with, say, the attractive force between two 1-kg masses separated by 1e9 ly due to regular gravity, that's around  7.052632e-36 N .   For our ruler, take  a 1-kg chunk at one end and perform a crude integral of the remainder, using  <span class="math-container">$F = \frac{Gm_1m_2}{R^2}$</span> , I get roughly 0.11 N total force exerted on that chunk (again, assuming that <span class="math-container">$\Lambda$</span> force hasn't ripped the ruler apart, and that we somehow managed to wait long enough for the gravitons to propogate a billion ly).</p>

CC BY-SA 4.0
2022-09-09T11:51:36.790Z
2022-09-09T11:51:36.790Z
Carl Witthoft
<p>Imagine a measuring stick (a rod or ruler) made of ordinary matter which has a length of a billion lightyears. The space surrounding the ruler expands. Will the ruler keep the same length by the accumulating electromagnetic forces opposing expansion? Or will expansion tear the ruler apart?</p>
<p>I have read somewhere here, in an accepted answer (I have no intention exposing the answer so you have to trust me on this) that the ruler expands along but I have my doubts. Will it not oppose expansion like the gravity of galaxies in a cluster opposes expansion?</p>

CC BY-SA 4.0
2022-09-08T21:49:55.940Z
should we suppose the ruler is made of perfectly rigid unobtanium
James K
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2022-09-08T21:57:18.230Z
@JamesK If you consider that ordinary matter... .if the ruler is made of wood, will there not be a difference with a one-billion ly row of disconnected atoms? I mean, isn't electromagnetism strong enough to oppose expansion?
Gerald
CC BY-SA 4.0
2022-09-08T22:07:03.023Z
tbh I don't know. I don't think it is obvious.  The electromagnetic forces aren't additive. There may be strong forces between neighbouring atoms, but these are short range. There is essentially no electromagnetic force between two parts of the ruler that are more than a few nanometers apart. This is unlike gravity, which is additive.
James K
CC BY-SA 4.0
2022-09-08T22:34:28.293Z
@JamesK I'm not sure either. If locally expansion is opposed through the global ruler, it seems globally expansion is opposed too. Which would mean you would get accelerated holding the ruler at one end.
Gerald
CC BY-SA 4.0
2022-09-08T22:56:18.257Z
@Gerald I've never been able to get answers to local expansion questions that I found truly satisfying. They usually end up in one of three categories; 1) "No, because of (insert short phrase to advanced concept without explanation)" 2) "No, because of (insert ballon with galaxies drawn on it or a piece of raisin cake)", 3) "No, because of (insert long, detailed GR answer indicating that there are unanswered questions or multiple formulations)". It seems that any experiment (thought, or real) that tries to see local expansion, can't. I still don't know why. :-)
uhoh
CC BY-SA 4.0
2022-09-08T22:58:24.803Z
[Would order 1AU metric expansion of space begin if the solar system were not inside a galaxy?](https://physics.stackexchange.com/q/268353/83380) and [Is the metric expansion of space relatively uniform on different length scales?](https://physics.stackexchange.com/q/251088/83380) (you may find the answer here of interest) and [Local expansion measured, near zero via Lunar Ranging - what about deep space probes?](https://space.stackexchange.com/q/14921/12102) There are many, many more in Physics SE.
uhoh
CC BY-SA 4.0
2022-09-08T23:06:55.050Z
@uhoh It confuses me uhoh.... Gravity between local cluster galaxies keeps them from expanding (for which a ruler obviously isn't enough, if it was "tied" between the galaxies and attractive gravity shut off). But what about the ruler itself? How big will the pulling force be without huge masses attached? Will it be like pulling a short ruler and will the long ruler just elongate along with space? Uhoh... :-)
Gerald
CC BY-SA 4.0
2022-09-08T23:08:58.020Z
It puzzles me too; good luck!
uhoh
CC BY-SA 4.0
2022-09-09T03:47:58.610Z
Very Controversial Comment Incoming: Space is literal emptiness so what, precisely, is expanding? There's nothing tangible *to* stretch the ruler, is there?
White Prime
CC BY-SA 4.0
2022-09-09T06:56:11.983Z
@WhitePrime I think that's exactly what I find difficult to imagine. Bus suppose we are at both ends of the ruler. Space between us stretches. If the ruler doesn't stretch along the ruler ends move from us. Which wouldn't be problematic but what if the ruler is 100 billion ly?
Gerald
CC BY-SA 4.0
2022-09-09T11:16:58.957Z
To me, this smacks of the "immovable object vs.  irresistable force" paradox.  I'm going to claim, without any basis in theory, that universe expansion means such a ruler could never be built or exist.
Carl Witthoft
CC BY-SA 4.0
2022-09-09T16:58:38.903Z
Do stars expand in an expanding universe? Would you notice?
Wastrel
CC BY-SA 4.0
2022-09-09T20:50:37.160Z
The Friedmann equations describing expansion of the universe were derived under the assumption that the mass density is the same everywhere. If you add a measuring stick (or any object with mass) you are violating that assumption.  General Relativity is nonlinear, so there is no easy way to determine the effects of that addition precisely. Any combination of effects that you propose will be an approximation at best.
D. Halsey
CC BY-SA 4.0
2022-09-09T22:40:04.447Z
@WhitePrime I can't consider space empty (or a void) since it is full of fields, vacuum energy, etc.
Michael
CC BY-SA 4.0
2022-09-10T06:20:50.053Z
Could you specify the initial conditions? Initially, is every point on the ruler at rest in the CMB reference system? Has it been like this for at least a billion years, so that we can assume that gravity has had time to propagate? Without initial conditions, answers may vary, depending on their interpretation of the initial state
Prallax
CC BY-SA 4.0
2022-09-10T12:21:19.547Z
@Michael With respect, man: saying space is "full" of fields, vacuum energy etc is very hyperbolic language. Aside from one or two atoms, space is literal emptiness and not tangible in any way.
White Prime
<expansion>
Does a measuring stick with a size of a billion lightyears in intergalactic space keep the same length in expanding space?
Gerald
# Does a measuring stick with a size of a billion lightyears in intergalactic space keep the same length in expanding space?

(Asked by: username_0 on 2022-09-08)

Imagine a measuring stick (a rod or ruler) made of ordinary matter which has a length of a billion lightyears. The space surrounding the ruler expands. Will the ruler keep the same length by the accumulating electromagnetic forces opposing expansion? Or will expansion tear the ruler apart?

I have read somewhere here, in an accepted answer (I have no intention exposing the answer so you have to trust me on this) that the ruler expands along but I have my doubts. Will it not oppose expansion like the gravity of galaxies in a cluster opposes expansion?

**Tags:** expansion

> **Comments:**
> **username_1**: should we suppose the ruler is made of perfectly rigid unobtanium  
> **username_0**: @username_1 If you consider that ordinary matter... .if the ruler is made of wood, will there not be a difference with a one-billion ly row of disconnected atoms? I mean, isn't electromagnetism strong enough to oppose expansion?  
> **username_1**: tbh I don't know. I don't think it is obvious.  The electromagnetic forces aren't additive. There may be strong forces between neighbouring atoms, but these are short range. There is essentially no electromagnetic force between two parts of the ruler that are more than a few nanometers apart. This is unlike gravity, which is additive.  
> **username_0**: @username_1 I'm not sure either. If locally expansion is opposed through the global ruler, it seems globally expansion is opposed too. Which would mean you would get accelerated holding the ruler at one end.  
> **username_2**: @username_0 I've never been able to get answers to local expansion questions that I found truly satisfying. They usually end up in one of three categories; 1) "No, because of (insert short phrase to advanced concept without explanation)" 2) "No, because of (insert ballon with galaxies drawn on it or a piece of raisin cake)", 3) "No, because of (insert long, detailed GR answer indicating that there are unanswered questions or multiple formulations)". It seems that any experiment (thought, or real) that tries to see local expansion, can't. I still don't know why. :-)  
> **username_2**: [Would order 1AU metric expansion of space begin if the solar system were not inside a galaxy?](https://physics.stackexchange.com/q/268353/83380) and [Is the metric expansion of space relatively uniform on different length scales?](https://physics.stackexchange.com/q/251088/83380) (you may find the answer here of interest) and [Local expansion measured, near zero via Lunar Ranging - what about deep space probes?](https://space.stackexchange.com/q/14921/12102) There are many, many more in Physics SE.  
> **username_0**: @username_2 It confuses me username_2.... Gravity between local cluster galaxies keeps them from expanding (for which a ruler obviously isn't enough, if it was "tied" between the galaxies and attractive gravity shut off). But what about the ruler itself? How big will the pulling force be without huge masses attached? Will it be like pulling a short ruler and will the long ruler just elongate along with space? username_2... :-)  
> **username_2**: It puzzles me too; good luck!  
> **username_3**: Very Controversial Comment Incoming: Space is literal emptiness so what, precisely, is expanding? There's nothing tangible *to* stretch the ruler, is there?  
> **username_0**: @username_3 I think that's exactly what I find difficult to imagine. Bus suppose we are at both ends of the ruler. Space between us stretches. If the ruler doesn't stretch along the ruler ends move from us. Which wouldn't be problematic but what if the ruler is 100 billion ly?  
> **username_4**: To me, this smacks of the "immovable object vs.  irresistable force" paradox.  I'm going to claim, without any basis in theory, that universe expansion means such a ruler could never be built or exist.  
> **username_5**: Do stars expand in an expanding universe? Would you notice?  
> **username_6**: The Friedmann equations describing expansion of the universe were derived under the assumption that the mass density is the same everywhere. If you add a measuring stick (or any object with mass) you are violating that assumption.  General Relativity is nonlinear, so there is no easy way to determine the effects of that addition precisely. Any combination of effects that you propose will be an approximation at best.  
> **username_7**: @username_3 I can't consider space empty (or a void) since it is full of fields, vacuum energy, etc.  
> **username_8**: Could you specify the initial conditions? Initially, is every point on the ruler at rest in the CMB reference system? Has it been like this for at least a billion years, so that we can assume that gravity has had time to propagate? Without initial conditions, answers may vary, depending on their interpretation of the initial state  
> **username_3**: @username_7 With respect, man: saying space is "full" of fields, vacuum energy etc is very hyperbolic language. Aside from one or two atoms, space is literal emptiness and not tangible in any way.  

---

## 3 answers

### Answer by username_9 on 2022-09-09. Score: 13

There is no Aristotelian/frictional force that drags things along with expanding space. The motion of objects is governed by

* inertia
* ordinary gravitation, which is an inward acceleration (note: acceleration, not velocity)
* the cosmological constant, which is an outward acceleration (if $Λ>0$)
* nongravitational forces

and nothing else. The reason matter has expanded for most of the universe's history is that it began with outward inertia (for unknown reasons – perhaps related to inflation), and the attractive force of gravity was too small to slow it to a stop. In the present era, attractive gravity has become weak enough that the repulsive $Λ$ force dominates.

If you plop a very long rod (of negligible mass) into this system, in the present era, there will be a net outward force on the ends, so it will be under tension. The result will be the same as if you were tugging on both ends: if the rod is too weak to survive the tension, then it will break; otherwise, it will be slightly longer than it would have been if it weren't under tension, but it won't grow over time (except see below).

If it's an earlier era, before $Λ$ dominance, then there will be a net inward force, and the rod will be under pressure. This is basically the same but with everything negated: either it will break or it will be slightly shorter than if it weren't under pressure.

If you leave the rod for a very long time, the matter responsible for the attractive force will continue to expand and thin, so the attractive force will decrease over time, and the rod will get slightly longer. Eventually, the matter density goes to zero, while the repulsive force remains constant, and that's the longest the rod will ever get (unless it breaks or deforms plastically). The repulsive force over a distance of 1 billion light years, given the measured $Λ$, is an acceleration of under $10^{-10}\text{ m/s}^2$.

> **Comments:**
> **username_10**: I think we should also take into account the fact that masses in space tend to take up a certain shape for a reason. For example, the planets ended up as spheres, and galaxies ended up as spirals etc. I'm pretty sure there would be more forces at play on the ruler.  
> **username_4**: Good thing you added "of negligeable mass" there, since, if I read your answer correctly, a sufficiently massive ruler would be in the regime of "ordinary gravity applies," in which case we might have to invoke the "mole of moles" rule (what-if.xkcd) and watch the ruler collapse into a spherical object, probably quickly into a black hole  
> **username_9**: @username_4 (and Julia) The whole thing is ridiculous, I'll grant you, but OP asked for a billion-light-year long stick and I was trying to honor that in some way.  
> **username_11**: This is what I always thought, but there are a lot of people who write narratives that are "in a zillion years, galaxies don't hold together, then stars don't hold together, then planets don't hold together, then atoms hold together, and eventually everything is lonely subatomic particles". I take it the real version is clumps of matter with ever larger and emptier expanses of nothing between them.  
> **username_9**: @username_11 There are two ways they might be correct. First, if they're talking about a [Big Rip](https://en.wikipedia.org/wiki/Big_Rip), which is not standard cosmology. Second, even in standard cosmology, everything will eventually fall apart because of thermal and quantum fluctuations, but it takes an extremely long time. At time scales of tens of billions of years from now, it's clumps of matter separated by huge voids as you said, and some clumps are very large (superclusters).  
> **username_12**: If we drop the stick at, say, the time of recombination, where only matter dominated, then as I understand you answer there is no "drag" (and in fact the inward pressure would make the rod collapse). Yet, the expansion velocity at this time at the distance of 1 Glyr is ~1600 times the speed of light. So if my end is stationary wrt. me, it seems to me that your answer imples that an alien 1 Glyr away would see the end move by them at 1600 c? What am I missing?  
> **username_9**: @username_12 You can't really have an object during recombination that is much larger than $c$ times the duration of recombination. Cosmological time and comoving distance are like latitude and longitude respectively. The recombination epoch is an annulus encircling the pole. You can't fit a straight (geodesic) line segment much longer than the width of the annulus into it. Farther from the pole, you can fit longer geodesics, and a constant-latitude curve (of length $a$ times the comoving distance) better approximates a geodesic.  
> **username_9**: @username_12 This is an interesting metric so I worked it out: in the approximation that $\dot a$ is constant over the time interval, the longest geodesic has length $c\sqrt{8\Delta t/H}$ where $H = H((t_i+t_f)/2)$. For the recombination epoch this is $\sim 10^6\text{ ly}$. For significantly shorter lengths you can probably come up with a reasonable heuristic definition of length remaining constant over time. For significantly longer lengths there's such a difference between constant-latitude and geodesic distance, and space along the geodesic is so nonuniform, that it's probably meaningless.  
> **username_12**: I suspect the problem is that you really *cannot* drop a long rod out of nowhere, but even if the rod is only 1 Myr at z ~ 1100 (where H ~ 1.6e6 km/s/Mpc), this would still mean the far end of the rod would be moving at v = 1.6c in local coordinates.  
> **username_9**: @username_12 I got distracted. What I should have said is that in the constant-$\dot a$ approximation that I used, the annulus (actually a 1+1 slice of it) is flat, so you can think of it as a region of 1+1 Minkowski space (that's how I calculated the maximum geodesic length), so any rod small enough to fit can manifestly stay the same length without exceeding $c$. It's just a worldtube with parallel edges. The comoving-length-times-$a$ of the rod changes with time because it's measured along a hyperbolic arc of time-varying radius, but the Minkowskian proper length is constant.  
> **username_12**: Wait, isn't "comoving-length-times-" the physical length? What is "Minkowskian proper length"? I think you're describing a 2+1 spacetime diagram with conformal time as a function of comoving distance, is that correct? Although a rod's worldtube wouldn't have parallel edges there. Sorry, I'm confused; the rod doesn't change its physical length, right?  
> **username_9**: @username_12 "Comoving length times $a$" is like distance measured along one of the curved parallels of a map like [this one](https://commons.wikimedia.org/wiki/File:1935_base_rev2.png). "Minkowskian proper length" is like the straight-line distance between two straight and parallel lines on that map. That map is a Lambert conformal conic projection, and I had in mind the analogous projection in 1+1 dimensions with hyperbolic parallels, though I was considering only a region of it where the conformal factor is constant so it's isometric in Minkowski space (which happens when $\ddot a=0$).  
> **username_12**: I'm not quite sure I understand this, but it seems this is a way of saying that dropping a rod so long that its end would be traveling a $v>c$ isn't possible, not because $v>c$ isn't possible, and not because dropping a long rod isn't possible, but because you can't send the "information of dropping" far enough to have a long enough rod. That would sort of make sense, although it feels a little unsatisfactory…  
> **username_9**: @username_12 I was talking about the difficulty of defining "maintain a constant length", not implementing it (you could use preprogrammed rockets all along the rod's length). I can draw a constant-width strip on a plane/sphere (Minkowski/de Sitter), but what does it mean to draw one on a more complicated shape? That problem vanishes in a small, roughly flat/spherical region, but the second issue is that on a sphere, constant $RΔλ\cos\phi$ won't produce a constant-width strip in the natural geometric sense. The FLRW equivalent is wrong enough to suggest you need $v>c$ when you really don't.  
> **username_12**: Okay, thanks. Did you by any chance write any papers on this where I can dive more into it?  

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### Answer by username_13 on 2022-09-09. Score: 2

I'll give it a try, and I'm perfectly willing to take my answer back if it's wrong beyond repair.

The first thing I take issue with is your wording "the space surrounding the ruler expands". I would rather say "the space the ruler is embedded in expands". The ruler doesn't "know" anything about it.

Does the expanding space change the material properties of the ruler, e.g. by weakening or stressing the electromagnetic forces holding its atoms together? I don't think so. I have not heard that star physics or chemistry have changed during the past phase of "normal" (past the inflationary period) expansion. (Of course it's possible that I just missed it, I'm open for additions and corrections.)

That the space in and around clusters of galaxies doesn't expand is owed to the fact that *gravity works on space proper,* "holding it together", if you want. Electromagnetism doesn't (apart from the comparably tiny amount of energy which electromagnetic fields contain).

Lastly you establish a nice paradox: How do we measure the expansion of the space a ruler is embedded in? It's a paradox because in the typical *gedankenexperiment* the ruler is *used* to measure distances, but now we want to gauge our gauge.

The only way to measure the expansion of space, I would think, is to measure the time needed to traverse it with a given fraction of the speed of light. If it takes longer than before, space has expanded; one could say "new space" has been mixed in.

In this sense the ruler would lengthen, yes. But it is not an elastic stretching ("against" the space it is embedded in); instead, the space itself has lengthened.

> **Comments:**
> **username_14**: +1 just for "That the space in and around clusters of galaxies doesn't expand is owed to the fact that gravity works on space proper, "holding it together", if you want. Electromagnetism doesn't", a key concept I hadn't understood before (I thought space expanded but the galaxies clung together in it) simply explained.  But I'm still a bit unclear on your conclusion, "the ruler would lengthen ... [because] the space itself has lengthened."  If I've understood, you mean, the ruler (like everything) contains space "between" its fundamental particles, and since the ruler's gravity over that ...  
> **username_14**: ... length is too small to hold its interior space together, that space will expand and so the ruler's constituent particles will move further apart?  [I'm assuming fundamental particles don't get bigger in diameter as space expands since the other forces prevail, and so we can say the ruler got longer measured against, e.g., width of one of its atoms.  Using the model of a particle can be said to have a defined diameter for simplicity]  

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### Answer by username_4 on 2022-09-09. Score: 1

Let's see.. suppose the ruler masses 0.001 g per meter. 1E9 ly \* 0.001 g $\approxeq$ 9.5E18 kg , roughly equal to 324 Bamberga.  
Leaving aside the obvious questions about rigidity in the presence of galactic gravity fields, if we take Benrig's value $\Lambda\_{accel} = 10^{-10} \frac{m}{s^2}$ , and then compare with, say, the attractive force between two 1-kg masses separated by 1e9 ly due to regular gravity, that's around 7.052632e-36 N . For our ruler, take a 1-kg chunk at one end and perform a crude integral of the remainder, using $F = \frac{Gm\_1m\_2}{R^2}$ , I get roughly 0.11 N total force exerted on that chunk (again, assuming that $\Lambda$ force hasn't ripped the ruler apart, and that we somehow managed to wait long enough for the gravitons to propogate a billion ly).


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