<p>Excepting a <a href="https://en.wikipedia.org/wiki/Big_Rip" rel="nofollow">Big Rip</a> scenario, there is no eventual 'clash'. </p>

<p>Consider a <a href="https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric" rel="nofollow">Friedmann–Lemaître–Robertson–Walker universe</a>:
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\left[\frac{\mathrm{d}r^2}{1-kr^2} + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\right]\text{,}$$
where $a(t)$ is the scale factor and $k\in\{-1,0,+1\}$ corresponds to a spatially open, flat, or closed cases, respectively. In a local orthonormal frame, the nonzero Riemann curvature components are, up to symmetries:
$$\begin{eqnarray*}
R_{\hat{t}\hat{r}\hat{t}\hat{r}} = R_{\hat{t}\hat{\theta}\hat{t}\hat{\theta}} = R_{\hat{t}\hat{\phi}\hat{t}\hat{\phi}} =&amp; -\frac{\ddot{a}}{a} &amp;= \frac{4\pi G}{3}\left(\rho + 3p\right)\text{,} \\
R_{\hat{r}\hat{\theta}\hat{r}\hat{\theta}} = R_{\hat{r}\hat{\phi}\hat{r}\hat{\phi}} = R_{\hat{\theta}\hat{\phi}\hat{\theta}\hat{\phi}} =&amp; \frac{k+\dot{a}^2}{a^2} &amp;= \frac{8\pi G}{3}\rho\text{,}
\end{eqnarray*}$$
where overdot denotes differentiation with respect to coordinate time $t$ and the <a href="https://en.wikipedia.org/wiki/Friedmann_equations" rel="nofollow">Friedmann equations</a> were used to rewrite them in terms of density $\rho$ and pressure $p$. From the link, note in particular that $\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right)$.</p>

<p>If dark energy is described by a cosmological constant $\Lambda$, as it is in the standard <a href="https://en.wikipedia.org/wiki/Lambda-CDM_model" rel="nofollow">ΛCDM</a> model, then it contributes a constant density and pressure $\rho = -p = \Lambda/8\pi G$, and so no amount of cosmic expansion would change things. Locally, things look just the same as they ever did: for a universe dominated by dark energy, the curvature stays the same, so the gravitational tidal forces do too. Through those tiny tidal forces, the dark energy provides some immeasurably tiny perturbation on the behavior of objects, including atomic nuclei, forcing a slightly different equilibrium size than would be otherwise. But it is so small that that it has no relevance to anything on that scale, nor do those equilibrium sizes change in time. The cosmological constant holds true to its name.</p>

<p>On the other hand, if dark energy has an equation of state $p = w\rho$, then a flat expanding universe dominated by dark energy has
$$\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right) = -3\sqrt{\frac{8\pi G}{3}}\left(1+w\right)\rho^{3/2}\text{,}$$
and immediately one can see that there is something special about $w&lt;-1$, leading to an accumulation of dark energy, while the cosmological constant's $w = -1$ leads to no change. This leads to a Big Rip more generally, as <strong>zibadawa timmy</strong>'s answer explains.</p>

<hr>

<blockquote>
  <p>? If the metric expands surely objects get further away from one another and that would include the stars inside galaxaies as well as the galaxies themselves?</p>
</blockquote>

<p>Not at all. It wouldn't even make any sense: if you have an object like an atom or a star in gravitational freefall, by the equivalence principle only tidal forces across it are relevant. The tidal forces stretch the object until the internal forces balance them. But for a Λ-driven accelerated expansion, dark energy contribution to tidal forces is <em>constant</em>. Hence, an object already in equilibrium has no reason to further change its size, no matter how long the cosmic acceleration occurs. This also applies to galaxies, only that the internal forces are also gravitational and balance the dark energy contribution.</p>

<p>Looking at this in more detail, imagine a test particle with four-velocity $u$, and a nearby one with the same four-velocity, separated by some vector $n$ connecting their worldlines. If they're both in gravitational freefall, then their relative acceleration is given by the <a href="https://en.wikipedia.org/wiki/Geodesic_deviation" rel="nofollow">geodesic deviation equation</a>, $\frac{D^2n^a}{d\tau^2} = -R^\alpha{}_{\mu\beta\nu}u^\mu u^\nu n^\beta$. The <a href="https://en.wikipedia.org/wiki/Bel_decomposition" rel="nofollow">gravitoelectric</a> part of the Riemann tensor, $\mathbb{E}^\alpha{}_\beta = R^\alpha{}_{\mu\beta\nu}u^\mu u^\nu$, represents the static tidal forces in a local inertial frame comoving with $u$, which will drive those particles apart (or together, depending). Hence, keeping those particles at the same distance would require a force between them, but it's not necessary for this force to change unless the tidal forces <em>also</em> change.</p>

<p>Galaxies don't change size through cosmological expansion. Stars don't either, nor do atoms. Not for Λ-driven expansion, at least. It would take an increase in tidal forces, such as those provided by a Big Rip, for them to do so.</p>

<p>A related way of looking at the issue is this: according to the Einstein field equation, the initial acceleration of a small ball of initially comoving test particles with some four-velocity $u$ (in the comoving inertial frame), which is given by the Ricci tensor, turns out to be:
$$\lim_{V\to 0}\left.\frac{\ddot{V}}{V}\right|_{t=0}\!\!\!\!= -\underbrace{R_{\mu\nu}u^\mu u^\nu}_{\mathbb{E}^\alpha{}_\alpha} = -4\pi G(\rho + 3p)\text{,}$$
where $\rho$ is density and $p$ is the average of the principal stresses in the local inertial frame. For a positive cosmological constant, $\rho = -p&gt;0$, and correspondingly a ball of test particles will expand. That's cosmic expansion on a local scale; because of uniformity of the FLRW universe, the same proportionality works on the large scale as well, as we can think of distant galaxies as themselves test particles if their interactions are too minute to make a difference.</p>

<p>Again, we are led to the same conclusion: if the ball has internal forces that prevent expansion, then those forces don't need to change in time unless the dark energy density also changes, which doesn't happen for cosmological constant. If they're not 'clashing' <em>now</em>, then they won't need to in the future either.</p>

CC BY-SA 3.0
2014-11-13T00:06:21.450Z
You seem to have a typo in the FLRW metric. It's singular at $k=\pm1$ as is, unless I've got a weird display error. Otherwise, where's my +10 button?
zibadawa timmy
CC BY-SA 3.0
2014-11-13T00:19:49.990Z
@zibadawatimmy thanks; somehow $kr^2$ was mistyped into $k^2$.
Stan Liou
CC BY-SA 3.0
2014-11-12T02:52:54.843Z
2014-11-13T17:19:26.560Z
2014-11-13T17:19:26.560Z
Stan Liou
<p>Definitely too small to measure. The expansion isn't even enough to measurably impact galaxy superclusters.</p>

<p>What you are otherwise thinking of is known as <a href="http://en.m.wikipedia.org/wiki/Big_Rip" rel="nofollow">the Big Rip</a>. This is not guaranteed to happen, though. It depends on a constant describing the density of dark energy over time, usually denoted $w$. </p>

<p>When $w&lt;-1$ the Big Rip occurs in finite time, separating all parts of a structure in the sense that lightspeed signals can no longer be transmitted between the parts. This includes hadrons like protons, so in the very last moments the quarks will no longer be able to communicate and will thus cease to be bound together.</p>

<p>Disclaimer: That ignores quantum gravity and QCD in highly curved spacetime. We can barely manage QCD in the flat case, and quantum gravity has been a holy grail for decades. Since a definitive answer to your question requires both of these things, the only definitive answer is "nobody knows for sure".</p>

<p>When $w=-1$ we have constant density and structures smaller than galaxies survive. Our solar system, assuming it survives a very long time, will remain bound together. Star clusters will remain together.</p>

<p>Note, however, that we would still, technically, be able to see everything we could see before. Telescopes can already see galaxies which are currently receding at superluminal speeds. We just see those last moments before such speeds were hit get smeared over ever longer time frames, and heavily redshifted. But we will never receive new signals from the moment it hit superluminal speeds. Only ever the faint remnants of those last moments. The redshifts will eventually be so extreme that everything at galactic distances and beyond becomes effectively impossible to detect these remnants from. At which point it's just you and your closest neighbors.</p>

<p>Current observations put $w$ close to -1, but we cannot currently distinguish between $w&lt;-1$, $w=-1$, and $w&gt;-1$</p>

CC BY-SA 3.0
2014-11-12T01:14:10.100Z
+1, though I disagree with the implication that a full theory of quantum gravity is relevant for this question. The QCD scale is large enough to say that we *do* understand quantum gravity there, and it's just GTR with very tiny corrections.
Stan Liou
CC BY-SA 3.0
2014-11-12T01:23:09.150Z
@StanLiou Perhaps. The question is explicitly about an event immediately preceding a singularity of GR. My usual presumption is that quantum gravity is inherently required to correctly describe singularities. But this is just my presumption, and may be incorrect/unnecessary in this instance.
zibadawa timmy
CC BY-SA 3.0
2014-11-12T01:36:14.560Z
I share your presumption; what I disagree about is whether this question is explicitly (or even implicitly) about an event immediately preceding (a would-be) singularity, even in the case of a Big Rip...  QCD would clash with gravity on scales above where full quantum gravity would become relevant.
Stan Liou
CC BY-SA 3.0
2014-11-11T21:54:25.080Z
2014-11-11T23:34:35.020Z
2014-11-11T23:34:35.020Z
zibadawa timmy
<p>I think there's several points that need addressing.</p>

<p>Firstly how expansion operates on scales below the intergalactic is very much an open question. The approximation of spatial homogeneity and isotropy fails before you reach these scales. Whilst it may be tempting to think that cosmic expansion could be modeled as something akin to a small repulsive force between objects on these scales that isn't necessarily the case. For example in the, admittedly not very realistic, Swiss-cheese model the regions around stars experience no expansion. This is before we even get to scales when we would expect to need more than general relativity to describe what is going on.</p>

<p>In de Sitter space w = -1 and it is essentially models a Universe that purely contains dark energy in the form of a cosmological constant and so it acts as a sort of limiting case for Universe with dark energy in the form of a cosmological constant. However de Sitter space, whilst it has accelerated expansion, is also static and there is no essential difference between how test particles behave in it at any given time. Consequently if a structure can exist in de Sitter space at a given time it can exist at all times, and as it is the limiting case, if a structure (like an atom for example) can exist in a Universe with a dark energy described by a cosmological constant at a given time, it can also exist at all later times too. Therefore you need to go into the realms of phantom energy for expansion that can rip apart atoms at later times in the Universe.</p>

<p>An additional point I would make though is that even in a Universe where dark energy is described by a cosmological constant structures such as galaxies will come apart. However this is due to effects such galactic evaporation rather than as a direct result of expansion. </p>

CC BY-SA 3.0
2014-11-13T00:33:35.597Z
2014-11-13T01:00:48.887Z
2014-11-13T01:00:48.887Z
John Davis
<p>If the expansion of the universe (the metric) continues (and perhaps is accelerating), in a very large but finite time the expansion of the metric will clash with the effects of the strong nuclear force. What happens then? Would the energy required halt the expansion and if so what is the slowing effect of the "stretch" on the strong force today - too small to be measured?</p>

<p><em>Update</em>: Thanks for the answers so far - I hadn't even considered the big rip scenario. I was merely thinking that as the metric expanded the strong force inside baryons would require unlimited resources of energy to be overcome - but I now understand that is wrong however it still looks to me like a potential perpetual motion machine - continually creating matter/anti-matter pairs. Perhaps that should be a separate question though...</p>

CC BY-SA 3.0
2014-11-11T22:11:42.080Z
How would it clash with the strong nuclear force?
HDE 226868
CC BY-SA 3.0
2014-11-11T23:39:20.007Z
@HDE226868 The energy to pull a quark out of a hadron vastly exceeds pair production energies, and would result in a stream of quarks and anti-quarks. A big rip would immediately disconnect these, effectively repeating the process. The simplest explanation, as in my answer, is that they are simply free quarks and then the singularity hits and nothing more can be said. The longer explanation is that we have no clue how gravity and QCD behave in these conditions.
zibadawa timmy
CC BY-SA 3.0
2014-11-12T02:28:28.010Z
I think that you're largely making the same mistake you made [here](http://astronomy.stackexchange.com/questions/1473/is-it-possible-that-we-see-our-solar-system-as-a-star) in thinking that cosmic expansion means objects such as galaxies or atoms expand. Excepting phantom energy, that's simply not the case.
Stan Liou
CC BY-SA 3.0
2014-11-12T12:44:55.073Z
@StanLiou ? If the metric expands surely objects get further away from one another and that would include the stars inside galaxaies as well as the galaxies themselves?
adrianmcmenamin
CC BY-SA 3.0
2014-11-12T21:15:28.527Z
@adrianmcmenamin The gravitational attraction between them will overpower the dark energy repulsion, unless we are in a big rip scenario.
zibadawa timmy
CC BY-SA 3.0
2014-11-12T22:04:05.767Z
@adrianmcmenamin No. Because I think that this confusion relates to the reason for this question, I've edited in a somewhat lengthy response below.
Stan Liou
<cosmology><general-relativity>
Strong force and metric expansion
adrianmcmenamin
# Strong force and metric expansion

(Asked by: username_0 on 2014-11-11)

If the expansion of the universe (the metric) continues (and perhaps is accelerating), in a very large but finite time the expansion of the metric will clash with the effects of the strong nuclear force. What happens then? Would the energy required halt the expansion and if so what is the slowing effect of the "stretch" on the strong force today - too small to be measured?

*Update*: Thanks for the answers so far - I hadn't even considered the big rip scenario. I was merely thinking that as the metric expanded the strong force inside baryons would require unlimited resources of energy to be overcome - but I now understand that is wrong however it still looks to me like a potential perpetual motion machine - continually creating matter/anti-matter pairs. Perhaps that should be a separate question though...

**Tags:** cosmology, general-relativity
> **Comments:**
> **username_1**: How would it clash with the strong nuclear force?  
> **username_2**: @HDE226868 The energy to pull a quark out of a hadron vastly exceeds pair production energies, and would result in a stream of quarks and anti-quarks. A big rip would immediately disconnect these, effectively repeating the process. The simplest explanation, as in my answer, is that they are simply free quarks and then the singularity hits and nothing more can be said. The longer explanation is that we have no clue how gravity and QCD behave in these conditions.  
> **username_3**: I think that you're largely making the same mistake you made [here](http://astronomy.stackexchange.com/questions/1473/is-it-possible-that-we-see-our-solar-system-as-a-star) in thinking that cosmic expansion means objects such as galaxies or atoms expand. Excepting phantom energy, that's simply not the case.  
> **username_0**: @StanLiou ? If the metric expands surely objects get further away from one another and that would include the stars inside galaxaies as well as the galaxies themselves?  
> **username_2**: @adrianmcmenamin The gravitational attraction between them will overpower the dark energy repulsion, unless we are in a big rip scenario.  
> **username_3**: @adrianmcmenamin No. Because I think that this confusion relates to the reason for this question, I've edited in a somewhat lengthy response below.  

---

## 3 answers

### Answer by username_3 on 2014-11-12. Score: 4 | Accepted answer

Excepting a [Big Rip](https://en.wikipedia.org/wiki/Big_Rip) scenario, there is no eventual 'clash'.

Consider a [Friedmann–Lemaître–Robertson–Walker universe](https://en.wikipedia.org/wiki/Friedmann%E2%80%93Lema%C3%AEtre%E2%80%93Robertson%E2%80%93Walker_metric):
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\left[\frac{\mathrm{d}r^2}{1-kr^2} + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\right]\text{,}$$
where $a(t)$ is the scale factor and $k\in\{-1,0,+1\}$ corresponds to a spatially open, flat, or closed cases, respectively. In a local orthonormal frame, the nonzero Riemann curvature components are, up to symmetries:
$$\begin{eqnarray\*}
R\_{\hat{t}\hat{r}\hat{t}\hat{r}} = R\_{\hat{t}\hat{\theta}\hat{t}\hat{\theta}} = R\_{\hat{t}\hat{\phi}\hat{t}\hat{\phi}} =& -\frac{\ddot{a}}{a} &= \frac{4\pi G}{3}\left(\rho + 3p\right)\text{,} \\
R\_{\hat{r}\hat{\theta}\hat{r}\hat{\theta}} = R\_{\hat{r}\hat{\phi}\hat{r}\hat{\phi}} = R\_{\hat{\theta}\hat{\phi}\hat{\theta}\hat{\phi}} =& \frac{k+\dot{a}^2}{a^2} &= \frac{8\pi G}{3}\rho\text{,}
\end{eqnarray\*}$$
where overdot denotes differentiation with respect to coordinate time $t$ and the [Friedmann equations](https://en.wikipedia.org/wiki/Friedmann_equations) were used to rewrite them in terms of density $\rho$ and pressure $p$. From the link, note in particular that $\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right)$.

If dark energy is described by a cosmological constant $\Lambda$, as it is in the standard [ΛCDM](https://en.wikipedia.org/wiki/Lambda-CDM_model) model, then it contributes a constant density and pressure $\rho = -p = \Lambda/8\pi G$, and so no amount of cosmic expansion would change things. Locally, things look just the same as they ever did: for a universe dominated by dark energy, the curvature stays the same, so the gravitational tidal forces do too. Through those tiny tidal forces, the dark energy provides some immeasurably tiny perturbation on the behavior of objects, including atomic nuclei, forcing a slightly different equilibrium size than would be otherwise. But it is so small that that it has no relevance to anything on that scale, nor do those equilibrium sizes change in time. The cosmological constant holds true to its name.

On the other hand, if dark energy has an equation of state $p = w\rho$, then a flat expanding universe dominated by dark energy has
$$\dot{\rho} = -3\frac{\dot{a}}{a}\left(\rho+p\right) = -3\sqrt{\frac{8\pi G}{3}}\left(1+w\right)\rho^{3/2}\text{,}$$
and immediately one can see that there is something special about $w<-1$, leading to an accumulation of dark energy, while the cosmological constant's $w = -1$ leads to no change. This leads to a Big Rip more generally, as **zibadawa timmy**'s answer explains.

---

> ? If the metric expands surely objects get further away from one another and that would include the stars inside galaxaies as well as the galaxies themselves?

Not at all. It wouldn't even make any sense: if you have an object like an atom or a star in gravitational freefall, by the equivalence principle only tidal forces across it are relevant. The tidal forces stretch the object until the internal forces balance them. But for a Λ-driven accelerated expansion, dark energy contribution to tidal forces is *constant*. Hence, an object already in equilibrium has no reason to further change its size, no matter how long the cosmic acceleration occurs. This also applies to galaxies, only that the internal forces are also gravitational and balance the dark energy contribution.

Looking at this in more detail, imagine a test particle with four-velocity $u$, and a nearby one with the same four-velocity, separated by some vector $n$ connecting their worldlines. If they're both in gravitational freefall, then their relative acceleration is given by the [geodesic deviation equation](https://en.wikipedia.org/wiki/Geodesic_deviation), $\frac{D^2n^a}{d\tau^2} = -R^\alpha{}\_{\mu\beta\nu}u^\mu u^\nu n^\beta$. The [gravitoelectric](https://en.wikipedia.org/wiki/Bel_decomposition) part of the Riemann tensor, $\mathbb{E}^\alpha{}\_\beta = R^\alpha{}\_{\mu\beta\nu}u^\mu u^\nu$, represents the static tidal forces in a local inertial frame comoving with $u$, which will drive those particles apart (or together, depending). Hence, keeping those particles at the same distance would require a force between them, but it's not necessary for this force to change unless the tidal forces *also* change.

Galaxies don't change size through cosmological expansion. Stars don't either, nor do atoms. Not for Λ-driven expansion, at least. It would take an increase in tidal forces, such as those provided by a Big Rip, for them to do so.

A related way of looking at the issue is this: according to the Einstein field equation, the initial acceleration of a small ball of initially comoving test particles with some four-velocity $u$ (in the comoving inertial frame), which is given by the Ricci tensor, turns out to be:
$$\lim\_{V\to 0}\left.\frac{\ddot{V}}{V}\right|\_{t=0}\!\!\!\!= -\underbrace{R\_{\mu\nu}u^\mu u^\nu}\_{\mathbb{E}^\alpha{}\_\alpha} = -4\pi G(\rho + 3p)\text{,}$$
where $\rho$ is density and $p$ is the average of the principal stresses in the local inertial frame. For a positive cosmological constant, $\rho = -p>0$, and correspondingly a ball of test particles will expand. That's cosmic expansion on a local scale; because of uniformity of the FLRW universe, the same proportionality works on the large scale as well, as we can think of distant galaxies as themselves test particles if their interactions are too minute to make a difference.

Again, we are led to the same conclusion: if the ball has internal forces that prevent expansion, then those forces don't need to change in time unless the dark energy density also changes, which doesn't happen for cosmological constant. If they're not 'clashing' *now*, then they won't need to in the future either.

> **Comments:**
> **username_2**: You seem to have a typo in the FLRW metric. It's singular at $k=\pm1$ as is, unless I've got a weird display error. Otherwise, where's my +10 button?  
> **username_3**: @zibadawatimmy thanks; somehow $kr^2$ was mistyped into $k^2$.  

---

### Answer by username_2 on 2014-11-11. Score: 3

Definitely too small to measure. The expansion isn't even enough to measurably impact galaxy superclusters.

What you are otherwise thinking of is known as [the Big Rip](http://en.m.wikipedia.org/wiki/Big_Rip). This is not guaranteed to happen, though. It depends on a constant describing the density of dark energy over time, usually denoted $w$.

When $w<-1$ the Big Rip occurs in finite time, separating all parts of a structure in the sense that lightspeed signals can no longer be transmitted between the parts. This includes hadrons like protons, so in the very last moments the quarks will no longer be able to communicate and will thus cease to be bound together.

Disclaimer: That ignores quantum gravity and QCD in highly curved spacetime. We can barely manage QCD in the flat case, and quantum gravity has been a holy grail for decades. Since a definitive answer to your question requires both of these things, the only definitive answer is "nobody knows for sure".

When $w=-1$ we have constant density and structures smaller than galaxies survive. Our solar system, assuming it survives a very long time, will remain bound together. Star clusters will remain together.

Note, however, that we would still, technically, be able to see everything we could see before. Telescopes can already see galaxies which are currently receding at superluminal speeds. We just see those last moments before such speeds were hit get smeared over ever longer time frames, and heavily redshifted. But we will never receive new signals from the moment it hit superluminal speeds. Only ever the faint remnants of those last moments. The redshifts will eventually be so extreme that everything at galactic distances and beyond becomes effectively impossible to detect these remnants from. At which point it's just you and your closest neighbors.

Current observations put $w$ close to -1, but we cannot currently distinguish between $w<-1$, $w=-1$, and $w>-1$

> **Comments:**
> **username_3**: +1, though I disagree with the implication that a full theory of quantum gravity is relevant for this question. The QCD scale is large enough to say that we *do* understand quantum gravity there, and it's just GTR with very tiny corrections.  
> **username_2**: @StanLiou Perhaps. The question is explicitly about an event immediately preceding a singularity of GR. My usual presumption is that quantum gravity is inherently required to correctly describe singularities. But this is just my presumption, and may be incorrect/unnecessary in this instance.  
> **username_3**: I share your presumption; what I disagree about is whether this question is explicitly (or even implicitly) about an event immediately preceding (a would-be) singularity, even in the case of a Big Rip...  QCD would clash with gravity on scales above where full quantum gravity would become relevant.  

---

### Answer by username_4 on 2014-11-13. Score: 2

I think there's several points that need addressing.

Firstly how expansion operates on scales below the intergalactic is very much an open question. The approximation of spatial homogeneity and isotropy fails before you reach these scales. Whilst it may be tempting to think that cosmic expansion could be modeled as something akin to a small repulsive force between objects on these scales that isn't necessarily the case. For example in the, admittedly not very realistic, Swiss-cheese model the regions around stars experience no expansion. This is before we even get to scales when we would expect to need more than general relativity to describe what is going on.

In de Sitter space w = -1 and it is essentially models a Universe that purely contains dark energy in the form of a cosmological constant and so it acts as a sort of limiting case for Universe with dark energy in the form of a cosmological constant. However de Sitter space, whilst it has accelerated expansion, is also static and there is no essential difference between how test particles behave in it at any given time. Consequently if a structure can exist in de Sitter space at a given time it can exist at all times, and as it is the limiting case, if a structure (like an atom for example) can exist in a Universe with a dark energy described by a cosmological constant at a given time, it can also exist at all later times too. Therefore you need to go into the realms of phantom energy for expansion that can rip apart atoms at later times in the Universe.

An additional point I would make though is that even in a Universe where dark energy is described by a cosmological constant structures such as galaxies will come apart. However this is due to effects such galactic evaporation rather than as a direct result of expansion.


---

