<p>In general relativity, there is no such thing as intrinsic curvature of spacetime that is not tied to the energy content. The Einstein field equations directly tie the curvature tensor to the stress-energy tensor.</p>
<p>Now, you are thinking about curvature of space and not spacetime. The situation is perhaps even worse there because the curvature of space is subjective. Space is a 3D surface in 4D spacetime, and there are infinitely many possible choices of spatial surface. You could always define coordinates so that your spatial surfaces are flat, or of negative curvature, or whatever. That's just a coordinate choice and does not change the physics.</p>
<p>However, in cosmology there is a family of preferred spatial surfaces. Roughly speaking, we define these spatial surfaces so that on any one surface, all the matter in the universe reports the same amount of elapsed time since the Big Bang. (More technically, we think of idealized comoving observers instead of real matter.) These are the surfaces cosmologists are talking about when we say the universe is spatially flat. And the curvature of these spatial surfaces is completely determined by the energy content and the expansion rate. You can't independently specify it. That is to say, there is also no such thing as intrinsic cosmological spatial curvature that is independent of the contents of the universe.</p>
<p>You can think about the curvature of these preferred spatial surfaces intuitively as a competition between gravity and kinematic time dilation. Light moves in straight lines on these surfaces, but gravity converges light. So gravity &quot;wants&quot; to give space positive curvature, which makes parallel lines converge. But the universe is expanding, which means distant objects are moving away and more distant objects are moving away faster. If they are moving faster, they are more time dilated, so less time has passed for them since the Big Bang. So you need to curve your spatial surfaces &quot;into the future&quot; at large distances in order for everyone on that surface to agree on how much time has elapsed. In the future, everything is farther apart. So this effect &quot;wants&quot; to make parallel lines diverge, corresponding to negative spatial curvature. Thus, more energy means more positive curvature, while a faster expansion rate means more negative curvature.</p>
<p>(But I'll caveat that this is an intuitive picture and some care would be needed to translate it into calculations. For example, you can't directly apply the kinematic time dilation equations in curved spacetimes.)</p>

CC BY-SA 4.0
2023-03-30T17:14:04.337Z
I agree that there is no such thing as intrinsic spatial curvature in standard general relativity but in the framework I've been investigating there actually can be. I tried my best to avoid mentioning the specific model though in case I was merely misunderstanding the meaning behind the parameters in the standard Friedmann equations. But from your answer it sounds like my understanding is sound and that my question might be one that has questionable meaning in standard general relativity. In any case, thanks for the answer!
Scott
CC BY-SA 4.0
2023-03-30T16:51:59.077Z
2023-03-30T17:04:24.340Z
2023-03-30T17:04:24.340Z
Sten
<p><span class="math-container">$H^2 = \frac{8 \pi G}{3}\rho-\frac{kc^2}{a^2}$</span></p>
<p>Shown above is the first Friedmann equation. I understand that the curvature parameter, k, represents the spatial curvature of the universe which is dictated by the energy density of the universe. The critical density is normally defined as the density required to halt the expansion after infinite time, or equivalently, the density that gives a flat universe (k=0). From this you can quite easily find the explicit expression for the critical density using the first friedmann equation.</p>
<p><span class="math-container">$\rho_c = \frac{3H^2}{8 \pi G}$</span></p>
<p>However, I've recently been considering hypothetical cosmological models where the universe itself intrinsically has negative spatial curvature INDEPENDENT of its matter-energy content. In this type of model, k=-1 is always the case (even in an empty universe) which is shown below:</p>
<p><span class="math-container">$H^2 = \frac{8 \pi G}{3}\rho+\frac{c^2}{a^2}$</span></p>
<p>From the work I've seen in this area, it's often assumed that the normal formula for critical density holds in this context but I'm not entirely convinced that is valid. In models where space itself has intrinsic negative curvature, and thus k=-1 is always the case, wouldn't the normal procedure for obtaining an expression for the critical density be impossible? Consequently, wouldn't the normal formula be unapplicable in models like these? Are my suspicions warranted or am I missing something here?</p>
<p><strong>If they are warranted then does the concept of a &quot;critical density&quot; even make sense in this context? My best guess would be that it's the density of energy required to &quot;counteract&quot; the intrinsic negative curvature such that the universe is now flat... however, I'm not entirely sure how to derive an expression for this critical density using the above modified Friedmann equations. Is it even possible?</strong></p>
<p>Apologies in advance if I'm misunderstanding something elementary here - I'm relatively new to the field of cosmology. Any help improving my understanding would be greatly appreciated!</p>

CC BY-SA 4.0
2023-03-31T07:13:41.650Z
The first Friedmann equation isn't a postulate of FLRW cosmology. The postulates are general relativity, homogeneity and isotropy, from which you can derive the first Friedmann equation – the standard one, not your variation. You would have to specify which of the underlying assumptions of standard cosmology are violated before the question would really be answerable, Sten's attempt notwithstanding.
benrg
CC BY-SA 4.0
2023-03-31T15:46:57.970Z
By the way, by "INDEPENDENT of its matter-energy content", do you mean matter and radiation specifically? That is, would it be acceptable to make spatial curvature by adding something that contributes to the stress-energy but isn't matter or radiation?
Sten
<cosmology><universe><expansion><fate-of-universe><friedmann-equations>
Critical density of universe with intrinsic spatial curvature
Scott
# Critical density of universe with intrinsic spatial curvature

(Asked by: username_0 on 2023-03-30)

$H^2 = \frac{8 \pi G}{3}\rho-\frac{kc^2}{a^2}$

Shown above is the first Friedmann equation. I understand that the curvature parameter, k, represents the spatial curvature of the universe which is dictated by the energy density of the universe. The critical density is normally defined as the density required to halt the expansion after infinite time, or equivalently, the density that gives a flat universe (k=0). From this you can quite easily find the explicit expression for the critical density using the first friedmann equation.

$\rho\_c = \frac{3H^2}{8 \pi G}$

However, I've recently been considering hypothetical cosmological models where the universe itself intrinsically has negative spatial curvature INDEPENDENT of its matter-energy content. In this type of model, k=-1 is always the case (even in an empty universe) which is shown below:

$H^2 = \frac{8 \pi G}{3}\rho+\frac{c^2}{a^2}$

From the work I've seen in this area, it's often assumed that the normal formula for critical density holds in this context but I'm not entirely convinced that is valid. In models where space itself has intrinsic negative curvature, and thus k=-1 is always the case, wouldn't the normal procedure for obtaining an expression for the critical density be impossible? Consequently, wouldn't the normal formula be unapplicable in models like these? Are my suspicions warranted or am I missing something here?

**If they are warranted then does the concept of a "critical density" even make sense in this context? My best guess would be that it's the density of energy required to "counteract" the intrinsic negative curvature such that the universe is now flat... however, I'm not entirely sure how to derive an expression for this critical density using the above modified Friedmann equations. Is it even possible?**

Apologies in advance if I'm misunderstanding something elementary here - I'm relatively new to the field of cosmology. Any help improving my understanding would be greatly appreciated!

**Tags:** cosmology, universe, expansion, fate-of-universe, friedmann-equations

> **Comments:**
> **username_1**: The first Friedmann equation isn't a postulate of FLRW cosmology. The postulates are general relativity, homogeneity and isotropy, from which you can derive the first Friedmann equation – the standard one, not your variation. You would have to specify which of the underlying assumptions of standard cosmology are violated before the question would really be answerable, username_2's attempt notwithstanding.  
> **username_2**: By the way, by "INDEPENDENT of its matter-energy content", do you mean matter and radiation specifically? That is, would it be acceptable to make spatial curvature by adding something that contributes to the stress-energy but isn't matter or radiation?  

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## 1 answers

### Answer by username_2 on 2023-03-30. Score: 2

In general relativity, there is no such thing as intrinsic curvature of spacetime that is not tied to the energy content. The Einstein field equations directly tie the curvature tensor to the stress-energy tensor.

Now, you are thinking about curvature of space and not spacetime. The situation is perhaps even worse there because the curvature of space is subjective. Space is a 3D surface in 4D spacetime, and there are infinitely many possible choices of spatial surface. You could always define coordinates so that your spatial surfaces are flat, or of negative curvature, or whatever. That's just a coordinate choice and does not change the physics.

However, in cosmology there is a family of preferred spatial surfaces. Roughly speaking, we define these spatial surfaces so that on any one surface, all the matter in the universe reports the same amount of elapsed time since the Big Bang. (More technically, we think of idealized comoving observers instead of real matter.) These are the surfaces cosmologists are talking about when we say the universe is spatially flat. And the curvature of these spatial surfaces is completely determined by the energy content and the expansion rate. You can't independently specify it. That is to say, there is also no such thing as intrinsic cosmological spatial curvature that is independent of the contents of the universe.

You can think about the curvature of these preferred spatial surfaces intuitively as a competition between gravity and kinematic time dilation. Light moves in straight lines on these surfaces, but gravity converges light. So gravity "wants" to give space positive curvature, which makes parallel lines converge. But the universe is expanding, which means distant objects are moving away and more distant objects are moving away faster. If they are moving faster, they are more time dilated, so less time has passed for them since the Big Bang. So you need to curve your spatial surfaces "into the future" at large distances in order for everyone on that surface to agree on how much time has elapsed. In the future, everything is farther apart. So this effect "wants" to make parallel lines diverge, corresponding to negative spatial curvature. Thus, more energy means more positive curvature, while a faster expansion rate means more negative curvature.

(But I'll caveat that this is an intuitive picture and some care would be needed to translate it into calculations. For example, you can't directly apply the kinematic time dilation equations in curved spacetimes.)

> **Comments:**
> **username_0**: I agree that there is no such thing as intrinsic spatial curvature in standard general relativity but in the framework I've been investigating there actually can be. I tried my best to avoid mentioning the specific model though in case I was merely misunderstanding the meaning behind the parameters in the standard Friedmann equations. But from your answer it sounds like my understanding is sound and that my question might be one that has questionable meaning in standard general relativity. In any case, thanks for the answer!  

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