Relationship between Energy density and Curvature

[Arman777]

I don't know GR so while answering the question if you prefer not to use that, I would be happy.

In the Friedmann Equations, is energy density has an effect on curvature or vice versa?
Or they are separate things and they don't affect each other?

For example can we have an energy density $\rho_0$ such that its less then a critical density $\rho_c\,,\,\,(\rho_0<\rho_c)$ in a positive curvature universe ?

Or a hyperbolic universe with $\rho_0>\rho_c$. In general, it seems it should affect, but I couldn't be sure.

 

[kimbyd]

Matter/energy density and curvature are intrinsically linked, yes.

In the FLRW universe, stuff is moving apart. The relationship between how dense stuff is and how fast it is moving gives the total space-time curvature. The expansion itself is a sort of space-time curvature, and the rest is spatial curvature. So if you have a lot of expansion, more than the total space-time curvature that the matter/energy causes, then you have negative spatial curvature. If you don't have a lot of expansion, so that the curvature from the expansion is smaller than the total, then you have positive spatial curvature.

This situation is exactly analogous to a situation where you can throw a baseball hard enough that it could leave orbit. If you do throw it that hard, it will never return. This is similar to the case where expansion is really fast compared to the matter density, where the curvature is negative. If you have only matter, then that universe expands forever.

If, however, you don't throw the baseball hard enough, then it will come back down to the ground. Similarly, if you have a matter-only universe with slow expansion compared to the matter density and positive curvature, that universe will collapse back on itself.

Dark energy does throw a wrench into the mix. Basically it's a constant density, which adds a constant space-time curvature. The gravity from stuff that is a constant density like this acts to push things away from one another.

 

[phyzguy]

I don't know GR so while answering the question if you prefer not to use that, I would be happy.

In the Friedmann Equations, is energy density has an effect on curvature or vice versa?
Or they are separate things and they don't affect each other?

For example can we have an energy density $\rho_0$ such that its less then a critical density $\rho_c\,,\,\,(\rho_0<\rho_c)$ in a positive curvature universe ?

Or a hyperbolic universe with $\rho_0>\rho_c$. In general, it seems it should affect, but I couldn't be sure.

The Friedmann equations are derived by applying Einstein's equation of General Relativity to the whole universe. How can we possibly answer these questions without referring to GR?

 

[kimbyd]

The Friedmann equations are derived by applying Einstein's equation of General Relativity to the whole universe. How can we possibly answer these questions without referring to GR?

I took it as more of a, "Please don't go into details about GR, as I won't understand them."

Edit:
By the way, to the OP, if you would prefer answers that are easier to understand because you worry about not having enough background, I suggest labeling the post as "beginner".

The post label doesn't restrict the kinds of questions you can ask, but acts as a signal to us for how we should try to answer. An intermediate label usually means we can feel free to get into the math a bit, but not too much.

 

[phyzguy]

I took it as more of a, "Please don't go into details about GR, as I won't understand them."

Sorry if I came off negatively. I just wanted to make sure that the OP understood that GR is where these results come from.

 

[Arman777]

Well you guys can go into math without using GR math. I know that in GR energy and curvature are related, but I have no idea about how the math works in there.

I think I understand the idea, so we cannot have a positive curvature with a density lower than the critical density or vice versa. The idea kind of confused me because in the friedmann equation we take curvature and energy density seperatly. But I guess when we try to solve them it turns that curvature can have some certain values or etc. (For example the empty universe case which I asked earlier cannot have hyperbolic geometry, scale factor does not make sense).

 

[Arman777]

I took it as more of a, "Please don't go into details about GR, as I won't understand them."

Edit:
By the way, to the OP, if you would prefer answers that are easier to understand because you worry about not having enough background, I suggest labeling the post as "beginner".

The post label doesn't restrict the kinds of questions you can ask, but acts as a signal to us for how we should try to answer. An intermediate label usually means we can feel free to get into the math a bit, but not too much.

I ll keep in my mind, thanks.

 

[kimbyd]

Well you guys can go into math without using GR math. I know that in GR energy and curvature are related, but I have no idea about how the math works in there.

I think I understand the idea, so we cannot have a positive curvature with a density lower than the critical density or vice versa. The idea kind of confused me because in the friedmann equation we take curvature and energy density seperatly. But I guess when we try to solve them it turns that curvature can have some certain values or etc. (For example the universe case which I asked earlier cannot have hyperbolic geometry, scale factor does not make sense).

I think the issue is that the way the Friedmann equations are usually written, they're written to be useful in doing calculations for cosmology. They're not written to make the physical meanings clear.

It'd be perhaps a little bit more understandable, for instance, if the first Friedmann equation was written as follows:
$$H^2 + {k c^2 \over a^2} = {8 \pi G \over 3} \rho$$

Here all the curvature-related terms are on the left, while all the density-related terms are on the right. I've also wrapped the cosmological constant term into $\rho$, though that can be done differently. The "critical density" is just the density that makes it so that parameter $k=0$.

This way of writing things is directly related to how the Einstein Field Equations can be written (again assuming $\Lambda$ is included in the matter/energy terms):
$$R_{\mu\nu} - {1\over 2}Rg_{\mu\nu} = {8\pi G\over c^4}T_{\mu\nu}$$

Here $R$ is a function of the curvature of space-time, while $T$ is the stress-energy tensor which describes matter (including energy density, pressure, momentum, and twisting forces).

 

[Arman777]

I think the issue is that the way the Friedmann equations are usually written, they're written to be useful in doing calculations for cosmology. They're not written to make the physical meanings clear.

It'd be perhaps a little bit more understandable, for instance, if the first Friedmann equation was written as follows:
$$H^2 + {k c^2 \over a^2} = {8 \pi G \over 3} \rho$$

Here all the curvature-related terms are on the left, while all the density-related terms are on the right. I've also wrapped the cosmological constant term into $\rho$, though that can be done differently. The "critical density" is just the density that makes it so that parameter $k=0$.

This way of writing things is directly related to how the Einstein Field Equations can be written (again assuming $\Lambda$ is included in the matter/energy terms):
$$R_{\mu\nu} - {1\over 2}Rg_{\mu\nu} = {8\pi G\over c^4}T_{\mu\nu}$$

Here $R$ is a function of the curvature of space-time, while $T$ is the stress-energy tensor which describes matter (including energy density, pressure, momentum, and twisting forces).

This seems nice so they are related as GR predicts.

 

[kimbyd]

This seems nice so they are related as GR predicts.

Well, the Friedmann Equations are derived directly from the Einstein Field Equations, so yes, they have to be!

To delve a tiny bit more into the details, the Einstein Field Equations are, when fully-expanded, a set of ten independent second-order differential equations. For the uniformly-expanding universe, however, we can use some symmetries to significantly reduce the problem. First, the universe is homogeneous and isotropic, so any momentum or twisting forces are out (both have a direction associated, and isotropic means no direction is special). And it can't have different pressure in different directions (again, because isotropic), so you're left with just two terms to describe the stress-energy tensor: energy density and pressure. Assuming that the space-time also follows that same symmetry, you are reduced to just two equations.

When you massage the equations in a certain way, those two equations become the two Friedmann equations. The first relates the rate of expansion (first derivative) to the energy density. The second relates the rate of change in the expansion to the energy density and pressure.

Of course, the way to write the two equations is a choice. But the symmetries of the system guarantee only two independent equations. The Friedmann equations are just how we usually choose to represent them.

 

[Arman777]

The first relates the rate of expansion (first derivative) to the energy density. The second relates the rate of change in the expansion to the energy density and pressure.

The fluid equation and the Friedmann equation I guess ?

But the symmetries of the system guarantee only two independent

And the Acceleration equation can be derived using the Friedmann Equation and the Fluid Equation. So..2 independent and 1 dependent equation...

First, the universe is homogeneous and isotropic, so any momentum or twisting forces are out.

Are these forces ever investigated as dark energy ? I am sure there has been some non-cosmological principle GR models. Are they good enough ?

 

[kimbyd]

The fluid equation and the Friedmann equation I guess ?

Not exactly. The symmetry of the situation guarantees that there can be only two independent equations. That doesn't say anything about which two equations they are. The choice of equations is entirely a matter of choice and convention.

Yes, you can go through the full GR calculation to derive both the fluid equation and the first Friedmann equation as your two equations. Or you can work through the GR calculation differently and select two different equations.

Are these forces ever investigated as dark energy ? I am sure there has been some non-cosmological principle GR models. Are they good enough ?

No. You need solid matter to support twisting forces. Fluids just flow, preventing any sort of twisting force from being maintained. Plus you can get rid of twisting forces by changing your coordinate system anyway (they show up as pressure in a different set of coordinates).

As an aside, a neat way of seeing which coordinate system makes the twisting force look like pressure can be visualized by twisting a piece of chalk: the chalk breaks in a spiral pattern. The twisting resolves to a pulling pressure perpendicular to the spiral break, and compression tangential to it.

The simplest explanation for dark energy is the cosmological constant, which is a parameter in General Relativity. It can be seen as either an integration constant for the curvature, which is another way of saying that the baseline space-time curvature is some constant but non-zero value, or as a form of energy which has a constant density. In string theory, for instance, the cosmological constant is the vacuum expectation value of a particular field.

 

[Arman777]

Not exactly. The symmetry of the situation guarantees that there can be only two independent equations. That doesn't say anything about which two equations they are. The choice of equations is entirely a matter of choice and convention.

Yes, you can go through the full GR calculation to derive both the fluid equation and the first Friedmann equation as your two equations. Or you can work through the GR calculation differently and select two different equations.No. You need solid matter to support twisting forces. Fluids just flow, preventing any sort of twisting force from being maintained. Plus you can get rid of twisting forces by changing your coordinate system anyway (they show up as pressure in a different set of coordinates).

As an aside, a neat way of seeing which coordinate system makes the twisting force look like pressure can be visualized by twisting a piece of chalk: the chalk breaks in a spiral pattern. The twisting resolves to a pulling pressure perpendicular to the spiral break, and compression tangential to it.

The simplest explanation for dark energy is the cosmological constant, which is a parameter in General Relativity. It can be seen as either an integration constant for the curvature, which is another way of saying that the baseline space-time curvature is some constant but non-zero value, or as a form of energy which has a constant density. In string theory, for instance, the cosmological constant is the vacuum expectation value of a particular field.

I see now. In QM cosmological constant seen as vacuum energy but calculations give huge discrepancy between observational and theoretical values..Is it not the same as the string theory I guess. This problem is not solved I think ? Well if it would we would now what dark energy is I guess.

 

[kimbyd]

I see now. In QM cosmological constant seen as vacuum energy but calculations give huge discrepancy between observational and theoretical values..Is it not the same as the string theory I guess. This problem is not solved I think ? Well if it would we would now what dark energy is I guess.

The cosmological constant is just a parameter in GR. It is unknown whether it relates to an energy density or not. It could. Certainly that's an appealing possibility.

Ideally if we can manage to nail down the correct theory of quantum gravity, that will tell us precisely where the parameter comes from. Sadly, nailing down the correct theory of quantum gravity may not happen any time soon.

 

[Arman777]

Thanks a lot

 

[substitute materials]

I think I have a question which is very similar, so I won't start a new thread: Dark Energy does 2 things in lambdaCDM as I understand it. First it provides a repulsive force that is accelerating the expansion of the universe, and secondly it constitutes the missing energy to bring the density of the observable universe up to the critical density (to give us zero spatial curvature). Are these 2 features linked quantitatively? Does setting the density of dark energy to fulfill the critical density yield the observed acceleration? Or are these 2 functions essentially independent of one another?

 

[Arman777]

Does setting the density of dark energy to fulfill the critical density yield the observed acceleration?

Yes, indeed. Thats actually what we are doing. Making measurements and then giving values to the density paramteres, then "running" the The friedmann Equations and see which denisty parameters would fit best to the observational data.

 

[kimbyd]

I think I have a question which is very similar, so I won't start a new thread: Dark Energy does 2 things in lambdaCDM as I understand it. First it provides a repulsive force that is accelerating the expansion of the universe, and secondly it constitutes the missing energy to bring the density of the observable universe up to the critical density (to give us zero spatial curvature). Are these 2 features linked quantitatively? Does setting the density of dark energy to fulfill the critical density yield the observed acceleration? Or are these 2 functions essentially independent of one another?

Absolutely.

One way to look at it is that impact of the spatial curvature of the universe scales as $1/a^2$. The spatial curvature, in turn, is a result of initial conditions. Lots of density and little expansion = positive curvature. Not much density and a lot of expansion = negative curvature. In a matter dominated universe, where the density scales as $1/a^3$, the curvature eventually becomes the dominant feature and determines the future fate: positive curvature means recollapse, while negative curvature means eternal expansion.

If you have something whose effect on the expansion scales more slowly than $1/a^2$, then it has the ability to overturn this. As long as the expansion is allowed to continue far enough, it is the slowly-diluting stuff that will determine the eventual fate of the universe, not the curvature. And it does so by causing acceleration. Furthermore, because it dilutes more slowly than $1/a^2$, it drives the universe towards flatness.

 

[PeterDonis]

you guys can go into math without using GR math.

The math for this problem is the Friedmann equations; those are "GR math". So the math that has been discussed here is "GR math".

 

[substitute materials]

Absolutely.

One way to look at it is that impact of the spatial curvature of the universe scales as $1/a^2$. The spatial curvature, in turn, is a result of initial conditions. Lots of density and little expansion = positive curvature. Not much density and a lot of expansion = negative curvature. In a matter dominated universe, where the density scales as $1/a^3$, the curvature eventually becomes the dominant feature and determines the future fate: positive curvature means recollapse, while negative curvature means eternal expansion.

I think I get this, I understand the balancing act of rate of expansion and gravitation in a matter dominated world. As you have said elsewhere on the site Kimbyd, it is a good analogy to escape velocity. I can see how this would only be stable, remaining flat, if the mass/energy were perfectly balanced with the rate of expansion.

But I guess I don't understand the nature of Dark Energy's repulsive contribution. Since Dark energy must gravitate in GR, if we believe that it makes up the difference to get us to the critical density, we can say that it would have to be ~6E-10 joules/m^3, right? My question is, is this the exact right amount of dark energy to produce the observed acceleration of expansion? Or is it unclear the "efficacy' of dark energy's repulsive force per unit energy?

 

[kimbyd]

I think I get this, I understand the balancing act of rate of expansion and gravitation in a matter dominated world. As you have said elsewhere on the site Kimbyd, it is a good analogy to escape velocity. I can see how this would only be stable, remaining flat, if the mass/energy were perfectly balanced with the rate of expansion.

But I guess I don't understand the nature of Dark Energy's repulsive contribution. Since Dark energy must gravitate in GR, if we believe that it makes up the difference to get us to the critical density, we can say that it would have to be ~6E-10 joules/m^3, right? My question is, is this the exact right amount of dark energy to produce the observed acceleration of expansion? Or is it unclear the "efficacy' of dark energy's repulsive force per unit energy?

Sean Carroll had a really good blog post about this, which talks about how this question is usually answered, and offers what I think is a far more intuitive answer:
http://www.preposterousuniverse.com...oes-dark-energy-make-the-universe-accelerate/

The short version is this: the amount of space-time curvature in the universe is a consequence of how much stuff is in the universe. Specifically, space-time curvature is a function of some combination of density and pressure. Exactly how they relate isn't at all important for the argument. What is important is recognizing that if the amount of stuff is constant (such as the cosmological constant), then the amount of space-time curvature will also be constant.

In an expanding universe, the space-time curvature is broken into two components: expansion and spatial curvature. If we take the extreme case where all we have is the cosmological constant, then the spatial curvature has to be zero, so a constant rate of expansion is implied by a cosmological constant-dominated universe.

A constant rate of expansion implies exponential increase in distances between test particles over time, because of the definition of $H(t)$:

$$H(t) = {1 \over a(t)} {da(t) \over dt} = H_0$$

The solution to this differential equation is an exponential:

$$a(t) = a(t=0) e^{H_0 t}$$

This situation describes the exponential future state of a universe dominated by a cosmological constant.

This argument won't tell you the exact conditions under which the universe crosses over into an accelerating state, but it at least motivates how it can be possible that slowly changing density causes gravity to become repulsive.

 

[kimbyd]

The math for this problem is the Friedmann equations; those are "GR math". So the math that has been discussed here is "GR math".

I would interpret this as asking us to avoid doing differential geometry calculations or providing explanations that rely on that level of understanding, such as killing vectors.

 

[lomidrevo]

But I guess I don't understand the nature of Dark Energy's repulsive contribution.

I agree with @kimbyd that explanation in her post is more intuitive.

But it might also be instructive to see why dark energy generates negative pressure. To get a picture of how this negative pressure "push things apart" pls read the "The Wrong Way" part of the Carroll's blog (you will also see in what aspect he consider it "wrong").

Consider a hypothetical container of volume V containing only dark energy (with constant energy density $\Lambda$) and let's try to squeeze the container by changing its volume by $dV$. The first law of thermodynamics says

$dE = TdS - PdV \ \ \ (1)$​

but as we are not increasing entropy, we have

$dE = - PdV \ \ \ (2)$​

As the energy density remains constant (no matter how we change volume):

$E = \Lambda \ V \ \ \ (3)$,​

then as $dV < 0$ (remember, we are squeezing the container), also $dE$ must be negative: $dE < 0$. But if in the equation (2) both $dE, dV$ are negative, $P$ must be also negative. Indeed you can see that $P = - \Lambda$, just by differentiating (3) (with respect to volume) and comparing with (2).

This calculation is maybe a bit heuristic but very easy to understand (however I am not sure that considering $dS = 0$ is fully correct). There are of course more involved derivations, which give you the same result.

 

[substitute materials]

If we take the extreme case where all we have is the cosmological constant, then the spatial curvature has to be zero

interesting, I didn't realize that this would be true.

I realize you are trying to explain this without referring to pressure Kimbyd, but I'm getting stuck on it. I think I understand that the matter in our particular universe has a pressure dictated by H- a sort of momentum in no direction. But why does dark energy have momentum, and how is its specific value observed?

I see pressure in the Friedmann equations, but I don't pretend to understand how it comes out of the field equations, they are still a black box to me, like the OP.Here is one quote by someone named Vmarko in the comments on the Sean Carroll article. "The point is that the “dark energy” also carries along its own “dark momentum” (usually called pressure), which is three times as strong [as its gravitational attraction] and contributes a repulsive gravitational effect. So one attractive energy plus three repulsive pressures give a net repulsive force of gravity. People who haven’t studied the details of GR have a hard time wrapping their heads around this"

This commenter is suggesting that Dark Energy has three times the pressure versus its density. Is this ratio determined by the 3 in the second Friedmann equation? Or could dark energy have ten times the pressure, or half the pressure, etc?

 

[substitute materials]

Consider a hypothetical container of volume V containing only dark energy (with constant energy density $\Lambda$) and let's try to squeeze the container by changing its volume by $dV$. The first law of thermodynamics says

$dE = TdS - PdV \ \ \ (1)$​

but as we are not increasing entropy, we have

$dE = - PdV \ \ \ (2)$​

As the energy density remains constant (no matter how we change volume):

$E = \Lambda \ V \ \ \ (3)$,​

then as $dV < 0$ (remember, we are squeezing the container), also $dE$ must be negative: $dE < 0$. But if in the equation (2) both $dE, dV$ are negative, $P$ must be also negative. Indeed you can see that $P = - \Lambda$, just by differentiating (3) (with respect to volume) and comparing with (2).

This calculation is maybe a bit heuristic but very easy to understand (however I am not sure that considering $dS = 0$ is fully correct). There are of course more involved derivations, which give you the same result.

What you both seem to be saying is that any form of energy that doesn't dilute as volume changes will exert negative pressure, yes? Shrinking a container with matter in it will do work on it, but not so with a container filled with dark energy. So this means that work is done on the universe by expanding it, because of the presence of dark energy? I'm weirded out by this violation of conservation of energy, even though I have seen many people suggest it is allowed.

 

[kimbyd]

I see pressure in the Friedmann equations, but I don't pretend to understand how it comes out of the field equations, they are still a black box to me, like the OP.

The source of Gravity in General Relativity is the stress-energy tensor. At any point in space, there is a set of coordinates where it will contain only four components: energy density and three pressure components along three different axes.

In the homogeneous, isotropic universe, those three pressure components have to be the same (different pressure in different directions would not be isotropic).

So the pressure occurs in the second Friedmann equation because the source of gravity is a combination of pressure and energy density. In fact, the second Friedmann equation is just the trace of the Einstein equations: the trace is the sum of the diagonal elements: $\rho + 3p$, with the three coming from pressure in the three directions. The rest of the second Friedmann equation comes from the very complicated math that is a function of the space-time curvature.

Here is one quote by someone named Vmarko in the comments on the Sean Carroll article. "The point is that the “dark energy” also carries along its own “dark momentum” (usually called pressure), which is three times as strong [as its gravitational attraction] and contributes a repulsive gravitational effect. So one attractive energy plus three repulsive pressures give a net repulsive force of gravity. People who haven’t studied the details of GR have a hard time wrapping their heads around this"

This commenter is suggesting that Dark Energy has three times the pressure versus its density. Is this ratio determined by the 3 in the second Friedmann equation? Or could dark energy have ten times the pressure, or half the pressure, etc?

Hopefully the above answers this: pressure in three directions, and the the formula for the second equation comes from the trace of the tensor.

 

[lomidrevo]

any form of energy that doesn't dilute as volume changes will exert negative pressure, yes?

Yes, for positive energy density the pressure is negative.

So this means that work is done on the universe by expanding it, because of the presence of dark energy?

Well, this seems to be a reasonable conclusion. But I think it just better to say that energy is not conserved in a universe dominated by dark energy.

 

[substitute materials]

Ok think I've got it. Thanks all!