I Relationship between Energy density and Curvature (1 Viewer)

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Arman777

Gold Member
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.

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kimbyd

Gold Member
2018 Award
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

Gold Member
2018 Award
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

Gold Member
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 beacause 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).

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Arman777

Gold Member
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

Gold Member
2018 Award
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 beacause 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

Gold Member
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

Gold Member
2018 Award
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

Gold Member
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

Gold Member
2018 Award
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

Gold Member
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 theoritical 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

Gold Member
2018 Award
I see now. In QM cosmological constant seen as vacuum energy but calculations give huge discrepancy between observational and theoritical 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.

Gold Member
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

Gold Member
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

Gold Member
2018 Award
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

Mentor
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?

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