# Form of Friedman eqn with Λ curvature constant

1. May 12, 2013

### marcus

The basic equation of GR has a curvature constant Λ on the lefthand (geometric) side.
The Friedman equation is derived from the Einstein Field Equation by making a simplifying assumption of uniformity. As a spacetime curvature Λ can be written either in units of reciprocal area or reciprocal time-squared. So in the Friedman equation you might expect to see Λ ( with units of reciprocal time2 or length2) again appearing on the lefthand side, and matter appearing on the right.

But that doesn't happen, so the question naturally comes up as to what the Friedman equation would look like if it were so. Well the Friedman is basically an equation governing the evolution of the expansion rate H(t). For simplicity I'll assume spatial flatness (measurements show largescale mean spatial curvature is nearly, if not precisely, zero). The Friedman actually tells us about the evolution of the SQUARE of the expansion rate, which in standard metric terms would be in units of seconds-2. And Λ as reciprocal time2 can also be expressed in that unit.

H2 - Λ/3 = (8πG/3c2) ρ

where ρ is the combined average energy density (ordinary and dark matter, electromagnetic radiation).

The righthand side goes to zero in the longterm future, due to expansion. So we can solve for the longterm Hubble expansion rate. Obviously it is (Λ/3).5 and therefore the Hubble time is (Λ/3)-.5

It turns out that the best-fit value of Λ, according to most recent Planck report, is
1.007 x 10-35 s-2

If you use that value, the longterm Hubble time (Λ/3)-.5 comes out to 17.3 billion years.

If you then take the present-day Hubble time to be 14.4 billion years and evaluate the LHS of the Friedman equation for the present-day:
H2 - Λ/3 = (8πG/3c2) ρ
at the mantissa level it is just 4.843 - 3.357 = 1.486
and you get 1.486 x 10-36 s-2
This can then be multiplied by 3c2/8πG to get the energy density corresponding to exact spatial flatness (energy equivalent of ordinary and dark matter etc).
It comes out to 0.2389 nanojoules per cubic meter. Google calculator gives this in the algebraically equivalent form 0.2389 nanopascal.

That is the critical density when what we are counting is just matter and energy that we actually know is energy, omitting a possibly fictitious "dark energy", which was already accounted for on the LHS by the curvature constant Lambda.

Last edited: May 12, 2013
2. May 13, 2013

### Jorrie

I am starting to buy into a spacetime curvature constant to replace dark energy with - which is what I understood you are saying. I also agree with the mechanistic conversions that you have done, but I still have problems with viewing energy density as equivalent to a pressure. Yes, there are units of length that cancel out (which is what the Google calculator does), but does it not leave a conceptual problem? As an engineer, I would be appalled if a student expresses energy density in pascals...

3. May 13, 2013

### Whitefire

Some time ago I tried to imagine what is going on inside a black hole. So I imagined matter squeezed... barions being packed so tightly you can't pack them any tighter... and then squeezing them further.

Then I got stuck.

Some time later i realized that perhaps the only answer would be for the elements to adapt by taking a form of what is sometimes referred to in quantum physics as a 'probability field'. 'Probability field' is what you have *before* you measure something (before 'now'). You could say, then, that by force of gravity matter is being pushed along the dimension of time. It actually forms future.

I will not defend this concept, I did not really think it over, but when I noticed you talking about pressure and energy, I couldn't help but mention it.

4. May 13, 2013

### marcus

I agree completely that an energy density is a different physical quantity from a pressure. Even though the units you measure them with boil down to the same primitive M,L,T terms, the same combination of mass, length, and time.

So maybe the metric SI people, the BIPM or CIPM should name the energy density unit after somebody else. The first two names that come to mind are Ringo Starr and Marilyn Monroe.

In the meantime I guess we'll just have to be patient and call the unnamed energy density unit "J m-3"

Point taken.

In case anyone is coming in new to the discussion, the metric pressure unit pascal (or Pa) boils down (in M, L, T = kilogram, meter, second terms) to ML-1T-2
and so does the energy density unit, the Marilyn or Ma as I would suggest it be called.

In sum, Pa and Ma are physically different but algebraically constitute one unit.

Last edited: May 13, 2013
5. May 13, 2013

### George Jones

Staff Emeritus
Something similar happens for torque and energy. Torque and energy are different concepts, but are dimensionally the same. It is standard to use (Newton)(metre) for torque and Joule for energy.

6. May 13, 2013

### Jorrie

7. May 13, 2013

### marcus

I should emphasize this is not meant as a contribution to Jorrie's project of a cosmic history tabulator, but is just something I'm trying out on the side: looking thru metric goggles at the Friedman equation.
To recapitulate, let's look back to that curvature constant form of the Friedman equation, namely:

H2 - Λ/3 = (8πG/3c2) ρ

where ρ is the combined average energy density (ordinary and dark matter, electromagnetic radiation). As a kind of mental exercise I was seeing what it would be like to explore that equation a little bit in metric terms. This isn't exactly conventional but I don't think it hurts to do it (and I WILL make the distinction Jorrie suggests, between Nm-2 and Jm-3 )

In light of the most recent Planck report we've chosen to work with convenient exact values of the current and future Hubble radii which are 14.4 Gly and 17.3 Gly.

This means that in metric the current Hubble growth rate H0 works out to
H0 = 2.2007 x 10-18 s-1
and the cosmological curvature constant works out to
Λ = 1.007 x 10-35 s-2

So, just to see what happens, if we evaluate the LHS of the Friedman equation as of the present day, squaring H0 to get
4.843 x 10-36 s-2 and dividing Λ by 3 to get
3.357 x 10-36 s-2 then the left hand side of Friedman comes out
1.486 x 10-36 s-2 so this then is the slight compensatory positive curvature which matter and electromagnetic radiation have to contribute in order to ensure exact spatial flatness.

And the way matter etc affects curvature is by having its equivalent energy density ρ multiplied by the Einstein equation's coefficient 8πG/3c2
or if you like to use the mass density version of ρ, then just multiply by 8πG/3
but here let's continue viewing the density in terms of energy per unit volume.
So it turns out that the matter etc density needed for exact spatial flatness is
0.2389 nanojoules per cubic meter or in abbreviated form
0.24 nJm-3
That is the energy density you need if you want to multiply by the coefficient 8πG/3c2
and get the desired curvature of 1.486 x 10-36 s-2.

Obviously to get that the only two numbers you need to know for starters are the metric values of H0 and Λ, namely 2.2007e-18 and 1.007e-35. The rest is just algebra with constants like pi, G, and c. Those metric values correspond to the choice of 14.4 and 17.3 billion years for the Hubble times. To repeat something from post#1, the righthand side of the Friedman equation goes to zero in the longterm future, due to expansion, so we can solve for the longterm Hubble expansion rate. H = (Λ/3).5 making the Hubble time (Λ/3)-.5 and that comes out to 17.3 billion years.

Last edited: May 13, 2013
8. May 13, 2013

### Jorrie

Back to the real topic. Do I have it correct that the cosmological constant represents a constant spacetime curvature and not a spatial curvature? I figured that for an Ω = 1 universe, with matter density diluted to insignificance in the far future, there remains only the cosmological constant, but zero spatial curvature - just (constant?) spacetime curvature.

Secondly, if this makes any sense, did the spacetime curvature "evolve" towards this long term constant from a different (matter dominated) curvature?

9. May 13, 2013

### marcus

I'd really like to hear how George Jones sorts these issues out. I'm not sure of my interpretation. Whatever I say is almost certain to be in some sense wrong. I'll think about it and try to offer as right an answer as I can, but then hope George or somebody like that weighs in.

I thought some, and I really don't know whether to call Lambda a constant spatial curvature or a spacetime curvature. Darn! Earlier in this thread I was referring to it as the latter. That may be wrong. Now I'm thinking spatial curvature.

Here's a way it might sort out: In context of FRIEDMAN equation there is an implied splitting into spatial slices and Lambda actually is a spatial curvature constant! That is how we should think and talk of it. But in the original Einstein equation it is a constant that is allowed by diffeomorphism invariance (general covariance) that therefore is unchanged no matter how you decide to slice things. So I have trouble thinking of it as a specifically spatial curvature in that context. Maybe in the original GR context Lambda is just a constant with units of reciprocal area, and that's all you can say---a constant which has to appear in the general form of the equation because it is one of just two gravitational constants that are allowed by the symmetries of the theory.

Last edited: May 13, 2013
10. May 13, 2013

### George Jones

Staff Emeritus
Here is what I know. A useful form of Einstein's equation is

$$R_{\mu \nu} = \Lambda g_{\mu \nu} + \kappa \left( T_{\mu \nu} - \frac{1}{2} T g_{\mu \nu} \right).$$

If anyone wants, I can show how to go from the standard form to this form.

A non-zero cosmological constant and zero energy-momentum tensor gives

$$R_{\mu \nu} = \Lambda g_{\mu \nu},$$

i.e., non-zero spacetime Ricci curvature.

Is this correct??? As I say, I don't know.

This is de Sitter, which admits all three types of spatial slicings.

Now I will stick my neck out and talk about what I don't know.

Three types of spatial slices of are possible for the same source (cosmological constant)! The three slicing correspond to three families of (geodesic) fundamental observers, with three different sets of "initial" conditions.

11. May 13, 2013

### marcus

Hi George, thanks for responding! I just now went back and added something to my post#9 without noticing that you had posted. Would this be a reasonable way to say it?
In Friedman equation context Lambda, a reciprocal area (or s^-2) quantity, can be thought of as an actual spatial curvature that contributes to the total spatial curvature. Because we have well-defined spatial slices.
But in the Einstein equation context instead of picturing it as a curvature you think of it as a coefficient that you multiply the metric tensor by to get a contribution to the Ricci curvature.​

12. May 13, 2013

### Mordred

I would be interested in how to go from the FLRW to the form you have.
It would help me with a related

http://arxiv.org/abs/1205.3855

its related to the discussion but Im still puzzling out aspects of it

Last edited by a moderator: May 14, 2013
13. May 14, 2013