# Lambda-CDM model and geometric units

1. Nov 21, 2007

### stevebd1

I'm currently putting together a basic summary of the Lambda-CDM model and I have a slight issue with the fact that the equation to calculate lambda (which includes factors to convert physical units into geometric units) is incorporated into the omega_lambda calculation (which incorporates the critical density equation that appears to be simply expressed in physical units). Is this the norm? I'd appreciate it if someone could confirm that the following is correct (I've added footnotes in bold)-

The equation for lambda is taken from Einstein's modified field equation for general relativity.

Einstein's field equation G = 8 pi G T/C^4

G is the Einstein tensor of curvature (space time).
T is the energy tensor of matter (matter energy).
8 pi is the concentration factor.

C and G are introduced to convert the quantity T (which is expressed in physical units) to geometric units (G/C^4 is used to convert units of energy into geometric units while G/C^2 is used to convert units of mass, when density is used instead of energy, the C^4 can be replaced with C^2). {evidence of converting to geometric units.}

Lambda = 8 pi G x vacuum energy/C^4 = 8 pi G x vacuum density/C^2

= 8 x 3.14159 x 6.6742x10^-11 x 0.67 x 10^-26/(3x10^8)^2

= 1.252 x 10^-52 m^-2 {geometric units?}

Based on the vacuum density being 0.67 x 10^-26 kg/m^3, the figure above is 1.252 x 10^-52 m^-2 which is expressed in geometric units. This divided by G/C^2 (7.43 x 10^-28) converts it to mass SI units of kg/m^3, providing a figure of 1.685 x 10^-25 kg/m^3 (or 1.685 x 10^-28 g/cm^3) in physical units for the cosmological constant.

Critical density = 3H^2/8 pi G

= 3 x (2.26x10^-18)^2/8 x 3.14159 x 6.6742x10^-11

= 0.918 x 10^-26 kg/m^3
{no evidence of converting to geometric units!}

Omega = actual density/critical density

-Normalized matter density (incorporating the critical density equation. Omega = density of universe/critical density, if we put the above critical density equation into this simple formula, we get the equation below).

Baryonic matter (b)

Omega b = 8 pi G x density of b/3H^2

= 8 x 3.14159 x 6.6742x10^-11x 0.4x10^-27/3 x (2.26x10^-18)^2

Omega b = 0.044 for baryonic matter (matter composed of protons, neutrons and electrons)

Dark matter (dm)

Omega dm = 8 pi G x density of dm/3H^2

= 8 x 3.14159 x 6.6742x10^-11 x 0.202x10^-26/3 x (2.26x10^-18)^2

Omega dm = 0.222 for dark matter

Omega b + omega dm = 0.044 + 0.222

-Normalized vacuum energy density (incorporating lambda equation. By altering the lambda equation shown above slightly, we get lambda C^2 = 8 pi G x vacuum density, therefore lambda C^2 replaces 8 pi G x vacuum density in the omega equation).

Dark energy (l)

Omega l = lambda C^2/3H^2

= 1.252x10^-52 x (3x10^8)^2/3 x (2.26x10^-18)^2
{equation based on combining lambda equation- geometric units and critical density equation- physical units?}

Omega l = 0.732 for dark energy

Hopefully this is clear. I would appreciate any feedback.

regards
Steve

2. Nov 22, 2007

### Chris Hillman

This is hard to read, which is probably why no response until now. Can you use the PF latex markup feature to reformat? Also, can you clarify: is this a school essay you want feedback on? Is the point to compute a numerical value for some physical quantity? If so, can you follow standard practice and put $G=c=1$ until you have computed the quantity you want in "geometrized units" and then insert the numerical factors to convert to cgs units as per this WP article (in the version cited, the last version I have read)? If you are using a particular textbook as a guide, it would also be a good idea to say which one.

3. Nov 22, 2007

### stevebd1

Thanks for your reply. I actually tried to write the equations with the figures top and bottom of a dividing line but when I previewed, it didn't appear as expected. I looked at the latex mark up but I'm not familiar with it (I'll see if there's some info about how to use it else where in the forum). Basically, by trade, I'm not a physicist though I have some understanding of maths and geometric units are something relatively new to me. What I was trying to put forward in my post is that I'm currently looking at the Lambda-CDM model and how various omega values were calculated based on the critical density.

omega = 'density' divided by 'critical density'
therefore
omega = 'density' divided by '('3 H^2' divided by '8 pi G')

omega = '8 pi G x density' divided by '3 H^2'

This makes clear sense to me. I was relatively happy with the results, then I realised that the omega_dark energy could be altered to include the actual figure of lambda-

lambda = '8 pi G x density of l' divided by 'C^2'
therefore
lambda C^2 = 8 pi G x density of l

which meant, lambda C^2 could replace 8 pi G x density of l in the omega_dark energy equation

omega_dark energy = 'lambda C^2' divided by '3 H^2'

This seems to combined geometric units and physical units (lambda being in geometric units and the critical density equation giving physical units).

Omega_baryonic and omega_dark matter equations don't incorporate geometric units, just the critical density equation (which I'm assuming is in physical units) yet the omega_lambda equation looks like it might incorporate geometric units also which seems at odds with the other omega equations.

Hopefully this makes more sense.

regards
Steve

Last edited: Nov 22, 2007
4. Nov 22, 2007

### Chris Hillman

Latex, please!!!

Try hitting the "reply" button under this post, but don't actually reply. In the windowpane, you should see the text with the VB/latex markup.

Einstein's equation:
$$G^{ab} = 8 \, \pi \, T^{ab}$$
Term on RHS corresponding to $\Lambda = 8 \, \pi \, \varepsilon$:
$$T\left[{\rm dark} \; {\rm energy}\right]_{ab} = \varepsilon \, \operatorname{diag} (1,-1,-1,-1 )$$

Can you figure out how to fix the format now?

5. Nov 23, 2007

### stevebd1

6. Nov 23, 2007

### Chris Hillman

Sorry, Steve, I avoid untrusted websites. However, there are others here at PF who have more patience for ill-formatted posts who will be able to help you.

7. Nov 23, 2007

### jonmtkisco

Hi Steve,

I'm not quite sure what you're saying. However, when vacuum density is stated as 0.67E-26 kg/m^3 (which I would state as 6.7E-27 kg/m^3), you do not divide it by C^2. You would divide it by C^2 only if it is stated in units of joules, that is, as energy rather than mass.

Jon

8. Nov 24, 2007

### stevebd1

Thanks for your response. From what I understand, the standard equation for lambda is 'lambda = 8 x pi x G x vacuum density / C^2' which I've seen expressed a number of times and gave the answer 1.252 x 10^-52 m^-2. As I had seen lambda expressed in kg/m^3 I enquired on the PF about how this had been achieved. I was informed that the G/C^2 had been introduced to the lambda equation in order to convert the units of mass into geometric figures (this tallied with the table on wiki). This explained why I had also seen the equation for lambda expressed as 'lambda = 8 x pi x vacuum density' which gives the answer in SI units of 1.685 x 10^-25 kg/m^3 which was reasonably close to other figures of Lambda I had seen expressed in SI units.

My issue here is that the lambda = 8 pi G x vacuum density / C^2 provides geometric units.
(which can be re-written as lambda C^2 = 8 pi G x vacuum density)

The critical density equation provides physical units and is used to find the density parameters, omega-

omega = density / critical density

omega = density / (3 H^2 / 8 pi G)

omega = 8 pi G x density / 3 H^2

I understand that omega isn't expressed in any unit of measurement or in geometric units for that matter, but I have seen it incorporate part of the equation for lambda when calculating omega_dark energy-

omega_dark energy = 8 pi G x density / 3 H^2

Due to the variation of the lambda equation, lambda C^2 = 8 pi G x vacuum density, the equation for omega_dark energy can be re-written as-

omega_dark energy = lambda C^2 / 3 H^2

It just seemed strange to combine units from an equation that expresses its answer in geometric units and an equation that expresses its answer in no units.

regards
Steve

Last edited: Nov 24, 2007
9. Nov 25, 2007

### jonmtkisco

Hi Steve,

I don't think I can answer your question. However, I typically see Lambda expressed as 10^-35-s^2; or 6.7E-26 kg/m^3. But I don't recall seeing it expressed in the form you gave using units of m^-2.

Jon

10. Nov 26, 2007

### stevebd1

Thanks for your reply, Jon.

I've seen a couple of examples where Lambda has been expressed in m^-2 (or km^-2), one being on wikipedia under 'cosmological constant' as shown below-

The cosmological constant Λ appears in Einstein's modified field equation in the form-

R - $$\frac{1}{2}$$Rg + $$\Lambda$$g = $$\frac{8 \pi G}{C^4}$$ T

where R and g pertain to the structure of spacetime, T pertains to matter (thought of as affecting that structure), and G and c are conversion factors which arise from using traditional units of measurement. When Λ is zero, this reduces to the original field equation of general relativity. When T is zero, the field equation describes empty space (the vacuum). Astronomical observations imply that the constant cannot exceed 10^-46 km-2.

Link
http://en.wikipedia.org/wiki/Cosmological_constant

I've also seen a couple of examples where the equation for lambda is expressed with C^2

$$\Lambda$$ = $$\frac{8 \pi G}{C^2}$$ x vacuum density

Example-

The notion that the cosmological constant is nonzero is based on the idea that the vacuum can contain energy. In the vacuum, for example, we know that quantum fluctuations lead to the appearance and disappearance of virtual pairs of particles which continuously pop into and then go out of existence. These cannot be measured directly, but they affect the curvature of space. These sum effect of the fluctuations could be positive, negative, or zero depending upon the details of our current theories. They are a major contribution to the background energy of the Universe. We parametrize their strength as:

Lambda = (8 pi G / c**4) (vacuum energy) = 8 pi G rho(vacuum) / c**2

Link
http://zebu.uoregon.edu/~imamura/209/may1/lambda.html [Broken]

Steve

Last edited by a moderator: May 3, 2017
11. Nov 27, 2007

### stevebd1

In addition to my above post regarding Lambda being expressed in m$$^{-2}$$, the following is an extract from the Hyperphysics web site -

Cosmological Constant

Einstein proposed a modification of the Friedmann equation which models the expanding universe. He added a term which he called the cosmological constant, which puts the Friedmann equation in the form

H$$^{2}$$ = $$\frac{8 \pi G \rho}{3 c^2}$$ - $$\frac{k}{R^2}$$ + $$\frac{\Lambda c^2}{3}$$ where $$\Lambda$$ = cosmological constant

The original motivation for the cosmological constant was to make possible a static universe which was isotropic and homogeneous. When the expansion of the universe was established without doubt, Einstein reportedly viewed the cosmological constant as the "worst mistake I ever made". But the idea of a cosmological constant is still under active discussion. Rohlf suggests that the physical interpretation of the cosmological constant was that vacuum fluctuations affected space time. A non-zero value for the cosmological constant could be implied from measurements of the volume densities of distant galaxies, but such measurements give a negative result, showing an upper bound of

|$$\Lambda$$| < 3 x 10$$^{-52}$$ m$$^{-2}$$

This implies that on the scale of the whole universe, vacuum fluctuation effects cancel out. This assessment comes at a time when theoretical calculations suggest vacuum fluctuation contributions from quarks on the order of 10-6 m-2.

Link-
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

12. Nov 28, 2007

### stevebd1

I thought the following was worth posting as it's another example of lambda being expressed in m^-2. (edited)

Below is an extract from 'Results and predictions of scale relativity' by Laurent Nottale, Director of Research at CNRS (National Centre for Scientific Research)-

'*Value of the cosmological constant: it is predicted to be $$\Lambda$$ = 1.36x10$$^{-56}$$ cm$$^{-2}$$, i.e. $$\Omega_{\Lambda}$$=0.36 h$$^{-2}$$ (=0.7±0.2 for H=70±10 km/s.Mpc) under the assumption that the fractal-nonfractal transition for the vacuum energy density occurs at about 70 MeV. This is the scale of the classical radius of the electron (70.02 MeV), but also the 6 flavor QCD scale (66±10 MeV) and even better, it is nothing else but the effective Compton length of quarks in the lightest meson (m$$\pi$$/2=69.78 MeV): this value could therefore arise from the quark-hadron transition in the primeval universe. This prediction is now confirmed: the reduced cosmological constant has been recently found to be $$\Omega_{\Lambda}$$=0.7±0.2 by several independant and complementary measurements (SNe's, Boomerang, gravitational lensing).'

Link- http://luth2.obspm.fr/~luthier/nottale/ukpredic.htm

I also find the following PF thread useful in understanding the units of the cosmological constant-

https://www.physicsforums.com/showthread.php?t=54718

Extract by Hellfire from the above thread-
'The cosmological constant is ... a geometric modification of the action of gravity (a modification of the Einstein-Hilbert action). This means that the way spacetime reacts against energy and momentum is postulated to be different whether a cosmological constant is introduced in the theory or not. In general relativity, it describes how spacetime is influenced by matter. The Einstein equations contain geometric terms on the left hand side (as a function of the metric of spacetime) and a description of matter on the right hand side (energy and momentum of matter). ... the cosmological constant is a term on the left hand side..'

Steve

Last edited by a moderator: Apr 23, 2017
13. Nov 28, 2007

### George Jones

Staff Emeritus
At least for, it's not clear that this sheds any light on what $\Lambda$ actually is.

14. Nov 29, 2007

### stevebd1

On a slight change of subject, Im currently looking at how Lambda is calculated in terms of quantum field theory and while looking at how to calculate the vacuum density with QFT, I happened upon 2 equations-

$$\rho$$ = $$\frac{m^4 c^3}{h^3}$$

where mass is calculated by 2 x $$\pi$$ x G x m$$^{2}$$ = h x c

therefore m = $$\sqrt{\frac{hc}{2 \pi G}}$$ = 1.2209x10$$^{19}$$ GeV/c$$^{2}$$ = 2.176x10$$^{-8}$$ kg

h = planck's constant
c = speed of light
G = Gravitaional constant

this gives an answer of 2.08x10$$^{91}$$ kg/m$$^{3}$$ (which tallies with the fact that the cosmological constant when calculated via QFT is 10$$^{120}$$ larger than the cc calculated by cosmology).

Link- http://www.astro.ucla.edu/~wright/cosmo_04.htm second para

This is the other equation-

$$\rho$$vc$$^{3}$$ = $$\frac{m^4}{(hc)^3}$$

which supposedly gives the answer-

$$\rho$$vc$$^{3}$$ ~ 3x10$$^{126}$$ eV cm$$^{-3}$$ (~ 10$$^{93}$$ gm cm$$^{-3}$$)

Link- http://www.astro.virginia.edu/class/whittle/astr553_old/Topic16/Lecture_16.html
(about 1/4 of the way down)

I have no problems with the first equation. With the second equation, I'm unable to achieve the answer provided and would appreciate any feedback on how to achieve this.

regards
Steve

Last edited: Nov 29, 2007
15. Dec 19, 2007

### stevebd1

I'd be grateful if someone could explain the difference between the vacuum density (an assumed 0.670x10^-26 kg/m^3) and the cosmological constant which can also be expressed in kg/m^3 (1.685x10^-25 kg/m^3). I'm assuming that the density is non-relativistic while the cc is the density multiplied by 8$$\pi$$ (the concentration factor), making it relativistic? I just find it a bit odd that they can be expressed using the same units but mean different things and would like to fully understand the difference between the two.

Steve

Last edited: Dec 19, 2007
16. Dec 19, 2007

### Garth

A subtle and common confusion Steve.

In Einstein's field equation:

$$R_{\mu \nu} - \frac{1}{2}g_{\mu \nu}R + \Lambda g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}$$

The LHS describes the curvature of space-time, the RHS describes the distribution of stress, mass and energy that cause that space-time curvature. ('Matter' tells space-time how to curve, curved space-time tells 'matter' how to move)

The cosmological constant on the LHS is simply an undetermined factor in the description of the curvature of space-time. If present it would mean that at 'short' ranges gravitation is Newtonian-like and attractive, but at great ranges it becomes replusive. No further explanation required - it is just that is how it would behave.

The false vacuum density, on the other hand, is entered in on the RHS of the equation as a component of $T_{\mu \nu}$, with the property that it has exactly the same equation of state as simulated by the CC.
($p = - \rho$) You can move the term from one side of the equation to the other.

Positive pressure adds energy and therefore greater attraction to a gravitational field.

In this case as this component has negative pressure the gravitation it causes is repulsive in nature, just like the CC.

So if you are dealing with cosmic acceleration, which is it to be, CC, or false vacuum energy? How can you tell? Does it matter what you call it, as both have the same effect?

Perhaps the only clue that they ought to be dealt with separately is the fact that the observed CC from distant SNe Ia data is a factor ~10120 smaller than that expected from an expected false vacuum energy density.

We await a quantum gravity theory to resolve the quandary!

Garth

Last edited: Dec 19, 2007
17. Dec 19, 2007

### marcus

nice bit of writing as usual Garth----don't know if I ever complimented you on your style which is often admirably concise and clear

Steve, you set off an interesting thread. got a lot of good comments.

I don't fully understand what you are getting at, or find confusing, but i will share a private perspective with you. I think of the cosmological constant as an ENERGY density

(just take the mass density you have for it and multiply by c^2 and it changes from kilogram per cubic into joules per cubic.)

And when I calculated it out one time I got that it was 0.6 joules per cubic kilometer.
I find that easy to remember.

And you can always change it back into joules per cubic meter, and then convert energy back into mass units and recover the kilograms per cubic meter you had before.
But 0.6 joules per cubic kilometer is somehow easier for me to visualize as this very dilute energy filling space.

And you can also (as Garth said shift it over to the LEFTHAND side and) express it as a basic small intrinsic CURVATURE that the world has. And then the units are one over length squared or reciprocal of an area. Because that is a curvature unit.
========================

This is not significant science---I'm just sharing with you how I personally keep what Garth was talking about in my mind. I ask myself what physical quantity do you multiply a curvature by to get an energy density?

We are just playing with units. An energy density is the same as a PRESSURE (force per unit area is the same, dimensionally, as forcedistance per unit volume)

If you multiply a FORCE times the reciprocal of area you get a pressure which is dimensionally equivalent to an energy density.

Here is a very natural unit of force where I write G for the newton gravity constant.

c^4/(8 pi G)

Let me flip that factor (a naturally defined force) over to the lefthand side of the einsteinequation that Garth wrote:

$$\frac{c^4}{8 \pi G}(R_{\mu \nu} - \frac{1}{2}g_{\mu \nu}R + \Lambda g_{\mu \nu}) = T_{\mu \nu}$$

Then you have PRESSURE OR ENERGY DENSITY on both sides of the equation.
the lefthand side is a force multiplied by a curvature---which is the same as a force divided by an area---
and on the righthand side you explicitly have an energy density.

If I remember this force is about 5x1038 tons of force but it probably doesnt matter what amount it is.
what the einstein equation seems to say is this force is what mediates between curvature and energy density----or another way to say is that it tells you how much a certain energydensity will bend spacetime.

there is no physical content to what i'm saying. I'm just saying how I look at it in order to make it all easier to remember

Last edited: Dec 19, 2007
18. Dec 19, 2007

### marcus

That is so right! Above all a quantum gravity theory should explain what cc is. And a couple of other things that Christine said----like explain what inertia is. (how matter and geometry interact)

Garth I would be grateful if you would look at Aldrovandi and Pereira most recent paper, where they CALCULATE a value for the CC energy density. It sounds too good to be true but they are respectable Brazilian physicists, no crackpots. If you could just explain why they are wrong it would be a comfort. I can't stop suspecting that they might actually be on the right track.

19. Dec 19, 2007

### stevebd1

Thanks for your replies Garth and Marcus. I understand that the cc is a description of the curvature of space-time and my understanding of EFE is a work in progress. What I have issue with is the fact that that the equation for lambda can be reduced to the following-

$$\Lambda$$ = $$8\pi\rho_{vac}$$

which provides 1.685x10^-25 kg/m^3

I prefer the answer in m^-2 when G and c^2 are introduced to the equation which I can relate to as a description of the curvature of space-time, especially when put into the 1 over the length square equation.

$$\Lambda$$ = $$\frac{8\pi G}{c^2} \rho_{vac}$$ = 1.252x10^-52 m^-2 (which is then put into the following equation)-

$$\Lambda$$ = $$\frac{1}{L^2}$$ therefore L = $$\sqrt{\frac{1}{\Lambda}}$$ = 8.937x10^-25 metres = 9.44675x10^9 Lys (9.45 billion Lys)

I understand the G and c are introduced to convert the density into geometric units but I fail to see how expressing the cc in kg/m^3 (as in the first equation) is of any benefit and how it might be recognised as units of curvature. Is it simply a different way of expressing the cc which might be applied in a different equation?

Thanks again
Steve

Last edited: Dec 19, 2007
20. Dec 19, 2007

### marcus

you never know when an alternative expression for something will come in handy. people in the mathematical sciences have an ingrained tendency to keep alternative equivalent expressions for stuff in their mind. it is a superstition. you never know when you will see something---one of its faces will pop up and grin at you in some unexpected circumstance-----then if you recognize it you make a discovery, perhaps.

so you should memorize what your girlfriend looked like at her 21st birthday party but also what she looked like at age 11 dressed as a tiger for halloween. Just to be safe.
She might, you know, dress up as a tiger some day and you would want to recognize who it was.

BTW compliments on your posts. you are doing a good job of hammering out all this stuff.

Some years back I calculated that L myself and it came out around 10 billion LY, just like with you. But I don't remember exactly. Anyway the squareroot of 1/Lambda is an important length, which one should know (at least approximately).

Maybe YOU should look at Aldrovandi and Pereira's paper. you wont understand but that's all right I don't either You could get a taste. they set up what they call DE SITTER general RELATIVITY instead of Lorentz general relativity, and then they say where the cc comes from, and (by working out a simplified case) why it is the value that it is. this is a tough paper to read, because they are not exactly crackpots!
they are basically good people (incidentally Brazilians) so it puts a strain on the mind to see what they do. In case you want to risk taking a look, I will give the link

Go here and click on PDF to download:
http://arxiv.org/abs/0711.2274

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