Value of lambda

1. Feb 2, 2007

stevebd1

I'm currently looking into the values for the 'critical density' and 'cosmological constant', I managed to calculate a figure for the critical density which was close to the generally accepted figure, with lambda I came up with an astronomically small number which I later realised after searching the web was close to the accepted value (my figure being 1.67 E-55 cm^-2). The most common way I've seen Lambda expressed is 10^-29 g/cm^3 which are the units used when calculating omega lambda. I'd be interested to know how the first figure in cm^-2 converts into g/cm^3. Any feedback would be welcome.

cheers
Steve

On a side note, any info regarding changing the value into planck units would be appreciated also.

Last edited: Feb 2, 2007
2. Feb 2, 2007

hellfire

It is expressed in geometric units that are often used in general relativity. The conversion factor for mass is G/c^2 (to convert to geometric units). You can read about this unit system here:
http://en.wikipedia.org/wiki/Geometrized_unit_system
The conversion factors are in the first table.

3. Feb 2, 2007

stevebd1

Thanks for that. I'm currently putting lambda into the vacuum energy density equation-

omega_lambda = lambda C^2/3H^2

and I'm getting 0.0009. I'd be grateful if some light could be shed on why the figure appears to be out by about 3 decimal places (or is the whole process suppose to be done backwards in order to calculate the actual density of the universe?).

Regarding the matter density equation for omega_baryonic and omega_dark matter, in order to achieve a total end figure of hopefully approx. 0.266, it appears the density figure needs to be lower than the estimated density of the universe (due to the fact that the other figures in the equation are constants). Is this relevent because of some kind of matter to volume ratio? Any feedback would be appreciated.

regards
Steve

Last edited: Feb 2, 2007
4. Feb 3, 2007

stevebd1

I've approached the Lambda CDM model as something of a layman with some knowledge of maths and would appreciate if someone could let me know if the following is more or less correct-

(the reason the figure mentioned above was out by three decimal places is because the density I used was in g/cm^3, I simply converted this to kg/m^3). I think I understand that the Lambda CDM model is based on a few things-

Based on info from the WMAP, which has placed the density of the universe within 2% of the critical density, the universe is more or less flat and that omega should equal 1. This is also supported by observations of type 1a supernovae and large red shift surveys.

The density of the observed 'luminous' matter in the universe is estimated by looking at a volume of space, working out the mass occupied by stars, nebula and free hydrogen and helium and dividing this by the volume, giving a density in the region of 0.5 x 10^-27 kg/m^2 (approx. 4.5% of the critical density)

This gave rise to the question 'if the universe is flat, where's the other mass that makes up the critical density?'. A figure of 95% for unseen matter was initially established.

While looking for this dark matter, another observation was made, that the universe was not only expanding but accelerating. This led to dark matter being split into 2 catagories- dark matter and dark energy. Using gravitational lensing and studying the speed of galaxies within clusters, a figure of 22% of the critical density was established for dark matter, approx. 0.202 x 10^-26 kg/m^3.

This left an assumed figure of 73% of the critical density for dark energy density, approx 0.644 x 10^-26 kg/m^3.

These are the density values used in the Lambda CDM model for omega_baryonic, omega_dark matter and lambda in order that omega_total more or less equals 1.

Steve

Last edited: Feb 4, 2007
5. Feb 3, 2007

hellfire

What you write seams correct to me.

By the way, the value of Omega_Lambda you want to obtain from that formula can be calculated also converting to geometric units (G = c = 1) which makes things easier. You only have to compute the value of the Hubble parameter in geometric units (convert H to 1/s and multiply with 1/c).

6. Feb 3, 2007

stevebd1

Thanks again. One last question, the answer I get now for Lambda using a density in kg/m^3 is 1.192 x 10^-52 m^-2. My question is, does this need to be converted into cm^-2 before dividing it by the G/c^2 in order to get the correct SI unit figure?

7. Feb 3, 2007

hellfire

If you take the value of the table in the wikipedia article then no, because there G/c2 is given in m/kg.

8. Feb 3, 2007

stevebd1

My knowledge of gaussian units is pretty limited at the most. The reason I ask is that the gaussian units seem to be predominantly shown in grams and centimetres while the SI units are shown in kilograms and metres. I thought maybe there was some relevance here.

9. Feb 3, 2007

marcus

I haven't time right now to read the thread but I gather you want a value for Lambda

I will try to give you one both as an inverse area
and as an energy density in ordinary metric units.

The CRITICAL DENSITY rho crit, calculated from current best value of Hubble parameter (71) is 0.85 joules per cubic kilometer

I calculated this several years back to be 0.83 and used that value for a long time, but the recommended G changed and so on, so when I recalculated a few weeks ago it came out 0.85, actually 0.850... with some spurious accuracy.

So you can simply take the current best value for Omega_Lambda which is 0.73 and multiply by 0.85 and you get 0.65,
so that means that as an energy density what you want is
0.65 joules per cubic kilometer.

Or 0.65 nanojoules per cubic meter.

Somewhere I have a "geometric" version-----Lambda as an inverse area.
I will look for it.

10. Feb 3, 2007

marcus

I think for a lot of calculations, the handiest thing is to have H as the inverse of the Hubble time, expressed either as
4.3 x 1017 seconds

or if you want to postpone rounding off, 4.346 x 1017 seconds

or 13.8 billion years.

and likewise Hubble distance as 1.303 x 1026 meters

or 13.8 billion light years

Then when you have some formula like yours you can read off the answer directly.

Because c2/H2 is obviously (13.8 billion lightyears)2

Your formula says Lambda = 3 Omega_Lambda H2/c2

= 3(0.73)/(13.8 billion LY)2

2.16 divided by the square of the Hubble distance.

I don't know whether you are interested. Let me know if you see I've made a mistake. the upshot is that the geometrical form of Lambda, a curvature, is the reciprocal of a huge area which is very roughly (order magnitude) equal to a square 10 billion LY on a side.

Last edited: Feb 3, 2007
11. Feb 3, 2007

stevebd1

Thanks for the reply. I've calculated lambda to be 1.192 x 10^-52 m^-2 (which seems to be in the right ball park) and as most articles I've seen express the figure in kg/m^3, I was looking to see how to convert lambda into kg/m^2. As hellfire pointed out, the figure is geometric and needs to be converted into SI units (kg/m^3), dividing the number by G/c^2 (as in the tables on wiki). Because all the gaussian figures are shown in grams and centimeters, I thought maybe the initial lambda figure of 1.192 x 10^-52 m^-2 might need converting into cm^-2 before dividing it by G/c^2 in order to obtain a correct value for the kg/m^3. Currently, when I do the conversion, the figure comes out as 1.76 x 10^-28 kg/m^3 which seems a little high (though converting the geometric number in cm^-2 would make the figure even higher).

Incidently, the critical density I calculated converts to 0.8226 j/km^3.

regards
Steve

Last edited: Feb 3, 2007
12. Feb 3, 2007

marcus

I am glad someone is interested in knowing the values of basic stuff about the universe like rho_Lambda etc.

Also it is nice that you got 0.8226 joules per cubic kilometer. As I said, for years I was using the value 0.83.
(although most recently got a slightly different 0.85)

One thing I would like if we could share and agree on would be a certain basic LENGTH call it L, defined by

Lambda = 1/L2

it would be handy to know this length and convenient for back-of-envelope calculations. It is just Lambda but packaged differently. Lambda is the reciprocal of a certain area which is the area of a square of side L.

I think it is order 10 billion LY. Do you know it in metric?

I guess it is 1.303 x 1026 meters divided by 1.4697 or whatever is the sqrt of 2.16, that would make it 0.8865 x 1026 meters.

I had something else going on here earlier and made a calculator error but that seems like a good figure for L now, and if you squared it you would get an area which would be like the reciprocal of what you calculated!

Last edited: Feb 3, 2007
13. Feb 4, 2007

stevebd1

The figure I get is-

0.916 x 10^26 metres, approx 9.68 billion Lys.

Last edited: Feb 4, 2007
14. Feb 4, 2007

marcus

we are in remarkably close agreement given that we are calculating it from different (approximate) data and by different methods.

that is very reassuring. thanks.

I would like to think of this length L as possibly a fundamental constant (like the Planck length except at the opposite end of the spectrum) and one which might play several other roles that we may not have figured out yet

but for now it is just the length L that we get the cosmo constant from by
Lambda = 1/L2

15. Feb 4, 2007

marcus

I see that hellfire already gave a wikipedia link for "geometrized units".

The NIST.gov site has a section on universal physical constants
http://www.physics.nist.gov/cgi-bin/cuu/Category?view=html&Universal.x=55&Universal.y=7
this has metric conversion factors for Planck mass, length, time, temperature.

and Wikipedia has an article specifically about Planck units.
http://en.wikipedia.org/wiki/Planck_units
It cites the NIST as source.

Last edited: Feb 4, 2007
16. Feb 5, 2007

stevebd1

Great, thanks. I'm still at something of a loss of how to convert to planck units. I know the figure is in the region of between 10^-120 and 3.2 x 10^-122. If I take a planck length- (1.616 x 10^-35 m), square this (2.612 x 10^-70 m^2) then multiply it by Lambda (1.2517 x 10^-52 m^-2) I get something very close- 3.269 x 10^-122. Otherwise, I don't see how its possible to get the large decimal place count.

regards
Steve

17. Feb 5, 2007

marcus

I dont have time to check in detail now, but it looks right.

Several years ago when I first calculated Lambda in Planck units I got something around E-123

I was calculating rho_Lambda, the ENERGY DENSITY that corresponds to Lambda and there is this rogue factor of 3 lurking around.

Then I started hearing people say E-120 and i think that may be simply that they dont want to be too accurate. It is more likely really E-122 or so but they are embarrrassed to sound like they are overstating the accuracy and they say 120 when they mean 122.

Another thing that confuses the situation is that some cosmologists use Planck-like units where you set 8 pi G = 1 instead of simply G = 1.

and then there is a factor of sqrt 8 pi G that can cause confusion.

So all I can say is trust your calculation and dont worry if they say it is "around E-120" because you are probably righter than they are.