# Cosmological Constant in mks units

Hey folks,

I'm writing a pop-sci article for a journal and one of the referees wants me to give the cosmological const in mks units instead of GeV. I just wanted to check with anyone in the know if the following looks correct.

First, the accepted value of lambda (order of mag):
$$\Lambda=10^{-47}(GeV)^4$$

I want to convert this to $J/m^3$

The following website http://hep.uchicago.edu/~dkrop/Natural_Units_Conversions.html [Broken] gives the conversion from meters to GeV

$$1m=5.07\times 10^{15}GeV$$

so I can convert lambda to GeV/m^3

$$\Lambda=\frac{10^{-47}GeV^4}{(5\times 10^{-15}GeV)^3}=8\times 10^{-2}GeV/m^3$$

Finally the website gives conversion from GeV to Joules

$$1 J=7 \times 10^{9}GeV$$

So,

$$\Lambda=\frac{8\times 10^{-2}GeV/m^3}{1 \times 10^{9}GeV}\approx 10^{-10} J/m^3$$

Has anyone performed this conversion before?
Does this figure of $\Lambda$ look correct???

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marcus
Gold Member
Dearly Missed
back in 2004 I calculated it to be between 0.60 and 0.65 joules per cubic kilometer

I think that is right and comes out ROUGHLY the same as what you found

I don't start from the GeV figure you do which is only a crude order of magnitude. I start with the current best estimate for the Hubble parameter----at that time it was 71 km/s per Mpc

from that you can get a precise conversion to SI units and calculate the critical density
as I recall the critical density works out to be around 0.85 joules per cubic kilometer

the dark energy density is estimated to be about 75 percent of rho-crit
you just get the best current estimate you can, say it is 75 percent, and you take that percentage of 0.85

I think your orderofmagnitude figure is 0.1 joules per cubic kilometer and that seems pretty sloppy to me. You can easily do better with widely available information. You should be getting 0.6 (if you try for two-place accuracy and then round down to one significant figure)

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Ok, thanks Marcus. I'll see if I can't do a little better. My gross rounding is a consequence of me working with QFT and higher dimensional models to try and obtain a prediction with the right order of magnitude instead of the factor 10^119 or so error one usually sees when summing the zero point energy. I'm not really looking for accuracy (something I should have mentioned so apologies). :)

Anyhow, your similar figure has given me additional confidence in my calculations so thanks for the input.

marcus
Gold Member
Dearly Missed
Robousy you are right, physicists like order-of-magnitude estimates. The peer-reviewer most likely wants an estimate like 10-9 or 10-10.

What I calculated was right inbetween I wouldn't know which way to round, up or down.
I guess I'd prefer to call 0.6 x 10-9 equal to 10-9

but supposing someone actually wants to know the energydensity version of the cosmological constant to one significant figure intead of just as order of magnitude?

as an educational exercise, how do we do it?

One way is rather fun. Just use the google calculator to find rho-crit and then that 75 percent of it.

To get rho-crit just put this into the google window and press search:
(3/8pi)*((c*71 km/s/Mpc)^2)/G

When you put that in the google window you get 8.5 x 10^-10 pascals
same as 0.85 x 10^-9 joules per cubic meter

You can see what I'm saying goes into the google window is just the usual formula for rho-crit
using 71 km/s/Mpc for the Hubble parameter

the rest is just the speed of light c and Newton constant G, which the calculator knows so it supplies the values and we don't worry about them.

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Robousy you are right, physicists like order-of-magnitude estimates. The peer-reviewer most likely wants an estimate like 10-9 or 10-10.

What I calculated was right inbetween I wouldn't know which way to round, up or down.
I guess I'd prefer to call 0.6 x 10-9 equal to 10-9

...

the rest is just the speed of light c and Newton constant G, which the calculator knows so it supplies the values and we don't worry about them.

Hey, thats pretty cool. I just stick in the formula and it gave me the answer! I didn't realize google could do that. Very impressed. Thanks!

I've just been playing around and it recognises the unit 'yards per month' ha ha. I'm going to try furlongs per week next.

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...update after moments of playing...google calculator understands furlongs, cubits and get this..spans/fortnight. (A span of a mans hand and a fortnight is 2 weeks in the uk)

marcus
Gold Member
Dearly Missed
you are finding out stuff about google calculator that I didn't know

you may already have discovered that it also can handle

earth mass
solar mass
electron mass
electron charge

whereever it sees those terms it puts in the handbook MKS values in kilograms or coulombs

I think I mentioned, where you type k it puts in the value of Boltzmann constant, which is a great help