# Units of cosmological constant

jimjohnson
The cosmological constant (Ʌ ) is equated to dark energy and has units of energy/volume. Why have I read that Ʌ equals the Hubble constant squared (H2) which has units of 1/sec2? There must be an obvious explanation.

## Answers and Replies

Gold Member
Dearly Missed
For most of its history since around 1920, Ʌ has been a curvature constant (reciprocal area) naturally appearing on lefthand side of Einstein GR equation along with other curvature terms.

Area is length squared so you could say the units were 1/length2

But if, as is often done in GR, you set c = 1 and measure length and time in same units, the unit for curvature, including the cosmo constant, would of course be 1/time2. Just as you have seen.

As far as we know "dark energy" is fictional--an artificial convention. I think it became a fashionable idea after 1998 when it was found that instead of being zero (as most people had assumed) the Ʌ constant in GR equation was a small positive curvature.

The GR equation, as simplified by Friedmann for cosmology, relates curvature on LHS to energymomentum density on RHS by way of conversion factor 8πG/(3c2)
or if c = 1 then the factor is simply 8πG/3
So you can always drag a curvature on the LHS over to the RHS, and multiply by 3/(8πG)
and treat it as an energy density. But is this a valid thing to do?
It is questionable whether dragging Ʌ across to RHS is physically appropriate. There may not be anything we would normally think of as an energy that corresponds to the vacuum curvature constant Ʌ.

It excites the public, always interested to hear about energy especially if mysterious e.g. "dark" and seemingly available in large amounts!
So this stimulates discussion and gets people fascinated and mystified. But so far that seems a bit artificial: all the evidence suggests that it is simply a vacuum curvature constant, the same everywhere, as per Einstein. No evidence of an associated energy field.

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Gold Member
Dearly Missed
That said, I should acknowledge that the corresponding energy density is a very convenient handle on Lambda! Maybe it's NOT fictional! Maybe we'll discover there IS this constant vacuum energy pervading all space for all time that is around 0.5 or 0.6 nanojoules per cubic meter.

And even if there isn't, and it is just a fictional energy density, it is such a handy easy way to remember the constant.

I'll check the calculation but what I recall is that the latest (Planck satellite report) figures would put the "dark energy density" at 0.54 nanojoule per cubic meter. So something around 0.5 or 0.6.

One nice thing about that is how easy it is to picture. It's the same as 0.54 joule per cubic kilometer. And a joule is the amount of energy released when you raise a one kilogram physics textbook 10 centimeters off your desk and drop it with a thud. So a joule per cubic kilometer is a very easy energy density to imagine and remember.

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Gold Member
Dearly Missed
...Why have I read that Ʌ equals the Hubble constant squared (H2) which has units of 1/sec2? ...

This is the really interesting part of your question. I think you are comfortable with the idea that the Hubble rate is a reciprocal time.
And in fact 1/H is called "Hubble time".
According to Planck report figures the current Hubble time is 14.4 billion years.

This translates into a percentage growth rate, for distances, of 1/144 percent per million years.

This percentage distance growth rate is slowly declining and is expected to level out at 1/173 percent per million years, when the U has expanded so much that matter density is negligible.

then the only constituent of curvature will be the vacuum curvature. So the Hubble constant squared WILL be equal to Lambda. It isn't yet (so what you said is not quite true yet) but it will be.

I want to write the current and future Hubble constant squared in a way that the google calculator likes so we can do a simple calculation.
H2 is now 1/(14.4 billion years)^2
In future when matter density is negligible H2 will be 1/(17.3 billion years)^2

The corresponding energy density is what you get when you open a google window and paste this into the box:

3c^2/(8 pi G)/(17.3 billion years)^2

I'm curious to know what you get when you open google and put that in. You may have to press return. It should give "5.4 x 10-10 pascal"
But pascal=N/m2 is also J/m3
so as an energy density that is 0.54 nJ/m3

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jimjohnson
Marcus, thanks for the explanation relating a curvature constant to energy density.
A number I have referenced for the value of dark energy is 5.81 x 10-30 gm/cm3 (assuming it exists) which is very close to what you quoted (5.2 joule/km3 to your .54 joule/km3).
Is there an online reference that shows the Friedmann and Einstein equations with units (setting c = 1 is always confuseing)?
Is the calculator you referenced the Ned Wright calculator?

Gold Member
Dearly Missed
Marcus, thanks for the explanation relating a curvature constant to energy density.
A number I have referenced for the value of dark energy is 5.81 x 10-30 gm/cm3 (assuming it exists) which is very close to what you quoted (5.2 joule/km3 to your .54 joule/km3).
Is there an online reference that shows the Friedmann and Einstein equations with units (setting c = 1 is always confuseing)?
Is the calculator you referenced the Ned Wright calculator?

The calculator I was using... well it's really two calculators. I like Jorrie's better than Ned Wright's:
http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7/LightCone.html
It is called Lightcone 7. Instead of being just a one-shot calculator, it makes tables where you specify the range and step-size. It also plots graphs of various curves that you specify. Fun to use.
You can always use it as a one-shot. it uses S = z+1, the factor by which the wavelength, or the distance, is enlarged. z=1 means the lengths get doubled, i.e. S=2. So you can set S and ask for a table with just one step, and it serves as a one-shot. Hovering over the blue info dots gives you information about the quantities---so fairly self-explanatory. I really like it.
=====================

BUT THE CALCULATOR I WAS JUST NOW USING in this thread was the GOOGLE calculator.
A lot of people do not know about this. Besides its "search" function you can put stuff in the google box that you want to calculate. And it knows the mass of the electron, the mass of the sun, the radius of the earth. So it is quicker to use for a lot of things because you don't have to look up constants or type in constants, that it knows. You can say "hbar" and "pi" and "G" and "c"

For exponents you use "^" and for multipying you use "*" if necessary.

You just go to google as if you wanted to search, and you type or paste in what you want calculated.

Like suppose you want to convert 5.2 kilograms to energy in joules.

"5.2 kg*c^2" goes into the google box (without the quote-marks) press "return" and what comes out is
4.67 x 1017 joules

Or if "5.81 kg*c^2" goes into the google box what comes out is
5.22 x 1017 joules
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So when I suggested that you go to google and paste 3c^2/(8 pi G)/(17.3 billion years)^2
into the window and press "return"
what I was suggesting is that you take the longterm value of H to be 1/17.3 billion years,
and square that to get H2 = 1/(17.3 billion years)^2
and multiply that value of Hubble rate squared, by this combination of physical constants:
3c2/(8πG)

That or its reciprocal (8πG/(3c2) is what appears in the Friedman equation relating the square of hubble rate on LHS to energy density terms on the RHS.

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jimjohnson
Marcus, thanks again. I just accessed Wikipedia on the Einstein's equation, quoting "The mathematics of general relativity are very complex. In Newton's theories of motions, an object's mass and length remain constant as it changes speed, and the rate of passage of time also remains unchanged. As a result, many problems in Newtonian mechanics can be solved with algebra alone. In relativity, on the other hand, mass, length, and the passage of time all change as an object's speed approaches the speed of light. The additional variables greatly complicate calculations of an object's motion. As a result, relativity requires the use of vectors, tensors, pseudotensors, curvilinear coordinates and many other complicated mathematical concepts.
Thus, my question on scale is beyond my math ability.

Gold Member
Dearly Missed
Thus, my question on scale is beyond my math ability.
Oh no! You sound discouraged. I've failed to understand what you were curious about, and proceeded in the wrong way.

Cosmology is much easier than full GR. It assumes the universe is uniform---same average density everywhere---and uses a simplified form of the GR equation called "Friedman equation".

Chances are you can handle the Friedman equation, and a lot of the math involved in cosmology.
So my advice would be to not let that passage you quoted about GR discourage you and just proceed in small steps to gradually understand basic cosmology.

But it's your choice, you may have other things that are higher priority right now.

jimjohnson
I've failed to understand what you were curious about, and proceeded in the wrong way.
No, you answered my conern on the units which I needed. I am researching dimensionless constants. Thanks

Devils
As far as we know "dark energy" is fictional--an artificial convention. I think it became a fashionable idea after 1998 when it was found that instead of being zero (as most people had assumed) the Ʌ constant in GR equation was a small positive curvature.

So is your explanation the accepted one? Didn't you say dark energy was not the correct explanation then go on to show how dark energy solves the problem.

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