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How to derive dark energy density from Jorrie's Hubble radius limit

  1. May 3, 2013 #1


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    How to derive "dark energy" density from Jorrie's Hubble radius limit

    (c^4/(8 pi G))*3*(17.3e9 light years)^(-2)

    Paste that in the google window, and see what you get.

    When I paste that into google I get 0.5393 nanopascal, which is equivalent to
    0.5393 nanojoule per cubic meter.

    If the cosmological curvature constant actually arose from a real "dark energy" that would be the density of that energy. I don't know of any observational evidence that it represents any sort of real energy. Observation is entirely consistent with it simply being a very small constant vacuum curvature. But real or fictitious, that's a way to get the energy density.

    A curvature is the reciprocal of an area. small curvature ↔ large area.
    The most readily available handle on the cosmo curvature constant is this distance:
    17.3 billion lightyears the longterm limiting value of the Hubble radius.

    When you open Jorrie's calculator you immediately see that the Hubble radius is tending to a longterm limit of 17.3 billion lightyears.

    The conventional Lambda in the Einstein gr equation is actually equal to 3*(17.3e9 light years)^(-2)
    That is what you'd put in the Einstein equation, in some form or another, if you were using the equation in its traditional form. Aside from that factor of 3, the area is just the square of 17.3 billion lightyears.

    To continue deriving the possibly fictitious "energy" density---the factor (c^4/(8 pi G)) is just a force. People familiar with "Planck units" will recognize it (up to the 8 pi factor) as the unit force belonging to that system of "natural units". If you multiply a force by a reciprocal area you get a pressure, or equivalently an energy density---IOW a nanojoules per cubic meter quantity, or nanopascals.

    Here's the calculator if you want to see what I mean about limiting value of Hubble radius:
    Last edited: May 3, 2013
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  3. May 3, 2013 #2
    This is good info I'm glad to get confirmation on my estimate of the energy density of the vacuum energy. Your estimate is more accurate than mine, but at least mine was in the same order of magnitude. The difference is just in the rounding off

    Like you I am getting addicted to Jorrie's calculator, its flexibility is amazing particularly some of the new changes
  4. May 3, 2013 #3


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    Delighted to hear this! It really is addictive :biggrin:

    I'm glad our estimates agree. Also I'm curious. Would you try pasting in what I wrote and see if the google calculator works for you? It has been a bit cranky lately, for me. Sometimes it just draws a blank and does not go immediately into calculator mode. I have to prod it by first doing some trivial calculation and then pasting in the real thing I want to evaluate.
    Does it work for you in the sense that if you paste in
    (c^4/(8 pi G))*3*(17.3e9 light years)^(-2)
    then you do actually get back something like 0.54 nanopascal?

    I just went back and tried: this seems to work,
    you paste it in the google window and then press "return"

    alternatively, you paste it in and then press "space" and then "equals sign",
    and then it knows to go into calculator mode and do the arithmetic for you.

    But does this work for you? If you wouldn't mind I'd like the reassurance that what I'm talking about works for others as well.
    Last edited: May 3, 2013
  5. May 3, 2013 #4
    yeah it works I had no problem with my google on it
  6. May 3, 2013 #5


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    Thanks for checking that it works. I've tried streamlining a bit and this seems to work:

    (c^4/(8 pi G))*3(17.3e9 light years)^-2

    This gets rid of some unnecessary symbols and also is more conceptual in the following sense:

    what is to the left of the * is a natural unit of force that comes up in GR, made only of universal physics constants.

    what is on the right of the * is a curvature and in case anyone is interested the METRIC unit for curvature is inverse square meter, namely m^-2
    But google calculator hasn't learned metric curvature units yet, so you have to prompt it to give you the answer in those terms, if you want it expressed as a curvature in metric.

    You have to paste in
    3(17.3e9 light years)^-2 in m^-2
    where you tack on "in m^-2" so it knows you want the answer expressed in those units. Then it evaluates to the authentic cosmological constant as it appears in Einstein GR equation, namely the constant curvature term required to be there by general covariance.
    Last edited: May 3, 2013
  7. May 3, 2013 #6
    yeah the second method is clearer coincidentally my original estimate was 6.0*10-10.

    that was the value I had on the expansion and redshift article I wrote for the forum usage. So from my end it was good to see a confirmation of that value
  8. May 3, 2013 #7


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    Yes! I got that same value, namely 0.60 nanopascal, equivalently 0.60 nanojoule per cubic meter, a couple of years ago when I was using the WMAP model parameters.

    Back then we were using something like 71, 0.73, 0.27
    and the Planck mission report suggests 68, 0.69, 0.31

    I'm rounding the numbers, I could have said 67.8, 0.693, 0.307. You know what I mean.
    the key thing is that .69 is a lot lower than .73 and so whereas before we calculated 0.60 nanopascal, now we calculate 0.53 nanopascal.

    But as far as we know there is no observational evidence that an energy exists, the vacuum curvature may have a quantum geometry explanation, so far, objectively speaking, it is just a curvature constant. So the 0.53 nanopascal is a fictitious energy density which people find is a convenient handle on the curvature quantity.

    Another handle I've seen people use is using inverse seconds squared and then it comes out by surprising coincidence to very close to 10-35 s-2

    A second can serve as a length, think "light second". So second squared can be thought of as a unit of area. And thus s-2 is itself a legitimate unit of curvature!

    When I paste this in
    3(17.3e9 years)^-2 in s^-2
    I get 1.00656 × 10-35 s-2

    This is the most convincing evidence I know of that God has 10 fingers. The cosmological curvature constant Λ is after all an important constant of nature. General covariance, the underlying space-time symmetry requires exactly TWO gravitational constants, G and Λ.
    These are the two constants allowed by the symmetry of the theory. Newton only knew about G, but he didn't understand general covariance so he missed the second gravitational constant.
    By around 1917 folks realized there was this other constant, which might or might not be exactly zero but in any case belonged in the equation and was to be determined by observation. Finally, after any twists and turns, as we all know a value was determined by observation.

    And it turns out to be a power of ten?, but no, that's meaningless because the second is not a natural unit. If it means anything it's that God has 10 fingers and wears a Rolex with a second hand.
    Last edited: May 3, 2013
  9. May 3, 2013 #8


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    here's something nice.

    What we have is that the cosmological constant Λ = 1.00656e-35 s^-2
    This is by far the cleanest way to write it in metric as a constant

    Now the longterm Hubble radius, essentially by definition, is (Λ/3)-.5

    So we can see if the google calculator will get us from the correct value of Λ to the R Hubble radius. Paste in this:
    (1.00656e-35 s^-2/3)^-.5

    Wow. It says exactly 1.73 × 1010 years
    Last edited: May 3, 2013
  10. May 3, 2013 #9
    that is a handy method to explain the future Hubble limit. I saw that on Jorrie's calculator as well
  11. May 3, 2013 #10


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    When you say "handy" method do you mean convenient, or do you mean the Creator having ten fingers? :wink:

    It came as something of a shock when I realized that according to latest (Planck mission) data the cosmological curvature constant was a power of ten. Why should the universe work in decimal numbers?

    Λ = 1.00656 x 10-35 seconds-2

    But perhaps in honor of the Great Secret Agent we should round that to

    Λ = 1.007 x 10-35 seconds-2
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