Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Berman's zero energy condition for universe

  1. Jul 31, 2012 #1
    I've been looking over the quantum gravity papers that were posted on another thread, and I've got a question about Berman's calculations

    I'm not seeing a cosmological constant in that paper, and the identities look to me as if they won't work if you add "dark energy". Is that the case?
  2. jcsd
  3. Jul 31, 2012 #2
    No, it works the same with a cosmological constant. Since a CC would also contribute a negative pressure, you can subtract this, leaving the equations the same. See Berman's comment about this in that paper:

  4. Aug 1, 2012 #3
    The problem is that the equation is for the matter density of the universe. If you model dark energy as a field then that field is going to have energy, and it's not going to zero as t -> infinity.

    On the other hand, one thing that I've seen with some quantum gravity papers is that they don't model the cosmological constant as a independent field, but as a characteristic of how gravity behaves. If Berman has that picture in mind, then I see how he gets this result.
  5. Aug 2, 2012 #4


    User Avatar
    Science Advisor
    Gold Member

    Berman has some rather fringish ideas, but, he is not demure.
  6. Aug 2, 2012 #5
    I don't think it matters. The equation of state for dark energy is [itex] w = -1[/itex]. So, [itex] \rho = -p[/itex]. Therefore, the right side of the equation vanishes, since [itex] \rho + p = 0[/itex] for a cosmological constant. So, it makes no contribution, and can be ignored.
    Last edited: Aug 2, 2012
  7. Aug 2, 2012 #6
    Still not sure.....

    In the section 2) where he derives the quantities from GR, he doesn't include the cosmological constant in any obvious way.

    Now in section 4), in equation 24) he derives the result that in a flat universe that the energy density of the universe with a cosmological constant is constant. That's fine, but in order to then get to result that the energy density of the universe is null he has to argue that the energy of "dark energy field" is zero rather than a positive constant.

    Alternatively what you could do define the "energy of the cosmological constant" as the zero level, and then argue that the energy of everything that isn't dark energy is zero relativity to that level.

    But I think that what might be happening is that QM and GR views are colliding. In QM, you have a vacuum and then you put fields into that vacuum. Dark energy is thought of as another field, and so it has an energy just like an electron field. In GR, everything is geometry, and asking about the energy of the dark energy field is like asking about the energy of the ether, it doesn't make any sense.

    I'm finding it amusing that one of the difficulties is that GR and QM seem to have different views of what is "nothing".
  8. Aug 3, 2012 #7
    In the first step, he contracts a pseudo-tensor with the stress-energy tensor. Since dark energy can be modeled as just a regular fluid with an equation of state w = -1, the stress-energy tensor should include any contribution from a cosmological constant in some region of space.

    Though, I'm not sure of this.

    Isn't it? Due to the negative pressure it contributes, it leaves the energy density the same (which is the key equation of his derivation). As I mentioned above, this is because, for dark energy, ##p = -\rho##.

    Does that really work, though? It seems a bit more like a 'slight-of-hand' rather than an actual derivation of a zero energy universe to just use different values for the same physics.

    Even though they differ conceptually, they are physically equivalent. For some cosmological constant ## \Lambda ##, there is an associated vacuum energy given by

    [tex] \rho_{vacuum} = \frac {\Lambda c^{2}} {8 \pi G} [/tex]

    So, I don't think that should pose too much of a problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook