 Quote by twofish-quant
In the section 2) where he derives the quantities from GR, he doesn't include the cosmological constant in any obvious way.
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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.
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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.
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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##.
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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.
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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.
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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.
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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.