The quantum based calculation of the cosmological constantby salvestrom Tags: based, calculation, constant, cosmological, quantum 

#1
Jan3012, 12:21 AM

P: 226

I nearly referred to it as infamous in the title. Unfair?
My question is this: why did they add up the combined energy of all the fields when only the Higgs field would be active in void space (I use this term to refer to the spaces between filaments, does it have its own name or is it usually just called intergalactic space?)? An extension of this question is why do they consider each force to have its own energy field. Isn't it simpler to assume they use a single source of energy, which reacts depending on how it is interacted with? I see a similarity here with String Theory duality. Should this be in the Quantum Mechanics forum? lol. If dark energy is consider to also come from the vaccum, wouldn't that tie in with the Higgs Field having a nonzero expectation value? I understand the purpose of the Higgs Field. Breaking symmetry, imparting mass. Usually, descriptions of it talk in terms as if the Higgs field is still operating today, i.e. that we and our planet gain their mass as a sort of ongoing process. I tend to end up thinking about what all this "excess" energy is doing out in deep space with no matter to wrestle with. I appreciate that when confronted with something that seems obvious, ones first reaction should be to determine if one hasn't simply misunderstood something along the way. That's where you lot come in ;). Something beyond "no" and before "wallofmathscritsyouforINFINITEdamage". (That one's for Drakkith. He's going to glare at the screen, annoyed when he reads this post, I can feel it.) 



#2
Jan3012, 07:24 AM

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P: 1,565

It's important to make the distinction between the Higgs field, which has a nonvanishing vacuum expectation value, [itex]\langle \phi \rangle = v[/itex] but has vanishing vacuum energy, and the type of scalar field that would be a dark energy candidate. Dark energy candidates must have nonzero vacuum energy, and this energy must dominate the stressenergy tensor for accelerated expansion to take place. Since the Higgs has effectively zero vacuum energy, it does not contribute (as a zero mode) to the stressenergy tensor.
With regards to summing up all particle species in the computation of the cosmological constant, this is done because all fields have vacuum fluctuations. The only difference between the Higgs, and say the electron field, is that the Higgs fluctuates about [itex]\langle \phi \rangle = v[/itex] whereas the electron field fluctuates about [itex]\langle \psi \rangle= 0[/itex]. 



#3
Jan3012, 02:56 PM

P: 226





#4
Jan3112, 04:04 PM

P: 514

The quantum based calculation of the cosmological constantThe only way that we are going to get a solution to this problem is by getting a theory of quantum gravity. Let's see what the problem is. Take a bosonic field. Its vacuum energy is [itex]E = g\sum_{\mathbf{k}} \frac12\omega_{\mathbf{k}}[/itex] A fermionic field has the sign reversed. g = degeneracy, k is a mode momentum, w is the mode energy. Taking the continuum limit, this formula gives the energy density: [itex]\rho = g\int \frac{d^3 k}{(2\pi)^3} \frac12\omega_k[/itex] Thus, [itex]\rho \sim E_{max}^4[/itex], where E_{max} is the maximum energy integrated over. But what is E_{max}? For quantum gravity, the appropriate energy scale is the Planck mass, and it would make the Universe's size the Planck length. But we don't live in that kind of Universe, so something must be wrong. The observed value is about 10[sup]120[/sub] this value, implying E_{max} ~ 0.01 eV. Supersymmetry seems to help. In supersymmetry, # bosonic modes = # fermionic modes, so we should get cancellation. But supersymmetry is broken, and a likely energy scale is a few TeV. That still produces an excessivelyhigh cosmological constant, by a factor of 10^{58} (3 TeV vs. 0.01 eV). So we don't know what's producing this almostbutnotquite cancellation. 



#5
Jan3112, 04:50 PM

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P: 1,565

You misread my post. I point out that it does not contribute "as a zero mode" to the stress tensor. In other words, its potential energy is zero in the vacuum state. Of course, Higgs fluctuations about this vacuum state contribute to the cosmological constant.
The distinction I was making in reponse to the OP was between a field having a nonzero VEV and a nonzero vacuum energy. The Higgs has the former, a dark energy candidate has the latter. 



#6
Jan3112, 05:08 PM

P: 514

The Higgs field could also contribute to the vacuum energy density.
I'll work out the integral above. It's for the noninteracting case, and I don't know how to do it for interactions. For large k, I find [itex]\rho \sim \frac{g}{4(2\pi)^3}\left(k^4 + k^2m^2 + \frac18m^4  \frac12m^4\log\frac{2k}{m}\right)[/itex] So to cancel, the sums of all these must cancel: P*g, m^{2}*P*g, m^{4}*P*g, m^{4}*log(m)*P*g, where the fermionic parity P is +1 for bosons and 1 for fermions. The supersymmetric sum rule is that the sum of P*g cancels, but supersymmetry supplies no constraints on the other sums. Let's now calculate the Higgsfield vacuum energy from its potential. Let [itex]V(\phi) = V_0 + \frac12V_2\phi^2 + \frac14V_4\phi^4[/itex] The minimum is at [itex]\phi = \sqrt{ \frac{V_2}{V_4}}[/itex] implying V_{2} < 0, and it yields [itex]V = V_0  \frac14\frac{V_2^2}{V_4}[/itex] 



#7
Jan3112, 05:09 PM

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#8
Jan3112, 07:41 PM

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#9
Jan3112, 07:44 PM

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#10
Jan3112, 07:58 PM

P: 226

Your second statememnt seems to be confirming what I said in the second quote. 



#11
Jan3112, 09:06 PM

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#12
Feb112, 05:16 AM

P: 514

[itex]\rho = V(\sqrt{V_2/V_4}) = V_0  \frac14\frac{V_2^2}{V_4}[/itex] Is there any justification within the Standard Model or the MSSM for fixing V_{0} so that this quantity is zero? I mean by the energy density the quantity that contributes to the energymomentum tensor T^{ij} in Einstein's equation G_{ij} = K*T_{ij} where G_{ij} is the Einstein curvature tensor. For a metric g_{tt} = 1, g_{tx} = 0, g_{xy} =  d_{xy} (Kronecker delta), and in a frame that moves with the matterenergy, the energymomentum tensor is T^{tt} = (energy density), T^{tx} = 0, T^{xy} = (pressure)*d_{xy} or in covariant notation, [itex]T^{ij} = (\rho + P)u^iu^j  (u_ku^k)Pg^{ij}[/itex] 



#13
Feb112, 07:19 AM

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P: 1,565

OK, I see your point. I guess I'm arguing that the Higgs has zero effective vacuum energy because we see no evidence for it. I make the distinction that the treelevel vacuum energy density of a scalar field gravitates in a way that we understand (e.g. inflation), while vacuum fluctuations might not (e.g. cosmological constant problem). Of course, the other option is that all vacuum energy gravitates the same and that we don't understand the cancellations.




#14
Feb112, 04:29 PM

P: 514

The Higgs vacuum value of the cosmological constant, without cancellations, is around 10^{54} times the observed value. So as you say, we don't understand where this cancellation might come from.
A dynamic effect could do it, like Quintessence (physics)  Wikipedia Many quintessence models have tracker behavior, where it tracks the matter density. Quintessence ("fifth stuff") or aether was the traditional celestial element, in additional to the traditional four terrestrial elements. I once saw this identification of the terrestrial ones with the major constituents of the Universe: Earth = Baryonic matter Water = Dark matter Air = Neutrinos Fire = Photons 


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