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

The quantum based calculation of the cosmological constant

  1. Jan 30, 2012 #1
    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 non-zero 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 "wall-of-maths-crits-you-for-INFINITE-damage". (That one's for Drakkith. He's going to glare at the screen, annoyed when he reads this post, I can feel it.)
     
  2. jcsd
  3. Jan 30, 2012 #2

    bapowell

    User Avatar
    Science Advisor

    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 stress-energy 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 stress-energy 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].
     
  4. Jan 30, 2012 #3
    That was way more straightforward than I expected. Thanks =D.
     
  5. Jan 31, 2012 #4
    It seems to me that that begs the question. Is it zero because its value is defined so that its minimum will be zero?

    The 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 Emax is the maximum energy integrated over.

    But what is Emax? 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 10-120[/sub] this value, implying Emax ~ 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 excessively-high cosmological constant, by a factor of 1058 (3 TeV vs. 0.01 eV).

    So we don't know what's producing this almost-but-not-quite cancellation.
     
  6. Jan 31, 2012 #5

    bapowell

    User Avatar
    Science Advisor

    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.
     
  7. Jan 31, 2012 #6
    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, m2*P*g, m4*P*g, m4*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 Higgs-field 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 V2 < 0, and it yields
    [itex]V = V_0 - \frac14\frac{V_2^2}{V_4}[/itex]
     
  8. Jan 31, 2012 #7
    Does this mean that the dark energy is producing more than it consumes, with the excess causing expansion, as where the other fields have a balanced diet? It seems very odd that the Higgs field should average out at non-zero. Is this predicted to be positive or negative? Actually, is this non-zero value precisely what gives the other particles mass? In other words if the VEV were 0 the field wouldn't actually do its job?
     
  9. Jan 31, 2012 #8

    bapowell

    User Avatar
    Science Advisor

    I don't know what you mean by this. Fields don't consume anything.
    The nonzero VEV is what gives other particles mass; the Higgs taking on a nonzero VEV is precisely how the Higgs breaks the symmetry of the theory.
     
  10. Jan 31, 2012 #9

    bapowell

    User Avatar
    Science Advisor

    Which is equal to what? I think you'll find that it vanishes in the Standard Model and MSSM.
     
  11. Jan 31, 2012 #10
    Fields create and annihilate virtual particles. They give out and take back. Produce and consume.

    Your second statememnt seems to be confirming what I said in the second quote.
     
  12. Jan 31, 2012 #11

    bapowell

    User Avatar
    Science Advisor

    Ah I see. The field driving the accelerated expansion is doing so on account of its nonzero vacuum energy, not through particle excitations. Its unique behavior stems from the bizarre nature of vacuum energy -- it has a constant density. As a comoving volume filled with dark energy expands, energy is created.
     
  13. Feb 1, 2012 #12
    Energy density for the Higgs vacuum value:
    [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 V0 so that this quantity is zero?

    I mean by the energy density the quantity that contributes to the energy-momentum tensor Tij in Einstein's equation

    Gij = K*Tij

    where Gij is the Einstein curvature tensor.

    For a metric gtt = 1, gtx = 0, gxy = - dxy (Kronecker delta), and in a frame that moves with the matter-energy, the energy-momentum tensor is
    Ttt = (energy density), Ttx = 0, Txy = (pressure)*dxy
    or in covariant notation,
    [itex]T^{ij} = (\rho + P)u^iu^j - (u_ku^k)Pg^{ij}[/itex]
     
  14. Feb 1, 2012 #13

    bapowell

    User Avatar
    Science Advisor

    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 tree-level 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.
     
  15. Feb 1, 2012 #14
    The Higgs vacuum value of the cosmological constant, without cancellations, is around 1054 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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The quantum based calculation of the cosmological constant
  1. Cosmological constant (Replies: 6)

  2. Cosmological constant (Replies: 3)

  3. Cosmological constant (Replies: 5)

Loading...