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Higgs vacuum energy-momentum

  1. Feb 1, 2010 #1
    Okay, these questions aren't completely about the Higgs, but it is a good starting point / explicit example.

    After spontaneous symmetry breaking occurs, such that the vacuum state itself no longer has the symmetry of the Lagrangian, will there always be something equivalent to a Higgs (a massive goldstone boson)?

    Because the vacuum state now has a non-zero expectation of these massive bosons, wouldn't that mean there is only one inertial frame in which the Higgs expectation energy-momentum four-vector has zero in the components for momentum? (ie. there exists a "higgs background rest frame"?) I assume the answer is no, but I'd like some explanation of why.

    Which leads to the third question, is it theoretically possible to have Lorentz symmetry be a symmetry of the Lagrangian, but have some expectation value of the vacuum violate this? Or can only gauge symmetries be affected in this manner, and Lorentz symmetry is not a gauge symmetry?
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  3. Feb 2, 2010 #2

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    Isn't the vev of the Higgs a Lorentz scalar? You seem to think it's the 0-th component of a four-vector.
  4. Feb 2, 2010 #3
    Excitations of the Higgs field can be interpretted as particles with energy and momentum. So I thought a non-zero expectation of this field meant there was a condensate of these particles in the ground state. Therefore I thought the vev was an energy and needed to transform like and energy ... and since I often see the vev listed as 246 GeV, that seemed to fit with my (mis)understanding. If this is somehow wrong, then can someone please give some explanation to help me understand?

    Anyway, if the answer to the second question is: it's a Lorentz scalar, I guess that is done. But I'd still enjoy insight on the other questions:

    What is necessary to get a massive vs massless goldstone boson?

    Is it theoretically possible to have Lorentz symmetry be a symmetry of the Lagrangian, but have some expectation value of the vacuum violate this? Or can only gauge symmetries be affected in this manner, and Lorentz symmetry is not a gauge symmetry?
  5. Feb 2, 2010 #4
    I wouldn't call the Higgs a massive goldstone boson. When the VEV of the Higgs breaks a global symmetry, one or more goldstone bosons appear. They correspond do the different deegrees of freedom of the Higgs, one for each of the directions in the symmetry group that was broken. If the symmetry was not exact to begin with, these would have a small mass instead.

    In the case of the EW gauge theory, these massless goldstone bosons combine with the EW gauge bosons to give them mass. So they are not seen elsewhere in the theory after the definition of the massive W,W,Z particles. Since one direction of the SU(2)xU(1) is not broken, one component of the Higgs scalar remains, and one gauge boson (photon) doesn't obtain a mass.

    The vacuum only has a non-zero expectation for the remaining components of the Higgs field (in EW, only one), not the components that appear as massless goldstone bosons/extra W,W,Z components. So the theory is "transformed" in a theory with only a U(1) symmetry, and a modified Higgs with a nonzero VEV, but which does not break the U(1) symmetry.

    There won't be such a rest frame, because of the fact that the Higgs is a Lorentz scalar.

    Any scalar field will contribute to the energy-momentum tensor of gravity, like this:

    but it doesn't define a preferred rest frame since it is a Lorentz scalar. No matter which Lorentz intertial frame you transform to, its scalar value will be the same.

    As far as I know, it is correct to say that the Lorentz symmetry actually is broken by the vacuum state in reality, otherwise spacetime would be completely homogeneous and isotropic, and not exist.

  6. Feb 2, 2010 #5
    I'm sorry, but I'm still confused.

    [tex]T^{\mu\nu} = \frac{\hbar^2}{m} (g^{\mu \alpha} g^{\nu \beta} + g^{\mu \beta} g^{\nu \alpha} - g^{\mu\nu} g^{\alpha \beta}) \partial_{\alpha}\bar\phi \partial_{\beta}\phi - g^{\mu\nu} m c^2 \bar\phi \phi[/tex]
    Which I can separate into the parts proportional to the metric (C) and parts (seemingly?) not (D):
    [tex]T^{\mu\nu} = - g^{\mu\nu} C + D^{\mu\nu}[/tex]
    [tex]C = \frac{\hbar^2}{m} \partial^{\beta}\bar\phi \partial_{\beta}\phi + m c^2 \bar\phi \phi[/tex]
    [tex]D^{\mu\nu} = \frac{\hbar^2}{m} (\partial^{\mu}\bar\phi \partial^{\nu}\phi + \partial^{\nu}\bar\phi \partial^{\mu}\phi)[/tex]

    C is a Lorentz scalar (scalar density?), and the metric is the only tensor whose components are the same in all inertial frames. So the C portion looks fine. But what about the D portion?

    It looks to me like the stress-energy tensor would still define a rest frame: the frame in which the momentum density and energy flux is zero.

    Is there someway to rewrite the D part to be proportional to the metric?

    Spacetime is not all vacuum. I'm talking just about the vacuum, which as far as I knew was homogenous and isotropic.

    Basically, the last question there boils down to this: the standard model is an example of a model in which the vacuum state breaks a gauge symmetry of the Lagrangian. While not the case in the standard model, are there known theoretical mechanisms whereby a theory has a vacuum state that breaks non-gauge symmetries of the theory (for example a parity invariant theory having a non-parity invariant vacuum, or a lorentz invariant theory not having a lorentz invariant vacuum, etc.)?
  7. Feb 2, 2010 #6
    The metric tensor components actually change when you change the reference frame. It transforms like any other 2-tensor.

    Btw, the D-term is not related/proportional to the metric tensor. For the vacuum state, [tex]D^{\mu\nu}=0[/tex], since [tex]\phi[/tex] is constant. So in this case, the total contribution to the energy-momentum tensor is just a number times the metric, so it appears just like a cosmological constant, which doesn't define a preferred frame.

    Your criticism of my answer to the question regarding breakage of Lorentz symmetry by the vacuum is valid :-) You can have spontaneous breaking of Lorent symmetry in a theory containing a 2-form field, although it can probably be done also with a vector field. It works just like in the Higgs mechanism, i.e. a potential that causes the field to have a nonzero VEV. The components of the field that correspond to unbroken symmetries appear as goldstone bosons just as in standard theory of symmetry breaking. Here is an article describing it for the antisymmetric field:


  8. Feb 3, 2010 #7
    Thanks for the article link. I'll check that out.

    Yes, in a general coordinate change. My point was merely that the vacuum stress energy tensor components need to look the same in all inertial coordinate systems, and the only tensors with this property are those proportional to the metric tensor.

    Why isn't there any time dependence? Shouldn't there be an exp(iEt/h) factor?

    Since the Klein-Gordon equation is
    [tex](\frac {m^2 c^2}{\hbar^2} - \partial^{\mu} \partial_{\mu}) \phi = 0[/tex]
    phi = a constant>0 is not a solution to this equation if m>0.
    Last edited: Feb 3, 2010
  9. Feb 3, 2010 #8
    Ok, I agree.

    You are forgetting the special potential of the Higgs here. The ordinary Klein-Gordon as you wrote it describes a scalar field in a potential of the form [tex]\phi^2[/tex]. In that case the solution [tex]\phi=0[/tex] is the only constant solution. Notice that this is at the bottom of the potential well.

    The Higgs potential uses a "Mexican Hat" potential of the form [tex]V\sim \phi^4-\phi^2[/tex]. For a general potential, the equation of motion is of the form

    \square \phi + V'(\phi) = 0

    In this case we have [tex]V'(\phi) = 0[/tex] at a nontrivial value of [tex]\phi[/tex], which becomes its VEV. This is a valid constant solution since both terms disappear at this value. So no time of space dependence.

  10. Feb 3, 2010 #9
    Ah, gotcha. That makes sense now. Thank you very much.
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