Spontaneous electroweak symmetry breaking (i.e. $SU(2)\times U(1)\to U(1)_{em}$ ) is at scale about 100 Gev. So, for Higgs mechanism, gauge bosons $Z$ & $W$ have masses about 100 GeV. But before this spontaneous symmetry breaking ( i.e. Energy > 100 GeV) the symmetry $SU(2)\times U(1)$ is not broken, and therefore gauge bosons are massless.

The same thing happens when we go around energy about $10^{16}$ GeV, where we have the Grand Unification between electroweak and strong interactions, in some bigger group ($SU(5)$, $SO(10)$ or others). So theoretically we should find gauge bosons $X$ and $Y$ with masses about $10^{16}$ GeV after GUT symmetry breaks into the Standard Model gauge group $SU(3)\times SU(2)\times U(1)$, and we should find massless X and Y bosons at bigger energies (where GUT isn't broken).

So this is what happened in the early universe: when temperature decreased, spontaneous symmetry breaking happened and firstly $X$ & $Y$ gauge bosons obtained mass and finally $Z$ & $W$ bosons obtained mass.

Now, I ask: have I understood this correctly? In other words, if we make experiments at energy above the electroweak scale (100 GeV) we are where $SU(2)\times U(1)$ isn't broken and then we should (experimentally) find $SU(2)$ and $U(1)$ massless gauge bosons, i.e. $W^1$, $W^2$, $W^3$ and $B$ with zero mass? But this is strange, because if I remember well in LHC we have just make experiments at energy about 1 TeV, but we haven't discovered any massless gauge bosons.


2 Answers 2


I think you have understood it almost well.

The masses do not change, they are what they are; at least at colliders. At high energy, it is true that the impact of masses and, more generally, of any soft term, becomes negligible. The theory for $E\gg v$ becomes very well described by a theory that respects the whole symmetry group.

Notice that to do so consistently in a theory of massive spin $-1$, you have to introduce the Higgs field as well at energies above the symmetry breaking scale. For the early universe, the story is slightly different because you are not in the Fock-like vacuum, and there are actual phase transitions (controlled by temperature and pressure) back to the symmetric phase where in fact the gauge bosons are massless (except perhaps for a thermal mass, not sure about it).


I'd like to edit a little further about the common misconception that above the symmetry breaking scale gauge bosons become massless. I am going to give you an explicit calculation for a simple toy mode: a $U(1)$ broken spontaneously by a charged Higgs field $\phi$ that picks vev $\langle\phi\rangle=v$. In this theory we also add two dirac fields $\psi$ and $\Psi$ with $m_\psi\ll m_\Psi$. In fact, I will take the limit $m_\psi\rightarrow 0$ in the following just for simplicity of the formulae. Let's imagine now to have a $\psi^{+}$ $\psi^-$ machine and increase the energy in the center of mass so that we can produce on-shell $\Psi^{+}$ $\Psi^{-}$ pairs via the exchange in s-channel of the massive gauge boson $A_\mu$. In the limit of $m_\psi\rightarrow 0$ the total cross-section for $\psi^-\psi^+\rightarrow \Psi^-\Psi^+$ is given (at tree-level) by $$ \sigma_{tot}(E)=\frac{16\alpha^2 \pi}{3(4E^2-M^2)^2}\sqrt{1-\frac{m_\Psi^2}{E^2}}\left(E^2+\frac{1}{2}m_\Psi^2\right) $$ where $M=gv$, the $A_\mu$-mass, is given in terms of the $U(1)$ charge $g$ of the Higgs field. In this formula $\alpha=q^2/(4\pi)$ where $\pm q$ are the charges of $\psi$ and $\Psi$. Let's increase the energy of the scattering $E$, well passed all mass scales in the problem, including $M$ $$ \sigma_{tot}(E\gg m_{i})=\frac{\pi\alpha^2}{3E^2}\left(1+\frac{M^2}{2E^2}+O(m_i^2/E^4)\right) $$ Now, the leading term in this formula is what you would get for a massless gauge boson, and as you can see it gets correction from the masses which are more irrelevant as $m_i/E$ is taken smaller and smaller by incrising the energy of the scattering. Now, this is a toy model but it shows the point: even for a realistic situation, say with a GUT group like $SU(5)$, if you scatter multiplets of $SU(5)$ at energy well above the unification scale, the masses of the gauge bosons will correct the result obtained by scattering massless gauge bosons only by $M/E$ to some power.

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    $\begingroup$ I am bit confused about your comment about the difference between early universe and LHC. Do you mean following - "We are stuck in a vacuum where the higgs field has a specific vacuum expectation value. We can't come out of this vacuum by feeding energy into electroweak fields only in a local region of spacetime. However if we do so globally at each point of space, then above the electroweak scale the gauge fields will be exactly massless". $\endgroup$
    – user10001
    Commented Jun 20, 2014 at 2:20
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    $\begingroup$ I agree with your way of putting it. Another way to say the same thing is looking at the Higgs potential in a thermal bath. Since the propagators are different than at zero temperature, the 1-loop potential gets a termal contribution. At high enough temperature it turn out that the symmetric phase can be restored in a thermal bath since the effective potential minimum is moved to zero by the thermal fluctuations $\endgroup$
    – TwoBs
    Commented Jun 20, 2014 at 7:21
  • $\begingroup$ @user10001 please see my edit in the beginning of my answer. There are two different "scale"s involved, the scale where broken weak SU(2) has to be used because the four fermion vertex diverges, called the electroweak scale. Before that electromagnetic interactions and weak interaction can be kept seperate phenomenologically, and the scale where the SU(2) symmetry was broken which is very high in energy as seen in the running coupling constants plot in my answer. Over that energy the gauge bosons are massles, actually there is one which at lowered energy break into the photon and the W+and W_ $\endgroup$
    – anna v
    Commented Jun 20, 2014 at 7:23
  • $\begingroup$ TwoBs I think, due to the locality property of quantum field theories, what I said can't be correct. Do you agree with what @anna v is saying? $\endgroup$
    – user10001
    Commented Jun 20, 2014 at 15:06
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    $\begingroup$ @user10001 Interactions and dymamics are local, but the thermal bath is spread all over with the same temperature at equilibrium. It is exactly the reasons why you can't change the vacuum in local experiments but you really need to do a global thing. And no, I do not agree at all with annav. Even at higher energies than GUT scale the bosons are not massless. It is just that the presence of a finite mass corrects the results for massless gauge boson only by $(m/E)^n$ with $n\geq1$ and $E\geq v_{GUT}$. I think it is in fact a common misconception. $\endgroup$
    – TwoBs
    Commented Jun 20, 2014 at 15:16

Technically, the gauge bosons are massless at high energy. But, technically, they're massless at low energy too.

All that really changes at low energy is that an effective theory with mass terms becomes accurate enough to be useful. At high energy, none of the interactions involved in that effective picture go away; they just become messier. It's common to say that particles "obtain" mass at low energy, but that seems very misleading. They don't really obtain anything they didn't already have.

Spontaneous symmetry breaking is sometimes analogized to a pencil that is initially balanced on its point and then falls in a random direction. That picture is wrong. At low energy the pencil does sit pointing in a particular direction, but at high energy it bounces randomly all over the place. The symmetry is unbroken at high energy because it spends as much time pointing in any direction as any other, not because it spends all of its time balanced on its point.

I'm pretty sure there is no energy regime in which all of the gauge bosons behave like the archetypal low-temperature massless particles: photons (or gravitons) traveling long distances at $c$ without interference. At high temperature, space is filled with a dense particle soup and the mean free path is short.


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