Temperature of electroweak phase transition

Question

How does one estimate the temperature at which electroweak phase transition (EWPT) occurred? Somewhere I have read it is around 100GeV but the reason was not explained.

Answer 1

We calculate the free energy (density) for the Higgs field $\phi$ at finite temperature. In the Standard Model, this looks like

$\mathcal{F}_{SM}(\phi,T) = -\frac{\pi^2}{90}g_* T^4+V_{SM}(\phi, T) \ ,$

where $g_*$ is the number of degrees of freedom in the SM ($g_*=106.75$).

The potential has the form

$V_{SM}(\phi,T) = D(T^2-T_0^2)\phi^2 - ET\phi^3+\frac{\lambda_T}{4}\phi^4\ ,$

with $D,E,T_0^2,\lambda_T$ some factors depending on particle masses, coupling constants, the Higgs v.e.v. and temperature.

At the phase transition (PT), there are two degenerate minima of the potential. One sits at $\phi=0$, where we are in the symmetric phase, the other is at $\phi=\phi_0$, where we are in the broken phase. If my quick calculation is correct, this leads to a critical temperature $T_c\approx 163 GeV$.

Note that in this case, the order parameter of the PT $\phi_c/T_c$ is very small. This means for one thing, that the PT is only very weakly first order and else, that perturbation theory is no longer reliable and we need to do non-perturbative calculations.

(Though the procedure is standard, I took this paper from Carena, Megevand, Quirós and Wagner as reference, just because it was the closest at hand, not because I particularly like it, which I don't btw.)

Answer 2

Let's define $T_{EW}$ the temperature where the coefficient $m^2_H(T)$ of the operator $H^2$ in the SM lagrangian vanishes: $$ m_H^2(T=T_{EW})=0\,. $$ For $T>T_{EW}$ the Higgs vev is vanishing, the EW symmetry in unbroken, and the elementary particles are massless. For $T<T_{EW}$ the the vev is non-vanishing, $v_T\propto -m_H^2/\lambda\neq 0$, the EW is broken spontaneously, and the various elementary particles acquire masses $m_i$.

Let's now determine $T_{EW}$ quantitatively. Since $m_i\simeq 0$ for $T\simeq T_{EW}$ is OK to make an expansion in small $m/T$. In this regime the 1-loop corrections to the Higgs potential from the thermal propagators are, expanded at leading order in $m/T$, given by \begin{equation} m^2(\tilde{T}_{EW})=m^2_{T=0}+\tilde{T}_{EW}^2\left[\frac{y_t^2}{4}+\frac{\lambda}{2}+\frac{g^{\prime\,2}}{16}+\frac{3g_2^2}{16}\right] \end{equation} where $m^2_{T=0}=m_H(T=0)=-\lambda v_{T=0}^2$, with $v_{T=0}=246$ GeV and $2\lambda v_{T=0}=m_h^2=(125\mathrm{GeV})^2$. The negative mass-squared receive positive thermal mass contributions form the loops of the particles the Higgs couples to. The leading contribution to $m_H^2(T)$ is, non-sorprisingly, coming from the largest couploing, top quark yukawa $y_t=\sqrt{2}m_t/v_{T=0}$. Neglecting the gauge couplings and the higg-self coupling, and solving $m^2(\tilde{T}_{EW})=0$ for $T_{EW}$ we get \begin{equation} \tilde{T}^2_{EW}\simeq m_h^2 v^2/m_t^2\simeq (178\mathrm{GeV})^2\,. \end{equation}

Answer 3

higgs field causing electroweak transition. before higgs field w,z boson photon all are massless. but higgs field gives masses to them and symmetry spontaneously break. because if particles having masses then su(2)u(1) symmetry break. all gauge symmetry break if we give mass to the particles.so transition must have to have below 125.6 gev that is the mass of higgs boson.

peoples are thinking about there may be another background field which causes strong and electroweak seperation.so that field can break the underlying symmetry.

In the standard model, the Higgs field is an SU(2) doublet, a complex scalar with four real components (or equivalently with two complex components). Its (weak hypercharge) U(1) charge is 1. That means that it transforms as a spinor under SU(2). Under U(1) rotations, it is multiplied by a phase, which thus mixes the real and imaginary parts of the complex spinor into each other—so this is not the same as two complex spinors mixing under U(1) (which would have eight real components between them), but instead is the spinor representation of the group U(2).

The Higgs field, through the interactions specified (summarized, represented, or even simulated) by its potential, induces spontaneous breaking of three out of the four generators ("directions") of the gauge group SU(2) × U(1): three out of its four components would ordinarily amount to Goldstone bosons, if they were not coupled to gauge fields.

However, after symmetry breaking, these three of the four degrees of freedom in the Higgs field mix with the three W and Z bosons (W+, W− and Z), and are only observable as spin components of these weak bosons, which are now massive; while the one remaining degree of freedom becomes the Higgs boson—a new scalar particle.

The photon as the part that remains massless

Answer 4

I don't think that an unambiguous justification can be given because the dynamic of the electroweak symmetry breaking (EWSB) is still unknown. We don't have a well established theory describing how the Higgs scalar potential evolves with the temperature. When people talk about the scale of the EWSB, they usually refer to two possible things:

1. before EWSB, the weak bosons are massless. After EWSB, they get a mass (91 GeV for $Z^0$ and 80 GeV for $W^\pm$). The scale is therefore of the order of the mass of the weak bosons, roughly 100 GeV.
2. before EWSB, the Higgs vacuum expectation value (v.e.v.) is 0, the field is symmetric. After EWSB, the v.e.v. is about 246 GeV. So again, the v.e.v. value is representative of the scale of the EWSB, still of the order of 100 GeV.