In an environment with high enough energy, the electromagnetic and weak interactions merge into a single electroweak interaction with 4 massless bosons; W1, W2, W3, and B. Would the three W bosons behave similar to the W± and Z bosons, with the exception that they can propagate indefinitely as opposed to quickly decaying into some pair of particles, or do they interact in completely different patterns? Similarly, is the B boson a 1 to 1 match with the photon considering (I believe) both come from U(1)?
$\begingroup$
$\endgroup$
4
-
2$\begingroup$ What do you mean by "behave similarly"? You know what their couplings to fermions and each other are. Likewise, why should the B be in correspondence to the photon, but not $W_3$, or a linear combination thereof? $\endgroup$– Cosmas ZachosCommented Aug 22, 2021 at 21:28
-
$\begingroup$ ‘Behave similar’ as in they change the flavor of particles like the W± and have a chiral preference. I don't understand what ‘coupling to fermions and each other’ means, I can only assume you imply coupling constant, which I still lack the sufficient background in mathematics and/or physics to understand. Finally, I asked about the correspondence between photon and B because if I recall correctly, they both come from U(1), whereas the W3 is from SU(2). $\endgroup$– eaeaa1232Commented Aug 22, 2021 at 21:35
-
1$\begingroup$ $B$ is not the photon because electroweak symmetry $SU(2)_L\times U(1)_Y$ is broken down not to $U(1)_Y$ but to another (electromagnetic) $U(1)$ which is a linear combination of $U(1)_Y$ and $U(1)$ subgroup of $SU(2)_L$. One combination of $B$ and $W_3$ gives the photon, and another gives $Z^0$ $\endgroup$– KosmCommented Aug 22, 2021 at 21:48
-
$\begingroup$ Are the W± bosons also a combination of the electroweak bosons then? Also, how would interactions mediated by electroweak bosons differ from those mediated by weak and electromagnetic bosons (besides the obvious difference that weak bosons are massive)? $\endgroup$– eaeaa1232Commented Aug 22, 2021 at 22:03
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Well, yes, in the limit of $E\gg v$, the weak mixing angle cosine remains the same, $$ \cos \theta_W= \frac{g}{\sqrt{g^2+ g'^2}}= \frac{m_W}{m_Z}, $$ where the last term is never 0/0: as v never really vanished, only becoming tiny. All interaction terms to fermions, and among gauge bosons, remain the same, so all weak interactions keep at it, except unhampered by mass suppressions, which have ceased favoring or disfavoring channels.
So, typically, in the EW soupy plasma, up-type quarks, all of them!, decay to down-like ones emitting $W^+$s, and vice versa, in thermal equilibrium, etc..., with the same couplings g and g', etc...
It is strictly up to you if you choose to use the $B, W_3$ basis or the equivalent $\gamma, Z$ basis, but all interactions of both the "before" and "after" SSB-Higgs bases work validly and equivalently.
Since Electromagnetism, in this framework, is defined by e = g sin θ = g′ cos θ, I would stick to the "after" basis, to compare to the low-energy limits of the amplitudes. 
So, for instance, the photon still does not interact with neutrinos, but only with the charged leptons we know, etc..., and up-like quarks still have charge +2/3, etc...
The Z still couples to neutrinos with pure V-A couplings, while the charged lepton neutral currents are still accidentally pure axial!! 95% of the magic of the SM persists, and especially the chirality features of the theory.
The only victim of this limit is the Higgs mechanism, that is, the Ws and the Z lose their longitudinal components, and it is best to think of these components as Higgs degrees of freedom, and use the "equivalence theorem" (Cornwall-Tiktopoulos) to describe the relevant scattering channels; but this is so technical it outranges the scope of the cartoon...
answered Aug 22, 2021 at 22:10
$\begingroup$
$\endgroup$
Prof. Zachos has given a very nice answer to this question, but I would like to add some additional information. In the electroweak theory, we start with the bosons $W^a$, $a = 1,2,3$ associated with the $SU(2)$ gauge interaction and the boson $B$ associated with the $U(1)$ gauge interactions. After the Higgs boson has its influence, these particles mix. Three of these bosons become the massive bosons $W^\pm$ and $Z$, and there is one massless boson left over that represents the photon.
It seems logical that, when the energy in a reaction is much greater than the masses of the $W$ and $Z$ bosons, the symmetry would be restored and the result would be the same as if we had the original $SU(2)\times U(1)$ symmetry.
A nice process to try this out on is the reaction $e^+e^-\to \mu^+\mu^-$, where the Feynman diagrams are simple to evaluate. Actually, to show what happens, I did this calcuation explicitly in a set of lectures that I gave at one of the CERN summer schools. The notes are available as arXiv:1708.09043. The original process goes through two diagrams, one with a virtual photon, one with a virtual $Z$, which have interference. The high-energy limit is worked out in eqs. (78)-(80),and, indeed, the result is exactly what one might have expected from unbroken $SU(2)\times U(1)$. This is a very nice exercise for students learning the structure of the particle physics Standard Model. Figure 7 of the notes shows the data from the LEP 2 accelerator at CERN (I thank Prof. Michael Hildreth for collecting this data set), which shows the that high-energy limit is actually reached rather quickly at energies just above the mass of the $Z$ boson.
As Prof. Zachos wrote, there is still a question: If the $W$ and $Z$ had to eat Higgs bosons to become massive, what happens to these Higgs bosons when the symmetry is restored? The answer is given in Section 5 of these lecture notes.
