Hello, I have read that above a certain temperature, the weak bosons become massless and become indistinguishable to the photon. Is the idea simply that at high enough energies, the Higgs field can sit on top of the peak in the mexican hat potential? I.e. at high enough energies, it's vacuum expectation value can be 0 (even if that were the case, surely it would still "prefer" to sit in the brim)? Would the leptons then also become massless? That's my main confusion, but as a sub-question I was also wondering what exactly "electroweak" implies. Does it suggest that above certain energies the electromagnetic and weak interactions are literally the same? Does it just mean that photons can mediate radioactive decay (and indeed are indistinguishable from the weak bosons), or does the unified force behave differently to the sum of its parts? Thanks Edit: I am trying to think of how they would behave as a single force in a similar way to electromagnetism, but I am not really sure how exactly EM behaves either. Electricity and Magnetism are two aspects of this one force, but they affect charges completely differently and beyond getting the photon I am not sure what unifying them does for us. I have a reasonable physics background but nowhere near the amount of maths or physics to be able to rigorously understand the standard model.
That's exactly right. The Higgs potential contains temperature-dependent terms that "restore" the symmetry at temperatures above the critical temperature of the phase transition: see
No, not exactly right. The OP thinks that the Higgs field would be sitting at the top of the hat - that is in a meta stable state. That's not what happens. As your own figure shows, at a temperature above the critical temperature, there is no hat.
Also do we know what happens with the λ coupling? From what I saw today, the λ changes by energy (thus by temperature) and in fact it tends to be zero (or even negative) for high energies (~10^10 GeV) which instead of having that "image" you posted would imply the total break down of the system with no stable point. Although I am not totally sure about the whole diagram... what makes it change like that? It must not be μ otherwise the Higg's mass wouldn't be a constant (it is given by sqrt[-2 μ^{2} ] ) And λ drops by energy :/
http://www.google.de/imgres?espv=21...w=127&start=0&ndsp=24&ved=1t:429,r:0,s:0,i:81 I am sorry, it was the fastest search, of course if you search more about it you'll find better sites... As for the |H|^6 terms it says I am not quiet sure...I asked my professor today about it, and he told me it would destroy the renormalizabilty of the theory :/ Though I need to look deeper in it to understand it Also if the answer lies in Beyond the SM physics, I don't know it (Today we saw that thing in class), so let's not care about that limit (of zero or negative) of high energies, but in lesser ones that of course SM should "hold" where lambda still drops
I don't see the problem with a mildly negative lambda. At high temperatures, the finite temperature terms (that go as [itex]\sim T^2\phi^2[/itex]) tend to dominate unless lambda is strongly negative. I don't know how strongly negative off-hand, but I recall that the theory goes non-perturbative anyway before this happens.
Maybe you are right. But if the temperature dependence is on ΦΦ term, wouldn't that imply that the Higg's mass is not constant? as I stated above... But on what you said- even the slightiest negative lambda would make the system unstable (your Vhiggs curve drops because of the ΦΦ term and grows because of the ΦΦΦΦ term)... that's why I said I wouldn't like to speak about the limit of negative lambdas, but the region where it still fell and SM is a nice theory...:P *correct myself- it would be metastable since the lifetime to "break out" would be extremely large even compared to the Universe's lifetime*