Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In this video, Leonard Susskind does a good job trying to explain succinctly the Higgs field and exactly how it gives elementary fermions mass, except for one point he seems to skip a few things. The weak hypercharge is endearingly called 'zilch' (0:46:00). He goes on to explain how (e.g.) an accelerating electron (or other zilch-charged particle) can emit a $Z$ boson. He then postulates (0:48:16) the existence of a field (none other than a 'ziggs boson condensate'), which (in a similar manner to a uniform electric field giving electric dipoles a potential energy - and therefore an extra mass - that depends on their orientation) gives fermions mass by the Dirac mechanism (explained earlier in the video). He adds how the ('zilch'-less) $Z$ boson also interacts with the 'ziggs' field to intermittently acquire 'zilch' (and becoming itself a 'ziggs'), and lose 'zilch', thereby acquiring mass as well.

At that point, he jumps to the Higgs boson without apparently any connection to the mechanism just explained (he employs different names for the fields). He explains that the Higgs boson is a particular excitation mode of the Higgs field, but what is then the (general) 'ziggs' boson? Are Higgs and ziggs actually the same? The $Z$ boson has been known experimentally for decades, what about its weakly hypercharged product? Does the Higgs boson actually have nothing to do with the Higgs phenomenon, being merely a consequence of the theory that was begging to be discovered experimentally?

share|improve this question
    
en.wikipedia.org/wiki/… is the only related link I could find, seems that Higgs symmetry breaks in four ways, three of which are coupled with electroweak bosons W+, W- and Z, the fourth then imposing itself as the Higgs boson. –  HarT Oct 20 '13 at 19:21
add comment

1 Answer

He explains that the Higgs boson is a particular excitation mode of the Higgs field, but what is then the (general) 'ziggs' boson? Are Higgs and ziggs actually the same?

If there was only a massive $Z$ boson, and nothing else, then this 'ziggs' mechanism would be sufficient and the slightly more complicated 'Higgs' mechanism would not be required. I would look at the 'ziggs' as what the 'Higgs' would look like in this scenario.

The Z boson has been known experimentally for decades, what about its weakly hypercharged product? Does the Higgs boson actually have nothing to do with the Higgs phenomenon, being merely a consequence of the theory that was begging to be discovered experimentally?

I presume by 'hypercharged product' you mean the charged $W^+$ and $W^-$ bosons. These are not products of the $Z$ boson, instead they are all a part of the same family of particles. Well its actually a bit more complicated than that... let's start from the top. The Higgs mechanism 'starts' with four massless bosons ($B^0$,$W^0$,$W^1$,$W^2$), which after some interacting with the Higgs field, will produce three massive bosons ($Z$,$W^+$,$W^-$) and one massless boson (which is the photon, $\gamma$).

As there are four original bosons, without going into the heavier details, this means that the Higgs-field should be constructed from four components (or degrees of freedom). After interaction with the four components of the field, the four original bosons acquire a mass. However - as mentioned earlier - the photon is massless. In order to add this condition into our theory, we mix two of the original four bosons (the $B^0$ and $W^0$, which have no electric charge) to produce the the observed $Z$ and $\gamma$ bosons in such an exact way as to leave the photon massless. The repercussion of this is that we are left with one free component of our Higgs-field, remaining from our original four-component Higgs-field. This spare component, analogously to the 'ziggs' mechanism, manifests itself as the Higgs boson.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.