Decoupling from the Higgs field?

[MigL]

If a fermion was accelerated to energy levels above electroweak symmetry breaking, would it stop interacting with the Higgs field and lose that property we call mass. Being a massless particle means it would then move at c , and possibly ( depending on other properties ) become its own antiparticle ?

[Strange]

If a fermion was accelerated to energy levels above electroweak symmetry breaking,...

 

The trouble is, energy is frame (observer) dependent. So while we might think the particle has a huge amount of energy, it its own frame of reference it has zero kinetic energy. (This is similar to wondering if an object would become a black hole if it is accelerated to high enough speed.)

[MigL]

The two are quite different.

While an object accelerated to high enough speeds cannot become a BH as its mass is 'intrinsic' ( I don't know if that's the right word ) to its frame, its interaction with the Higgs field is 'extrinsic', i.e. it crosses frames.

At electroweak energy levels ( with respect to the Higgs field frame ) would the interaction not cease and the fermion lose the property of 'rest' mass, along with a jump to lightspeed ?

 

I'm not asserting this Strange, rather I'm asking if its a possibility or if my thinking is at fault.

[Strange]

While an object accelerated to high enough speeds cannot become a BH as its mass is 'intrinsic' ( I don't know if that's the right word ) to its frame, its interaction with the Higgs field is 'extrinsic', i.e. it crosses frames.

 

Accelerating a particle only increases its energy in your frame of reference, not in its own. If you think you can define energy or velocity relative to the Higgs field, then that would imply that it is some sort of absolute frame of reference.

 

 

I'm not asserting this Strange, rather I'm asking if its a possibility or if my thinking is at fault.

 

I understand that. But I think you are wrong with regard this specific point (regarding the energy imparted to the particle). The way fermions interact with the Higgs mechanism is way over my head!

Edited September 10, 2015 by Strange

[swansont]

 

At electroweak energy levels ( with respect to the Higgs field frame ) would the interaction not cease and the fermion lose the property of 'rest' mass, along with a jump to lightspeed ?

 

I don't think there is a "Higgs field frame"; it's Lorentz invariant. AFAIK, the interaction is always going to be there.

[timo]

Strange gave/indicated a very good reason why a particle would not stop interacting with the Higgs field (the paradox that every massive particle would always interact with the Higgs field and also not do so). On the other hand, there is no reason why it actually should stop interacting with the Higgs field (and even less reasons why it would suddenly lose other properties and become its own anti-particle). So the answer to the question looks like a solid "no" to me.

Edited September 10, 2015 by timo

[MigL]

Thanks for clarifying guys.

 

But I have to ask. When we talk about electroweak unification at about 125 GeV, where the weak and EM forces unify and the intermediate gauge bosons ( W and Z ) become massless and indistinguishable from photons, do they not need to decouple from the Higgs field to become massless ? If that is the case, doesn't the W then become its own antiparticle, hence a majorana ?

 

What 125 GeV energy are we talking about, then ?

Edited September 10, 2015 by MigL

[Strange]

The W boson would still have charge and could still have an anti-particle (I assume).

[imatfaal]

Thanks for clarifying guys.

 

But I have to ask. When we talk about electroweak unification at about 125 GeV, where the weak and EM forces unify and the intermediate gauge bosons ( W and Z ) become massless and indistinguishable from photons, do they not need to decouple from the Higgs field to become massless ? If that is the case, doesn't the W then become its own antiparticle, hence a majorana ?

 

What 125 GeV energy are we talking about, then ?

 

Surely it is the Z boson that would be a candidate for being its own anti-particle - Majorana fermions are (speculated to be) neutral spin half particles; the W of either charge does not fit this model

[timo]

I have to ask. When we talk about electroweak unification at about 125 GeV, where the weak and EM forces unify and the intermediate gauge bosons ( W and Z ) become massless and indistinguishable from photons, do they not need to decouple from the Higgs field to become massless ? If that is the case, doesn't the W then become its own antiparticle, hence a majorana ?

 

What 125 GeV energy are we talking about, then ?

I find it hard to formulate a proper answer here. There are a lot of misconceptions in what you say. And I am afraid there is a reason why teaching the basics of the Standard Model takes a full semester when being taught to advanced physics students. Well, and on top of that I am not qualified to teach particle physics at university. But I don't expect anyone here to write a full explanation so I can as well give it a shot ...

 

I'll start with pointing out a few flaws:

- W and Z do not become indistinguishable from photons at any point.

- The only time I heard about Majorana particles is in the context of Majorana fermions (and only then in exotic particle physics, not in the Standard Model). The W is a boson, so it definitely is not a Majorana fermion.

- No one except you talked about 125 GeV (could not resist this comment ). Presumably, that is the mass of the Higgs Boson.

 

Statements such as "at the xxx GeV scale" refer to the center-of-mass energy of particle interactions. They are often depicted via Feynman diagrams, in which a number of particles go in, a possibly different set of particles goes out, and an in-principle arbitrary number of intermediate lines connect the in-coming and out-going particles. These intermediate lines are, in the language of Fenyman diagrams, associated to intermediate particles (or "virtual particles", because they are just lines in a diagram and do not appear physically). Now, if a particle has a mass higher than the total energy of the process it cannot be an outer line (that part is simple: you simply don't have enough energy to create the amount of mass). Also, diagrams that contain such a particle as an intermediate line are unlikely to contribute much to the outcome of the process (that part is less simple to understand and I won't even try explaining it - the slang to look for is "off-shellness").

 

So in conclusion from the previous paragraph: If your process has significantly less than 125 GeV of energy you can ignore the existence of Higgs Bosons (or to put it the other way round: if you want Higgs Bosons to have any effect at all you have to burn a lot of money on a huge device that puts lots of energy into tiny particles ). Sadly, the part explained so far was the easy one. For the second part of the story I will have to refer to analogies and imprecise language, and also have to speculate a bit since I don't want to dig up my old physics books. So don't take every statement at face value but rather consider it a nice bedtime story I just made up:

 

Particles are excitations from so-called fields. And a weird feature of these excitations is that they do not appear at any value but only as multiples of a minimum excitation (a single particle). The default for almost all fields is a value of zero. So if there is way too little energy to excite the field the effect of interaction with it is almost zero. The Higgs field is an exception, as its default value is not zero but a constant number. So if you have way too little energy to excite the Higgs field the effect is some constant one. This effect is the same as a mass of the particle. The excitation of the Higgs field is the Higgs Boson. If you have loads of energy to excite the Higgs fields around its default value (create lots of Higgs bosons) then at some point you should not consider your particles interacting with a constant background Higgs field but with a highly dynamic one, instead. And you can forget about the default value. Then, the term formerly associated to the mass of your particles does not exist in your picture, anymore. In that sense, the particles lose their mass. They do not lose their interactions with the Higgs field.

 

 

[MigL]

Thanks again for taking the time to explain, timo ( and others ).

I appreciate it.

Edited September 11, 2015 by MigL