# Tag Info

3

The Higgs interaction is an interaction between quantum fields not between particles. To take electrons as an example: it is as a result of the interaction between the Higgs field and the electron field that the excitations of the electron field, i.e. electrons, have mass. So all electrons will have the same mass because they are all excitations of the same ...

4

You are slightly misinterpreting some words by Prof Matt Strassler. He says that the force mediated by the exchange of the Higgs bosons – the "Higgs force" – is attractive, much like gravity between two ordinary positive-mass objects. But that doesn't mean that "everything" in the presence of a Higgs field is attractive. Indeed, the Higgs potential ...

1

The hypercharge of a doublet cannot be "deduced". When one builds a gauge theory, the first step is to define the particle content of your theory and to postulate the representation of all particle multiplets. In particular, if the gauge group is abelian, then we have to assign numbers usually called charges. So, I reformulate your question: Why do we ...

3

The Higgs field is not giving mass to all other particles. There are particles that acquire mass differently - for example neutrinos. Or the yet undiscovered particles of dark matter probably don't get mass from the Higgs field. Also please notice the difference between the Higgs field and the Higgs boson. The Higgs field is giving mass to some particles, ...

6

Gluons are bosons, they have spin one and are mediators of the strong force. They have mass 0 and there is no weak interaction vertex with a gluon. The extract is talking of the Higgs field, not the Higgs boson. A field in physics is A field is a physical quantity that has a value for each point in space and time at that point. The Higgs field ...

1

Re: "what about its weakly hypercharged product?" Just listened to LS explaining this. (Using his terms) "when the Z boson picks up its quantum (quanta?) of zilch, it becomes a ziggs". The point he makes is that this is the famous "Brout-Englert-Higgs" mechanism. Re Higgs boson: As I understand it, Peter's name got attached (rather than Brout, Englert ...

2

If I'm understanding, you're asking: How does the Higgs field "give" particles mass? This is an explanation given by Henry of MinutePhysics (To be clear, we're talking about the Higgs field and NOT the Higgs Boson, which is merely an excitation leftover after the process we're about to explain. But I digress…back to mass!) To begin, we need to know what ...

8

Does a particle enter/interact with the Higgs Field when created, or at some other time? After reading your question a couple of times as well as your comments, it occurs to me that you're picturing something like this: a massless particle is created, interacts once with the Higgs field to acquire a permanent classical like mass which it then ...

7

I don't think you understand QFT. To be fair, I'm no expert myself, but I can certainly point out where you're going wrong here. A particle does not enter the Higgs field. However, the particle field that gets mass from the Higgs field does interact with the Higgs field. What this means is that in the Lagrangian of your model, there exists a term that will ...

2

It is not the Higgs boson, but rather the Higgs field, that gives mass to the elementary particles, Higgs boson included. In fact, even in Higgsless theories, e.g. such as a technicolor, the W and Z get mass but there is no Higgs boson (although there is a composite Higgs field made of techniquarks). Said this, the Higgs boson has a finite mass because its ...

2

No. The mass of most particles is not a problem. But for the force carriers, i.e. gauge bosons like the gluon or the W and Z bosons, it is a theorem that they must be (naively) massless. But we find that the W and Z bosons act as if they have a mass! The mechanism by which this mass arises is the Higgs mechanism, but we can have masses without it - just not ...

0

One argument could be the Yukawa coupling, which is responsible for the coupling to the fermions. In the Yukawa coupling term in the Lagrangian, $\mathcal{L}_{\text{Yukawa}}$ , there are no terms that contain a $\gamma^5$ matrix, defined as $$\gamma^5 := i\gamma^0 \gamma^1 \gamma^2 \gamma^3$$ This publication states how terms in the Lagrangian transform ...

2

It is not an assumption; both $0^+$ and $0^-$ were considered as possible Higgs states. The angular distribution of decay products (like in $h\to ZZ$, $h\to f\bar{f}$, $h\to \gamma\gamma$ or in Higgstrahlung) is dependent on the parity of the Higgs particle. Alternatively, you can measure the helicities of the outgoing photons (in the $h\to\gamma\gamma$ ...

1

The notion of relativistic mass is an archaic one that most physicists don't use anymore. A particle's mass is a Lorentz invariant, meaning a constant property of the particle, regardless of its energy. Instead, energy and momentum are seen to be non-Lorentz invariant, such that the invariant mass remains, well, invariant: $$E^2-p^2c^2=m^2c^4.$$ Also, the ...

4

No. What happens to an object at high speed has nothing to do with the Higgs field. There are two errors in the argument you've made to come to this conclusion: Energy is not directly equivalent to mass. An object has a certain amount of energy simply by virtue of its mass; or equivalently, an object's mass is the amount of energy it would have in the ...

2

The "Mexican Hat Potential" (although now more politically correctly called the "Champagne Bottle Potential" after the punt at the base) is the potential energy curve for the vacuum expectation value (VEV) of the Higgs field. Think of the blue dot as being "the vacuum", and the radial direction as turning up the strength of a background field that permeates ...

4

The Higgs is inferred from its decays - no Higgs bosons themselves are actually detected. As such, an experiment can count the number of Higgs boson decays to a particular final state, if it can measure the particles in that final state. For that reason, it is impossible to measure $\sigma$ or the branching ratios independently, you can only measure \$\sigma ...

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