So the top quark is pure mass, with no internal structure, as heavy as a gold nuclei?

[bananan]

The gold nucleus has an atomic mass of 197 with 79 protons and 118N and 79 electrons.

The top quark has about the same mass, but no internal structure. It's just a particle with pure mass.

Why would a structure of pure mass and nothing internal decay?

 

[RandallB]

Why sould a Top Quark decay and not be stable?
(question edited to make clear)

I don’t see how your question can be answered without clarifying many assumptions already included within the question.

You say that unlike the gold nucleus the Top Quark is “pure mass” – other than “mass” what is the gold nucleus made of then??

You question why it is not stable on its own;
Since it is a Quark what about it would you expect to be different as compared to any other quark.
Or do you have some information about a Up, Down, or any other quark able to remain stable on it own, without needing at least one or more quarks with it to establish Quark Confinement in the form of a Baryon or Meson?

Seems to me the Tau Quark is following the same pattern as any other Quark.

 

[Rade]

The top quark decays into less massive quarks (the Bottom quark and Strange quark) via the weak force, see this link:http://www.eurekalert.org/pub_releases/2005-02/uor-nmu021005.php
So the Top quark is not really a mass with "nothing internal", the Bottom quark and/or Strange quark (plus other particles) are internal and released during Top quark decay.

 

[bananan]

The top quark decays into less massive quarks (the Bottom quark and Strange quark) via the weak force, see this link:http://www.eurekalert.org/pub_releases/2005-02/uor-nmu021005.php
So the Top quark is not really a mass with "nothing internal", the Bottom quark and/or Strange quark (plus other particles) are internal and released during Top quark decay.

So the second and third generation fermions, particularly quarks, are just composite structures of first generation fermions as bound states?

Is there a reason you can't get a fourth generation quark as bound states?

Then all a preon theory has to do is describe the first generation fermions.

 

[Rade]

So the second and third generation fermions, particularly quarks, are just composite structures of first generation fermions as bound states? Is there a reason you can't get a fourth generation quark as bound states? Then all a preon theory has to do is describe the first generation fermions.

Well, see here for information about particle "generations" and lack of evidence for 4th generation fermions within the Standard Model:http://en.wikipedia.org/wiki/Generation_(particle_physics). So yes, 2nd and 3rd generation quarks can be seen as composite structures that ultimately decay into Up <---> Down first generation transformations. I do not know preon theory.

 

[vanesch]

The gold nucleus has an atomic mass of 197 with 79 protons and 118N and 79 electrons.
The top quark has about the same mass, but no internal structure. It's just a particle with pure mass.
Why would a structure of pure mass and nothing internal decay?

Worse! The top quark, according to the standard model, has no mass at all, but only mimicks it because of its interaction with the Higgs field.

Within the standard model (which is the domain I'm going to limit myself to), the top quark is a quantum excitation of a massless Dirac field, the "field of the top quark". However, this field also couples to the Higgs field, which has a non-zero vacuum expectation value, and this coupling term looks exactly as would a mass term in the Dirac equation. Turns out that the equivalent mass is of the order of 170 GeV.

http://en.wikipedia.org/wiki/Top_quark

As to why it can "decay" into lower-mass quarks, the only reason for that is that in the standard model, there are pathways for this reaction, and the reaction is energetically possible.
But it is not because something decays, that it is "made of" its decay products, as in chemistry. You essentially have to consider that the top quark disappears, and that a new set of particles appears.

This is better understandable by considering that the top quark (as is any particle in QFT) is just an "excitation of a field". Due to couplings between different fields, an "excitation in one field" can de-excite, and can excite other fields.

In atomic physics, you could have that an electron state "decays" and excites another electron state. So one state "disappears" and other states "appear". This doesn't mean that the first excited state is "made of" the other excited states. It is in a similar way that one has to see particle decay in QFT.

 

[Rade]

A recent summary of top quark physics:http://ej.iop.org/links/re0Y0CoX9/oJ2HOGR12xGiS5Bqav5vpA/jpconf5_18_003.pdf

 

[arivero]

Worse! The top quark, according to the standard model, has no mass at all, but only mimicks it because of its interaction with the Higgs field.

... on the other side, the nucleus gets its mass mostly from the gluons, not from the quarks.

 

[LURCH]

Worse! The top quark, according to the standard model, has no mass at all, but only mimicks it because of its interaction with the Higgs field.

Maybe I've been missunderstanding something for a while. I thought that, in the standard model, interaction with the Higgs field is what gives all things their mass, that thiis interaction is mass. Is that incorrect? (My knowledge of QM is quite limitted)

 

[selfAdjoint]

Maybe I've been missunderstanding something for a while. I thought that, in the standard model, interaction with the Higgs field is what gives all things their mass, that thiis interaction is mass. Is that incorrect? (My knowledge of QM is quite limitted)

The Higgs interaction, as described by Vanesch, gives the quarks their mass; they are intrinsically zero-mass in QCD without Higgs. But quark mass is only a minor part of the mass of ponderable matter as we experience it. The majority of that is the mass equivalent of the binding energy of the gluons which confine the up and down quarks withing the proton and neutron. It is only in high energy physics that quark mass becomes the significant factor that it is.

 

[ValenceE]

Hello to all,

Dear vanesch, you posted


Within the standard model (which is the domain I'm going to limit myself to), the top quark is a quantum excitation of a massless Dirac field, the "field of the top quark". However, this field also couples to the Higgs field, which has a non-zero vacuum expectation value, and this coupling term looks exactly as would a mass term in the Dirac equation.

Can you elaborate on how the quantum excitation takes place.

Regards,

VE

 

[ValenceE]

Edit : ... actually, not taking anything away from vanesh, I guess anyone who can elaborate on this, please do.

 

[vanesch]

Maybe I've been missunderstanding something for a while. I thought that, in the standard model, interaction with the Higgs field is what gives all things their mass, that thiis interaction is mass. Is that incorrect? (My knowledge of QM is quite limitted)

You have to know that a Dirac field has "mass" when, in the lagrangian, a certain term is present: it is the "m" in http://en.wikipedia.org/wiki/Dirac_field

However, that m would ruin a certain property of the Lagrangian in the standard model (chiral symmetry), so we simply cannot write it that way. The trick is that an interaction with the Higgs field, in the following sense:

g.psi-bar_L.psi_R.phi_0

looks like the mass term: m psi-bar_L.psi_R

because phi_0 has a non-zero vacuum value (g.phi_0 plays the role of m here). However, the form of this term is not a mass term, but an interaction (there are 3 field factors present). It is an interaction between the left-handed part, the right-handed part of a massless Dirac field, with the Higgs field, and if you consider the higgs field "at rest", then this looks exactly like the normal mass term of a Dirac field. Of course, at very high energies, the interaction will be different from a true mass term, but at low energies, it will behave similarly (because the Higgs field is essentially "at rest" and hence a constant value). This interaction is called the Yukawa coupling.

But as such, the problem of the genuine mass term introducing an unwanted coupling between L and R terms has been solved, while nevertheless accounting for the apparent appearance of a mass term.

So the trick of introducing 1) a non-zero higgs field 2) taking away the mass terms from the Dirac equation and 3) introducing an interaction between the higgs field and the dirac fields gives us a way to avoid having an explicit dirac mass term, while nevertheless micking one at low enough energies.

 

[vanesch]

Hello to all,
Can you elaborate on how the quantum excitation takes place.

A quantum excitation is simply an interaction which makes a quantum system evolve from its ground state into a higher-energy state. Strictly speaking, this can only come about when we are considering quantum systems as approximations to the true overall system. This is also the case here: we are taking in fact the *free* top quark field, while it is of course part of a bigger system with a lot of interactions. This approximation, of considering free systems, which then are in interaction with other systems (instead of viewing the entire overall system without approximation) then leads to changes of states in the free system due to interactions: the state of the free system evolves through the interactions. So it can go from its ground state to an exicted state, or back, for instance. The ground state of the top quark field is: "no top quarks are around" and its excited states correspond to different possible configurations, like "there's one top quark flying to the left, and three to the right" or so. The simplest excitations are those with one top quark with a certain momentum k. But these are states of the free top quark field. Interactions can bring hence the top quark field from the ground state "there's no top quark" into the (excited) state "there's a top quark flying to the left"... or vice versa.

These interactions are exactly what are described by Feynman diagrams (which allow you then to calculate the probability that such a thing will happen).