# Weak nuclear force: Charge

1. Jul 18, 2011

### Schreiberdk

Hi PF

I was wondering today about the origin of the fundamental forces. Now gravity comes from mass (lets call this massive charge), electromagnetism comes from electrical charge and the strong force comes from color charge. These are all kinds of "charges", but what kind of "charge" gives origin to the weak nuclear force, in the same analogy as the rest of the forces gets (or is this really nonsense in terms of the weak force)?

Schreiber

2. Jul 18, 2011

### Staff: Mentor

The weak force has been combined with the Electromagnetic force to produce the Electroweak force. It is generally talked about as a separate force still however, as until you get to high energies it appears very different.

Edit: Also, be careful with what you mean by "origin" of the forces. All things with mass produce and are affected by gravity, but so are photons, which have no mass, only momentum.

3. Jul 19, 2011

### Schreiberdk

Yes, sorry about that. I have been looking around, and found the term "weak isospin". Could that be put on foot with mass, electrical and color charge?

4. Jul 19, 2011

### Staff: Mentor

I don't really know. I know the situation is a bit complicated, just from a little reading on wikipedia about it. But I don't know anything else beyond that really.

5. Jul 19, 2011

### kurros

In terms of particle physics, charges aren't really the "origin" of forces, instead they describe the ways in which various sorts of particles interact. If you are a particle, having electric charge just means that you can interact with photons, and having colour charge just means you can interact with gluons. So in this sense yes weak isospin is the same, if you have weak isospin then you can interact with W and Z bosons. Right handed fermions have a weak isospin of zero so they don't participate in weak interactions, which makes the Standard Model the strange chiral theory that it is.

It is all much cooler and more interesting than just that, but you have to dig into the group theory to see where it all comes from. I was going to write a bunch of stuff about it, but after I wrote a bit I realised that I don't know it well enough to explain it properly and I would just make it worse for you, so I'll skip it :).

edit: The chiral thing is pretty strange, it makes it seem like left and right handed particles should really be conceptualised as separate physical entities, but then one remembers that particles don't propagate the same way as they interact. An electron propagating along is a mixture of left and right handed interaction-basis electrons (because they have mass, interestingly), yet only the left handed 'piece' of it interacts via weak interactions. It's a bit of a crazy universe.

Last edited: Jul 19, 2011
6. Jul 19, 2011

### Schreiberdk

Well.. What then determines if, for example, a nucleus is unstable and makes a beta decay? How does this connect to weak isospin? Does unstable nuclei have larger weak isospin than stable ones, or how does it work? :) Thank for you :D

And by the way: It actually makes quite good sense to me, that charge makes you able to interact with the forces, rather than being the origin of the forces (because we do fx have photons comming from other sources than electrical charges

7. Jul 19, 2011

### kurros

Err, well hmm I am not sure. I was going to say I don't think that is the case (unstable nuclei have larger weak isospin than stable ones) but now that I think about it maybe it is. Ok lets think; we are really thinking about up and down quarks here. Up and down (left handed) quarks have opposite weak isospin (+1/2 and -1/2, I forget which is which). So I suppose that means that protons and neutrons also have opposite weak isospin. Which in turn means that if a nucleus has an excess of protons or neutrons then the nucleus does have an overall excess of weak isospin proportional to the excess of protons or neutrons.

Despite this I am not quite game to say that this is why proton/neutron imbalances make beta decay more likely, despite how tempting it seems. I think it must play a role, but it is the energy considerations that are more important, which comes down to strong interactions and nuclear shell structures and coulomb repulsions and "liquid drop" models of nuclei and other horrible things. The weak interactions just provide a mechanism through which these energy problems can be relieved. Besides, stable heavy nuclei have many more neutrons than protons, further screwing this weak isospin line of thinking.

In summary I don't rightly know.

8. Jul 19, 2011

### Staff: Mentor

The different decay methods are a result of a combination of forces and interactions, not simply the weak force.

9. Jul 19, 2011

### Dickfore

The Wikipedia article about Electroweak Interaction states that it belongs to the expanded gauge group $SU(2) \times U(1)$. The generators of the $SU(2)$ are called weak isospin and the gauge bosons associated with it are $W^{+}, W^{-}, W^{0}$, while the generator of $U(1)_{Y}$ is the weak hypercharge and the gauge boson associated with it is $B^{0}$.

Next, there is spontaneous symmetry breaking under the Higgs mechanism down to $U(1)_{em}$, which is a different copy of the $U(1)$ gauge group (and hence the different subscript) which is the gauge group of the electromagnetic intercation. The generator of this group is the electric charge Q, which is related to the previous ones according to:
$$Q = I_{3} + \frac{Y_{W}}{2}$$
and the associated gauge boson is the photon $\gamma$. Since this boson is a Nambu-Goldstone boson, it is massless, while the other three ($W^{+}, W^{-}, W^{0}$) gain mass and correspond to the weak interaction. I would say that the generator associated with the weak interaction are $I_{1}$, $I_{2}$ and another linear combination of $I_{3}$ and $Y_{W}$. The word generator used here is the same as what you call 'charge'. Because the gauge bosons associated with it are massive, the weak force has a very short range.

10. Jul 20, 2011

### daschaich

Dickfore, that's all correct except for:

The photon is massless because it is a gauge boson; the three Nambu--Goldstone bosons resulting from the $SU(2) \times U(1) \to U(1)$ symmetry breaking are "eaten" by the W and Z becoming massive.

The electric charge (i.e., electromagnetic coupling) $e$ can be written in terms of the two charges associated with the weak isospin and weak hypercharge $SU(2) \times U(1)$. These are often called $g$ and $g'$, but I prefer to use $g_2$ and $g_1$ to make it easier to remember which is which. Then $e = \frac{g_1 g_2}{\sqrt{g_1^2 + g_2^2}}$; if anyone wants to see where that comes from, Halzen and Martin's Quarks and Leptons is a reasonable place to start.

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