18
$\begingroup$

A question was asked over at EE.SE recently which I tried to answer, but much of my answer was speculative. I'm hoping someone here can help my ignorance.

In electronics design, there are four physical quantities of interest: voltage, flux, charge, and current. If you have four things and want to pick two, order not mattering, there are 4C2 = 6 ways to do that. Two of the physical quantities are defined in terms of the other two. (Current is change in charge over time. Voltage is change in flux over time.) That leaves four possible relationships: resistance, inductance, capacitance, and memristance. These are the values we use in every-day electronics design and analysis. (Well, not memristance, but one can dream.)

enter image description here

If you want another fundamental component, you need another physical quantity to relate to these four. And while there are many physical quantities one might measure, none seem so tightly coupled as these. I'd suppose this is because electricity and magnetism are two aspects of the same force. I'd further suppose that since electromagnetism is now understood to be part of the electroweak force, one might be able to posit some relationships between the weak nuclear interaction and our four quantities of voltage, current, charge, and flux.

I haven't the first clue how this would be physically manifested, especially given the relative weakness of the weak nuclear force at anything short of intranuclear distances. Perhaps in the presence of strong magnetic or electrical fields affecting the rates of radioactive decay? Or in precipitating or preventing nuclear fusion? I'd yet further suppose (I'm on a roll) that the field strengths required would be phenomenal, which is why they're not practical for everyday engineering.

But that's a lot of supposition. I am a mere engineer, and unqualified to comment intelligently on such things. Am I in the ballpark? Are there theoretical or demonstrable mathematical relationships between the weak nuclear interaction and charge, voltage, flux, and/or current which would correspond to my above suppositions? Or am I just looking for symmetry where none exists?

$\endgroup$
  • $\begingroup$ Hehe, I was to ask the question here but I don't know if physicists are familiar with memristors and such stuff. $\endgroup$ – jinawee Jan 29 '14 at 21:38
  • 1
    $\begingroup$ I almost put a bounty on this. Would love if someone tried to make an answer using graphs to represent the couplings. $\endgroup$ – WalyKu Feb 27 '15 at 16:24
  • $\begingroup$ I'm just going to gibber here for a minute, by way of expounding on my question without knowledge. The weak force is responsible for beta decay, which involves movement of charge (current) in and out of a nucleus. So the rate of beta decay would be somewhat analogous to voltage. Change in magnetic field induces voltage. So in this ignorant analogy, exposing a nucleus to extreme deltas in magnetic field strength would alter its rate of beta decay, potentially causing it to emit either electrons or positrons, and changing its atomic number. Now, what does reality actually do? $\endgroup$ – Stephen Collings Feb 27 '15 at 17:11
  • $\begingroup$ I now actually put a bounty on it :-D. I hope someone will help us understand it. $\endgroup$ – WalyKu Aug 14 '15 at 7:15
  • $\begingroup$ The electrical charge has to be replaced with weak isospin in the context of the weak force. The quantity which generalizes charge and weak isospin is weak hypercharge. One thing you might be interested in is Feynmann diagrams involving the weak interaction. You will see that there are nodes and paths, much like electrical diagrams, for which conservation laws or selection rules apply. $\endgroup$ – vosov Aug 14 '15 at 10:09
12
+50
$\begingroup$

There is no analogy of "voltage", "current", "charge" or "flux" to electromagnetism for the weak force, at least none that would be helpful.

The reason for this is that all of these are classical concepts, while the notion of the weak force is completely quantum. Taking the classical limit just makes it vanish because the classical force law of forces with massive bosons is exponentially suppressed by their mass, and the W and Z bosons are quite heavy. You don't notice it as an actual classical force at scales where you could meaningfully talk about "flux" or "current".

However, the concepts of "charge" and "voltage", in some way, generalize. Voltage is just the (difference in the) potential of the electric field, and thus, relativistically, the time-component of the four-potential of electrodynamics. The concept of four-potential immediately generalizes to the weak force, where it then is no longer real-number-valued, but $\mathrm{SU}(2)$-matrix-valued. Much like the electromagnetic four-potential contains the creation operators for photons, the weak four-potential contains the creation operators for W and Z bosons.

"Charge", instead of being a positive or negative multiple of the elementary charge of quarks, would be more akin to spin (since both spin and weak numbers arise from the representation theory of the symmetry group $\mathrm{SU}(2)$), and hence you get particles with "half-integer total weak charge" and "directional weak charges" which can, in any spatial direction, take a value between + and - their total weak charge.

However, much like color is not an observable, this directional weak charge would likewise be unobservable because the weak gauge transformations freely change it.

There's little influence of electromagnetic fields on radioactive decay because, at our scales, the electroweak interaction is broken into the electromagnetic and weak force. The electroweak scale where they begin to merge again lies somewhere on the order of hundreds of $\textrm{GeV}$, and I'm pretty sure you don't reach an energy scale like that just by applying "strong" magnetic fields - afaik, the only processes this highly energetic are incoming cosmic rays and our own particle colliders.

In conclusion, I think there is, on the level of the quantities you mention, no useful analogy between the electromagnetic and the weak or electroweak force.

$\endgroup$
-6
$\begingroup$

Are there analogs to resistance, inductance, capacitance, and memristance connecting the weak force to electromagnetism?

Yes, but not the way you're thinking. Take a look at weak interaction on Wikipedia and note this: "The first type is called the 'charged-current interaction' because it is mediated by particles that carry an electric charge (the W+ or W− bosons), and is responsible for the beta decay phenomenon. The second type is called the 'neutral-current interaction' because it is mediated by a neutral particle, the Z boson". The electromagnetic interaction features the charged current whilst the weak interaction features a neutral current. And like I said in my comment, conduction current is not the only current.

In electronics design, there are four physical quantities of interest: voltage, flux, charge, and current. If you have four things and want to pick two, order not mattering, there are 4C2 = 6 ways to do that. Two of the physical quantities are defined in terms of the other two.

Which ought to tell you that you're going round and round in circles instead of getting down to the fundamentals.

If you want another fundamental component, you need another physical quantity to relate to these four.

Since electronics involves electrons, and since we make electrons (and positrons) out of photons in pair production, you need to look at light, including displacement current, taking note of c = √(1/ε0μ0) where ε0 is vacuum permittivity and μ0 is vacuum permeability.

And while there are many physical quantities one might measure, none seem so tightly coupled as these.

Because they're all derivative.

I'd suppose this is because electricity and magnetism are two aspects of the same force.

Not quite. See what Minkowski said in Space and Time:

"In the description of the field caused by the electron itself, then it will appear that the division of the field into electric and magnetic forces is a relative one with respect to the time-axis assumed; the two forces considered together can most vividly be described by a certain analogy to the force-screw in mechanics; the analogy is, however, imperfect".

It's one field and two forces. The electron has an electromagnetic field. Electromagnetic field interactions between charged particles result in linear electric force and/or rotational magnetic force. When we only see the linear force we talk about an electric field, when we only see the rotational force we talk about a magnetic field. Understanding why we see linear and/or rotational force is crucial.

I'd further suppose that since electromagnetism is now understood to be part of the electroweak force, one might be able to posit some relationships between the weak nuclear interaction and our four quantities of voltage, current, charge, and flux.

You're going too far with this. IMHO you need to pull back and understand the electron first. We can make electrons and positrons. Only then do we have the thing we call charge. Then when we move the electrons we call it conduction current. And we move them using other charged particles, contrived in a box that we call a battery. Or contrived with rotational motion as the thing we call a magnet.

I haven't the first clue how this would be physically manifested...

The first clue comes from light and electrons. That's what QED is all about. My advice is to focus on that first.

But that's a lot of supposition. I am a mere engineer

Then put your engineering hat on, and ask yourself how you make an electron and a positron out of photons in pair production. And while you're at it, have a think about this:

$$Z_0 = \frac{E}{H} = \mu_0 c_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} = \frac{1}{\varepsilon_0 c_0}$$

Am I in the ballpark? Are there theoretical or demonstrable mathematical relationships between the weak nuclear interaction and charge, voltage, flux, and/or current which would correspond to my above suppositions? Or am I just looking for symmetry where none exists?

1) No. 2) Yes but not the way you're thinking. 3) Yes, but you are thinking, so good for you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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