The screw nature of electromagnetism

Farsight seems to think that electric charge is some winding number or something similar. It isn't. It's an interaction strength. That is evident from the Standard-Model Lagrangian, a function that summarizes equations of motion.

Let's now consider charge quantization, something that might be considered evidence that charge is some winding number.

In fairness, I think there is a way that quantised charge could, in an implausible theory, arise directly from topological concerns.

[sketch of crank theory]
You'd start with a Yang-Mills gauge theory set up so that after symmetry breaking you have a U(1) with a large coupling constant g >> 1, but in which there were no particles which actually coupled to the U(1) gauge field. The theory has 't Hooft-Polyakov monopoles, and you find that their magnetic charges are determined by the winding number of maps from the sphere at infinity to U(1).

You now have single type of source of electromagnetic fields (namely, magnetic monopoles) with a small coupling constant ~1/g << 1. Switch your "magnetic" and "electric" terminology, and you're done.
[/sketch of crank theory]

It is somewhat pointless, probably, as the theory we started with could (as you mentioned) give you charge quantisation anyway, in a more direct fashion. :)
 
I don't have an in-detail account to hand I'm afraid.

I think it would be a good idea for you to work this stuff out in some detail. My questions regarding pions and baryons were really intended to try and explore that stuff in some depth - to find out exactly how spin causes charge in the picture of things you've been describing. I'd like to be able to take a case you haven't considered yet and use this picture to figure out the answer myself, ideally.

No worries for now, though. It seems like there are a lot of different topics being discussed in this thread, and to be honest the other strand of conversation regarding spirals is enough for the time being!

One small point, though, on your statement about W-bosons:

Don't take anything for granted. Think about free neutron decay. It is said to occur because of the weak interaction, mediated by the W boson. How can an 80GeV particle be involved in the decay of a 939 MeV particle?

As you can see by your energy conservation argument, there is no real W-boson produced in when a free neutron undergoes beta decay. There is definitely a kind of activity in the W-field which is commonly described with evocative words like "virtual particle", but nothing happens which violates energy conservation.

That aside, we've been producing bona fide, non-virtual Ws and Zs in particle accelerators since as long ago as November 1982 (the date of the first W sighting in the UA1 experiment at CERN). Despite their seemingly short lifetime, we've been able to measure several of their properties to great accuracy and, truly amazingly, we find that they are in agreement with the predictions of electroweak theory (together with data from observations of weak interaction processes) - including, first and foremost, the prediction that these particles would exist in the first place!
 
A Universe of magnetic monopoles. That's a good one.

You heard it here first! Possibly. :)

I made a bit of a silly error: rather than maps from S[sup]2[/sup] --> U(1), I meant maps from S[sup]2[/sup] --> G/U(1), where G is the unbroken symmetry group (this is a simple theory where U(1) is all that's left after SSB). These maps fall into distinct homotopy types, and so the monopole charge is quantised. Serves me right for posting in a hurry...
 
Belated welcome lpetrich & btr. Great to see more good folks thrashing the cranks. Not that Farsight has any clue what you're even talking about. You have to speak in gross oversimplifications, suitable for a child, with lots of metaphors. Of course the average kid will ask questions. Farsight is of course one of those "exceptional kids". ;)
 
Sure. What I meant was, what physical system would give rise to the particular field you depicted in the spiral diagram? If it is supposed to be illustrative of what happens with a point charge, I think you should really replace the concentric circles with dipole field lines, and make whatever changes are necessary on the right hand side of the equation. Could you do that? You might be surprised by the result. ;)
It isn't supposed to be illustrative of a point charge. There are no point charges. It's supposed to be illustrative of a "vorton" with "intrinsic" spin. I'll have a look at replacing the concentric circles with dipole field lines. As I said I avoided that to keep things simple and to steer clear of the toroidal electron and having to inflate the torus to yield spherical symmetry.

I have a follow-up question: I get that it is only meant to be illustrative, but if the idea has potential (so to speak) it should be possible to find a mathematical procedure which leads to similar diagrams, while also relating in a clear mathematical way to the fields. On the other hand, if it isn't possible to relate the diagrams to the fields in some mathematical way, that would tell you something too. Have you tried to test it in this way?
No. It's a thought.

I'll give you an example of the sort of thing I'm talking about. For ordinary vector fields, the integral lines (a.k.a. field lines or streamlines) are generated by a simple mathematical procedure; you just integrate the first-order differential equation which equates each line's tangent vector to the field value. You can also do a partial inverse, at least in theory; given all of the field lines, you can reconstruct the original vector field up to an overall scaling. Thereby, the field lines are rigorously proved to contain actual information about the field itself.
Which lands us right back with the central issue: that E isn't "the" field, it denotes the linear force resulting from the Fμv field interactions.

Sorry, I worded that ambiguously. When I said "a single value at each point of space" I didn't mean to imply a that the value was necessarily a scalar, just that it was a single field rather than multiple fields. What your diagram actually suggests to me is a vector field of some sort (with the lines being the integral lines - a.k.a. stream lines or field lines - of that field).
It's a single field, but it's more than just a vector field. You know that "field strength" diminishes with distance, which starts you off with a scalar. Then the spiral lines have a direction, which steps you up to a vector. But it doesn't stop there. Each line is "curled" in the third dimension rather just curved on the flat, it's more than just a straight line with a direction. You need more information, so it's a tensor field.

Be careful; you can have also rotational electric fields! Whenever B is varying with time such things are predicted by the Maxwell-Faraday equation:

$$ \nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t} $$​
The moot point is that you don't create an electric field by varying a magnetic field, because they're two aspects of the greater whole.

This is how electric generators work, in fact; a varying magnetic field creates circulating lines of E, which in turn create electromotive forces in circuits, and hence a flowing electric current is created.
Please can you try to be cautious with the word create? And please note that those lines don't actually exist. I suppose I can't really complain about creating forces and current, but again my central point is that E denotes the force - which moves the electrons in the electric conduction current. When you pedal a bicycle with a dynamo you can feel a slight resistance when you turn it on. You are "pedalling against the electrons" which you are moving. For an analogy imagine your pedals are connected to a propeller which you dip into the water.

The only real difference between what lines of E do and what lines of B do is most readily seen in their divergences when there are charges around:

$$ \text{div}\mathbf{E} = \rho / \epsilon_0 ; \qquad\qquad \text{div}\mathbf{B} = 0 . $$​

Pictorially, these say that E can have integral lines which terminate on charges, while the lines of B can never terminate (unless it turns out that magnetic monopoles exist, in which case we would have to alter our equations).
Lines of E are lines of force, but lines of B aren't. And there are charges around! A charged particle like an electron has an electromagnetic field. It doesn't have an electric field, so "electric charge" is a misnomer. Therefore "magnetic charge" is misguided. Think in terms of frame-dragging and thing of the electromagnetic field as a "twist field". If you were moving through it but didn't know it, you would think of it as a "turn field". A magnetic field. But because the frame-dragging is there you know that there can be no region of space that is turning like a roller-bearing disconnected from the surrounding space. Hence magnetic monopoles cannot exist. But do note this.

It is a laudible goal to try to find ways of picturing the electromagnetic field, but I think the idea of using "field lines" to represent a tensor is fundamentally problematic. One issue I see with this strategy is that the field's value at each point is a skew-symmetric rank 2 tensor (a total of six numbers per point), while the spiral diagram implies that the electromagnetic field can be represented by streamlines, which then very intuitively leads one to think of a 3-vector with only three numbers per point. To see what sort of ambiguities this can lead to, it might be instructive if you tried drawing the electromagnetic field using your method for the following two cases: (i) an electric dipole; (ii) a magnetic dipole.
I'll think about the magnetic dipole, but life is a constant battle to keep things simple enough for the modern attention span. Maybe I'll do another separate essay which talks about the toroidal dipole and how one "inflates" it to yield a spindle-sphere torus, like this. Topologically it's a torus, hence the electron's magnetic dipole, but geometrically it's a sphere, hence the electron's electric monopole.

That said, there are pictorial ways of thinking about skew-symmetric tensor fields (a.k.a. differential forms). There is one involving surfaces, tubes, boxes and so forth, which Misner, Thorne and Wheeler describe in their book Gravitation (whether they invented it, I do not know). I see that you have already been learning about that from Markus Hanke at thephysicsforum.com (at least, I assume it's you; the thread is recent, and called Questions re: Covariant Electromagnetism), so I won't repeat that material here. I will point you at this document, though, which describes another approach (as well as providing a reasonably good introduction to differential forms):

arxiv.org/pdf/math/0306194v1.pdf
Thanks. Yes, that's me. Markus is a good egg. He is sincere.
 
I think it would be a good idea for you to work this stuff out in some detail. My questions regarding pions and baryons were really intended to try and explore that stuff in some depth - to find out exactly how spin causes charge in the picture of things you've been describing. I'd like to be able to take a case you haven't considered yet and use this picture to figure out the answer myself, ideally.
I'd say start with the photon, then examine pair production* and the electron, and then the proton. Relegate short-lived particles to the back of the queue, along with any particles whose existence has been inferred or which have never actually been seen, but when you do study them look hard at what they decay into and play around with paper strips. Meanwhile take a look at this Topological Quantum Field Theory Club webpage. See those blue trefoil knots at the top? Pick one, start at the bottom left, and trace around it anticlockwise calling out the crossing-over directions: up down up. Ring any bells? Imagine the knot is elastic like the bag model. When you pull so hard that you break it, you don't get a mess of quarks and gluons spilling out like beans from a bag, you get pions, like this: 8. Only they aren't symmetrical so they aren't balanced so they aren't stable.

As you can see by your energy conservation argument, there is no real W-boson produced in when a free neutron undergoes beta decay. There is definitely a kind of activity in the W-field which is commonly described with evocative words like "virtual particle", but nothing happens which violates energy conservation. That aside, we've been producing bona fide, non-virtual Ws and Zs in particle accelerators since as long ago as November 1982 (the date of the first W sighting in the UA1 experiment at CERN). Despite their seemingly short lifetime, we've been able to measure several of their properties to great accuracy and, truly amazingly, we find that they are in agreement with the predictions of electroweak theory (together with data from observations of weak interaction processes) - including, first and foremost, the prediction that these particles would exist in the first place!
Let's talk about that another time.

* there's a photon-photon interaction. So much to discover. So much low hanging fruit.
 
Belated welcome lpetrich & btr. Great to see more good folks thrashing the cranks. Not that Farsight has any clue what you're even talking about. You have to speak in gross oversimplifications, suitable for a child, with lots of metaphors. Of course the average kid will ask questions. Farsight is of course one of those "exceptional kids". ;)

Do us a favour and tone it down eh?
 
It isn't supposed to be illustrative of a point charge. There are no point charges. It's supposed to be illustrative of a "vorton" with "intrinsic" spin. I'll have a look at replacing the concentric circles with dipole field lines. As I said I avoided that to keep things simple and to steer clear of the toroidal electron and having to inflate the torus to yield spherical symmetry.

No. It's a thought.

OK, good on both points.

Which lands us right back with the central issue: that E isn't "the" field, it denotes the linear force resulting from the Fμν field interactions.

It's a single field, but it's more than just a vector field. You know that "field strength" diminishes with distance, which starts you off with a scalar. Then the spiral lines have a direction, which steps you up to a vector. But it doesn't stop there. Each line is "curled" in the third dimension rather just curved on the flat, it's more than just a straight line with a direction. You need more information, so it's a tensor field.

More correctly, E is the force a stationary unit test charge feels; it isn't always equal (or proportional) to the force in general! But anyway, all I wanted to illustrate was how the "traditional" electric and magnetic field lines carry actual information about the "traditional" 3-vector fields E and B.

You are correct to say that it is more in keeping with relativity to speak in terms of the tensor F, and that brings us to the very issue under discussion: how do you draw F and get a similar preservation of information? You could just draw both sets of field lines, and that technically works, but it feels a bit like a cop-out.

The moot point is that you don't create an electric field by varying a magnetic field, because they're two aspects of the greater whole.

Absolutely. However, do note that it is inaccurate to say that what we mean by electric fields is somehow a linear part of F, while magnetic fields are somehow a rotation part. Like I was saying, you can have rotational parts of the field which people would still call "electric".

A better way of stating it is to say that when F has non-zero space/time components for some observer, that observer will call that an electric field; when F has non-zero space/space components for some observer, that observer will call that a magnetic field. A more Lorentz-invariant way might be this: if u[sup]μ[/sup] are the 4-velocity components of a "traditional" observer (one who does not know about the covariant formulation of electromagnetism), and F[sub]μν[/sub] are the electromagnetic field components, then:


F[sub]μν[/sub]u[sup]ν[/sup] = the bit of F the observer will think of as an electric field;
*F[sub]μν[/sub]u[sup]ν[/sup] = the bit of F the observer will think of as a magnetic field.

I've introduced the quantities *F[sub]μν[/sub] = (1/2)ε[sub]μνλρ[/sub]F[sup]λρ[/sup], which are the components of the so-called dual tensor to F. They contain exactly the same information as F, but with the components shuffled around.

Please can you try to be cautious with the word create? And please note that those lines don't actually exist. I suppose I can't really complain about creating forces and current, but again my central point is that E denotes the force - which moves the electrons in the electric conduction current. When you pedal a bicycle with a dynamo you can feel a slight resistance when you turn it on. You are "pedalling against the electrons" which you are moving. For an analogy imagine your pedals are connected to a propeller which you dip into the water.

I think it is pretty common to speak loosely and say things like "varying magnetic fields create EMFs", but I agree that it is rather imprecise. That's a good illustration of why mathematics is useful in physics! And I hope I didn't suggest that field lines actually exist; they are definitely just mathematical abstractions which are sometimes useful for visualising some apsects of a field. I hope my meaning was clear, all the same.

Lines of E are lines of force, but lines of B aren't.

Is this what you mean: in covariant language, the Lorentz force is alway given by qF[sub]μν[/sub]u[sup]ν[/sup], but because u = (1,0,0,0) in the rest frame, the force (in that rest frame) on the particle only depends on the time/space components of F, which is what a "traditional" observer ignorant of relativity would call the electric field. In the (3+1) language of that kind of observer, if you look at things in a charged particle's instantaneous rest frame, you discover that the particle only ever "feels" an electric field; in other words, because it is at rest in its own reference frame (i.e. v = 0), the Lorentz force reduces to F = qE. It's interesting stuff.

And there are charges around!

Oh, most definitely. I only meant that I was interested in regions of space where the charge density was non-zero, as opposed to a vacuum somewhere. Did the rest of it make sense?

A charged particle like an electron has an electromagnetic field. It doesn't have an electric field, so "electric charge" is a misnomer. Therefore "magnetic charge" is misguided.

I'm not sure I agree with the "therefore" here, although I don't have a problem with the conclusion that monopoles don't exist, provided that Maxwell's equations are correct of course (as normally written, with div B = 0, or in the covariant formulation d*F = 0).

Think in terms of frame-dragging and thing of the electromagnetic field as a "twist field". If you were moving through it but didn't know it, you would think of it as a "turn field". A magnetic field. But because the frame-dragging is there you know that there can be no region of space that is turning like a roller-bearing disconnected from the surrounding space. Hence magnetic monopoles cannot exist. But do note this.

I'm afraid I didn't grasp your analogy there, but no worries.

I'll think about the magnetic dipole, but life is a constant battle to keep things simple enough for the modern attention span.

In all honesty, my request for the diagrams for electric and magnetic dipoles were more for your benefit; I wanted you to see the ambiguity I mentioned for yourself, rather than expect you to simply take my word for it. Wolfram Alpha (wolframalpha.com) can do streamline plots, incidentally, if you don't have plotting software. You give it a query like this:

Code:
 StreamPlot[ {x/(x^2 + y^2), y/(x^2 + y^2)}, {x, -1, 1}, {y, -1, 1}]

(that's an inverse square vector field) or like this:

Code:
 StreamPlot[ {x*y/(x^2 + y^2)^(5/2), ((2/3)*y^2 - (1/3)*x^2)/(x^2 + y^2)^(5/2)}, {x, -1, 1}, {y, -1, 1}]

(that's a dipole vector field).

Maybe I'll do another separate essay which talks about the toroidal dipole and how one "inflates" it to yield a spindle-sphere torus, like this. Topologically it's a torus, hence the electron's magnetic dipole, but geometrically it's a sphere, hence the electron's electric monopole.

That sounds intriguing. Is this a novel model, or something equivalent (in terms of observables) to the standard one?


You're welcome.
 
A better way of stating it is to say that when F has non-zero space/time components for some observer, that observer will call that an electric field; when F has non-zero space/space components for some observer, that observer will call that a magnetic field. A more Lorentz-invariant way might be this: if u[sup]μ[/sup] are the 4-velocity components of a "traditional" observer (one who does not know about the covariant formulation of electromagnetism), and F[sub]μν[/sub] are the electromagnetic field components, then:


F[sub]μν[/sub]u[sup]ν[/sup] = the bit of F the observer will think of as an electric field;
*F[sub]μν[/sub]u[sup]ν[/sup] = the bit of F the observer will think of as a magnetic field.

I've introduced the quantities *F[sub]μν[/sub] = (1/2)ε[sub]μνλρ[/sub]F[sup]λρ[/sup], which are the components of the so-called dual tensor to F. They contain exactly the same information as F, but with the components shuffled around.

The electromagnetic dual tensor $$*F_{\mu\nu} = \frac{1}{2} \varepsilon_{\mu\nu\alpha\beta} \eta^{\alpha\gamma} \eta^{\beta\delta} F_{\gamma\delta} = \frac{1}{2} \varepsilon_{\mu\nu\alpha\beta} F^{\alpha\beta} $$ is also an anti-symmetric rank-2 tensor that looks like electric and magnetic components swapped places.

I think you might have been expressing the special case where $$v^{\mu} \propto \begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ in which case $$F_{\mu\nu} v^{\nu} $$ and $$*F_{\mu\nu} v^{\nu} $$ have spatial components proportional to $$\vec{E}$$ and $$\vec{B}$$, respectively. This is not generally the case for other velocity 4-vectors since you get a 4-force where the spatial components are not proportional to the 3-force.
 
Oops, for an inverse-square vector field, it's
Code:
StreamPlot[ {x/(x^2 + y^2)^(3/2), y/(x^2 + y^2)^(3/2)}, {x, -1, 1}, {y, -1, 1}]

Farsight's pictures suggest that he seems to think that the electromagnetic tensor F contains E and B mixed in together, when F contains E and B side by side as separate components. Sort of like Ex, Ey, Ez for E and Bx, By, Bz for B. His main argument seems to be overly literal interpretation of certain analogies, rather than an attempt to explain the appropriate math in nonmathematical terms, as I have tried to do.

Farsight physics also includes total incomprehension of space-time unity. This despite the fact that it was first described by someone whose writings Farsight thumps: Hermann Minkowski.

As to "math dumps", Maxwell's and Einstein's and Minkowski's writings are full of them, and those "math dumps" are important parts of their theories.
 
I think you might have been expressing the special case where $$v^{\mu} \propto \begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ in which case $$F_{\mu\nu} v^{\nu} $$ and $$*F_{\mu\nu} v^{\nu} $$ have spatial components proportional to $$\vec{E}$$ and $$\vec{B}$$, respectively.

Not quite, but very nearly. What I was actually doing was constructing a pair of spacelike 4-vectors with components

$$ E_\mu = F_{\mu\nu} u^\nu \qquad ; \qquad\qquad B_\mu = *F_{\mu\nu} u^\nu $$​

such that an observer with 4-velocity matching u, in their rest frame, sees that the spatial components of the 4-vectors E and B are the same as the components of the electric and magnetic 3-vectors E and B that they would extract from F.

In other words, once given a 4-velocity u (which we can imagine corresponds to some observer, but it needn't) there is a Lorentz-invariant way of writing the decomposition of the electromagnetic field into "electric" and "magnetic" spacelike vectors, without having to work in a special frame where u has components (1,0,0,0).
 
Oops, for an inverse-square vector field, it's
Code:
StreamPlot[ {x/(x^2 + y^2)^(3/2), y/(x^2 + y^2)^(3/2)}, {x, -1, 1}, {y, -1, 1}]

Farsight's pictures suggest that he seems to think that the electromagnetic tensor F contains E and B mixed in together, when F contains E and B side by side as separate components. Sort of like Ex, Ey, Ez for E and Bx, By, Bz for B. His main argument seems to be overly literal interpretation of certain analogies, rather than an attempt to explain the appropriate math in nonmathematical terms, as I have tried to do.

Farsight physics also includes total incomprehension of space-time unity. This despite the fact that it was first described by someone whose writings Farsight thumps: Hermann Minkowski.

As to "math dumps", Maxwell's and Einstein's and Minkowski's writings are full of them, and those "math dumps" are important parts of their theories.

He also thinks the Schwarzschild bookkeeper coordinates are preferred. Another reason to believe "Farsight physics also includes total incomprehension of space-time unity." The key word 'incomprehension'.
 
He also thinks the Schwarzschild bookkeeper coordinates are preferred. Another reason to believe "Farsight physics also includes total incomprehension of space-time unity." The key word 'incomprehension'.

And was it he or the other one that also claims, time stops at a BH's EH, from the local FoR, when in fact it is never seen to stop from any FoR.
 
Farsight seems to jump from "topological charge" to "charge is topological" -- "electric charge is proportional to a topological invariant". Seems like carelessness with words and an overly literal-minded interpretation of the "charge" of "topological charge".

He also thinks the Schwarzschild bookkeeper coordinates are preferred. Another reason to believe "Farsight physics also includes total incomprehension of space-time unity." The key word 'incomprehension'.
I haven't followed very much what Farsight claims about black holes, but he does not seem to like Kruskal-Szekeres coordinates.

Coordinate systems are often selected for convenience, like making problems easier to solve or making certain features readily manifest. A good way to do that is to associate coordinates with symmetries or approximate symmetries. That is rather obvious for rectangular coordinates in a Euclidean space or a pseudo-Euclidean space like a Minkowskian space. The coordinates are associated with translation symmetries.

The Schwarzschild solution is static and spherically symmetric. It has translation and reflection symmetry in time, and spherical symmetry in directions from the center, specified with two coordinates. The remaining coordinate is the radius. It has a coordinate singularity at the event horizon, so that's why K-S coordinates were invented and halfway coordinates were invented: Eddington-Finkelstein. K-S coordinates are smooth through the event horizon, enabling them to clarify some features of the Schwarzschild solution.
 
Not quite, but very nearly. What I was actually doing was constructing a pair of spacelike 4-vectors with components

$$ E_\mu = F_{\mu\nu} u^\nu \qquad ; \qquad\qquad B_\mu = *F_{\mu\nu} u^\nu $$​

such that an observer with 4-velocity matching u, in their rest frame, sees that the spatial components of the 4-vectors E and B are the same as the components of the electric and magnetic 3-vectors E and B that they would extract from F.

In other words, once given a 4-velocity u (which we can imagine corresponds to some observer, but it needn't) there is a Lorentz-invariant way of writing the decomposition of the electromagnetic field into "electric" and "magnetic" spacelike vectors, without having to work in a special frame where u has components (1,0,0,0).

I'm pretty sure if you want that then you have to double-apply the Lorentz transform that takes $$v^{\mu}$$ to (1,0,0,0) on the rank-two electromagnetic tensor.

$$f^{\mu} = q F_{\nu}^{\mu} v^{\nu} \quad \Rightarrow \quad \Lambda_{\mu}^{\alpha} f^{\mu} = q \Lambda_{\mu}^{\alpha} F_{\nu}^{\mu} \left( \Lambda^{\tiny -1} \right)_{\beta}^{\nu} \Lambda_{\gamma}^{\beta} v^{\gamma} = q \Lambda_{\mu}^{\alpha} F_{\nu}^{\mu} v^{\nu} $$
So if we take $$\Lambda_{\gamma}^{\beta} v^{\gamma} \propto \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ then it follows that $$ v^{\nu} \propto \left( \Lambda^{\tiny -1} \right)_{\beta}^{\nu} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}^{\beta} $$ and thus E in coordinate frame where $$v^{\nu}$$ is our state of rest that we can read the E components directly off $$F'_{\beta}^{\alpha} = \Lambda_{\mu}^{\alpha} F_{\nu}^{\mu} \left( \Lambda^{\tiny -1} \right)_{\beta}^{\nu}$$.
 
More correctly, E is the force a stationary unit test charge feels; it isn't always equal (or proportional) to the force in general! But anyway, all I wanted to illustrate was how the "traditional" electric and magnetic field lines carry actual information about the "traditional" 3-vector fields E and B.
Understood.

You are correct to say that it is more in keeping with relativity to speak in terms of the tensor F, and that brings us to the very issue under discussion: how do you draw F and get a similar preservation of information? You could just draw both sets of field lines, and that technically works, but it feels a bit like a cop-out.
A field is a state of space. You draw it like you draw the frame-dragging of geomagnetism. But to draw it properly you draw it with chirality in three dimensions. Maybe a fly-through animation would depict that well. The spindle sphere torus is in the middle surrounded by the "spiral". Only there's no surface to the spindle sphere torus. And F isn't fundamental. Which is why I said you start with a photon. Which takes many paths like a seismic wave takes many paths.

Absolutely. However, do note that it is inaccurate to say that what we mean by electric fields is somehow a linear part of F, while magnetic fields are somehow a rotation part. Like I was saying, you can have rotational parts of the field which people would still call "electric".
Noted. It's difficult to combine precision clarity and brevity. You will note that I didn't mention that two magnets move linearly together.

A better way of stating it is to say that when F has non-zero space/time components for some observer, that observer will call that an electric field; when F has non-zero space/space components for some observer, that observer will call that a magnetic field.
I'm something of an advocate for relativity, but actually I don't agree with that. The field is what it is. As I said, it's a "state of space". It doesn't suddenly acquire space/space components just because the observer started to move. It didn't change one jot. You might think of that as something that goes against the grain of relativity, but please don't. It's a "big picture" interpretation, that's all.

A more Lorentz-invariant way might be this: if u[sup]μ[/sup] are the 4-velocity components of a "traditional" observer (one who does not know about the covariant formulation of electromagnetism), and F[sub]μν[/sub] are the electromagnetic field components, then:


F[sub]μν[/sub]u[sup]ν[/sup] = the bit of F the observer will think of as an electric field;
*F[sub]μν[/sub]u[sup]ν[/sup] = the bit of F the observer will think of as a magnetic field.

I've introduced the quantities *F[sub]μν[/sub] = (1/2)ε[sub]μνλρ[/sub]F[sup]λρ[/sup], which are the components of the so-called dual tensor to F. They contain exactly the same information as F, but with the components shuffled around.
If you say so. But there's still something vital missing - the way that forces only result from field interactions. The way you're describing it suggests the test particle is passive participant when it's not. I think the best place to start is with two particles only.

I think it is pretty common to speak loosely and say things like "varying magnetic fields create EMFs", but I agree that it is rather imprecise. That's a good illustration of why mathematics is useful in physics! And I hope I didn't suggest that field lines actually exist; they are definitely just mathematical abstractions which are sometimes useful for visualising some apsects of a field. I hope my meaning was clear, all the same.
It was. And as I suggested above, it isn't always easy to stay precise.

Is this what you mean: in covariant language, the Lorentz force is alway given by qF[sub]μν[/sub]u[sup]ν[/sup], but because u = (1,0,0,0) in the rest frame, the force (in that rest frame) on the particle only depends on the time/space components of F, which is what a "traditional" observer ignorant of relativity would call the electric field. In the (3+1) language of that kind of observer, if you look at things in a charged particle's instantaneous rest frame, you discover that the particle only ever "feels" an electric field; in other words, because it is at rest in its own reference frame (i.e. v = 0), the Lorentz force reduces to F = qE. It's interesting stuff.
LOL, no! I meant the test particle moves linearly along the electric field lines and around the magnetic field lines. The electric field lines are "lines of force" but the magnetic field lines aren't.

To be continued.
 
I'm pretty sure if you want that then you have to double-apply the Lorentz transform that takes $$v^{\mu}$$ to (1,0,0,0) on the rank-two electromagnetic tensor.

$$f^{\mu} = q F_{\nu}^{\mu} v^{\nu} \quad \Rightarrow \quad \Lambda_{\mu}^{\alpha} f^{\mu} = q \Lambda_{\mu}^{\alpha} F_{\nu}^{\mu} \left( \Lambda^{\tiny -1} \right)_{\beta}^{\nu} \Lambda_{\gamma}^{\beta} v^{\gamma} = q \Lambda_{\mu}^{\alpha} F_{\nu}^{\mu} v^{\nu} $$
So if we take $$\Lambda_{\gamma}^{\beta} v^{\gamma} \propto \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ then it follows that $$ v^{\nu} \propto \left( \Lambda^{\tiny -1} \right)_{\beta}^{\nu} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}^{\beta} $$ and thus E in coordinate frame where $$v^{\nu}$$ is our state of rest that we can read the E components directly off $$F'_{\beta}^{\alpha} = \Lambda_{\mu}^{\alpha} F_{\nu}^{\mu} \left( \Lambda^{\tiny -1} \right)_{\beta}^{\nu}$$.

We're trying to achieve subtley different, and extremely closely related things. :)

Let's forget entirely about coordinate systems temporarily: if you are given an electromagnetic field described by the 2-form F at some event in spacetime, and any (fixed) timelike vector u at that same event, normalised so that u[sub]μ[/sub]u[sup]μ[/sup] = 1 (I use the "mostly minus" sign convention for the metric), you can construct a perfectly good Lorentz 4-vector B using the prescription B[sub]μ[/sub] = *F[sub]μν[/sub]u[sup]ν[/sup]. All observers, if given the same two geometric objects F and u, will construct this same 4-vector B. Agreed?

I now claim that this 4-vector B has the following properties:

1. B is spacelike and orthogonal to u (this follows from the skew-symmetry of F).
2. If you did introduce an inertial observer with 4-velocity u at the event in question, they would detect a magnetic 3-vector field B with components equal to the spatial components of B. They would also find that B[sub]0[/sub] = 0 in their reference frame.

Similar remarks apply to my construction of the 4-vector E.
 
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