The screw nature of electromagnetism

Discussion in 'Alternative Theories' started by Farsight, May 1, 2014.

  1. btr Registered Member

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    93
    I'll speak about these points below...

    Yes, absolutely; that particular property is what we call "charge". A slightly deeper explanation of "charge" (without getting overly technical) is that it expresses how strongly the particle interacts with photons. It could well be that a particle's charge can also be deduced from some other property (or properties) of the particle. Then again, perhaps not; life isn't always accommodating!

    NOTE: I'm leaving well aside for now the fact that electroweak theory complicates this picture considerably. What we see as electric charge is really a certain combination of weak hypercharge and a component of weak isospin, and what we see as a photon is really a certain combination of the W[sup]0[/sup] and B bosons. We can come back to that if necessary...

    Now, it seems to me that what you are saying, simply, is that spin somehow determines charge, but only for the "elementary" particles (whatever those turn out to be in the future - we still don't know, and may never!), and in a way that seems not to depend on the actual value of the spin. This is counterintuitive, but let's probe the idea and see where it leads.

    As I hinted above, I'm not clear on why it should matter whether a particle is composite or elementary. I think it would be useful at this stage to probe the idea further, so could we consider some specific examples?

    1. How (in detail) does this view of things account for the fact that some pions are neutral (and spin 0), while others are charged (but still spin 0)?

    2. How (in detail) does it account for the fact that baryons are observed with <spin, charge> combinations such as <1/2, 0>, <1/2, +1>, <1/2, +2>, <3/2, 0>, <3/2, +1> and <3/2, +2>?

    3. How (in detail) does it account for the fact that electrons have charge -1, neutrinos/antineutrinos are neutral, and positrons have charge +1, despite the fact that all three are (apparently "elementary") spin-1/2 particles described by spinor-valued fields?

    4. I'd like to consider W bosons in more depth. While their half-life is certainly short (on the order of 10[sup]-25[/sup] seconds in the particle's rest frame), their existence is not in any serious doubt, and nor are their properties such as charge and spin. As such, I think they still remain as good counterexamples, at least for now.

    The point you raised about gravity sounds it might take us too far afield right now, but perhaps we can return to it later?

    And thanks for the welcome!
     
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  3. lpetrich Registered Senior Member

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    So the writings of Maxwell and Einstein and Minkowski are sacred scriptures?

    He was describing how to measure the electric field.
    What childish literal-mindedness. This is an attempt to explain something in nontechnical terms, and it's not surprising that it may be misleading.
    Book-thumping. So like a theologian.
    LIGO is for looking for waves in space-time. Yes, space-time, not space-motion. That's gravitational radiation, not electromagnetic waves / photons. There's a difference between the two.

    Dubious analogies snipped. Maxwell and Einstein and Minkowski and Feynman did something that you seem to find very difficult. They did the math, so they didn't have to make such analogies.

    That you have a lot to learn about the math involved in it.

    What an excuse for low standards.
    So like a theologian.

    Farsight, the electromagnetic interaction has nothing to do with spinorness. In fact, I don't think that you understand what a spinor is.
    So you are in favor of dumbing down physics.

    It does. I've worked it out mathemtically.
    So like a theologian. "I've read the sacred scriptures".

    Do you understand the mathematics of the divergence operator?

    Irrelevant.

    Farsight, since you are such an expert on spinors, I want for you to write down the spinor angular-momentum operator. Also write down the vector one and the orbital one. What mathematical features do they have in common? How are they related to generators of rotations?
     
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  5. Farsight

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    How can I forget that? A photon is comprised of displacement current and it moves linearly at c. We put it through pair production and we focus on the electron. It isn't moving linearly at c, it's not moving anywhere. But it's got charge. It's got spin, it's got magnetic dipole moment. An electron indeed behaves like a tiny bar magnet because of the displacement current going round and round. That's more fundamental than conduction current. Read Taming Light at the Nanoscale:

    "Look around, and you will probably see numerous electronic and optical gadgets, such as mobile phones, personal digital assistants, laptops, TVs and digital cameras. These may all do different things but they have one thing in common: in the electronic circuits that drive these devices, charged particles flow through components and impart power via what is known as the conduction current. But is the motion of charged particles the only current we have available?!

    I'm not with you. Because the static Ampere's law is said to violate conservation of charge for a time-varying field? But the electron indeed behaves like a tiny bar magnet. Even when it's just sitting there, not moving, and thus there's no conduction current. Because a time-varying electromagnetic field that was going linearly is now going round and round just so, and looks like a standing field. it looks static, but it isn't. The Einstein de-Haas effect should make that clear. So should annihilation. The electron is like a photon in a box of its own making, so it the positron. Open one box with another and two photons are off like a shot. At c. They can't do that from a standing start. They were always going at c.
     
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  7. Farsight

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    No problem, I mentioned this in post #38. I used the concentric-circle magnetic field lines because this is the very simplest depiction, and I wanted to avoid complication. I also wanted to avoid being accused of promoting the Williamson / van der Mark toroidal electron. And avoid getting bogged down inflating the torus to a spindle-sphere because the electron's field is spherically symmetrical.

    Sorry, can you restate this? There is no charge source per se. It's topological. The electron is essentially a "knot" of field-variation trapped in a Dirac's belt configuration.

    There's no exact mathematical procedure behind the combination of the lines. I'm merely depicting "electromagnetic field lines" in a fashion that matches the Minkowski and Maxwell comments about the screw, the concept of a spinor, attraction and repulsion, and the frame-dragging of electromagnetism. Field lines don't actually exist, electric field lines are actually lines of force, charged particles go round magnetic field lines, and forces only occur because two electromagnetic fields interact.

    It's clearly not a scalar field.

    You don't. There is no new field, and the existing mathematics is good. You just appreciate what Minkowski was saying and realise that electromagnetic field interactions result in force, and when it's linear you talk of an electric field, and when it's rotational you talk of a magnetic field.

    It hasn't created a new field, it's merely depicting the "greater whole" Fuv electromagnetic field which is said to be a tensor field.
     
  8. Farsight

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    3,492
    All points noted przyk.
     
  9. PhysBang Valued Senior Member

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    We all knew that this wasn't about the theories of other people, this is all about Farsight's pet theory that electrons are made up of photons. Never mind that he knows of no way to show how this could possibly happen; we are to believe him on faith.
     
  10. Farsight

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    3,492
    Noted.

    Yes let's come back to it.

    OK.

    Yes. But apologies, the wife calls, and calls, I have to go. I'll respond to your points later.
     
  11. rpenner Fully Wired Valued Senior Member

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    That's all we asked for.
    That's what we said.
     
  12. CptBork Valued Senior Member

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    None of these responses from you answer the question, nor do you need to know anything about particle physics in order to answer it. Suppose I have an electric current density \(\vec{J}\) which is diverging away from a point. Why is it mathematically impossible to find any field \(\vec{B}\), whether corresponding to a real field or a purely fictitious one, such that \(\vec{\nabla}\times\vec{B}=\mu_o\vec{J}\) when \(\vec{J}\) is diverging? Even Maxwell himself could have answered this one in his sleep.
     
  13. rpenner Fully Wired Valued Senior Member

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    Simply explained because chapter 11 is the chapter that introduces the unification of space and time into space-time and the unification of electric phenomena and magnetic phenomena into electromagnetic phenomena. In a physics textbook, an author makes sure the readers have the background to work usefully and precisely with the material as it is given. This frequently takes the form of a history-of-physics approach of sneaking up on modern physics by discussing pre-modern physics.
    But it's not wrong to speak of the E and B fields just as long as you know that their particular values are dependent on the choice of coordinates.

    Well, first of all, Jackson's textbook is a bit dated in concepts and notation. Therefore it won't trivially square with certain modern sources until you take the differences of notation involved. Likewise it won't square with Maxwell at all until you apply many more differences of notation. I'll use Cartesian coordinates throughout and (-+++) sign conventions for \(\eta_{\mu\nu}\).

    Jackson deals with Classical Electrodynamics and in this context, this ignores quantum effects. So we are limited to Maxwell + Einstein.

    Relativistic dynamics is about the replacement of \(d\vec{p}/dt \; = \; \vec{f} \; = \; m \, \vec{a} \; = \; m \, d^2 \vec{r} / dt^2\) which comprises Newtonian dynamics for a classical particle. Relativity requires that we think about time as well as space, so we replace relative location (\(\vec{r}\)) with a 4-vector:
    \(R^{\mu} = \begin{pmatrix} c \Delta t \\ \vec{r} \end{pmatrix} = \begin{pmatrix} c \Delta t \\ \Delta r_x\\ \Delta r_y\\ \Delta r_z \end{pmatrix}\)
    For a particle moving slower than the speed of light, the inner product of the 4-position of the particle (relative to some point on the particle's world line) is negative.
    \(R^2 = R_{\mu} R^{\mu} = \eta_{\mu\nu} R^{\mu} R^{\nu} = -c^2 (\Delta t)^2 + (\Delta r_x)^2 + (\Delta r_y)^2 + (\Delta r_z)^2 = \vec{r}^2 -c^2 (\Delta t)^2 < 0\)
    This quantity \(R^2\) analogous to a (squared) Euclidean length in many ways, specifically it does not change value under a change of inertial coordinates (Rotations or Lorentz transforms or translations of space or time). So for a general slower-than light path we model the time experienced by the particle as:

    \(\Delta \tau = \int_{t_0}^{t_1} \sqrt{ - \frac{(d R)^2}{c^2 (dt)^2} } \, dt \; = \; \int_{t_0}^{t_1} \sqrt{ 1 - \left( \frac{ d\vec{r} }{c dt} \right)^2 } \, dt \; = \; \int_{t_0}^{t_1} \sqrt{ 1 - \left( \frac{\vec{v} }{ c} \right)^2 } \, dt \; = \; \int_{t_0}^{t_1} \frac{1}{\gamma(\vec{v})} \, dt \)
    So for unaccelerated motion we have the relations: \( \Delta t = \gamma(\vec{v}) \, \Delta \tau, \quad \left. R^2 \right|_{t_0}^{t_1} = (\Delta \vec{r})^2 - c^2 ( \Delta t)^2 = ( \vec{v}^2 - c^2 ) ( \Delta t)^2 = -c^2 ( \Delta \tau )^2\)


    The relativistic momentum for a massive particle is:
    \(P^{\mu} = \begin{pmatrix} E/c \\ \vec{p} \end{pmatrix} = \begin{pmatrix} \gamma(\vec{v}) m c \\ \gamma(\vec{v}) m \vec{v} \end{pmatrix} = m \begin{pmatrix} \gamma(\vec{v}) c \\ \gamma(\vec{v}) \vec{v} \end{pmatrix} = m \begin{pmatrix} \frac{c}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \\ \frac{v_x}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \\ \frac{v_y}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \\ \frac{v_z}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \end{pmatrix} = m V^{\mu}\)
    Because it is unchanged by the Lorentz transform, the inner product of 4-velocity (or 4-momentum) with itself is a scalar invariant of special relativity:
    \(V^2 = V_{\mu} V^{\mu} = \eta_{\mu\nu} V^{\mu} V^{\nu} = - \left( \frac{c}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \right)^2 + \left( \frac{v_x}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \right)^2 + \left( \frac{v_y}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \right)^2 + \left( \frac{v_z}{\sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}}} \right)^2 = \frac{- c^2 + v_x^2 + v_y^2 + v_z^2}{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}} = -c^2 P^2 = - m^2c^2\)
    Thus the two important equations of relativistic kinematics of a free particle are: \(E^2 = (mc^2)^2 + (\vec{p} c)^2 \\ E \vec{v} = c^2 \vec{p}\) or with 4-vectors: \(0 = c^2 + V^2 \\ m V^{\mu} = P^{\mu}\)

    From comparison of \(V^{\mu}\) and \(R^{\mu}\) we see that \(V^{\mu} = \gamma(\vec{v}) \frac{d R^{\mu}}{d t} = \frac{d R^{\mu}}{d \tau}\) which suggests why it transforms under the Lorentz transform like \(R^{\mu}\). It also suggests that the proper analogue of an acceleration vector is:
    \(\alpha^{\mu} = \frac{d V^{\mu}}{d \tau} = \gamma(\vec{v}) \frac{d V^{\mu}}{d t} = \gamma(\vec{v}) \begin{pmatrix} \frac{ d \gamma(\vec{v}) c}{dt} \\ \frac{ d \gamma(\vec{v}) \vec{v} }{dt} \end{pmatrix} = \gamma(\vec{v}) \begin{pmatrix} \frac{ d \gamma(\vec{v})}{dt} c \\ \frac{ d \gamma(\vec{v}) }{dt} \vec{v} + \gamma(\vec{v}) \frac{ d \vec{v} }{dt} \end{pmatrix} = \gamma(\vec{v}) \begin{pmatrix} \gamma^3(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c^2} c \\ \gamma^3(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c^2} \vec{v} + \gamma(\vec{v}) \vec{a} \end{pmatrix} = \begin{pmatrix} \gamma^4(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c} \\ \gamma^4(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c^2} \vec{v} + \gamma^2(\vec{v}) \vec{a} \end{pmatrix}\)

    Then
    \(\alpha^2 = - \gamma^8(\vec{v}) \frac{ \left( \vec{a} \cdot \vec{v} \right)^2 }{c^2} + \gamma^8 (\vec{v}) \frac{ \left( \vec{a} \cdot \vec{v}\right)^2 }{c^4} \vec{v}^2 + 2 \gamma^6(\vec{v}) \frac{ \left( \vec{a} \cdot \vec{v} \right)^2 }{c^2} + \gamma^4(\vec{v}) \vec{a}^2 = \gamma^6(\vec{v}) \frac{ \left( \vec{a} \cdot \vec{v} \right)^2 }{c^2} + \gamma^4(\vec{v}) \vec{a}^2 =\gamma^4(\vec{v}) \left( \vec{a}^2 + \frac{ \left( \vec{a} \cdot \vec{v} \right)^2 }{c^2 - \vec{v}^2} \right) \)
    \( \alpha^{\mu} V_{\mu} = -\gamma^5(\vec{v}) \left( \vec{a} \cdot \vec{v} \right) + \gamma^5(\vec{v}) \left( \vec{a} \cdot \vec{v} \right) \frac{ \vec{v}^2}{c^2} + \gamma^3(\vec{v}) \left( \vec{a} \cdot \vec{v} \right) = \gamma^3(\vec{v}) \left( \vec{a} \cdot \vec{v} \right) \left( -\gamma^2(\vec{v}) + \gamma^2(\vec{v}) \frac{ \vec{v}^2}{c^2} + 1 \right) = 0\) (for \(\alpha^{\mu}\) and \(V^{\mu}\) corresponding to the same event on the same particle's path.

    Geometrically, with the convention of the inner product we are using, \(\alpha^{\mu}\) and \(V^{\mu}\) are orthogonal and in many ways the acceleration 4-vector is analogous to the curvature vector of a path through Euclidean space while the velocity 4-vector is analogous to the tangent vector.

    So we have our analog with Newton's second law:

    \(\frac{d P^{\mu}}{d\tau} \; = \; f^{\mu} \; = \; m \, \alpha^{\mu} \; = \; m \, \frac{d^2 R^{\mu}}{ d\tau^2}\)

    So if we apply the relativistic force law for a massive particle of charge q, \( f^{\mu} = q F_{\nu}^{\mu} V^{\nu}\) we get:

    \(\gamma(\vec{v}) \begin{pmatrix} \frac{1}{c} \frac{d E}{dt} \\ \frac{d \vec{p}}{dt} \end{\pmatrix} = m \begin{pmatrix} \gamma^4(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c} \\ \gamma^4(\vec{v}) \frac{ \vec{a} \cdot \vec{v} }{c^2} \vec{v} + \gamma^2(\vec{v}) \vec{a} \end{pmatrix} = f^{\mu} = q F_{\nu}^{\mu} V^{\nu} = q \begin{pmatrix} 0 & \frac{1}{c} E_x & \frac{1}{c} E_y & \frac{1}{c} E_z \\ \frac{1}{c} E_x & 0 & B_z & -B_y \\ \frac{1}{c} E_y & -B_z & 0 & B_x \\ \frac{1}{c} E_z & B_y & -B_x & 0 \end{pmatrix} \begin{pmatrix} \gamma(\vec{v}) c \\ \gamma(\vec{v}) v_x \\ \gamma(\vec{v}) v_y \\ \gamma(\vec{v}) v_z \end{pmatrix} = q \gamma(\vec{v}) \begin{pmatrix} \frac{1}{c} \vec{E}\cdot \vec{v} \\ \vec{E} + \vec{v} \times \vec{B} \end{pmatrix}\)

    Or \(\frac{dE}{dt} = q \vec{E}\cdot \vec{v} \\ \vec{f} = \frac{d \vec{p}}{dt} = q \left( \vec{E} + \vec{v} \times \vec{B} \right)\)

    So the dynamics of energy and momentum exchange in terms of the electromagnetic tensor \(F_{\mu\nu} = \eta_{\mu \xi} F_{\nu}^{\xi}\) has exactly the same number of components (6) as \(\vec{E} \oplus \vec{B}\) (6), in detail they are the same components, and they lead to the same physical prediction of changes of momentum and changes of energy. Thus Farsight's discovery of the Minkowski quote was not a revelation about electromagnetic phenomena, but a unification of \(\vec{E} \oplus \vec{B}\) into a mathematical object that transforms under Lorentz boosts of the coordinate frame. This was analogous the "force-screw of mechanics" that Minkowski was talking about. Screw theory is a mathematical topic about 6-dimensional objects and one of these objects is called the "force-screw" for it's two 3-dimensional components. Similarly, the electromagnetic tensor permits dissection into two 3-dimensional components, the \(\vec{E}\) and \(\vec{B}\) vectors of Maxwell's theory.
     
  14. lpetrich Registered Senior Member

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    Whatever "displacement current" is supposed to be in Farsight physics. Because in "normal" physics, the displacement current is the time derivative of the electric field that gets added to the electric current in one of Maxwell's equations.
    A circling photon? However it's supposed to explain spin 1/2 and numerous other electron properties. Farsight, show that the circling-photon hypothesis reduces to the Dirac-field hypothesis, complete with the necessary mathematics.

    Seems like a lot of fumbling to me. Ampere's law is one of Maxwell's equations in the quasi-static limit, where one ignores time variation in the equations.

    Except that photons don't have self-interaction strong enough to confine themselves in that way.

    Farsight, you seem like you are running away from this hypothesis.

    Electric charge is not some winding number. It's a gauge-field interaction strength.

    They were making analogies to explain cross products. Their "screw" comments should not be considered revealed truth, and literal revealed truth at that.
    Give us spinor angular-momentum operators.
    A consequence of the interactions.
    It's gravity that does frame-dragging, not electromagnetism.
    Farsight, write it out in component-by-component fashion.
     
  15. lpetrich Registered Senior Member

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    117
    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 both classical and quantum electrodynamics, a particle's charge is a free parameter. Charge quantization is unexplained. After electroweak symmetry breaking, the Standard Model's particles' charges all follow this relation:
    charge = integer - (QCD triality)/3
    Triality is a QCD quantum number that adds modulo 3. Quarks are 1 and antiquarks 2. Colorless particles and gluons are all 0. Not surprisingly, hadrons are also 0.

    Before EWSB, the elementary fermions are divided up into left-handed and right-handed particles. In EWSB, the Higgs particle bridges that gap, producing the familiar EF's. The SM weak hypercharges are somewhat constrained by the Higgs-EF interaction. The electroweak interaction is two interactions, the weak-isospin one and the weak-hypercharge one. EWSB creates a mixture of them.
    Weak isospin = I
    Weak hypercharge = Y

    Gauge particles: I = 0 or 1, Y = 0
    Higgs particle: I = 1/2, Y = 1/2 (antiparticle has I = 1/2, Y = -1/2)
    Elementary fermions:
    I[sub]L[/sub] = 1/2
    I[sub]R[/sub] = 0
    Y[sub]R,up[/sub]= Y[sub]L[/sub] + 1/2
    Y[sub]R,down[/sub]= Y[sub]L[/sub] - 1/2

    We have:
    weak hypercharge = integer - (QCD triality)/3 + (weak isospin)
    Still unexplained.

    -

    Turning to Grand Unified Theories, the Georgi-Glashow SU(5) one fixes all the weak-hypercharge values. Other ones, like SO(10) and Pati-Salam SU(4)*SU(2)*SU(2) ~ SO(6)*SO(4), can add a factor proportional to B-L, (baryon number) - (lepton number). So GUT's may provide the solution to the quantization problem.
     
  16. Farsight

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    Who's moved my thread to alternative theories?
     
  17. Farsight

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    I don't have an in-detail account to hand I'm afraid. But think in terms of "wavefunction configuration" and TQFT. If you take a look at neutral pion decay the typical product is two gamma photons. The antiparticle is "self". It's akin to para-positronium which has spin 0. Note that ortho-positronium has spin 1, and the only difference between para and ortho positronium is that of "configuration". The lifetimes are very different of course, ortho lasting much longer, just as a charged pion lasts much longer than a neutral pion. But a charged pion isn't much like ortho-positronium. Again look to the decay and the typical π+ decay product is a muon neutrino and an antimuon which decays into a positron, a muon antineutrino, and an electron neutrino. Two neutrinos and an antineutrino don't resemble an electron. But neutrinos do have spin, so you can liken the charged pion to a backspinning positron. Do you know what I mean by that? Imagine you have a tornado rotating at 60rpm. Now rotate it bodily at 60rpm the other way.

    I don't have an in-detail account to hand I'm afraid. But can I say this: most of the particles with unequivocal mass are transient. The only particles that are totally stable are the electron and the proton and their antiparticles. The free neutron is almost stable, but not quite. All the rest are... ephemera.

    Again I don't have an in-detail account to hand. But I will say this: the neutrino is more like a photon than an electron.

    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?

    No problem.

    My pleasure. I'm sorry that we have had no effective moderation of the abuse, and yet the thread has been moved to "alternative theories". Sigh.
     
  18. Farsight

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    Why do you keep asking me irrelevant questions? Because B is the curl or rot or rotor or "vorticity" of the potential, and when you've got two charged particles moving apart, there isn't any. Press two electrons together and let go. They move apart, the net conduction current is zero, and they don't "generate" a magnetic field because they each have an electromagnetic field.
     
  19. PhysBang Valued Senior Member

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    A paragon amongst human beings.
     
  20. PhysBang Valued Senior Member

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    Could you please explain this using your diagrams?
     
  21. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Since you've acknowledged you didn't use any physics (math) to generate your picture and it has no mathematical meaning, you are lucky it didn't end up next to Victor's thread in Pseudoscience!

    More to the point, if Minkowski had drawn the picture, it would have been appropriate for the physics section. Since you drew it, it is at best "alternative".
     
  22. btr Registered Member

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    I apologise for the late reply. I've been a little busy.

    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.

    Please Register or Log in to view the hidden image!



    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?

    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.

    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 liens - of that 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:

    \( \nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t} \)​

    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.

    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).

    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.

    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
     
  23. btr Registered Member

    Messages:
    93
    No problem, and thanks.
     

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