The screw nature of electromagnetism

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

  1. Farsight

    Messages:
    3,492
    Yes. I have to say though that IMHO all the source-and-sink stuff is just so misleading. Next time you have a bath, study the water coming from the tap, then study the vortex as the water goes down the plughole. Then get yourself a plate and find a pond, and make a Falaco soliton. There is no source, and no sink either.

    OK noted.

    I think it's important to try to "grasp" it. Don't just shut up and calculate. Imagine space is a stiff lattice. Reach in with your right hand and grab it, then turn it clockwise by one full turn. Then reach in round the side with your left hand and turn it anticlockwise by half a turn. Now remove your hands, and the lattice stays hitched. That's what the electron electromagnetic field "looks" like. Try not to think of the electron as some little round thing in the middle. The whole thing is the electron, and it isn't static, it's dynamical, it's a wave going round and round such that a propagating field variation looks like standing field.

    I'll look into it. Thanks.

    I don't think it's novel, there's been a lot of "electron model" papers over the years, but they tend to go unnoticed. For example http://arxiv.org/abs/physics/0512265 isn't something that appeared in Nature. As for the standard one, there is no electron model. Which is why I'm forever saying the work required is "within the Standard Model", not "beyond the Standard Model".

    Here, for your amusement and delectation, take a look at the spindle-sphere torus again:

    Please Register or Log in to view the hidden image!



    You see a sphere. Whenever you see a sphere you think 4π. You also see a rotation, and knowing as you do about pair production, c springs to mind. But there's another orthogonal rotation at "half the rate" as per the Dirac's belt Moebius strip. You know that Planck length is l=√(ћG/c³). Replace √(ћG) with 4πn where n is a suitable value with the right dimensionality. You still have your Planck length l=4πn/√(c³). You can write that as l=4πn/c[sup]1½[/sup]. Now set n to 1, and work out 4πn/c[sup]1½[/sup]. It isn't a perfect answer because of the binding energy but it ought to be close enough to be intriguing. People complain about units, but when you dig down to the fundamentals you've got waves in space and that's it. You end up with harmonics and ratios. Speaking of which, try c[sup]½[/sup] / 3π. Forget about the dimensionality, imagine you've adjusted for that with some other n=1 artifice.

    c[sup]½[/sup] = 17314.5158177
    3π = 9.424778
    c[sup]½[/sup] / 3π = 17314.5158177 / 9.424778 = 1837.12717877

    Again not perfection, but close enough to be interesting.

    Doo doo doo doo, doo doo doo doo. Welcome. To the Twilight Zone!
     
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  3. Farsight

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    My main argument is that E and B denote the forces that result from Fμv interactions. Note that a tensor is a matrix. It isn't what the field is. And the map is not the territory.

    I utterly, utterly understand it. And around here, I'm probably the only one who does. The map is not the territory.

    You know what I mean. Rpenner has no counterargument, and he comes out with smoke and mirrors in an attempt to fool other readers. He tries to pull the Emperor's New Clothes trick in lieu of carefully-crafted counterargument backed by references and logic to counter mine. It never works.
     
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  5. PhysBang Valued Senior Member

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    Note that a tensor is not a matrix. This is a basic misunderstanding you have, Farsight.

    Please, Farsight, learn the mathematics. You have had over ten years of stubbornly refusing to start to learn mathematics and continued bannings from site after site. Why not try a new tactic and learn the physics?
     
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  7. rpenner Fully Wired Valued Senior Member

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    4,833
    I agree.
    I disagree. I think if \(v^{\nu}v_{\nu} =c^2\) and \(B_{\mu} = \frac{1}{c} \left(*F \right)_{\mu\nu} v^{\nu} = \begin{pmatrix} \gamma \vec{v} \cdot \vec{B}/c \\ \gamma \left( \vec{B} - \frac{\vec{v} \times \vec{E}}{c^2} \right) \end{pmatrix}\) but by the transformation laws the expected B-field for that observer is a vector in his rest coordinate system, \(\vec{B}' = \gamma \left( \vec{B} - \frac{\vec{v} \times \vec{E}}{c^2} \right) - \left(\gamma - 1\right) \frac{ \vec{v} \cdot \vec{B}}{\vec{v}^2} \vec{v}\) where I am optimistic that you might recover this last expression by Lorentz transformation of the components of the \(B_{\mu}\) 4-vector in the original coordinates.
    I agree. But I think this would be \(B'_{\mu} = \left(\Lambda^{\tiny -1} \right)_{\mu}^{\nu} \left(*F \right)_{\nu\xi} v^{\xi}\) that has \(B'_0 \propto \vec{B}' \cdot \vec{v}' = \vec{B}' \cdot 0 = 0\).
    I agree.
     
  8. rpenner Fully Wired Valued Senior Member

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    4,833
    As has been repeatedly explained to Farsight, \(K = \frac{c m_e^2}{9 \pi^2 m_p^2}\) just happens to be a physical quantity close to 1 m/s in SI units. (I get 1.00106176 m / s). But those SI units reflect purely human choices. What Farsight is asking us to do by ignore the units is really to cancel the units like this: \(\frac{1}{3 \pi} \sqrt{\frac{c}{K}} = \frac{m_p}{m_e}\) which is dimensionally correct and not mysterious. But the numerological trick of ignoring units ignores that c may be measured in a wide variety of units.

    In cgs \(\frac{1}{3 \pi} \sqrt{\frac{c}{1 \textrm{cm}/\textrm{s}}} = \sqrt{\frac{K}{1 \textrm{cm}/\textrm{s}}} \frac{m_p}{m_e} \approx 10 \frac{m_p}{m_e}\) and in the derived foot-pound-second system, \(\frac{1}{3 \pi} \sqrt{\frac{c}{1 \textrm{ft}/\textrm{s}}} = \sqrt{\frac{K}{1 \textrm{ft}/\textrm{s}}} \frac{m_p}{m_e} \approx 1.8 \frac{m_p}{m_e}\)

    In natural units, c = 1 and the trick fails utterly.
     
  9. OnlyMe Valued Senior Member

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    3,914
    Without any comment on the discussion generally...,

    I have never thought of c =1 as a natural unit. It has always seemed more a matter of convention and convenience. Can you explain?
     
  10. Russ_Watters Not a Trump supporter... Valued Senior Member

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    5,051
    The term "natural unit" has a specific definition:

    http://en.m.wikipedia.org/wiki/Natural_units
     
  11. lpetrich Registered Senior Member

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    117
    I've snipped a lot of hand-waving arguments, since that's a poor substitute for doing the math.
    So the Dirac equation is not good enough for you? There are oodles of evidence that electrons are plain Dirac fields with the interactions specified in the Standard Model. The previous occupant of the LHC's tunnels, the LEP, had produced lots of it -- collisions of electrons with energies up to about 110 GeV. No departures from the Standard Model were observed to within experimental accuracy.

    (attempts to find values of various fundamental constants with number manipulation...)
    Complete with carelessness about units.

    (me on Farsight's lack of understanding of space-time unification)
    That's a big laugh. Let's see you show us that you understand what a Lorentz boost is and how it differs from a Galilean boost.

    It's only "smoke and mirrors" if one does not understand the mathematics. I had no trouble doing so. Farsight, what is your excuse for not learning the mathematics?
     
  12. Farsight

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    3,492
    I won't be posting any further in this section of the forum.
     
  13. btr Registered Member

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    93
    To recap, mostly for my own benefit:

    I think where we differ is that you are comparing the spatial components of the 4-vector B in some other reference frame with the components of the 3-vector B in the observer's frame. I very deliberately only introduced a single reference frame in my statement: that of the intertial observer with 4-velocity u. When I introduced that observer and then started talking about components of the various fields, I meant specifically the components in their reference frame; as no other reference frame was involved in the problem, I allowed myself to be a bit over-terse, perhaps, so I have to apologise that I wasn't more explicit. Hopefully that clears things up!
     
  14. btr Registered Member

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    93
    I'm pretty certain I didn't bring up the issue of sources and sinks in fluid flows (the topic of your link), but OK.

    I appreciate a good analogy, but they can never truly be a substitute for mathematics. It is only the mathematical form of a theory which can be tested against experiment. I have to say, without wishing to sound rude, that I am doubtful that the analogy you present is helpful. Certainly to me, it doesn't seem to convey the content of Maxwell's equations in a clearer way than simply picturing lines of E and B threading through space (or tubes of F, if you prefer).

    Thanks, I'll read it. Is that representative of your model?

    That's not exactly true. There is a very good model; it is just very abstract in nature and difficult to put into ordinary words.

    Skipping ahead a bit...

    I think I see what you are driving at; it's numerically close to the proton-electron mass ratio. However, the dimensions are all over the place, and your hidden factor n will only numerically equal 1 (with some units) in S.I. units. It will have quite horrible values in other systems. Perhaps as an exercise, you could try expressing it in terms of pounds and inches?
     
  15. Farsight

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    3,492
    No. The n is then something other than 1. Ratios don't change. Thompson and Tait are still the guys who coined the phrase spherical harmonics. See above, I've decided I won't be posting in this section any further.
     
  16. lpetrich Registered Senior Member

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    117
    However, the value of n is to be justified.
    Seems like a prelude to more book-thumping. Farsight, can you explain the math behind spherical harmonics?
    As you wish.

    More seriously, lots of people try to find the proton-to-electron mass ratio, the fine structure constant, etc. using number manipulations, and some of them get rather close. But if one has a lot of numbers and does a lot of trial and error, one can get very good results. Furthermore, those quantities are not exactly fundamental. Nucleons' masses are mostly due to quarks and gluons' kinetic energies, and those in turn are due to color confinement. The electron's effective mass changes over energy as a result of vacuum-polarization effects. It's rather different at GUT energy scales, or more properly, its Higgs-particle coupling is rather different. The FSC is about 1/137 at low energies, and about 1/128 at the W particle's mass. Further up in energy, and it isn't very well-defined -- the electromagnetic field is a mixture of weak-isospin and weak-hypercharge fields.

    So one has to extrapolate to GUT energies and work from there.
     
  17. OnlyMe Valued Senior Member

    Messages:
    3,914
    O.K. That is understood...

    As I have mentioned, it has been a long time since I have worked with the math being presented. I accept rpenner's math skills and thus often don't go to the effort of working through some of it.., just accepting his skill.., and interpretation. Thus, not looking closely at the math in that post, I misunderstood the purpose of his examples... My bad!
     
  18. btr Registered Member

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    93
    If you mean under arbitrary changes of units, you are right if and only if the ratio is dimensionless. That's the issue.

    To fix up your formula you could (just like you hinted, and as treated in detail above by rpenner) write it as something like

    (c/n)[sup]1/2[/sup]/3π = m[sub]proton[/sub]/m[sub]electron[/sub]

    where n is 1/299474489[sup]th[/sup] of the speed of light. This makes both sides bona fide dimensionless quantities, and the equation is therefore true in all rational systems of units. However, it immediately raises the question what is so special about 1/299474489[sup]th[/sup] of the speed of light?

    That's a pity. I hope you take the time to look at the things I mentioned earlier on.
     
  19. btr Registered Member

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    93
    Sir Arthur "Adding-One" was a notable example, a respectable physicist who turned to the numerological dark side in his quest to derive the value of 1/α.
     
  20. btr Registered Member

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    93
    Farsight won't be returning to this thread, unfortunately, so I've taken the liberty of following up on my own request from a few pages back, on the subject of the spiral diagram:

    Here is what you get if you literally add a radial inverse-square electric field to a dipole magnetic field, in the spirit of the spiral diagram from earlier:

    View attachment 7092

    Code:
    StreamPlot[{x/(x^2 + y^2)^(3/2) + x*y/(x^2 + y^2)^(5/2), y/(x^2 + y^2)^(3/2) + ((2/3)*y^2 - (1/3)*x^2)/(x^2 + y^2)^(5/2)}, {x, -1, 1}, {y, -1, 1}]
    To be clear, I don't claim that this has any useful meaning. Adding E to B makes no physical sense, and immediately loses half of the data at each point in space. There is no way to calculate forces on particles from the combined field any more (neither magnitude nor direction), nor can you predict the future evolution of the fields given the current value and rate of change of the combined field at each point. As a result, combining the two fields with this sort of strategy cannot faithfully represent the true electromagnetic field.

    Features of the combined field:
    • At large distances, the dipole component is negligible in comparison with the inverse square component, and approximate spherical symmetry is seen in the combined field. The system looks very like an ordinary point source at this range.
    • At intermediate distances, the combined field begins to deviate markedly from spherical symmetry. It still has rotational symmetry about the dipole axis, however.
    • At very close distances, the dipole component dominates. On the dipole axis, the field still points away from the source (and its magnitude varies approximately as the inverse cube of the distance); however, in the horizontal plane, the field points "downwards" instead of away from the source.
    • As you approach the dipole along its axis from "below" the horizontal plane, there is a point where the field actually changes direction from outwards to inwards! In the attached plot, that happens at (x, y) = (0, -2/3).
     
  21. Beer w/Straw Transcendental Ignorance! Valued Senior Member

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    6,549
  22. btr Registered Member

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    93
    I've tried to post a reply in that thread (for some reason Farsight has addressed me there, in reply to my post here); I was informed by the forum software that my post (the second I have ever made on those forums) was pending approval. Odd, since I did not get that with my first post.

    Anyway, to Farsight, if he is reading this: I have already commented on your formula for the proton/electron mass ratio in this thread, as have lpetrich and rpenner (see above).
     
  23. rpenner Fully Wired Valued Senior Member

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    4,833
    Perhaps they have an anti-spam measure against posts with links from relatively new users.
     

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