Mysterious property of light.

Discussion in 'Physics & Math' started by ash64449, Nov 13, 2012.

  1. Walter L. Wagner Cosmic Truth Seeker Valued Senior Member

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    I took the original post to mean, as for an example, that a beam of electrons can be bent or deflected by an electric or magnetic field when the beam is passed through that field in a vacuum, but a beam of visible light would not exhibit such bending with the same electric or magnetic field. The various effects of light bending (because of an electric or magnetic field) when in the presence of a medium are of course well known, and I believed that that was not what the original post was speaking about.

    However, your link is quite interesting and I would appreciate more information about that.

    And it also appears that physics can make strange bedfellows. I'm not certain where RO comes off about credentials, but I have nearly 2,000 posts on this forum which evidence a degree of knowledge of science in various fields in which I post (biology/botany, linguistics, radioactive materials, etc.), acquired during various professional pursuits.
     
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  3. prometheus viva voce! Registered Senior Member

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    Mod note: 16 off topic posts have been moved here
     
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  5. Farsight

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    See for example this:

    "The beam is asymmetric, with one bright region at the center and a series of progressively dimmer patches on one side of the central spot. But rather than propagating in a straight line, the entire pattern of bright and dark patches curves toward one side. At the same time, the width and intensity of each patch remains essentially constant, even after an ordinary beam would have dropped to nearly half its original intensity and spread to several times its original width."

    I think it's important to remember this when looking at say two-photon physics on wikipedia:

    "From quantum electrodynamics it can be found that photons cannot couple directly to each other, since they carry no charge, but they can interact through higher-order processes. A photon can, within the bounds of the uncertainty principle, fluctuate into a charged fermion-antifermion pair, to either of which the other photon can couple."

    This is saying pair production occurs because pair production occurs. Because a photon, all on its own, spontaneously transforms itself into an electron and a positron, which then magically turn back into a single photon. This explanation is wrong, it's treating a calculation method as experimental fact, when it doesn't square with experimental fact. Put a magnetic field around a single laser beam, along with the modern equivalent of a cloud chamber, and you don't detect any electrons and positrons. You only get pair production with two crossing laser beams. Search on SLAC pair production for details. Experimental fact tells you light interacts with light.
     
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  7. rpenner Fully Wired Valued Senior Member

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    The article is about \(\vec{E}(t,x,y,z) = \begin{pmatrix} 0 \\ E_0 \; J_{\alpha} \left( \frac{\omega}{c} \sqrt{x^2 + z^2} \right) \; \cos( \phi_0 + \alpha \tan^{\tiny -1} \frac{z}{x} - \omega t ) \\ 0 \end{pmatrix}\) which requires a space-filling, time varying current density of:
    \(\frac{\partial \vec{J}(t,x,y,z)}{\partial t} = - \frac{1}{\mu_0}\nabla \times \nabla \times \vec{E}(t,x,y,z) - \frac{1}{\mu_0 c^2} \frac{\partial^2 \vec{E}(t,x,y,z)}{\partial t^2} = \begin{pmatrix} 0 \\ \frac{E_0}{\mu_0} \; \left( \frac{\omega^2}{4 c^2} \left( J_{\alpha-2} \left( \frac{\omega}{c} \sqrt{x^2 + z^2} \right) + J_{\alpha+2} \left( \frac{\omega}{c} \sqrt{x^2 + z^2} \right) \right) + \frac{\omega}{2 c \sqrt{x^2 + z^2} } \left( J_{\alpha-1} \left( \frac{\omega}{c} \sqrt{x^2 + z^2} \right) - J_{\alpha+1} \left( \frac{\omega}{c} \sqrt{x^2 + z^2} \right) \right) + \left( \frac{\omega^2}{2 c^2} - \frac{\alpha^2}{x^2 + z^2} \right) J_{\alpha} \left( \frac{\omega}{c} \sqrt{x^2 + z^2} \right) \right) \; \cos( \phi_0 + \alpha \tan^{\tiny -1} \frac{z}{x} - \omega t ) \\ 0 \end{pmatrix} \)
    which is peculiar for a purported vacuum solution.

    //Edit:
    Or are they claiming a vacuum solution? That's the easiest way to read "self-bending solution" but is it the only one?
     
  8. Walter L. Wagner Cosmic Truth Seeker Valued Senior Member

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    Quotes at the bottom of the article:

    "As with the Bessel beam’s diffraction-free “propagation,” light doesn’t actually propagate along the curved path. The beam is the pattern created by interference of light from the 500,000 carefully-phased pixels of the SLM. Still, Mordechai Segev of the Israel Institute of Technology in Haifa is impressed. “It’s a beautiful piece of work,” he says. Segev envisions using similar phasing tricks in situations where multiple light beams are needed to create a specific light effect at a precise distance from the source. “If you engineer the phase the right way, you will only have white light where the pulses meet. Imagine if you did something like that in the atmosphere,” where Airy-like beams could improve LIDAR, a type of laser-based radar used to study the atmosphere."

    In other words, it is a carefully constructed interference pattern that mimics the bending of light, if my reading of the article is correct. Rpenner, any comments?
     
  9. rpenner Fully Wired Valued Senior Member

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    Yes, I understand that about the 2007 article. I was responding to the claims about the 2012 article. http://physics.aps.org/articles/v5/44 (Link to free PDF in upper right corner.)
     
  10. Hellenologophobia Registered Senior Member

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    http://en.wikipedia.org/wiki/Electromagnetic_wave_equation#Multipole_expansion
     
  11. kevinalm Registered Senior Member

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    As to your first question, it can be explained as a fundamental property of mathematics. Specifically calculus. Maxwell's equations, which govern electromagnetism, are "differential" equations. It is a fundamental mathematical fact that the derivative of two funtions that differ only by a constant amount (f(x) = g(x) + k) have the same derivatives. (The derivative of f(x) = the derivative of g(x)) Therefore, static electric and magnetic fields can have no effect on the solution to the E/M wave equations, ie. on the E/M wave.
     
  12. Farsight

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    Think about what SLAC did. They arranged things so that light travelled through light. The result was an electron (and a positron). You can diffract an electron. It has a wave nature, magnetic moment, spin ½, and so on. People tend to think of it as some static thing with a static field, but it isn't, it's dynamical. The same is true of a standing wave in a cavity. It looks like it's static, but whip away one of the sides and it comes out at c from what looks like a standing start. It wasn't a standing start. It was always going at c, but in this direction → and this direction ← at the same time. It's something like water waves starting from each end of a tank. Where they pass each other in the middle they're out of phase, so the water there is flat. But they go right through each other and keep going. Now imagine a very short tank, so short that it consists only of the middle bit where the water is flat. Anyway, annihilate an electron with a positron and it's isn't all that different. Again see atomic orbitals and pay attention to "electrons exist as standing waves". People are happy with an electron in an atomic orbital existing as a standing wave, but for some reason they aren't happy with a free electron at rest existing as a standing wave. It has a wave nature, magnetic moment, spin ½, and so on. It's a standing wave all right. And the spin ½ tells you it's somehow like a moebius strip. Draw a line around a moebius strip, and think of that line as light. What have you got? Light travelling through light. That's all you ever had. Hence the Williamson / van der Mark electron. People tend to dismiss this, which I think is a great shame. Percy Hammond's The role of potentials in electromagnetism doesn't seem to get any attention either. Look at the bit near the end-note: "We conclude that the field describes the curvature which characterizes the electromagnetic interaction".
     
  13. rpenner Fully Wired Valued Senior Member

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    Aha.
    \(\frac{v^2}{4}\left( J_{u-2} (v) + J_{u+2} (v) \right) + \frac{v}{2} \left( J_{u-1} (v) - J_{u+1} (v) \right) + \frac{v^2}{2} J_{u} (v) = u^2 J_{u} (v)\)
    Yeah, good luck finding that identity spelled out.
    so \(\frac{\partial \vec{J}(t,x,y,z)}{\partial t} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)
    Bing, bang, boom -- vacuum conditions satisfied for arbitrary value of \(\alpha\)
    The condition \((\nabla^2 + k^2) \vec{E} = 0\) is equivalent to my imposing vacuum conditions (\(k \equiv \frac{\omega}{c}\))
     
  14. tesla2 Banned Banned

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    I think that light it can be many small ball
     
  15. Hellenologophobia Registered Senior Member

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    http://en.wikipedia.org/wiki/Particle_in_a_spherically_symmetric_potential#Vacuum_case

    http://en.wikipedia.org/wiki/Bessel_function#Spherical_Bessel_functions:_jn.2C_yn
     

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