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The diagram below shows that the propagation of light is perpendicular to the direction of the E and B fields. This explanation makes sense when looking at this two dimensional representation, shown below. However, with a little imagination the E and B fields can be viewed in three dimensions as spheres created from point sources. Once this is done the clear direction of the A and B fields becomes unclear. My question is: How does light travel perpendicular to a sphere?

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    $\begingroup$ Light propagates radially from a point source. And still does if happens that the point is on spherical surface. $\endgroup$ – Alchimista Jan 30 '18 at 15:17
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    $\begingroup$ @Lambda - Sure, but the point is that so long as the symmetries share a common origin (e.g., two concentric spheres), then all radial vectors will be locally orthogonal to the spherical surfaces. $\endgroup$ – honeste_vivere Jan 30 '18 at 15:49
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    $\begingroup$ If all vectors of the A and B fields are orthogonal then what is the direction of the propagation of the EM wave? $\endgroup$ – Lambda Jan 30 '18 at 16:08
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    $\begingroup$ Are you saying that there is no electric-monopole radiation? $\endgroup$ – JEB Jan 30 '18 at 16:09
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    $\begingroup$ In other words if the sphere is the source , each point on it radiates in all outpointing direction. Is your premise as in the title that doesn't hold, if I understood it correctly. $\endgroup$ – Alchimista Jan 30 '18 at 17:52
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  • Unless you're talking about light ray coming from one vibrating charge only, the so-called point source usually understood to be photons come from small region that can be approximated as spherical boundary. The EM waves go out incoherently and randomly polarized. See more on polarization and coherence.

    enter image description here

    All you observed is rapid switching of wave pulses emitted in all radial directions with random orientation of polarization governed by $\mathbf{k\times E=\omega B}$.

    See another question here.

  • By rotating polaroid sunglasses (or polarizer), you can see the change in brightness of the blue sky.

    enter image description here

    Air molecules scatter the unpolarized sunlight and the EM wave re-radiate spherically outwards but only the transverse direction can be transmitted.

    enter image description here

    enter image description here

  • In short antenna, the EM wave is transmitted efficiently along the direction perpendicular to the direction. Meanwhile, no energy is transmitted along the the axis of symmetry, i.e. the longitudinal direction.

    enter image description here

    See also the animation in another question here.

    The $E$ and $B$ fields produced by an oscillating dipole $\mathbf{p}=p_0 e^{-i\omega t} \mathbf{e}_z$ at the origin, in spherical polar coordinates, are given by:

    \begin{align} \begin{pmatrix} E_r \\ E_{\theta} \\ E_{\phi} \end{pmatrix} &= \begin{pmatrix} -\frac{p_0\cos \theta}{2\pi \epsilon_0 r} \left( \frac{ik}{r}-\frac{1}{r^2} \right) \\ -\frac{p_0\sin \theta}{4\pi \epsilon_0 r} \left( -k^2-\frac{ik}{r}+\frac{1}{r^2} \right) \\ 0 \end{pmatrix} e^{i(kr-\omega t)} \\[5pt] \begin{pmatrix} B_r \\ B_{\theta} \\ B_{\phi} \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \\ \frac{i\omega \mu_0}{4\pi} \frac{p_0\sin \theta}{r} \left( -ik-\frac{1}{r} \right) \end{pmatrix} e^{i(kr-\omega t)} \end{align}

    The leading terms are \begin{align} \mathbf{E} &= \frac{k^2}{4\pi \epsilon_0 r} p_0 \sin \theta \, e^{i(kr-\omega t)} \mathbf{e}_{\theta} \\ \mathbf{B} &= \frac{\mu_0 \omega k}{4\pi r} p_0 \sin \theta \, e^{i(kr-\omega t)} \mathbf{e}_{\phi} \end{align}

    enter image description here

    Loosely speaking, the anisotropic "spherical" wavefront decays as $\dfrac{1}{r}$ and the amplitude is maximal for $\theta=90^{\circ}$ and zero when $\theta=0^{\circ}$ or $180^{\circ}$ (i.e. the longitudinal direction). This is due to resolution of vector into the component perpendicular to the direction of propagation, viz., $\mathbf{k}=k\, \mathbf{e}_r$ (i.e. the direction towards the observer).

    N.B. For cylindrical (or circular) wave, $$\psi(r, \phi, z, t) \propto \frac{e^{i(kr-\omega t)}}{\sqrt{r}}$$

  • Miscellaneous pictures that may help

    enter image description here

    enter image description here

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  • $\begingroup$ What I am considering is not the wave as a whole but rather the structure of the wave. The rise and fall of the E and B fields that propel the wave. These fields are spherical. So how can the these rising and collapsing fields create a wave that is perpendicular to the field. Every where on a sphere is the same. The are many right angles to the sphere. What gives the wave motion in one direction over another? $\endgroup$ – Lambda Jan 30 '18 at 18:00
  • $\begingroup$ Say for light bulb, it's unpolarized (do test it with sunglasses). One light beam is mixed with light pulses with random polarization. What you observe is average. But for antenna, EM waves are polarized. You can't receive radio signal at the top of the transmission station (and radio antenna is always aligning vertically). Please refer to the link here. $\endgroup$ – Ng Chung Tak Jan 30 '18 at 18:14
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    $\begingroup$ Actually, the $E$ and $B$ fields emitted from attenna do have $\theta$ and $\phi$ dependence. $\endgroup$ – Ng Chung Tak Jan 30 '18 at 18:21
  • $\begingroup$ I guess there is something I am missing because none of the comments or answer seems to address the question. EM waves travel by leap frogging from E to B fields. Because the E and B fields are spheres why/how does the leap frogging travel in straight lines? Spheres have no direction. Any direction is perpendicular to the sphere. $\endgroup$ – Lambda Feb 1 '18 at 19:14
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    $\begingroup$ Strictly speaking, EM wave can only have plane or circular wave front but not spherical wave front. Ripples at water surface is circular. Laser beam is plane wave. You observe light from a point source transmitting radially is a macroscopic view. The point source contain many many oscillators. Say charge $A$ at $t=10^{-8}$ second oscillates in direction $(1,0,0)$, and then charge $B$ at $t=2\times 10^{-8}$ second oscillates in direction $(0,-0.8,-0.6)$, and then charge $C$ at $t=2.5\times 10^{-8}$ second oscillates in direction $(0.1,-0.2,-0.975)$ and so on. $\endgroup$ – Ng Chung Tak Feb 1 '18 at 19:56

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