The EM wave impedance of free space is said to be ~377 ohms and represents the ratio of Electric field strength (E) to magnetic field strength (H). So that: $$ \frac{E}{H} = ~377 \,\Omega $$ When considering the photon, what aspect of the photon reflects this $E/M$ ratio or wave impedance?
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$\begingroup$ Does this website help? If so I can try to find some of the references, etc. $\endgroup$– audenCommented Jul 14, 2016 at 16:06
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$\begingroup$ @heather I will read the paper to see if it helps. It may. $\endgroup$– K7PEHCommented Jul 14, 2016 at 16:13
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$\begingroup$ @heather My first fast read of the paper raises more questions. Further study would require that I investigate the basis of his arguments which seem fishy to me -- not sure I trust his paper. However, the paper does raise some other interesting questions about physical or quantum origins of the permittivity/permeability of free-space. $\endgroup$– K7PEHCommented Jul 14, 2016 at 16:31
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$\begingroup$ okay. That was really the only thing I found. I'll look for some of the sources online so if you want you can look to where he based his arguments. $\endgroup$– audenCommented Jul 14, 2016 at 16:33
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$\begingroup$ The impedance exists regardless of the attributes of a particular charged particle, so that wouldn't make a difference. Although i don't understand it enough to give any details, I believe you can use en.wikipedia.org/wiki/… as a starting point. Specifically, consider the portion of an electric field that contributes to the relativistic magnetic fields (it has to do with the speed of the particle carrying the field). $\endgroup$– DigiprocCommented Jul 14, 2016 at 16:33
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Sorry all, but the impedance of free space is set by the choice of units. If you use natural or Planck units where $c$ and $\hbar$ and other constants are 1, the Planck impedance comes out to be about 30. More to the point, Planck units sets Coulomb's constant $\frac{1}{4 \pi \epsilon_0} = 1$. This represents a choice of units.
To the point. If instead you set the permitivity $\epsilon_0 = 1$, along with $c=1$, then the permeability $\mu_0$ is also 1, and the impedance of free space is also 1.
It is simply units, there is no physics behind the 377 ohms.
See Wikipedia at https://en.m.wikipedia.org/wiki/Planck_units
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$\begingroup$ Of course that there is no physics behind the 377 ohms. But, that is not my question -- the units themselves are of zero interest to me. It is the interaction that my question was about but yesterday I realized that I did not ask the question specific enough. Still processing some of the other resources and other related questions posted here. $\endgroup$– K7PEHCommented Jul 15, 2016 at 14:20