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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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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