Is E/B = c for spherical EM Wave in Vacuum?

[LarryS]

TL;DR Summary

Is E/B = c for spherical EM Wave in Vacuum?

In classical EM, consider an EM plane wave traveling in free space. The ratio of the amplitude of the electric field to the amplitude of the magnetic field is the velocity of the wave, the speed of light.

Is the above also true if the wave is spherical, expanding from a point source, as in a light cone?

Thanks in advance.

 

[Baluncore]

Or is E/H related to the intrinsic impedance of free space ≈ 377 ohms?
https://en.wikipedia.org/wiki/Wave_impedance
https://en.wikipedia.org/wiki/Impedance_of_free_space

 

[alan123hk]

Summary: Is E/B = c for spherical EM Wave in Vacuum?

Is the above also true if the wave is spherical, expanding from a point source, as in a light cone?

I don't think this should matter.
Light travels in straight lines in free space, and spherical waves are also composed of countless straight lines propagating in different directions. As long as you zoom into see the detail anywhere and imagine that it's just one of the straight lines, the conclusion should be the same.

 

[Meir Achuz]

yes, E=cB (or E=B in Gaussian units) for spherical waves.

 

[Delta2]

I would like to add that in modern technology the transmitted waves (for radio, television, mobile phones)are generated from dipole antennas (or array of dipole antennas) and the fields of a dipole antenna kind of look like spherical waves but satisfy the relation $E=cB$ only in the far field.

 

[LarryS]

I would like to add that in modern technology the transmitted waves (for radio, television, mobile phones)are generated from dipole antennas (or array of dipole antennas) and the fields of a dipole antenna kind of look like spherical waves but satisfy the relation $E=cB$ only in the far field.

Are you saying that because in the far field limit the wave type approaches planar?

 

[Delta2]

Are you saying that because in the far field limit the wave type approaches planar?

The far field takes the form of a radiating spherical wave, but I said it because that's what the formulas for the far field of a dipole antenna tell us. You might find of use the following wikipedia article

https://en.wikipedia.org/wiki/Dipole#Dipole_radiation

 

[vanhees71]

The said relation only holds for plane waves, i.e. in the "far zone". This becomes clear from the multipole expansion.

 

[Meir Achuz]

The wave front is spherical, even in the far zone. Otherwise, there would be no angular distribution.

 

[vanhees71]

Even a spherical wave looks asymptotically like a plane wave. Just look at the multipole expansion in the far-field region, $r \rightarrow \infty$, assuming that the source (charge-current distribution) is localized around the origin.

 

[Meir Achuz]

"looks"
Dipole:
\begin{equation}
{\bf E}=-{\bf{\hat r}\times B}=
\frac{k^2e^{ikr}}{r}[{\bf p-( p\cdot{\hat r}){\hat r}}]
\end{equation}