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I was wondering what the near-field expression of the Electric field for a cylindrical wave is. I know that in the case of a spherical wave, as shown in Jackson (9.18), the far field has a 1/r dependance, as expected for the energy to be conserved, and the near field has a 1/r^3 dependance.

In the case of a cylindrical wave the Electric field in the far-field has a 1/sqrt(rho) dependance and I was wondering what the near-field dependance is.

I unsuccessfully attempted to reproduce Jackson's procedure for a cylindrical wave.

I then looked at a book by Roger Harrington considering an infinite fillament as a source. His fields are written in terms of the cylindrical Hankel functions of the first kind for both the infinite fillament case, as well as generally.

Based on his fields I get the Poynting vector from the corresponding Electric and Magnetic field expressions. By then considering the assymptotic form of the Hankel functions I get a 1/rho dependance, which is what one would expect for the far field (I get this dependance for both the assymptotic forms k * rho -->0 and k * rho-->infinity). It seems like by considering his procedure you cannot obtain the near-fields as I also followed and continued his case of a spherical wave and only got the 1/r far-field dependance.

While it seems like this should be a very simple problem I can't seem to find any literature mentioning it.

Any help/insight would be appreciated. Thanks :)

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In general, the near fields are given by the static/quasi static fields. In general you should not expect the same scaling for cylindrical waves and spherical waves.

For simple setups, you can calculate the waves analytically. For example, in this previous answer, I detailed the case of an infinite wire of oscillating current.

Hope this helps.

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  • $\begingroup$ Thanks, I believe this is what I wanted. So in the case you described, when taking the limit r --> 0 in your expressions and considering DLH rule, both the magnetic and the electric fields end up having a 1/r dependance, leading at a Poynting vector proportional to 1/r^2 in the near-field. $\endgroup$
    – arsenis
    Commented Oct 26, 2023 at 14:58
  • $\begingroup$ Actually, the Poynting vector rather diverges as $\frac{\ln r}{r}$, but you still get an infinite flux of energy around the wire. $\endgroup$
    – LPZ
    Commented Oct 27, 2023 at 23:06

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