This paper I was reading said: "At a high level, the propagation speed of EM waves through conductive materials is SLOWER than their speed in air".

  1. Don't conductive materials intrinsically have less impedance than air as propagation mediums, to be conductive in the first place. Wouldn't the ray flow through this conductor faster with less 'resistance'?

  2. Do EM waves only travel at the speed of light in a vacuum?

Or am I Stupid??

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    \$\begingroup\$ You are not stupid. If you were stupid you would not come here and ask this question. This confuses a lot of people. But the paper is not good. They should not say EM waves travel through conductive materials because that does not generally happen. \$\endgroup\$
    – user57037
    Commented Oct 12, 2019 at 1:10
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    \$\begingroup\$ It depends on the conductive material and in general since the medium will be dispersive on the frequency. In a diluted plasma you get reflection under a certain frequency (plasma frequency) and then the medium slowly becomes 'transparent'. In a perfect conductor the charges are so free that by following the field they generate so big a current that it will obliterate the electric and magnetic field inside the conductor and will add magnetic field outside. In imperfect conductors you usually get some propagation because the current is not able to completely cancel the field inside... \$\endgroup\$ Commented Oct 12, 2019 at 3:52
  • \$\begingroup\$ ...in lossy dielectrics you get a much bigger penetration, but still exponential attenuation. In all cases, if you are interested in the speed of propagation, you want to look at the dispersion relation. From there you can get an idea of the phase and group velocities at different frequencies. In vacuum the dispersion relation is linear and the speed is c. \$\endgroup\$ Commented Oct 12, 2019 at 3:56

2 Answers 2


You need to go much deeper into the physics in order to understand what is really going on. The equation of the wave inside a conductor can be derived from Maxwell’s equations

$$\mathbf{E} = \mathbf{E}_0 e^{j(\alpha z -\omega t)} e^{-\beta z},$$

where \$z\$ is the direction perpendicular to the conductor in our case and

$$\alpha = \omega \sqrt{\mu\varepsilon} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{1 + \frac{\sigma^2}{\omega^2\varepsilon^2}}}.$$

Here the first exponential is the wave and the second exponential is the attenuating factor. It can be shown that for a bad conductor, satisfying \$\sigma \ll \omega \varepsilon\$, the wave number is

$$\alpha \approx \omega \sqrt{\mu \varepsilon}.$$

It can also be shown that for a good conductor, satisfying \$\sigma \gg \omega \varepsilon\$, the wave number is

$$\alpha \approx \sqrt{\frac{\omega \mu \sigma}{2}} = \sqrt{\mu\varepsilon} \sqrt{\frac{\omega \sigma}{2\varepsilon}}.$$

Hence the speed of propagation for a bad conductor is

$$v_\mathrm{p} = \frac{\omega}{\alpha} \approx \frac{1}{\sqrt{\mu\varepsilon}}.$$

This is the same as the speed of light in non-conductive media. However in a good conductor

$$v_\mathrm{p} = \frac{\omega}{\alpha} \approx \frac{1}{\sqrt{\mu\varepsilon}} \sqrt{\frac{2\omega\varepsilon}{\sigma}}.$$

It can be seen that the speed of light you would expect to see is multiplied by a second factor, which is less than one due the fact \$\sigma \gg \omega \varepsilon\$. Thus we arrive at the counter intuitive result that the better the conductor the lower the propagation speed.

But wait, there’s more! We use conductors only as anchors for the wave. The wave propagates in the space around the conductor and only penetrates up to the skin depth. Note that \$z\$ is in the direction perpendicular to the conductor. Usually what we really care about is the propagation in the direction parallel to the conductor.

  • \$\begingroup\$ Would it be overly misleading to say that the motion of electrons or "current" flowing in the conductor is due to the wave "dipping down" inside the conductor? And that those electrons have a motion that follows the motion of the wave (let's say a sinusoid) but the z component is typically ignored while the component parallel to the conductor gets emphasis? \$\endgroup\$
    – DKNguyen
    Commented Oct 28, 2019 at 19:58
  • \$\begingroup\$ @DKNguyen We are getting pretty deep into quantum mechanics here. You actually don’t need any QM to describe the EM wave in and around the conductor. What is happening on the QM level is actually quite complex. Firstly, the EM wave is only moving forward along the conductor. This may seem counter intuitive as a wave has an alternating fields, but it is moving in one direction only. Secondly, without the presence of a field the electrons are moving in all directions but they cancel each other out. \$\endgroup\$
    – user110971
    Commented Oct 28, 2019 at 21:14
  • \$\begingroup\$ So the oscillation (as in an actual back and forth motion of something in the mechanical sense) of the wave is more of a macro-level conceptualization that doesn't actually happen at the quantum level? I guess that makes sense. \$\endgroup\$
    – DKNguyen
    Commented Oct 28, 2019 at 21:16
  • \$\begingroup\$ @DKNguyen Electrons are moving along the Fermi surface, which is a surface in momentum space, not physical space. Applying an EM field causes a shift in the Fermi surface. This results in the net current becoming positive in one direction. Although the real Fermi surface for a conductor is more complex, a good visualization can be found here: homepage.lnu.se/staff/pkumsi/1FY805/Fermi_sphere.png. \$\endgroup\$
    – user110971
    Commented Oct 28, 2019 at 21:17
  • \$\begingroup\$ @DKNguyen Even in the classical sense the wave is moving in one direction. To see this, just pick a peak and follow it along. A single point can experience oscillatory motion, but so can a fixed point in a conductor experience a varying EM field. But the electrons are not free to move in any direction. The atoms are in the way. \$\endgroup\$
    – user110971
    Commented Oct 28, 2019 at 21:26

Not at all a stupid question. It looks like you’re reading a confusing paper.

Electromagnetic waves don’t generally flow very far through conductors. Light reflects from metal, it doesn’t go through it. Radio doesn’t propagate in sea water. These things happen because the mobile charges in the conductor, well, move and that cancels the wave as it penetrates in.

Glass and other non-conductors have charges that can move, but not freely. The motion of their changes acts to slow light down, but not completely cancel it.

With no material, light moves at the speed of light in a vacuum because it is in a vacuum.

  • \$\begingroup\$ radio does penetrate sea water, if the frequency is << 1MHz. Thus the US Navy uses 20,000 Hertz carrier, from a huge transmitter in Michigan, to communicate with the missle subs. \$\endgroup\$ Commented Oct 12, 2019 at 3:27
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    \$\begingroup\$ With a 15km wavelength, that’s radio EM wave, but once it’s in the water it’s an “evanescent” wave. That’s the same non-1/r^2 phenomenon that determines the skin depth in a metal. \$\endgroup\$ Commented Oct 12, 2019 at 3:51

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