This seems like such a simple question but there is just something that is not making sense to me. I'm a 4th year ECE student so I think I should know this answer. Here's the question:

Recently I've been doing a lot of work on RF stuff at my internship. I've gotten a good grasp on the fundamentals and can work on satellite ground station equipment and know what I am doing. But I started thinking tonight, and I am not sure why energy that we send into an antenna is radiated away, instead of appearing as reactive power.

In my mind, I'm thinking of it like this. Let's say we have a simple dipole antenna. Let's apply a negative voltage to one end of the dipole. This is going to effectively "cram" the electrons into the far end of the dipole. This change in the electric field is radiated away at the speed of light and can be picked up. If we then instantly removed the voltage from the dipole and assumed it was a lossless conductor, the electrons would try to restore themselves to a neutral distribution, but they are going to overshoot and oscillate. During this transition, they will generate a magnetic field. Afterwards, they will equalize with an electric field that was opposite to the one we applied. This is going to continue on forever provided no loss. I know that real antennas have loss but I am assuming that this is not actually where the power is going in antennas, as then they would be heating up by insane amounts.

This is my mind, constitutes an antenna that should be radiating power away, but I am not sure where it is going. I could understand that the power is transferred if there was another dipole nearby, but the radiation energy loss occurs whether or not there is a receiver for the radiated power. Why does the above example radiate power away even if it was left alone in an empty universe? Does it not and I'm just misunderstanding something? I was reading something about far vs near fields that might have to do with my question but I was still confused. Thanks for your help.

  • \$\begingroup\$ simply an E field of driver to radiate it's pattern usually 1/2wave and then there are Friis losses 1/r^2 \$\endgroup\$ Aug 26, 2021 at 2:12
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    \$\begingroup\$ The way the voltage and current is distributed in the antenna gives rise to a propagating EM wave. Why? Well, I could say because that's what Maxell's equations dictate. But, of course, the equations describe reality. They are not the cause of any phenomena. So the real answer is, the universe works this way and nobody knows why. Why do planets orbit the sun? Because Newtoninan physics tells them they have to? No. Gravity is just a property of mass in the universe. Nobody can really say why it is that way. \$\endgroup\$
    – user57037
    Aug 26, 2021 at 6:09

2 Answers 2


The cause for the propagation of electromagnetic waves are the Maxwell equations or their solutions. Here my attempt at an intuitive explanation that does not require the solution of the wave equation. The two important equations for this are:

$$\mathrm{curl}\,\vec{E} = -\frac{\partial \vec{B}}{\partial t} \qquad \mathrm{and} \qquad \mathrm{curl}\,\vec{H}=\vec{J} + \frac{\partial \vec{D}}{\partial t}$$

This means that an electric field creates a magnetic one and vice versa, as shown here:

electric and magnetic field

If a harmonic current density is impressed, a harmonically changing magnetic field is created. This magnetic field in turn creates a harmonic electric field, which then generates a magnetic field itself. And now you're back to the beginning. Magnetic field -> electric field -> magnetic field -> ... and so on.

creation of new electric and magnetic fields

The way in which the fields are directed towards each other creates a wave that propagates away from the cause (current density). In theory, the whole thing goes on indefinitely as long as the derivative of the quantities does not become zero. In practical terms, you need a high frequency for this to work for a long time. Derivation means multiplication by jω which is greater the greater the frequency.

This image shows the wave propagation: creation of electromagnetic wave

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    \$\begingroup\$ In English math and physics textbooks the ∇× operator is usually written as curl. In my language it is correct to write rot, but English readers may not be used to that notation. \$\endgroup\$
    – Bart
    Aug 26, 2021 at 10:46
  • \$\begingroup\$ @Bart Thanks for the note, I have edited it. \$\endgroup\$
    – mais
    Aug 26, 2021 at 11:26

It's simple - some of the solutions to Maxwell's equations are radiating waves. It sounds like you don't understand these - have you covered them in class? You seem to understand mutual inductance, but wave propagation can send energy out into empty space. It is exactly the same as shining a torch up into the night sky - some of that radiation may well carry on forever and never meet anything.

It's actually quite hard to make unshielded electronic circuitry that doesn't excite radiating waves. Reducing unwanted radiated waves is one of the things that EMC engineering is about.

Antennas are designed to couple to these radiating modes, generally, but not always, as efficiently as possible. Solving the field equations around the dipole shows that there are both reactive fields and radiation fields generated. Radiation fields are the ones that carry power off into the distance.

Is there a simpler explanation - I'm not sure. It's like when you send an impulse down a garden hose by shaking it - if the medium supports wave propagation then waves are likely to carry energy from the source. Sound waves propagate in air. When you vibrate a speaker cone in an enclosed space then this is like a reactance and the power is somewhat trapped, but in open space - some power propagates away from the source, regardless of whether there is a listener.

  • \$\begingroup\$ Thanks for the response. I took emag when covid hit and my school focuses on transmission line theory first so I did not learn nearly as much as I should about Maxwell's equations. Looks like I'll need to go back and learn it again. Your comment that one of the solutions to the equations is a propagating wave was helpful to me and I think I just didn't remember the solutions to the equations and what they mean for EM radiation. \$\endgroup\$
    – cEEa
    Aug 26, 2021 at 12:06

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