Two dipoles radiating out of phase

I thought I’d write about one of my favorite problems from freshman year. It doesn’t require any math to understand, but it points out many of the risks and subtleties that can arise when physics problems make too many “ideal” assumptions:

Suppose that you have two simple antennae, each consisting of a single, straight length of copper wire through which a single frequency of alternating current is passing. The two antennae are positioned some fixed distance apart, and they are oriented in parallel. If a remote physicist operating the two antennae introduces an appropriate delay between their driving signals, causing the AC waveforms in the two antennae to be 90 degrees out-of-phase (but still at the same frequency), then the electric field in the region between the two antennae will vanish due to destructive interference. Yet the two antennae are still emitting radiation; they are still each drawing current, and presumably the power they consume to create this current must be transferred into the resulting fields they emit. So when the two antenna destructively interfere, where does the energy go?

The conventional response to this question (and the one my freshman lab TA insisted upon) is that the field cancels out in some regions—such as between two two antennae–but it increases by a compensatory amount in other regions where the waves constructively interfere, resulting in the net energy stored in the fields (throughout all of space) remaining constant. While this is certainly a satisfactory answer for most textbook treatments of dipole radiation, it remains troubling because one can easily envision a case in which there are no other regions in which the waves can constructively interfere—for example, if mirrors were used carefully. If, instead of antennae, one pictures two out-of-phase lasers pointed towards each other, then it becomes much less clear where the compensating region between the two lasers would be. However, there’s another way of looking at the problem that sheds light on this inconsistency:

Conventional electrodynamics tells us why the two copper antennae will generate radiation: the moving charges in each antenna beget changing magnetic fields, which in turn create electric fields via Faraday’s law, which then create new magnetic fields as they collapse. This cycle of electric and magnetic fields taking turns forming and collapsing gives rise to self-propagating electromagnetic waves—a collapsing electric field changes quickly, thus inducing a magnetic field which eventually collapses to produce a new electric field, and so on. The power transmitted by the wave is thus determined by the amplitude of the initial magnetic field generated by the antenna, which in turn is proportional to the current through the wire. This current is, in turn, determined by the resistance of the wire comprising the antenna—if the wire were a impossibly perfect conductor, then even the most minor voltage difference between the two ends of the antenna would generate an impossibly infinite current via Ohm’s law. Thus the power put into a single antenna is determined by the resistance of its wire, and this power exits the antenna as electromagnetic radiation—so far, energy has neither been created nor destroyed.

The subtlety of the problem arises because an additional effect that occurs when there are two antennae near each other. The electrons moving back and forth inside one antenna aren’t just limited in motion due to the resistance of the copper wire, but also by the electric field due to the other antenna. If the other antenna is in-phase (no delay), then the electrons will keep experiencing a Lorentz force in the opposite direction to the way that the antenna’s power source wants them to move, and so the power source will need to provide more power in order to generate waves of the same amplitude—the current, and therefore power, drawn by the antenna increases. In the case when the sources are out of phase, or the waves are destructively interfering, the opposite effect occurs: the field from the other antenna actually helps the electrons along, allowing a given electron to oscillate at a certain amplitude without requiring as much energy from the power source. In other words, placing the antennae out of phase reduces the effective resistance, or impedance, of the two antennae, and thus reduces their power consumption by an amount equivalent to the drop in the energy of the electromagnetic field due to their destructive interference.

In the laser formulation of this problem, this explanation would amount to the light from one laser damping excitations in the lasing medium of the other laser, resulting in less power drawn from the source.

What I like about this scenario is the manner in which a very common assumption used in physics problems—that power supplies are monoliths, steadily providing a fixed voltage and current to each component of a system–turns out to be the source of the ambiguity. An electrical engineer who places an ammenter in series with one of the antennas would immediately notice the drop in input power when the antennae are placed out-of-phase. But in the way that the problem is often presented, the power consumption of the antennae seems like a fixed quantity, giving rise to the supposed paradox.

Advertisements

One thought on “Two dipoles radiating out of phase

  1. I spent last two days and nights thinking about exactly these two scenarios, (two out of phase dipoles and two out of phase laser) for dipoles it is easy to see that driving filed of one dipole will interact with the driving field of other in such a way that they would strongly oppose each other, making it harder for the dipoles to radiate which amounts to destructive interference.

    But in the case of lasers, where both the laser beams are traveling in the same direction but out of phase, -= , how would one see that the laser beam which was having energy and momentum vanished altogether because of addition of more energy and momentum from another out of phase beam, I guess, I’m wondering where does this energy vanish ?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s