# How a dipole should behave in order to produce an electromagnetic field?

I am trying to understand what an electric dipole "has to do" to produce electromagnetic waves. I know that an oscillating electric dipole will produce EM waves and by oscillating electric dipole I mean two opposite charges $$q$$ that move with an harmonic motion approaching each other;

I also know that this is possible since every time we have a time-varying electric field or a time-varying magnetic field we see from Maxwell's equations that EM waves are produced so, back to the oscillating dipole, of course it will produce a time-varying electric field.

Now, hoping I have said everything right so far (let me know), the doubt:

I've read about Hertz's oscillator and that this kind of "oscillating" dipole can be reproduced with an RC circuit in AC current. What does it happen in an RC circuit in AC current? It happens that the capacitor's armatures will charge with a positive charge that will go from $$0$$ to a certain positive charge $$q$$, consequently the other armature will charge with a negative charge from $$0$$ to the same amount of charge $$-q$$;

then the current will change direction so that the first armature will lose gradually the positive charge $$q$$ unloading completely and will recharge with negative charge $$-q$$ and so will do the second armatur in the opposite way.

What we get with this circuit is: two areas in the space separated by a certain distance $$d$$, the distance between capacitor's armatures, which periodically have a positive charge and a negative charge that grow, decrease by passing through $$0$$, grow in the opposite direction decrease by passing through $$0$$ and so on.

So physically we have two different situations if in a laboratory we have an RC circuit in AC current or two opposite charges oscillating (like an atom in an EM field) that act in the same ways if we look at EM waves they produce.

And here is my point:

The definition I know of an electric dipole is:An electric dipole is a pair of equal and opposite point charges $$-q$$ and $$q$$ separated by a distance $$d$$, so whenever I look at the armatures of the capacitor I see a dipole but not an oscillating dipole since the distance $$d$$ is a constant, so we are producing EM waves without an oscillating dipole. Is it possible?

1. Is this system actually an oscillating dipole or is it simply a model that behaves as one?

2. This circuit produces EM waves since it is actually an oscillating dipole or in general because every time we have a time-varying electric field or a time-varying magnetic field we see from Maxwell's equations that EM waves are produced? and this is just a realistic and simple way to emulate an oscillating dipole but actually not properly an oscillating dipole?

As long as the dipole moment (a vector quantity) oscillates, the system radiates. If you know how the dipole moment changes with time, that's enough. The reason why the dipole moment changes doesn't matter. Radiation will be emitted.

There are actually only two ways in which the dipole moment can change. Either charged particles can move, which changes the displacement vectors between them and therefore the dipole moment. (If they manage to do a dance where the total dipole moment doesn't change, then radiation might not be emitted, although there is still the possibility of radiation from the quadrupole or higher moments.) Or, fundamental dipoles (and we are not sure whether any actually exist in nature) can rotate, changing the directions of their dipole moments.

Your scenario, involving an oscillating circuit, is actually the same scenario as oppositely charged particles oscillating; it just has more particles! The only way that one plate can become positively charged while the other becomes negatively charged is by the movement of negatively charged particles from the former to the latter, or the movement of positively charged particles from the latter to the former. A charge separation builds up, where none previously existed, because of the movement of charge carriers. This gives the capacitor a dipole moment. When the capacitor discharges, it is because the charge separation is winding down. Fundamentally, there isn't much difference between this and a single pair of oppositely charged particles that initially move away from each other, then toward each other.

• @Salmone I believe as long as the second derivative of the dipole moment is nonzero, there will be radiation. If the dipole moment oscillates, in the sense of varying in time and regularly returning to its original value, its second derivative will have to be nonzero, at least some of the time. Commented Jul 29, 2022 at 1:05
• @Salmone The charge on the plates on the capacitor can't just increase without charges moving. The only way that the charge of an object can change is through the movement of charged particles. Electrons, most likely, move through the circuit. When the capacitor is charging, it's because electrons are becoming separated from the protons they are normally bound to, and moving to somewhere else that has more electrons than protons. The charge separation causes the dipole moment. Commented Jul 29, 2022 at 1:06
• @Salmone 1. I believe the general condition is that the second derivative must be nonzero, and the radiated power will be proportional to the squared magnitude of the second derivative of the dipole moment. However there are tricky definitional issues when non-oscillatory motions are involved, leading to controversy. 2. Yes, your scenario is like that except that the two ends (the capacitor plates) are fairly close together rather than being on opposite ends. Commented Jul 29, 2022 at 14:59
• @Salmone For example, see here for a discussion of the controversy as applied to uniformly accelerating point charges. If you analyze a uniformly accelerating pair of opposite charges (so that the second derivative of the dipole moment is a nonzero, at least during the Newtonian regime) I suspect you might run into similar issues. When we confine ourselves to oscillating systems, these issues go away. Commented Jul 29, 2022 at 15:09
• @Salmone Every charged particle in the system counts. Usually, that just means protons and electrons. We can calculate the dipole moment by summing the vector positions of all the electrons, summing the vector positions of all the protons, and then taking the difference and finally multiplying by the charge of 1 electron. Anything that tends to systematically separate electrons from protons, sending the electrons all off to one side, tends to make this quantity deviate from zero. This is what happens when a capacitor charges up. Commented Jul 29, 2022 at 23:46

Suppose you had a dipole with a constant distance d between the equal charges, and the dipole rotated.

Then the charges would be accelerated and it would produce EM waves.

The EM waves would be linearly polarized in the plane of rotation, and would be circularly polarized normal to that plane.

So I don't see any problem if you choose to call a capacitor a dipole, and think of it as oscillating. Why not?

• @J Thomas Yes I don't see any problems either I was wondering just this, we THINK of it as oscillating, so the capacitor is a model representing a real oscillating dipole, is that what you're saying? That is my first question. Like: the capacitor so constructed and a real oscillating dipole behave in the same way, watching the EM waves they produce, so we can think the first being a good representation of the second? Commented Jul 28, 2022 at 22:34

Suppose you have a dipole moment consisting of two charges $$\pm q$$ and the length between them is $$d$$. But now suppose that the charge at "each end" of the dipole is time dependent such that $$q = q_0 \sin \omega t$$, then the dipole moment is $$p = q_0d \sin \omega t$$

But current is $$dq/dt$$, so $$I = \omega q_0 \cos\omega t = I_0 \cos \omega t\ ,$$ where we have now defined $$I$$ to be an AC current with amplitude $$I_0 = \omega q_0$$.

Using this definition, we can see that the electic dipole moment could be written as $$p = \frac{I_0 d}{\omega} \sin \omega t \ .$$

Thus the two situations are entirely equivalent, with the electric dipole moment amplitude replaced by $$I_{0}d/\omega$$ and the oscillating AC current produces an oscillating electric dipole moment.

• I agree with those formulas but I see two different situations with the same dipole moment. If we consider a charge $q$ and a charge $-q$ oscillating toward each other the dipole moment has the same formula you wrote where the time dependence arises from the harmonic motion in space; on the other hand, if we consider the first example you gave, the same dipole moment formula has a time dependence arises from the "harmonic" variation of the charges. Now the two systems act in the same manner in emitting EM waves but physically we have different situations. Can you explain this? Commented Jul 29, 2022 at 14:08
• @Salmone that isn;t your question or the scenario in your question. In your question $d$ is fixed. If you fix $q$ and change $d$ then obviously you have an oscillatory dipole moment. Commented Jul 29, 2022 at 15:55
• I understand but do I have an oscillatory dipole moment even if $d$ is fixed and charges change with $q(t)=q_osin(\omega t)$? Commented Jul 29, 2022 at 19:48
• @Salmone that is exactly what I have said in my answer. Commented Jul 29, 2022 at 20:29