Introduction
I'm sure a lot of us have seen the controversial video made by Veritasium: The Big Misconception About Electricity
Now what I want to ask in this post is not to argue about any of the claims made during the video, as I am pretty sure he cleared a lot of them up in his follow-up video. No, what I want to ask is regarding his animation representing the electromagnetic fields around an ac circuit:
Now I am pretty sure everyone would agree that the entire field turning on and off in phase with the battery is obviously the wrong picture here since that would most definitely violate the second postulate of SR (which is literally built on electromagnetism).
We all know that the true picture is of the circuit radiating EM waves as the electric field of the battery oscillates like that of an oscillating dipole (Oscillating Dipole by John Belcher). Now my question is then how exactly would the EM waves radiate and in what direction?
The following sections are just me trying to answer the question myself but obviously failing but is optional in answering this question (everything you need to know to answer this question is all in this introduction). But if you want to know why I have been troubled by this question, feel free to continue reading:
The Math
Now let me get into some maths first with the conservation of energy from the Poynting theorem. Now if we integrate over the entire space for the equation $\vec{J}\cdot\vec{E} + \frac{\partial u}{\partial t} = -\vec{\nabla}\cdot\vec{S}$, we get:
$\frac{d}{dt}(E_{Mechanical} + E_{Field}) = 0$
Where $E_{Mechanical}$ is contributed by work done by the field on charges ($\vec{J}\cdot\vec{E}$, i.e. energy dissipated in the resistor: $I^2 R$), and $E_{Field} = \iiint_{\mathbb{R}} (\frac{\epsilon_0}{2}|\vec{E}|^2 + \frac{1}{2\mu_0}|\vec{B}|^2)dV$ is the energy stored in the field, or more accurately, the potential energy of the system.
We are also missing one more term: $E_{Chemical}$ or $E_{Battery}$, which is the mysterious energy source that keeps replenishing the dipole of the battery, and so the full equation becomes:
$\frac{d}{dt}(-E_{Battery} - E_{Mechanical}) = \frac{d}{dt}E_{Field}$
Proposals
To make this simple, let us make a few assumptions first:
- The battery and the resistor are perfect dipoles (i.e. d -> 0)
- Surface charges can instantaneously rearrange themselves
Now here are a few proposals with some reasoning as to why I think they might be false:
Direction of Travel
1. Radially:
This may first seem reasonable as it simply just uses the principle of relativity to assume information (i.e. the battery oscillating) travels at the speed of light.
But then it can be realised that this propagating electromagnetic field can also influence the surface charges all along the wire of the circuit, which then themselves will also create its own "sphere of influence" (pun intended) propagating outwards at the speed of light, which then refutes this proposal.
In addition to that, the energy from the battery should all really be transferred to the battery, not to infinity.
2. Along Poynting vectors in the dc version of the circuit:
Now this would definitely seem very plausible at first, as it makes sense that the waves should all be propagating from the battery toward the resistor to transfer all its energy.
Yet the problem comes from the math of the problem:
$\frac{d}{dt}(-E_{Battery} - E_{Mechanical}) = \frac{d}{dt}E_{Field}$
Now from simple circuit analysis, we know that $E_{Battery} = -IV$, while $E_{Mechanical} = I^2 R$, resulting in:
$\frac{d}{dt}(IV - I^2 R) = \frac{d}{dt}E_{Field}$
Now usually, we can just say $IV - I^2 R = 0$, but that's only true because the speed of light is very fast for any normal circuit, and since we're doing physics, we can imagine a really long circuit, now the first problem presents itself:
Now if we agree with our proposal that the waves propagate along the Poynting vectors, this would result in multiple paths for the electromagnetic field to be "updated". But each of these paths has different lengths and so what exactly would be the electric field at the resistor, or more importantly, what is the time delay between $IV$ and $I^2 R$?
Even if we assume that the time delay is the time taken for information to travel in the shortest path (i.e. the straight line from the battery to the resistor), and take the length of the circuit to be $c/f$ ($f$ is the frequency of oscillation of the voltage), we will get $IV - I^2 R = 0$ as the phase difference at the battery and the resistor is $2π$, and so now we have:
$\frac{d}{dt}E_{Field} = 0$
But again, if we consider the multiple paths the EM waves take again, they are all out of phase with the shortest path and so will not result in a constant $E_{Field}$, contradicting the above equation.
How it Travels
1. The same field as a dc circuit but oscillating in magnitude spatially:
Although this is the basis of how transmission lines work, it just always seems so hand-wavy. Now although not a sound argument, I always imagined the propagation of an EM wave to be a complicated intermediate region between the new and old field (Decreasing Dipole by John Belcher) and so should be more complicated than just some form of variation of the "static/dc case".
Hopefully, someone can perhaps provide me with a simple animation of how an AC circuit actually radiates and clear my confusion regarding this problem.
