Light and radiation

Introduction: Scale Matters

One of the points which Richard Feynman, as a great physics teacher, does admirably well is to point out why scale matters. In fact, ‘old’ physics are not incorrect per se. It’s just that ‘new’ physics analyzes stuff at a much smaller scale.

For example, Snell’s Law, or Fermat’s Principle of Least Time, which were ‘discovered’ 500 years ago – and they are actually older, because they formalize something that the Greeks had already found out: refraction of light, as it travels from one medium (air, for example) into another (water, for example) – are still fine when studying focusing lenses and mirrors, i.e. geometrical optics. The dimensions of the analysis, or the equipment involved (i.e. the lenses or the mirrors), are huge as compared to the wavelength of the light and, hence, we can effectively look at light as a beam that travels from one point to another in a straight line, that bounces of a surface, or as a beam that gets refracted when it passes from one medium to another.

However, when we let the light pass through very narrow slits, it starts behaving like a wave. Geometrical optics does not help us, then, to understand its behavior: we will, effectively, analyze light as a wave-like thing at that scale, and analyze wave-like phenomena, such as interference, the Doppler effect and what have you. That level of analysis is referred to as the classical theory of electromagnetic radiation, and it’s what we’ll be introducing in this post.

The analysis of light as photons, i.e. as a bunch of ‘particles’ described by some kind of ‘wave function’ (which does not describe any real wave, but only some ‘probability amplitude’), is the third and final level of analysis, referred to as quantum mechanics or, to be more precise, as quantum electrodynamics (QED). [Note the terminology: quantum mechanics describes the behavior of matter particles, such as protons and electrons, while quantum electrodynamics (QED) describes the nature of photons, a force-carrying particle, and their interaction with matter particles.]

But so we’ll focus on the second level of analysis in this post.

Different mathematical approaches

One other thing which Feynman points out in his Lectures is that, even within a well-agreed level of analysis, there are different mathematical approaches to a problem. In fact, while, at any level of analysis, there’s (probably) only one fully mathematically correct analysis, approximate approaches may actually be easier to work with, not only because they actually allow us to solve a practical problem, but also because they help us to understand what’s going on. 

Feynman’s treatment of electromagnetic radiation (Volume I, Chapters 28 to 34) is a case in point. While he notes that Maxwell’s field equations are actually the ones to be used, he writes them in a mathematical form that we can understand more easily, and then simplifies that mathematical form even further, in order to derive all that a sophomore student is supposed to know about electromagnetic radiation (EMR), which, of course, not only includes what we call light but also radio waves, radar waves, infrared waves and, on the other side of the spectrum, x-rays and gamma rays. 

But let’s get down to business now.

The oscillating charge

Radiation is caused by some far-away electric charge (q) that’s moving in various directions in a non-uniform way, i.e. it is accelerating or decelerating, and perhaps reversing direction in the process. From our point of view (P), we draw a unit vector er’ in the direction of the charge. [If you want a drawing, there’s one further down.]

We write r’ (r prime), not r, because it is the retarded distance: when we look at the charge, we see where it was r’/c seconds ago: r’/c is indeed the time that’s needed for some influence to travel from the charge to the here and now, i.e. to P. So now we can write Coulomb’s Law:

E1 = –qer’/4πe0r’2

This formula can quickly be explained as follows:

  1. The minus sign makes the direction of the force come out alright: like charges do not attract but repel, unlike gravitation. [Indeed, for gravitation, there’s only one ‘charge’, a mass, and masses always attract. Hence, for gravitation, the force law is that like charges attract, but so that’s not the case here.]
  2. E and er’ and, hence, the electric force, are all directed along the line of sight.
  3. The Coulomb force is proportional to the amount of charge, and the factor of proportionality is 1/4πe0r’2.
  4. Finally, and most importantly in this context (study of EMR), the influence quickly diminishes with the distance: it varies inversely as the square of the distance (i.e. it varies as the inverse square).

Coulomb’s Law is not all that comes out of Maxwell’s field equations. Maxwell’s equations also cover electrodynamics. Fortunately, because we are, indeed, talking moving charges here, so electrostatics is only part of the picture and, in fact, the least important one in this case. 🙂 That’s why I wrote E1, with as subscript, above – not E.

So we have a second term, and I’ll actually be introducing a third term in a minute or so. But let’s first look at the second term. I am not sure how Feynman derives it from Maxwell’s equations – I am sure I’ll see the light 🙂 when reading Volume II – but, from Maxwell’s equations, he does, somehow, derive the following, secondary, effect:

Formula 1

This is a term I struggled with in a first read, and I still do. As mentioned above, I need to read Feynman’s Volume II, I guess. But, while I still don’t understand the why, I now understand what this expression catches. The term between brackets is the Coulomb effect, which we mentioned above already, and the time derivative is the rate of change. We multiply that with the time delay (i.e. r’/c). So what’s going on? As Feynman writes it: “Nature seems to be attempting to guess what the field at the present time is going to be, by taking the rate of change and multiplying by the time that is delayed.” 

OK. As said, I don’t really understand where this formula comes from but it makes sense, somehow. As for now, we just need to answer another question in order to understand what’s going on: in what direction is the Coulomb field changing?

It could be either: if the charge is moving along the direction of sight er’ won’t change but r’ will. However, if r’ does not change, then it’s er’ that changes direction, and that change will be perpendicular to the line of sight, or transverse (as opposed to radial), as Feynman puts it. Or, of course, it could be a combination of both. [Don’t worry too much if you’re not getting this: we will need this again in just a minute or so, and then I will also give you a drawing so you’ll see what I mean.]

The point is, these first two terms are actually not important because electromagnetic radiation is given by the third effect, which is written as:

Formula 3

Wow ! This looks even more complicated, doesn’t it? Let’s analyze it. The first thing to note is that there is no r’ or r’2 in this equation. However, that’s an optical illusion of sorts, because r’ does matter when looking at that second-order derivative. How? Well… Let’s go step by step and first look at that second-order derivative. It’s the acceleration (or deceleration) of er’. Indeed, visualize er’ wiggling about, trying to follow the charge by pointing at where the charge was r’/c seconds ago. Let me help you here by, finally, inserting hat drawing I promised you.

Capture

This acceleration will have a transverse as well as a radial component: we can imagine the end of er’ (i.e. the point of the arrow) being on the surface of a unit sphere indeed. So as it wiggles about, the tip of the arrow moves back a bit from the tangential line. That’s the radial component of the acceleration. It’s easy to see that it’s quite small as compared to the transverse component, which is the component along the line that’s tangent to the surface (i.e. perpendicular to er’).

Now, we need to watch out: we are not talking displacement or velocity here but acceleration. Hence, even if the displacement of the charge is very small, and even if velocities would not be phenomenal either (i.e. non-relativistic), the acceleration involved can take on any value really. Hence, even with small displacements, we can have large accelerations, so the radial component is small relative to the transverse component only, not in an absolute sense.

That being said, it’s easy to see that both the transverse as well as the radial component depend on the distance r’ but in a different way. I won’t bother you with the geometrical proof (it’s not that obvious). Just accept that the radial component varies, more or less as the inverse square of the distance. Hence, we will simplify and say that we’re considering large distances r’ only – i.e. large in comparison to the length of the unit vector, which just means large in comparison to one (1) – and then it’s only the transverse component of a that matters, which we’ll denote by ax.

However, if we drop that radial component, then we should drop E1 as well, because the Coulomb effect will be very small as compared to the radiation effect (i.e. E3). And, then, if we drop E1, we can drop the ‘correction’ E2 as well, of course. Indeed, that’s what Feynman does. He ends up with this third term only, which he terms the law of radiation:

Formula 4

So there we are. That’s all I wanted to introduce here. But let’s analyze it a bit more. Just to make sure we’re all getting it here.

The dipole radiator

All that simplification business above is tricky, you’ll say. First, why do we write t – r/c for the retarded time (t’)? It should be t – r’/c, no? You’re right. There’s another simplification here: we fix the delay time, assuming that the charge only moves very small distances at an effectively constant distance r. Think of some far-away antenna indeed.

Hmm… But then we have that 1/c2 factor, so that should reduce the effect to zilch, isn’t it? And then… Hey! Wait a minute! Where does that r suddenly come from? Well, we’ve replaced d2er’/dt2 by the lateral acceleration of the charge itself (i.e. its component perpendicular to the line of sight, denoted by ax) divided by r. That’s just similar triangles.

Phew! That’s a lot of simplifications and/or approximations indeed. How do we know this law really works? And, if it does, for what distance? When is that 1/r part (i.e. E3) so large as compared to the other two terms (E1 and E2) that the latter two don’t matter anymore? Well… That seems to depend on the wavelength of the radiation, but we haven’t introduced that concept yet. Let me conclude this first introduction by just noting this ‘law’ can easily be confirmed by experiment.

A so-called dipole oscillator or radiator can be constructed, as shown below: a generator drives electrons up and down in two wires (A and B). Why do we put the generator in the middle? That’s because we want a net effect: the radiation effect of the electrons in the wires connecting the generator with A and B will be neutral, because the electrons move right next to each other in opposite direction. With the generator in the middle, A and B form one antenna, which we’ll denote by G (for generator).

dipole radiator

Now, another antenna can act as a receiver, and we can amplify the signal to hear it. That’s the D (for detector) shown below. Now, one of the consequences of the above ‘law’ for electromagnetic radiation is, obviously, that the strength of the received signal should become weaker as we turn the detector. The strongest signal should be when D is parallel to G. At point 2, there is a projection effect and, hence, the strength of the field should be less. Indeed, remember that the strength of the field is proportional to the acceleration of the charge projected perpendicular to the line of sight. Hence, at point 3, it should be zero, because the projection is zero.

dipole radiator - field

Now, that’s what an experiment like this would indeed confirm. [I am tempted now to explain how a radio receiver works, but I will resist the temptation.]

I just need to make a last point here in order to make sure that we understand the formula above and – more importantly – that we can use in subsequent chapters without having to wonder where it comes from. The formula above implies that the direction of the field is at right angles to the line of sight. Now, if a charge is just accelerating up and down, in a motion of very small amplitude, i.e. like the motion in that antenna, then the magnitude (or strength let’s say) of the field will be given by the following formula:

Formula 5

θ, in this formula, is the angle between the axis of motion and the line of sight, as illustrated below:

Fig 29-1

So… That’s all we need to know for now. We’re done. As for now that is. This was quite technical, I guess, but I am afraid the next post will be even more technical. Sorry for that. I guess this is just a piece we need to get through.

Post scriptum:

You’ll remember that, with moving and accelerating charges, we should also have a magnetic field, usually denoted by B. That’s correct. If we have a changing electric field, then we will also have a magnetic field. There’s a formula for B:

B = –er’´E/c = –| er’||E|c–1sin(er’, En = –(E/c)·n

This is a vector cross-product. The angle between the unit vector er’ and E is π/2, so the sine is one. The vector n is the vector normal to both vectors as defined by the right-hand screw rule. [As for the minus sign, note that –a´b = b´a, so we could have reversed the vectors: the minus sign just reverses the direction of the normal vector.] In short, the magnetic field vector B is perpendicular to E, but its magnitude is tiny: E/c. That’s why Feynman neglects it, but we will come back on that in later posts.

One thought on “Light and radiation

  1. Pingback: Vector analysis (II) | Reading Penrose

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