The electric field in (and from) a conductor

This is just a quick post to answer a question of my 16-year old son, Vincent: why are we safe in a car when lightning strikes? What’s the Faraday effect really?

He wants to become an engineer, and so I told him what I knew: the electric charges reside at the surface of a conductor and, therefore, a fully-enclosed, all-metallic vehicle is safe. One should just not touch the interior metallic areas, surely not during the strike, but also not after the strike. Why? Because there may still be some residual charge left on the vehicle, even if the metal frame should direct all lightning currents to the ground.

Through the rubber of the tyres? Yes. In fact, it’s the rubber and other insulators that explain why some residual charge might be left. Indeed, the common assumption that, somehow, it’s the rubber that protects the occupants of a car (or that, somehow, rubber soles would insulate us in an electric storm and, hence, less likely to get hit) is ridiculous—completely false, really! The following quote from the US National Weather Service is clear enough on that:

“While rubber is an electric insulator, it’s only effective to a certain point. The average lightning bolt carries about 30,000 amps of charge, has 100 million volts of electric potential, and is about 50,000°F. These amounts are several orders of magnitude higher than what humans use on a daily basis and can burn through any insulator—even the ceramic insulators on power lines! Besides, the lightning bolt may just have traveled many miles through the atmosphere, which is a good insulator. Half an inch (or less) of rubber will make no difference.”

So that’s what I told him—sort of. However, I felt my answer (which I tried to get across as I was driving the car, in fact) was superficial and incomplete. So…

Vincent, here’s the full answer! I promise, no integrals or complex numbers. At the same time, it will be not so easy as the physics you learned in school, because I want to teach you something new. 🙂 Just try it. What I want to explain to you is Gauss’ Law. If you manage to go through it, you’ll know all you need to know about electrostatics, and it will make your first undergrad year a lot easier. [Especially that vector equation, as I always felt my math teacher never told me what a vector really was: it’s something physical. :-)]

Forces and fields

You’ve surely seen Coulomb’s Law:

F = ke·(q1q2)·(1/r212)

The ke factor is Coulomb’s constant: it is just a constant of proportionality, so it’s there to make the units come out alright. Indeed, Coulomb’s formula is simple enough: it says that the force is directly proportional to the amount of charge and inversely proportional to the square of the distance. That’s all. However, the units in which we measure stuff are not necessarily compatible: we measure distance in meter, electric charge in coulomb, and force in newton. So, if we’d define the newton as the force between two charges of one coulomb separated by a distance of one meter, then we wouldn’t need to put that kfactor there. But the newton has another definition: one newton is the force needed to accelerate 1 kg at a rate of 1 m/s per second.

Coulomb’s constant is usually written as k= 1/4πε0 factor in more serious textbooks. Why? Well… You can read my note at the end of this post, but it doesn’t matter right now. It’s much more important to try to understand the vector form of Coulomb’s Law, which is written as:

Coulomb's Law

I used boldface to denote F1 and F2 because they are force vectors. Vectors are physical ‘quantities’ with a magnitude (denoted by F1 and F2, so no boldface here) and a direction. That direction is given by the unit vector e12 in the equation: it’s a unit vector (so its length is one) from q2 to q1. Read again: from q2 to q1, not from q2 to q1. It’s important to get this one thing right, otherwise you’ll make a mess of the signs. Indeed, in the example below, q1 and q2 have the same sign (+) but their sign may differ (so we have a plus and a minus), and the formula above should still work. Check it yourself by doing the drawing for opposite charges.

Coulomb's LawIn fact, my drawing above has a small mistake: Fis the same as Fbut I forgot to put the minus sign: the force on q2 is F= –F1. It’s the action = reaction principle, really.

OK. That’s clear. Now you need to learn about the concept of a field: the field is the force per unit charge. So the field at q1, or the field at point (1), is the force on q1 divided by q1. For example, if q1 is three Coulomb, we divide by three. More in general, we write:

Field

So now you know what the field vector E stands for: it is the force on a unit charge we would place in the field. To be clear, a unit charge is +1 unit. We can measure it in coulomb, or the proton charge, or the charge of a quark, or in whatever unit we want, but we’ve been using coulomb so far so let’s stick to that. Just in case you wonder: one coulomb is the charge of approximately 6.241×1018 protons, so… Yes. That’s quite a lot. 🙂

OK. Next thing.

Gauss’ Law

The field is real. We don’t have to put any charge there. The field is there, and it has energy. [There’s a formula for the energy, but I won’t bother you with that here, because we don’t need it.] The magnitude of the electric field, i.e. the field strength E = |E|, is measured in newton (N) per coulomb (C), so in N/C. In physics, we’ll multiply the field strength with a surface area so we get the so-called flux of the field, which is measured in (N/C)·m2. The illustration below (which I took from Feynman’s Lectures) is just as good as any. In fact, we have several surfaces here: we have a closed surface S with several faces, including surface a and b, which are spherical surfaces. The other surfaces of this box are so-called radial faces. The E field coming out of the charge is like a flow, and so the flow going through face a is the same as the flow going through face b: the face is larger, but the field strength is less.

Flux

It is easy to show that the net flux is zero: Coulomb’s Law tells us that the magnitude of E decreases as 1/r2 while, from our geometry classes, we know that the surface area increases as r2, so their product is the same. So, if the surface area of a is Δa, and the surface area of b is Δb, then Ea·Δa = Eb·Δb and so the net flux through the box is equal to Eb·Δb − Ea·Δa = 0. So the flux of E into face a is just cancelled by the flux out of face b. Needless to say, there is no flux through the radial surfaces. Why? Because the electric force is a radial force.

OK. Let’s look at a more complicated situation:

Flux box

When calculating the flux through a surface, we need to take the component of E that is normal to the surface, so that’s En = E·n = |E|·|ncosθ = |Ecosθ. I am sure you’ve seen that much in your math classes: n is the so-called normal vector, so its length is one and it’s perpendicular to the surface. In any case, the point is: the net flux through this closed surface will still be zero.

Now it’s time for the Big Move. Look at the volume enclosed by the surface S below: we can think of it as completely made up of infinitesimal truncated cones and, for each of these cones, the flux of E from one end of each conical segment will be equal and opposite to the flux from the other end. So the total net flux from the surface S is still zero!

Flux any volume

So we have a very general result here:

The (net) flux out of a volume that has no charge(s) in it is zero, always!

You’ll say: so what? Well… It’s a most remarkable result, really. First, it’s not what you’d expect intuitively, and, second, we can now use a clever trick to calculate the flux out of a volume that has some charge(s) in it. Let’s be clever about it. Look at the surface S below: it’s got a point charge q in it. Now we imagine another surface S’ around it: we imagine a little sphere centered on the charge.

Flux with charge

From Coulomb’s Law, we know that, if the radius of our little sphere is equal to r, then the field strength E, everywhere on its surface, is equal to:

formula 1

From your geometry class, you also know that the surface of a sphere is equal to 4πr2, so the flux from the surface of our little sphere is just the product of the field and the surface, so we write:

formula 2

Now, the nice thing is that we can generalize this result for many charges, or for charge distributions, because we can simply add the fields for each of them: EE+ EE+ … That gives us Gauss’ Law:

The flux from any closed surface S = Qinside0

Qinside is, obviously, the sum of the charges inside the volume enclosed by the surface.

OK. That’s Gauss’ Law. Let’s go back to our car. 🙂

The field in (and from) a conductor

An electrical conductor is a solid that contains many free electrons. Free electrons can move freely around, but cannot leave the surface. When we charge a conductor, the electrons will move around until they have arranged themselves to produce a zero electric field everywhere inside the conductor. It’s the corollary of Gauss’ Law: the (net) flux out of a volume that has no charge(s) in it is zero, always! And so the electrons will arrange themselves in order to make sure that happens.

Think about the dynamics of the situation: as long as there’s some field inside, the charges will keep moving. Fortunately (especially if you’re in a car or a plane hit by lightning!), the re-arrangement happens in a fraction of a second. Hence, if we have some kind of shell, then the field everywhere inside of the shell will be zero, always. In addition, when we charge a conductor, the electrons will push each other away and try to spread as much as possible, so they will reside at the surface of the conductor. In fact, the excess charge of any conductor is, on the average, within one or two atomic layers of the surface only. The situation is illustrated below:

Flux out of a conductor

Let me sum up the main conclusions:

  1. The electric field inside the conductor (E1) is zero. In other words, if a cavity is completely enclosed by a conductor, no distribution of charges outside can ever produce any field inside. But no field is no force, so that’s how the shielding really works!
  2. The electric field just outside the surface of a conductor (E2) is normal to the surface. There can be no tangential component. If there were a tangential component, the electrons would move along the surface until it was gone.

To be fully complete, the formula for the field just outside the surface of the conductor is E = σ/ε0, where σ is the local surface charge density. That local surface charge density can be quite high, of course, especially when lightning is involved—but it works! You’re safe in a car!

There’s one more point. You may think that you’ve seen that E = σ/ε0 formula before: it’s the formula for the field from a charged sheet, which is easy to calculate from Gauss’ Law. Indeed, if we look at some imaginary rectangular box that cuts through the sheet, as shown below (it’s referred to as a Gaussian surface), then the total flux is, once again, the field times the area. Now, if the charge density (so the charge per unit area) is ρ, then the total charge enclosed in the box is σA. So the flux, on each side of the sheet, must be equal to E·A = σA/ε0, from which we get: E = σ/ε0. But so we have a field left and right. For our conductor, we only have the E = σ/ε0 field outside. So how does it work really?

Charged sheet

We only have a field outside the conductor – and, hence, no field inside – because the charges in the immediate neighborhood of a point P on the surface will arrange themselves in such a way so as to produce a field that neutralizes the E = σ/ε0 field we’d expect on the inside. So we have ‘other charges’ here that come into play. The mechanics behind are similar to the mechanics behind the polarization phenomenon. If we have a negative charge density on the surface, we’ll have a positive charge density in the layer below. However, it’s quite complicated and, to analyze it properly, we’d need to analyze the electric properties of matter in more detail, which we won’t do here.

So… When everything is said and done, the phenomenon of ‘shielding’ is extremely complex indeed: it’s all about charges arranging themselves in patterns, and the result is truly remarkable: the fields on the two sides of a closed conducting shell are completely independent—zero on the inside, and E = σ/ε0 on the outside, with σ the local surface charge density. And it also works the other way around: if we’d have some distribution of charges inside of a closed conductor, those charges would not produce any field outside. So shielding works both ways!

Some closing remarks

A car is not a sphere. Some surfaces may have points or sharp ends, like the object sketched below. Again, the charges will try to spread out as much as possible on the surface, and the tip of a sharp point is as far away as it is possible from most of the surface. Therefore, we should expect the surface density to be very high there. Now, a high charge density means a high field just outside. In fact, if the electric field is too great, air will break down, so we get a discharge. As Feynman explains it:

“Air will break down if the electric field is too great. What happens is that a loose charge (electron, or ion) somewhere in the air is accelerated by the field, and if the field is very great, the charge can pick up enough speed before it hits another atom to be able to knock an electron off that atom. As a result, more and more ions are produced. Their motion constitutes a discharge, or spark. If you want to charge an object to a high potential and not have it discharge itself by sparks in the air, you must be sure that the surface is smooth, so that there is no place where the field is abnormally large.”

Sharp tip

It explains why lightning is attracted to pointy objects, so you should stay away from them.

What about planes and lightning? Well… There’s a nice article on that on the Scientific American website. Let me quote a paragraph that sort of sums up what actually happens:

“Although passengers and crew may see a flash and hear a loud noise if lightning strikes their plane, nothing serious should happen because of the careful lightning protection engineered into the aircraft and its sensitive components. Initially, the lightning will attach to an extremity such as the nose or wing tip. The airplane then flies through the lightning flash, which reattaches itself to the fuselage at other locations while the airplane is in the electric “circuit” between the cloud regions of opposite polarity. The current will travel through the conductive exterior skin and structures of the aircraft and exit off some other extremity, such as the tail. Pilots occasionally report temporary flickering of lights or short-lived interference with instruments.”

One more thing perhaps: isn’t incredible that, even when lightning goes through a car or a plane, it’s only the surface that’s being affected? I mean… It’s fairly easy to see the equilibrium situation, which has the charges on the surface only. But what about the dynamics indeed? 30,000 amps, 100 million volts, and 25,000 to 30,000 degrees Celsius… As lightning strikes, that must go everywhere, no? Well… Yes and no. If there are pointy objects, lightning will effectively burn through them. For an example of the damage of lightning on the nose of an airplane, click this link. 🙂 But then… Well… Let me copy Feynman as he introduces the electric force:

“Consider a force like gravitation which varies predominantly inversely as the square of the distance, but which is about a billion-billion-billion-billion times stronger. And with another difference. There are two kinds of “matter,” which we can call positive and negative. Like kinds repel and unlike kinds attract—unlike gravity where there is only attraction. What would happen? A bunch of positives would repel with an enormous force and spread out in all directions. A bunch of negatives would do the same.”

So that’s what happens. The charges spread out, in a fraction of a second, all away from each other, and so they stay on the surface only, because that’s as far away as they can get from each other. As mentioned above, we’re talking atomic or molecular layers really, so they don’t penetrate, despite the incredible charges and voltages involved. Let me continue the quote—just to illustrate the strength of the forces involved:

“But an evenly mixed bunch of positives and negatives would do something completely different. The opposite pieces would be pulled together by the enormous attractions. The net result would be that the terrific forces would balance themselves out almost perfectly, by forming tight, fine mixtures of the positive and the negative, and between two separate bunches of such mixtures there would be practically no attraction or repulsion at all. […] There is such a force: the electrical force. And all matter is a mixture of positive protons and negative electrons which are attracting and repelling with this great force. So perfect is the balance, however, that when you stand near someone else you don’t feel any force at all. If there were even a little bit of unbalance you would know it. If you were standing at arm’s length from someone and each of you had one percent more electrons than protons, the repelling force would be incredible. How great? Enough to lift the Empire State Building? No! To lift Mount Everest? No! The repulsion would be enough to lift a “weight” equal to that of the entire earth!”

So… Well… That’s it. I’ll close this post with the promised note on Coulomb’s constant and the electric constant, but it’s just an addendum, so you don’t have to read it if you don’t feel like it, Vincent. 🙂

Addendum: Coulomb’s constant and the electric constant

The ke = 1/4πε0 factor in Coulomb’s Law is just a constant of proportionality. Coulomb’s formula is simple enough – it says that the force is directly proportional to the amount of charge and inversely proportional to the square of the distance – but it would be a miracle if the units came out alright, wouldn’t it? Indeed, we measure distance in meter, charge in coulomb, and force in newton. Now, we could re-define one of those units so as to get rid of the 1/4πε0 factor, but so that’s not what we’re going to do. Why not? First, the constant of proportionality depends on the medium. Indeed, εis the so-called permittivity in a vacuum, so that’s in empty space. The constant of proportionality will be different in a gas, and it will be different for different gases and different temperatures and at different pressure. You can check it online if you want – just click the link here for some examples – but I guess you’ll believe me. So, if we write 1/4πε instead of ke then we can put in a different ε for each medium and our formula is still OK.

Now, because you’re a smart kid, you’ll say that doesn’t quite answer the question: why do we write is as 1/4πε? Why don’t we simply write μ instead of 1/4πε, or just k or a or something? Well… There is an answer to that, but it’s complicated. First, the μ and μ0 symbols are already used for something else: it’s something similar as ε and εbut then for magnetic fields. To be precise, μ0 is referred to as the permeability of the vacuum (and μ is just the permeability of some non-vacuum medium, of course). Now, because electricity and magnetism are part of one and the same phenomenon in Nature (when you’re going for engineer, you’ll get one course on electromagnetism, not two separate ones), ε0 are μ0 related. In fact, they’re related through a marvelous formulas—a formula like E = mc2 in physics or, in math, eiπ+ 1 = 0. Don’t try to understand it. Just look at it:

c2ε0μ0 = (cε0)(cμ0) = 1

Amazing, isn’t it? The c here is the speed of light in a vacuum, obviously. So it’s a physical constant. In other words, unlike ε0 or μ0, it’s got nothing to do with proportionality or units: the speed of light is the speed of light no matter what units we use—meters or light-seconds or whatever. OK. Just swallow this and don’t pay too much attention. It’s just a digression, but let me finish it.

The equivalent of Coulomb’s Law in magnetism is Ampère’s Law, and it involves the circulation of a field, as illustrated below. So that’s why Ampère’s Law involves a 2π factor.

MagneticWireAttraction-2nd

In fact, because we’re talking two wires (or two conductors) with currents going through them (I1 and I2 respectively), the proportionality constant in Ampère’s Law is written as 2kA.

Ampere Law

Now, I won’t go too much into the detail but the thing about the circulation and that factor 2 in Ampère’s Law result in μbeing written as μ0 = 4π×10–7 N/A2. As for the units: N is newton and A is ampere obviously. And so that’s why we have the 4π in the proportionality constant for Coulomb’s Law as well. And, of course, the (cε0)(cμ0) = 1 equation makes it obvious that cε0 and cμ0 are reciprocal numbers, so that’s why we write 1/4πε0 for the proportionality constant in Coulomb’s Law, rather than kor a or whatever other simple thing. […] Well… Sort of. In any case, nothing to worry about. 🙂

Advertisements

Applied vector analysis (II)

We’ve covered a lot of ground in the previous post, but we’re not quite there yet. We need to look at a few more things in order to gain some kind of ‘physical’ understanding’ of Maxwell’s equations, as opposed to a merely ‘mathematical’ understanding only. That will probably disappoint you. In fact, you probably wonder why one needs to know about Gauss’ and Stokes’ Theorems if the only objective is to ‘understand’ Maxwell’s equations.

To some extent, your skepticism is justified. It’s already quite something to get some feel for those two new operators we’ve introduced in the previous post, i.e. the divergence (div) and curl operators, denoted by ∇• and × respectively. By now, you understand that these two operators act on a vector field, such as the electric field vector E, or the magnetic field vector B, or, in the example we used, the heat flow h, so we should write •(a vector) and ×(a vector. And, as for that del operator – i.e.  without the dot (•) or the cross (×) – if there’s one diagram you should be able to draw off the top of your head, it’s the one below, which shows:

  1. The heat flow vector h, whose magnitude is the thermal energy that passes, per unit time and per unit area, through an infinitesimally small isothermal surface, so we write: h = |h| = ΔJ/ΔA.
  2. The gradient vector T, whose direction is opposite to that of h, and whose magnitude is proportional to h, so we can write the so-called differential equation of heat flow: h = –κT.
  3. The components of the vector dot product ΔT = T•ΔR = |T|·ΔR·cosθ.

Temperature drop

You should also remember that we can re-write that ΔT = T•ΔR = |T|·ΔR·cosθ equation – which we can also write as ΔT/ΔR = |T|·cosθ – in a more general form:

Δψ/ΔR = |ψ|·cosθ

That equation says that the component of the gradient vector ψ along a small displacement ΔR is equal to the rate of change of ψ in the direction of ΔRAnd then we had three important theorems, but I can imagine you don’t want to hear about them anymore. So what can we do without them? Let’s have a look at Maxwell’s equations again and explore some linkages.

Curl-free and divergence-free fields

From what I wrote in my previous post, you should remember that:

  1. The curl of a vector field (i.e. ×C) represents its circulation, i.e. its (infinitesimal) rotation.
  2. Its divergence (i.e. ∇•C) represents the outward flux out of an (infinitesimal) volume around the point we’re considering.

Back to Maxwell’s equations:

Maxwell's equations-2

Let’s start at the bottom, i.e. with equation (4). It says that a changing electric field (i.e. ∂E/∂t ≠ 0) and/or a (steady) electric current (j0) will cause some circulation of B, i.e. the magnetic field. It’s important to note that (a) the electric field has to change and/or (b) that electric charges (positive or negative) have to move  in order to cause some circulation of B: a steady electric field will not result in any magnetic effects.

This brings us to the first and easiest of all the circumstances we can analyze: the static case. In that case, the time derivatives ∂E/∂t and ∂B/∂t are zero, and Maxwell’s equations reduce to:

  1. ∇•E = ρ/ε0. In this equation, we have ρ, which represents the so-called charge density, which describes the distribution of electric charges in space: ρ = ρ(x, y, z). To put it simply: ρ is the ‘amount of charge’ (which we’ll denote by Δq) per unit volume at a given point. Hence, if we  consider a small volume (ΔV) located at point (x, y, z) in space – an infinitesimally small volume, in fact (as usual) –then we can write: Δq =  ρ(x, y, z)ΔV. [As for ε0, you already know this is a constant which ensures all units are ‘compatible’.] This equation basically says we have some flux of E, the exact amount of which is determined by the charge density ρ or, more in general, by the charge distribution in space.  
  2. ×E = 0. That means that the curl of E is zero: everywhere, and always. So there’s no circulation of E. We call this a curl-free field.
  3. B = 0. That means that the divergence of B is zero: everywhere, and always. So there’s no flux of B. None. We call this a divergence-free field.
  4. c2∇×B = j0. So here we have steady current(s) causing some circulation of B, the exact amount of which is determined by the (total) current j. [What about that cfactor? Well… We talked about that before: magnetism is, basically, a relativistic effect, and so that’s where that factor comes from. I’ll just refer you to what Feynman writes about this in his Lectures, and warmly recommend to read it, because it’s really quite interesting: it gave me at least a much deeper understanding of what it’s all about, and so I hope it will help you as much.]

Now you’ll say: why bother with all these difficult mathematical constructs if we’re going to consider curl-free and divergence-free fields only. Well… B is not curl-free, and E is not divergence-free. To be precise:

  1. E is a field with zero curl and a given divergence, and
  2. B is a field with zero divergence and a given curl.

Yeah, but why can’t we analyze fields that have both curl and divergence? The answer is: we can, and we will, but we have to start somewhere, and so we start with an easier analysis first.

Electrostatics and magnetostatics

The first thing you should note is that, in the static case (i.e. when charges and currents are static), there is no interdependence between E and B. The two fields are not interconnected, so to say. Therefore, we can neatly separate them into two pairs:

  1. Electrostatics: (1) ∇•E = ρ/ε0 and (2) ×E = 0.
  2. Magnetostatics: (1) ∇×B = j/c2ε0 and (2) B = 0.

Now, I won’t go through all of the particularities involved. In fact, I’ll refer you to a real physics textbook on that (like Feynman’s Lectures indeed). My aim here is to use these equations to introduce some more math and to gain a better understanding of vector calculus – an understanding that goes, in fact, beyond the math (i.e. a ‘physical’ understanding, as Feynman terms it).

At this point, I have to introduce two additional theorems. They are nice and easy to understand (although not so easy to prove, and so I won’t):

Theorem 1: If we have a vector field – let’s denote it by C – and we find that its curl is zero everywhere, then C must be the gradient of something. In other words, there must be some scalar field ψ (psi) such that C is equal to the gradient of ψ. It’s easier to write this down as follows:

If ×= 0, there is a ψ such that C = ψ.

Theorem 2: If we have a vector field – let’s denote it by D, just to introduce yet another letter – and we find that its divergence is zero everywhere, then D must be the curl of some vector field A. So we can write:

If D = 0, there is an A such that D = ×A.

We can apply this to the situation at hand:

  1. For E, there is some scalar potential Φ such that E = –Φ. [Note that we could integrate the minus sign in Φ, but we leave it there as a reminder that the situation is similar to that of heat flow. It’s a matter of convention really: E ‘flows’ from higher to lower potential.]
  2. For B, there is a so-called vector potential A such that B = ×A.

The whole game is then to compute Φ and A everywhere. We can then take the gradient of Φ, and the curl of A, to find the electric and magnetic field respectively, at every single point in space. In fact, most of Feynman’s second Volume of his Lectures is devoted to that, so I’ll refer you that if you’d be interested. As said, my goal here is just to introduce the basics of vector calculus, so you gain a better understanding of physics, i.e. an understanding which goes beyond the math.

Electrodynamics

We’re almost done. Electrodynamics is, of course, much more complicated than the static case, but I don’t have the intention to go too much in detail here. The important thing is to see the linkages in Maxwell’s equations. I’ve highlighted them below:

Maxwell interaction

I know this looks messy, but it’s actually not so complicated. The interactions between the electric and magnetic field are governed by equation (2) and (4), so equation (1) and (3) is just ‘statics’. Something needs to trigger it all, of course. I assume it’s an electric current (that’s the arrow marked by [0]).

Indeed, equation (4), i.e. c2∇×B = ∂E/∂t + j0, implies that a changing electric current – an accelerating electric charge, for instance – will cause the circulation of B to change. More specifically, we can write: ∂[c2∇×B]/∂t = ∂[j0]∂t. However, as the circulation of B changes, the magnetic field B itself must be changing. Hence, we have a non-zero time derivative of B (∂B/∂t ≠ 0). But, then, according to equation (2), i.e. ∇×E = –∂B/∂t, we’ll have some circulation of E. That’s the dynamics marked by the red arrows [1].

Now, assuming that ∂B/∂t is not constant (because that electric charge accelerates and decelerates, for example), the time derivative ∂E/∂t will be non-zero too (∂E/∂t ≠ 0). But so that feeds back into equation (4), according to which a changing electric field will cause the circulation of B to change. That’s the dynamics marked by the yellow arrows [2].

The ‘feedback loop’ is closed now: I’ve just explained how an electromagnetic field (or radiation) actually propagates through space. Below you can see one of the fancier animations you can find on the Web. The blue oscillation is supposed to represent the oscillating magnetic vector, while the red oscillation is supposed to represent the electric field vector. Note how the effect travels through space.

emwave2

This is, of course, an extremely simplified view. To be precise, it assumes that the light wave (that’s what an electromagnetic wave actually is) is linearly (aka as plane) polarized, as the electric (and magnetic field) oscillate on a straight line. If we choose the direction of propagation as the z-axis of our reference frame, the electric field vector will oscillate in the xy-plane. In other words, the electric field will have an x- and a y-component, which we’ll denote as Ex and Erespectively, as shown in the diagrams below, which give various examples of linear polarization.

linear polarizationLight is, of course, not necessarily plane-polarized. The animation below shows circular polarization, which is a special case of the more general elliptical polarization condition.

Circular.Polarization.Circularly.Polarized.Light_Right.Handed.Animation.305x190.255Colors

The relativity of magnetic and electric fields

Allow me to make a small digression here, which has more to do with physics than with vector analysis. You’ll have noticed that we didn’t talk about the magnetic field vector anymore when discussing the polarization of light. Indeed, when discussing electromagnetic radiation, most – if not all – textbooks start by noting we have E and B vectors, but then proceed to discuss the E vector only. Where’s the magnetic field? We need to note two things here.

1. First, I need to remind you of the force on any electrically charged particle (and note we only have electric charge: there’s no such thing as a magnetic charge according to Maxwell’s third equation) consists of two components. Indeed, the total electromagnetic force (aka Lorentz force) on a charge q is:

F = q(E + v×B) = qE + q(v×B) = FE + FM

The velocity vector v is the velocity of the charge: if the charge is not moving, then there’s no magnetic force. The illustration below shows you the components of the vector cross product that, by now, you’re fully familiar with. Indeed, in my previous post, I gave you the expressions for the x, y and z coordinate of a cross product, but there’s a geometrical definition as well:

v×B = |v||B|sin(θ)n

magnetic force507px-Right_hand_rule_cross_product

The magnetic force FM is q(v×B) = qv×B q|v||B|sin(θ)n. The unit vector n determines the direction of the force, which is determined by that right-hand rule that, by now, you also are fully familiar with: it’s perpendicular to both v and B (cf. the two 90° angles in the illustration). Just to make sure, I’ve also added the right-hand rule illustration above: check it out, as it does involve a bit of arm-twisting in this case. 🙂

In any case, the point to note here is that there’s only one electromagnetic force on the particle. While we distinguish between an E and a B vector, the E and B vector depend on our reference frame. Huh? Yes. The velocity v is relative: we specify the magnetic field in a so-called inertial frame of reference here. If we’d be moving with the charge, the magnetic force would, quite simply, disappear, because we’d have a v equal to zero, so we’d have v×B = 0×B= 0. Of course, all other charges (i.e. all ‘stationary’ and ‘moving’ charges that were causing the field in the first place) would have different velocities as well and, hence, our E and B vector would look very different too: they would come in a ‘different mixture’, as Feynman puts it. [If you’d want to know in what mixture exactly, I’ll refer you Feynman: it’s a rather lengthy analysis (five rather dense pages, in fact), but I can warmly recommend it: in fact, you should go through it if only to test your knowledge at this point, I think.]

You’ll say: So what? That doesn’t answer the question above. Why do physicists leave out the magnetic field vector in all those illustrations?

You’re right. I haven’t answered the question. This first remark is more like a warning. Let me quote Feynman on it:

“Since electric and magnetic fields appear in different mixtures if we change our frame of reference, we must be careful about how we look at the fields E and B. […] The fields are our way of describing what goes on at a point in space. In particular, E and B tell us about the forces that will act on a moving particle. The question “What is the force on a charge from a moving magnetic field?” doesn’t mean anything precise. The force is given by the values of E and B at the charge, and the F = q(E + v×B) formula is not to be altered if the source of E or B is moving: it is the values of E and B that will be altered by the motion. Our mathematical description deals only with the fields as a function of xy, z, and t with respect to some inertial frame.”

If you allow me, I’ll take this opportunity to insert another warning, one that’s quite specific to how we should interpret this concept of an electromagnetic wave. When we say that an electromagnetic wave ‘travels’ through space, we often tend to think of a wave traveling on a string: we’re smart enough to understand that what is traveling is not the string itself (or some part of the string) but the amplitude of the oscillation: it’s the vertical displacement (i.e. the movement that’s perpendicular to the direction of ‘travel’) that appears first at one place and then at the next and so on and so on. It’s in that sense, and in that sense only, that the wave ‘travels’. However, the problem with this comparison to a wave traveling on a string is that we tend to think that an electromagnetic wave also occupies some space in the directions that are perpendicular to the direction of travel (i.e. the x and y directions in those illustrations on polarization). Now that’s a huge misconception! The electromagnetic field is something physical, for sure, but the E and B vectors do not occupy any physical space in the x and y direction as they ‘travel’ along the z direction!

Let me conclude this digression with Feynman’s conclusion on all of this:

“If we choose another coordinate system, we find another mixture of E and B fields. However, electric and magnetic forces are part of one physical phenomenon—the electromagnetic interactions of particles. While the separation of this interaction into electric and magnetic parts depends very much on the reference frame chosen for the description, the complete electromagnetic description is invariant: electricity and magnetism taken together are consistent with Einstein’s relativity.”

2. You’ll say: I don’t give a damn about other reference frames. Answer the question. Why are magnetic fields left out of the analysis when discussing electromagnetic radiation?

The answer to that question is very mundane. When we know E (in one or the other reference frame), we also know B, and, while B is as ‘essential’ as E when analyzing how an electromagnetic wave propagates through space, the truth is that the magnitude of B is only a very tiny fraction of that of E.

Huh? Yes. That animation with these oscillating blue and red vectors is very misleading in this regard. Let me be precise here and give you the formulas:

E vector of wave

B vector of a wave

I’ve analyzed these formulas in one of my other posts (see, for example, my first post on light and radiation), and so I won’t repeat myself too much here. However, let me recall the basics of it all. The eR′ vector is a unit vector pointing in the apparent direction of the charge. When I say ‘apparent’, I mean that this unit vector is not pointing towards the present position of the charge, but at where is was a little while ago, because this ‘signal’ can only travel from the charge to where we are now at the same speed of the wave, i.e. at the speed of light c. That’s why we prime the (radial) vector R also (so we write R′ instead of R). So that unit vector wiggles up and down and, as the formula makes clear, it’s the second-order derivative of that movement which determines the electric field. That second-order derivative is the acceleration vector, and it can be substituted for the vertical component of the acceleration of the charge that caused the radiation in the first place but, again, I’ll refer you my post on that, as it’s not the topic we want to cover here.

What we do want to look at here, is that formula for B: it’s the cross product of that eR′ vector (the minus sign just reverses the direction of the whole thing) and E divided by c. We also know that the E and eR′ vectors are at right angles to each, so the sine factor (sinθ) is 1 (or –1) too. In other words, the magnitude of B is |E|/c =  E/c, which is a very tiny fraction of E indeed (remember: c ≈ 3×108).

So… Yes, for all practical purposes, B doesn’t matter all that much when analyzing electromagnetic radiation, and so that’s why physicists will note it but then proceed and look at E only when discussing radiation. Poor BThat being said, the magnetic force may be tiny, but it’s quite interesting. Just look at its direction! Huh? Why? What’s so interesting about it?  I am not talking the direction of B here: I am talking the direction of the force. Oh… OK… Hmm… Well…

Let me spell it out. Take the force formula: F = q(E + v×B) = qE + q(v×B). When our electromagnetic wave hits something real (I mean anything real, like a wall, or some molecule of gas), it is likely to hit some electron, i.e. an actual electric charge. Hence, the electric and magnetic field should have some impact on it. Now, as we pointed here, the magnitude of the electric force will be the most important one – by far – and, hence, it’s the electric field that will ‘drive’ that charge and, in the process, give it some velocity v, as shown below. In what direction? Don’t ask stupid questions: look at the equation. FE = qE, so the electric force will have the same direction as E.

radiation pressure

But we’ve got a moving charge now and, therefore, the magnetic force comes into play as well! That force is FM  = q(v×B) and its direction is given by the right-hand rule: it’s the F above in the direction of the light beam itself. Admittedly, it’s a tiny force, as its magnitude is F = qvE/c only, but it’s there, and it’s what causes the so-called radiation pressure (or light pressure tout court). So, yes, you can start dreaming of fancy solar sailing ships (the illustration below shows one out of of Star Trek) but… Well… Good luck with it! The force is very tiny indeed and, of course, don’t forget there’s light coming from all directions in space!

solar sail

Jokes aside, it’s a real and interesting effect indeed, but I won’t say much more about it. Just note that we are really talking the momentum of light here, and it’s a ‘real’ as any momentum. In an interesting analysis, Feynman calculates this momentum and, rather unsurprisingly (but please do check out how he calculates these things, as it’s quite interesting), the same 1/c factor comes into play once: the momentum (p) that’s being delivered when light hits something real is equal to 1/c of the energy that’s being absorbed. So, if we denote the energy by W (in order to not create confusion with the E symbol we’ve used already), we can write: p = W/c.

Now I can’t resist one more digression. We’re, obviously, fully in classical physics here and, hence, we shouldn’t mention anything quantum-mechanical here. That being said, you already know that, in quantum physics, we’ll look at light as a stream of photons, i.e. ‘light particles’ that also have energy and momentum. The formula for the energy of a photon is given by the Planck relation: E = hf. The h factor is Planck’s constant here – also quite tiny, as you know – and f is the light frequency of course. Oh – and I am switching back to the symbol E to denote energy, as it’s clear from the context I am no longer talking about the electric field here.

Now, you may or may not remember that relativity theory yields the following relations between momentum and energy:  

E2 – p2c2 = m0cand/or pc = Ev/c

In this equations, mstands, obviously, for the rest mass of the particle, i.e. its mass at v = 0. Now, photons have zero rest mass, but their speed is c. Hence, both equations reduce to p = E/c, so that’s the same as what Feynman found out above: p = W/c.

Of course, you’ll say: that’s obvious. Well… No, it’s not obvious at all. We do find the same formula for the momentum of light (p) – which is great, of course –  but so we find the same thing coming from very different necks parts of the woods. The formula for the (relativistic) momentum and energy of particles comes from a very classical analysis of particles – ‘real-life’ objects with mass, a very definite position in space and whatever other properties you’d associate with billiard balls – while that other p = W/c formula comes out of a very long and tedious analysis of light as an electromagnetic wave. The two analytical frameworks couldn’t differ much more, could they? Yet, we come to the same conclusion indeed.

Physics is wonderful. 🙂

So what’s left?

Lots, of course! For starters, it would be nice to show how these formulas for E and B with eR′ in them can be derived from Maxwell’s equations. There’s no obvious relation, is there? You’re right. Yet, they do come out of the very same equations. However, for the details, I have to refer you to Feynman’s Lectures once again – to the second Volume to be precise. Indeed, besides calculating scalar and vector potentials in various situations, a lot of what he writes there is about how to calculate these wave equations from Maxwell’s equations. But so that’s not the topic of this post really. It’s, quite simply, impossible to ‘summarize’ all those arguments and derivations in a single post. The objective here was to give you some idea of what vector analysis really is in physics, and I hope you got the gist of it, because that’s what needed to proceed. 🙂

The other thing I left out is much more relevant to vector calculus. It’s about that del operator () again: you should note that it can be used in many more combinations. More in particular, it can be used in combinations involving second-order derivatives. Indeed, till now, we’ve limited ourselves to first-order derivatives only. I’ll spare you the details and just copy a table with some key results:

  1. •(T) = div(grad T) = T = ()T = ∇2T = ∂2T/∂x+ ∂2T/∂y+ ∂2T/∂z= a scalar field
  2. ()h = ∇2= a vector field
  3. (h) = grad(div h) = a vector field
  4. ×(×h) = curl(curl h) =(h) – ∇2h
  5. ∇•(×h) = div(curl h) = 0 (always)
  6. ×(T) = curl(grad T) = 0 (always)

So we have yet another set of operators here: not less than six, to be precise. You may think that we can have some more, like (×), for example. But… No. A (×) operator doesn’t make sense. Just write it out and think about it. Perhaps you’ll see why. You can try to invent some more but, if you manage, you’ll see they won’t make sense either. The combinations that do make sense are listed above, all of them.

Now, while of these combinations make (some) sense, it’s obvious that some of these combinations are more useful than others. More in particular, the first operator, ∇2, appears very often in physics and, hence, has a special name: it’s the Laplacian. As you can see, it’s the divergence of the gradient of a function.

Note that the Laplace operator (∇2) can be applied to both scalar as well as vector functions. If we operate with it on a vector, we’ll apply it to each component of the vector function. The Wikipedia article on the Laplace operator shows how and where it’s used in physics, and so I’ll refer to that if you’d want to know more. Below, I’ll just write out the operator itself, as well as how we apply it to a vector:

Laplacian

Laplacian-2

So that covers (1) and (2) above. What about the other ‘operators’?

Let me start at the bottom. Equations (5) and (6) are just what they are: two results that you can use in some mathematical argument or derivation. Equation (4) is… Well… Similar: it’s an identity that may or may not help one when doing some derivation.

What about (3), i.e. the gradient of the divergence of some vector function? Nothing special. As Feynman puts it: “It is a possible vector field, but there is nothing special to say about it. It’s just some vector field which may occasionally come up.”

So… That should conclude my little introduction to vector analysis, and so I’ll call it a day now. 🙂 I hope you enjoyed it.