Transcription
What you're looking at is the electric field produced by a charged particle that's accelerating to nearly the speed of light. With each burst of acceleration, the charge sends an electromagnetic wave radiating out to infinity, like the ripples of a stone dropped into a pond.
Needless to say, it's quite a bit more complicated than the electric field of a stationary charge described by Kulum's law, which is what you probably learned the first time you were ever taught about electricity. But Kulum's law is only the beginning. And if you stop there, you'll be missing out on the fascinating physics that's responsible for light itself. And that's why in this video, I'm going to show you how to go beyond that simple static picture of a charge sitting at rest, to understand the fields produced by moving and accelerating electric charges.
We'll start off with the basics, reviewing what an electric field is and the different ways that we can visualize it. We'll develop the geometric intuition behind Gaus's law and see how it quantifies the connection between the electric field and the charges that sourced it. From there, we'll allow our charged particles to start moving at first with constant velocity. And we'll discover that in order to be consistent with the laws of special relativity, the electric field produced by a moving charge is very different from the familiar coolum field as the particle approaches the speed of light. And finally, we'll return to the problem of an accelerating charge. The fields here can get quite complicated, but by focusing on the simplest possible example, we'll discover that an accelerating charge must produce electromagnetic waves that expand to infinity at the speed of light.
And as usual, I've written up notes to go along with this video that go into greater detail about all the topics we'll cover. And you can get those for free by following the link down in the description. Let's begin.
Let's start with a brief review of the fundamentals. Picture a particle of charge big Q sitting at rest. As you probably learned in a high school science class, if any other test charge little Q is placed nearby, it will experience an electric force along the line connecting the two charges. The force is proportional to the product of the two charges and one over the distance r between them squared. And if we write r hat for the unit vector that points radially away from the origin, we arrive at kum's law for the electric force exerted on little q, where k is a constant that sets the strength of the electric force.
Actually, it's more convenient to write it in the form 1 / 4 pi epsilon0, for reasons that'll become clear a little later, where epsilon0 is just another constant. If the two charges have the same sign, then their product is positive and so the coolum force points in the positive r direction, repelling the test charge away from the origin. If they have opposite signs, then the product is negative and the test charge is attracted toward the source. But let's say they're both positive. Then here the force points up and to the left. Here it's down and to the left. Here down and to the right, and so on. In fact, at every point in space, we can draw the arrow that represents the force that the test charge would feel were we to place it at that location.
Better yet, we can draw the arrows that represent the force per unit test charge, F divided by little Q. That way, the arbitrary test charge drops out entirely and the resulting collection of arrows is what we call the electric field E produced by the source. E is a vector field, meaning that it's a function that assigns a vector to each point in space. And it tells us the force that would be exerted on any test charge at that location, simply by multiplying by little Q. From now on, though, we'll mostly forget about the test charge and just focus on the electric field itself.
First of all, it's important to understand that although I've been drawing pictures in a two-dimensional plane here, the actual electric field fills the full three-dimensional space surrounding the particle. That's a bit overwhelming to try to visualize though. So, I'll mainly be illustrating a two-dimensional slice of the field to keep things simpler. Also, rather than drawing the full length of the arrows like this, it's often more convenient to show the strength of the field using color instead. That way, we can draw a lot more arrows without having to worry about them overlapping each other.
If we have multiple source charges, we can just add up the electric fields from each of them to get the total. And you might have seen before some of the pretty pictures that you can get from making different arrangements of positive and negative charges. Just remember that the arrows always lead away from the positive charges and toward the negative ones, unless they go out to infinity. As we'll see though, the field of a single charge is already going to be complicated enough once we allow it to start moving. Indeed, it's important to understand that Kulum's formula only describes the field of a charge that's sitting at rest. When we allow the charge to start moving close to the speed of light and in particular accelerating, the field that it produces becomes much more intricate. We'll come back to moving and accelerating charges later though.
For now, there's more for us to learn about the field produced by a single stationary charge, because next we need to talk about one of the fundamental equations of electromagnetism, Gaus's law. To motivate where it comes from though, let's first discuss another very useful way of visualizing the electric field. Rather than drawing the actual arrows at a smattering of sample points in space like this, we can connect them up with streamlines that trace out the direction of the field at each point they pass through. These are the particles electric field lines, and they're reminiscent of the jets of water that you might see spraying out from a sprinkler.
In fact, pushing that analogy a little further, if we wanted to measure the strength of a sprinkler, we could imagine surrounding the whole thing in a giant rubber balloon and measuring how much water per unit time is hitting its surface. In a similar way, let's picture an imaginary sphere surrounding our point charge. Then, the strength of the field lines passing through the surface should give us a measure of the amount of charge contained inside. To make that idea precise, imagine slicing the surface up into lots of tiny patches. For each patch, we can construct a vector da whose magnitude is the area of the patch and whose direction points perpendicularly away from the surface. And by taking the dot product between that area vector and the electric field vector, we get a number that quantifies the strength of the electric field that's passing outward through that tiny area. Finally, we'll add up the contributions from all the little patches making up the sphere by integrating the dot product over the entire surface. The result is a number called the electric flux passing through the surface. And it would correspond to the amount of water flowing outward per unit time.
Back in our sprinkler analogy, here in the electric case, the flux is supposed to measure for us the amount of charge contained within the surface. So, let's see what we get when we plug in our formula for the kum field. The integral may look a little complicated at first, but it's actually dead simple because we have spherical symmetry. First of all, let's pull all these constant factors out front. What's more, because we're integrating over the surface of a sphere, the radial coordinate r is also a constant, and we can pull that factor outside the integral as well. As for this dot product, remember that the area vector is defined to point perpendicularly away from the surface, which for a sphere just means that it points radially outward along the R hat direction. And when we take the dot product of the unit vector R hat with itself here, we just get one. So all that remains of that complicated looking integral is just the area da of each little patch summed up over all the little patches making up the entire surface. And that sum is simply the total area of the sphere, 4 pi r squared. So lo and behold, the factors of 4 pi r squared cancel. That's why it was convenient before to write kulum's constant in that funny way. And so, just as we'd hoped, the total electric flux passing through our surface has indeed told us the amount of charge Q contained inside of it.
And this is in fact a completely general property of the electric field known as Gaus's law. It holds for any configuration of electric charges and for any surface S surrounding them. It doesn't necessarily have to be a sphere. And it also holds not just for stationary charges like we've been considering so far, but also for charges that are moving with non-zero velocity. And that's the topic that we'll turn to next.
As mentioned before, Kulum's law describes the electric field of a stationary charged particle. The field of a rapidly moving charge can look quite different. To keep things simple to begin with, let's consider a charge that spent its whole life moving with constant speed V to the right. We'll deal with accelerating charges later on. Then what does the electric field produced by this particle look like? Well, your first guess might be that it looks about the same, except that the moving charge drags the usual coolum field along with it as it goes. In other words, instead of pointing in the radial direction away from the origin along little r hat, we'd guess that the field points radially away from the current position of the charge at time t. If we call that new vector big r hat and that new distance big r, then the natural guess would be that the electric field is the same as our old kum formula, but with all the little rs replaced by big rs that refer to the current moving position of the charge. That's a good guess. And in fact, it's an excellent approximation to the field, provided that the charge isn't moving too fast. But as the speed of the charge increases and approaches the speed of light, the shape of its electric field changes dramatically. The field gets squished along the direction of motion, becoming very weak in the forward and backward directions, and conversely [clears throat] very strong in the perpendicular direction.
And the easiest way to derive that field is to demand that it's consistent with the principles of special relativity. In other words, consider again the ordinary kulum field for a charged particle sitting at rest. If we hop on a train that's moving to the left at speed V and look out the window from the perspective of the train's frame of reference, it's the particle that's now moving to the right at speed V. And therefore, by transforming the original coolum field from the particle's arrest frame to the frame of the train, we can determine the electric field produced by a moving charge. And if you've studied a little special relativity before, the result is very similar to the effect of length contraction.
To review how that works, picture a box of side length L sitting at rest. If we now set it moving to the right with speed V and measure its length again, we'll find that the answer has been shortened along the direction of motion by a factor called gamma that depends on the speed of the box. That factor is given by 1 over the square root of 1 - v squared / c squared. And it's a function that shows up all over the place in special relativity. Here's what the plot of it looks like. When the speed v is equal to zero, gamma is equal to 1 and the length of the box is therefore unchanged. But as v increases and gets closer to the speed of light, the denominator here starts going to zero. And so gamma gets larger and larger. As a result, our box gets squished flatter and flatter along its direction of motion the faster and faster that it moves. And essentially the same effect leads to the squishing of the electric field of our moving particle when its speed is anywhere close to the speed of light.
Like we just discussed, the field is well approximated by Kulum's formula provided that V is small. It points radially away from the charge and it has the same magnitude proportional to 1 / r squared in all directions surrounding the particle. But as we increase the speed, the spherical symmetry breaks and the coolum approximation falls apart. Instead, the magnitude of the field is modified by an additional factor f that depends on the angle theta from the direction of motion to the point where we want to know the field. This is the generalization of Kulum's law for a particle moving with constant velocity V, where this additional factor F is given explicitly by 1 - v squared C squared / 1 - sin squared of theta times v squared C squared all raised to the 3/2 power. I'll show you where that comes from in detail in the notes, but for now I just want to explain how it's consistent with the squashed shape of the field that I've tried to motivate here.
First of all, when v is small compared to the speed of light, these factors of v squared / c squared are close to zero and the whole thing in brackets approaches 1. In that case, we therefore get back the approximate kum field and spherical symmetry has been restored because the factors of theta have disappeared. But now suppose that the particle is moving faster so that we can't ignore those corrections anymore. If we look at the field directly in front of the particle where theta is equal to 0, then the sin of theta vanishes and we simply get a factor of 1 - v squared / c squared, or in other words, 1 / gamma squared. Remembering that gamma gets really big when a particle approaches the speed of light. That means that the electric field has been reduced by a large factor compared to Kulum's formula directly in front of and also behind the moving particle in the perpendicular direction. On the other hand, where theta is equal to pi / 2, the sin is now equal to 1. And therefore, the factor in brackets becomes 1 / 1 - v squared C squared to the 1/2, which is just gamma again, meaning that the electric field at perpendicular angles with respect to the motion has been amplified by a factor of that large number gamma. And so this formula for the field is indeed consistent with what we might have expected for the squashing of the field lines along the direction of motion.
In fact, we can draw a simple picture of the magnitude of the field by making a polar plot of this function with respect to the angle theta. When the particle is sitting at rest, this function is equal to one and its polar plot simply traces out a circle, meaning that the magnitude of the electric field is the same in any direction surrounding the particle, as we would expect from Kulum's law. But as we begin to crank up the speed, the shape of the circle is deformed into something more like a peanut, showing again that the magnitude of the field is drastically reduced in the directions in front of and behind the particle, and simultaneously increased at right angles to the motion.
But spherical symmetry or no, Gaus's law nevertheless applies. Meaning that if we once again surround the particle by an imaginary sphere or any surface and compute the total flux of the electric field passing through it, the result is still given by the charge Q contained inside divided by epsilon0. It's just that most of that flux comes from the perpendicular directions where the field is strongest, rather than being uniformly distributed like it was for a stationary charge.
Finally, before we move on to discuss the electric field of an accelerating charge, it's important to understand that a moving charge produces not just an electric field, but a magnetic field as well. Whereas the electric field lines emanated radially away from the charge, the magnetic field lines circle around it. And taken together, the two fields look something like a wheel and its spokes. I'll continue to mainly discuss the electric field in this video so that we can focus on one thing at a time. But know that electricity is really just one half of the larger subject of electromagnetism.
At last we come to the most challenging, but also the most fascinating, variety of electric field: the field produced by an accelerating charge. We started off with the simple radial field of a charge sitting at rest. And from there we generalized to the electric field of a charge moving at constant velocity. But now we'll finally allow our charge to accelerate. And in doing so, we'll discover that the field must carry out waves of information that communicate the state of the charge to the rest of the universe.
To focus on perhaps the simplest example, suppose our charge is initially sitting at rest at the origin until somebody comes along and gives it a sharp kick at time t=0, rapidly accelerating it to a relativistic speed v. Then what does the electric field of the charge look like both before and after that abrupt burst of acceleration? Well, beforehand, the charge has just been sitting at rest at the origin. And so it sources the usual coolum field that we started with from the very beginning. A short time after the kick though, we'll find the particle traveling to the right with its new speed V. And given what we just talked about, you might guess that the charge is just going to drag along that same coolum field. And maybe it's squished a bit along the direction of motion depending on how fast that it's moving. But that's wrong. That picture was for a charge that's been moving at constant velocity forever, not one that's been suddenly accelerated up to speed v. And we can see that it's wrong for a very simple reason.
Suppose we go to draw what the field looks like at t=1 second after the initial kick. Picture yourself standing with your electric field meter at a point like this. That's a fair distance away from the origin. Let's say it's two light seconds away. Then at this moment, the news that the particle has suddenly started moving hasn't [clears throat] even reached you yet. It can't have, because a light signal emitted from the origin has only managed to travel half that distance in the intervening second. As far as you know, from so far away, the charge is still sitting at rest at the origin. And the electric field that you measure can't have changed at all yet. Instead, from the moment of the kick at t equals 0, the news that the particle has accelerated radiates away from the origin in a spherical shell that expands outward at the speed of light. At any point outside the shell, nothing's changed yet compared to the state of the field before the kick. The field lines still point radially away from where the charge was at the origin for all the long time before the clock struck t=0. Inside the shell, on the other hand, the news that the particle is now moving has arrived. And in this region, the field does look very much like what we just discussed for the field lines of a charge moving with constant velocity V.
All that we can deduce about the field of this accelerating charge solely on the basis of causality, that no signal can travel faster than the speed of light. What's even more interesting though is how the field behaves in between those two regions, in the vicinity of the expanding shell. The outer field lines of the charge that was at rest at the origin must smoothly connect to the inner field lines of the charge in motion. And there's really only one way that they can: in zigzagging arcs that run almost tangent to the shell. This is the electric field of a charge that's been rapidly accelerated from rest. Combined with the magnetic component, those expanding zigzags are the electromagnetic wave radiated out from the origin that communicates the news of the particles acceleration. In other words, an accelerating charge must produce an electromagnetic wave, a ripple in the electric and magnetic fields that expands outward at the speed of light.
To understand why the zigzagging lines take the shape that they do, consider just one line that emerges from the charge at some angle inside the shell. It extends radially away from the charge until it hits that boundary. And then it emerges at some new angle outside the shell, extending radially away from the origin. And again, in between, there's only one way the field line can go: tangentially along the arc of the circle. There's one thing causality alone doesn't tell us, though. Given the initial angle of the field line inside the shell, what angle does it wind up at when it emerges on the outside? To answer that question, we turn once more to Gaus's law with a clever choice for our surface S. Picking a starting point near the charge inside the shell. Imagine following the field line until we emerge at some point on the outside. By construction, then, the electric field points parallel to that path everywhere along it. Meanwhile, at the two ends of the segment, we can draw circular arcs that connect back down to the horizontal axis, the inner arc being centered at the current location of the charge while the outer arc is centered at the original location at the origin. Along those arcs, the electric field is then perpendicular to the path. Finally, by revolving that curve around the axis of motion, we obtain a closed surface S. But crucially, it's a surface that contains no charge anywhere inside of it. And Gaus's law therefore says that the total flux of the electric field passing through the surface must vanish.
The way we've constructed it though, the only contributions to that flux come from the inner and outer caps of the surface where the field is perpendicular to it. And therefore, the flux passing into the inner cap must cancel against the flux going out through the outer cap, so that the total adds up to zero. The outer flux is due to the original kulum field of the charge at rest at the origin. And so it's just some function of the angle of that outgoing field line. The inner flux on the other hand is due to the field of the moving charge. And so it's a function both of that inner angle as well as the speed v of the particle. Evaluating those two fluxes isn't too hard, but I don't want us to get bogged down computing integrals right now. So I'll once again leave the details for the notes. The upshot is that we get a simple relationship between the inner and outer angles of the field line and [clears throat] the speed of the charge. The tangent of the inner angle is equal to gamma times the tangent of the outer angle. The outer angles are of course uniformly distributed since they correspond to the usual kulum field lines. And if the particle isn't moving too fast compared to the speed of light, then gamma is close to one and the inner angles are almost identical to the outer angles, meaning that a field line that makes say a 30° angle outside the shell connects to one that's likewise at about 30° inside the shell. And if we draw a plot of the inner angle versus the outer angle, it's very nearly a straight line of slope one.
But as the particle speeds up, gamma gets bigger and bigger and the plot of this inverse tangent function begins to flatten out with a horizontal plateau at 90°. Meaning that all the outer field lines in that ever widening range emerge from the particle at very nearly a right angle. Which is exactly what we should expect because as we just learned, the electric field of a particle traveling at nearly light speed is squished along its direction of motion, very strong in the perpendicular direction, but conversely very weak elsewhere.
Remember that you can get the notes that I wrote to go along with this video for free at the link in the description. I want to say a special thank you to the Patreon donors whose support helped to make this video possible. You can join too at the link up in the corner. If you'd like to see more videos like this in the future, the visualizations I made for this video were only possible thanks to Manom, the animation library created by 3Blue1Brown. I'm thankful to Grant for creating and sharing such a powerful tool. And I'll also put a link down below to a beautiful video that he made about electromagnetic waves. Thank you so much for watching and I'll see you back here soon for another physics lesson.