Fundamentals of Physics II

PHYS 201 - Lecture 5 - The Electric Potential and Conservation of Energy

Chapter 1: Review of Electrostatics [00:00:00]

Professor Ramamurti Shankar: All right, guys, let’s start today with a new idea. But before proceeding, I should tell you where we stand. We can do the following problem. You give me any number of charges which are fixed, anywhere you want. So here they are. Here’s one guy q₁, here’s q₂, here’s q₃. And instead of these discrete charges, you may have a line charge too, maybe λ coulombs per meter. That guy is here. They are fixed. They are not dynamical. The object of interest to us is this charge q, which can actually move, allowed to move.

The question we ask is, do we know what this charge will do? And the answer is, yes, we know what it will do, because if it’s sitting here and we have already computed the electric field everywhere — you know how to compute electric field. Add the field due to this one, that one, that one, add that. That is some electric field sitting here. Then we know, the force on this charge we’re interested in, this q times the electric field at the location of the charge. So at every instant, we know what force is acting on it. Then from Newton’s law, that will give me the rate of change of velocity or the acceleration. You can write it either way. That’s all I need, because using that I can find out the new position, new velocity, right? The acceleration tells you, if you wait 1 nanosecond, you go to this place, because the initial velocity, I can predict where it will be, and because you know the acceleration, I know what velocity it will have. Then I keep on updating it and I get my trajectory.

Student: I think it should be dt squared.

Professor Ramamurti Shankar: Yes, thank you. So everyone understands this is how you proceed. This charge will of course push those guys, but we’re not worried, because we don’t let them move. They are just fixed. And if there are much such charges, we can find what they all do in this environment, which is the background electric field. That’s all we can do right now, this is electrostatics. In the real world, everybody is moving and all kinds of stuff happens, but you cannot use electrostatics for that. This asymmetry between the background and the test charge that we are studying is in the static situation. Guys that move and guys that don’t.

Now this is something you could have done on the first day, because I think on the very first day I told you about Coulomb’s law and the principle of superposition. If you put them together, you know the force on a charge q. Then you use F = ma, you predict its trajectory. But in principle, that’s all you need to do electrostatic problems, but in practice you want to clean up the machinery a little bit, make it more efficient. For example, we found Gauss’s law. Gauss’s law follows from Coulomb’s law, but we found it’s quite useful. For example, if we give you a ball of charge, spherical charge, going back to Coulomb’s law and first principles is very difficult. But if you use Gauss’s law and the symmetries, you’re able to find the electric field around the sphere, or an infinitely long cylinder, or an infinite plane or simple geometries like that. So it’s not enough to say “I’ve got the basic equations.” You always want to see if there are ways to speed up the computation to make it more efficient. So let me remind you what we did in the first part of the course.

Chapter 2: Review of Law of Conservation of Energy [00:03:50]

There, if you have the following question: here’s a spring and here’s a mass, the spring likes to have that length and I pulled it to way over there and I released it, and right now it happens to be here, going at some speed. I want to know what velocity it will have when it is here, so this position is x₁, this speed is v₁, this position is x₂. I want to know v₂. That’s a reasonable thing you can ask. Unlike the electric force, the force on this mass is due to the spring. Apart from that, it’s the same situation. So one way to do that is to say, I know F = ma, so the instant you start the problem here, I know the acceleration because the force of the spring is -k times x₁. Once I know the force, I know the acceleration. I can find the new velocity, new position, keep on doing it, or I can solve the equation, d²x/dt² is −kx/m, and I can eventually end up here. Then I can find the velocity here. But we all know that in this problem, there’s a shortcut, right? Yes? Anybody in the last row want to tell me what the shortcut is? Yes?

Student: You can use conservation of energy.

Professor Ramamurti Shankar: Conservation of energy. So conservation of energy is a consequence of Newton’s laws and certain properties of the force. I’m going to remind you how it’s done. One attitude I could take is, you’ve already done this before, but usually it’s better to make sure that you’re all on the same page when we do this. So let me remind you how it works there. You go to Newton’s laws and we write m dv/dt = F. It’s all in one dimension though. I’m moving just along the line. Then I multiply both sides by v. On the left hand side, if you know basic calculus, this is really d/dt of mv²/2, by the chain rule of calculus. The d/dt derivative is the v derivative, followed by the derivative of v with respect to t. v derivative will give you mv and then there is that. This becomes F dx/dt.

This means in every tiny instant of time, the force times the change in the position is equal to the change in this quantity, which we know is called kinetic energy. So if you multiply both sides by dt, then you find this relation, and if you add up all the changes, we get this relation, m — so let’s see what we are integrating from a starting point to an ending point and some starting time, ending time. We get mv₂²/2 - mv₁²/2 = integral f(x)dx from x₁ to x₂. And this is called the work energy theorem. The work energy theorem tells you something you may expect, namely, when a force acts on a body, it’s going to accelerate it. It’s going to accelerate it, it’s going to change the velocity. You’re going to ask, “How much velocity change do I get for a certain action of the force?” and the statement is, the force pushes the body from x₁ to x₂, then this is the change in this quantity called kinetic energy. This is true no matter what the force is. It’s true for friction, it’s true for spring, it’s true for everything, because Newton’s law is always valid. But if you want to extract from this a law of conservation of energy, you have to do a certain manipulation and that’s what I’m going to describe.

Chapter 3: Deriving the Work-Energy Theorem and the Law of Conservation of Energy [00:08:13]

So let’s go back here and try to extract from it a law of conservation of energy. So I’m going to write here k₂ - k₁ is integral F(x) dx, from x₁ to x₂. Now it’s a fundamental result in calculus that integral of any function of x from the point x₁ to the point x₂ can be written as G(x₂) - G(x₁), where G is any function whose derivative is the F I’m integrating. For example, if F = x, G = x²/2, because derivative x²/2 is x. In fact, it’s not just x²/2 . You can add x²/2 + a constant, but the constant is not important, because they’ll cancel between this and this. So you can pick the solution without a constant, okay?

Now what does that mean? Let’s rearrange the expression so I get K₂ − G(x₂) = K₁ − G(x₁). That tells you the combination K - G does not change as the particle moves, but this doesn’t look very nice. We want to get the old law of conservation of energy in the standard form. You define the function U(x), you simply - this G(x). It’s a very trivial change. Then you can say K₂ + U(x₂) = K₁ + U(x₁), and that is the energy that does not change with time. So the relation between U, which is called the potential energy, and the force, is the following. The force is - the derivative of the potential energy.

The - came because G is the integral of F, but U is -G, so that this formula, this means you have this relation. And you can also say U(x₂) − U(x₁) with an integral of F(x) dx from x₁ to x₂, there’s a - sign here. Again, the - sign is present because F is defined to be - the derivative of U. So you can go back and forth from potential to force and force to potential. If you knew the force, you can integrate it to get the potential. If you knew the potential, you can differentiate it to get the force, and more importantly, you have this great conservation law, which is very helpful in the spring problem, because if you know the potential energy, once and for all, it’s kx²/2, you balance the kinetic and the potential, you can find the velocity at one other point. Now why doesn’t this work for friction? Friction is the one dimensional force. Yes?

Student: It’s not conservative.

Professor Ramamurti Shankar: It’s not conservative. It’s just a definition. But what’s wrong with my argument? Where in this argument does it fail? It looks like any function — yes?

Student: Because the direction of the force depends on the direction that the particle’s moving.

Professor Ramamurti Shankar: Very good. The point is that frictional force is not a function just of your location. It depends on your velocity. In other words, whereas the spring will always exert a force -kx — if you are displaced to the right by an x, it exerts a force -kx — whether or not you’re moving to the right, away from the equilibrium or towards it. Thus given x, it gives you a given force, but friction is not like that. Friction does not have a definite direction. The only thing friction wants to do is to slow you down. If you want to go right, it wants to act left. If you want to go left, it wants to act right. I will resist all political analogies. I’ll just point out to you that friction does not have a definite sign, therefore it’s not a function of x, but as she said, it’s a function of x and of x dot, which is the velocity. So it doesn’t have a unique answer.

That does not mean you cannot find the work due to friction. If you know it’s moving to the right, then the force is to the left. You can integrate that force, as long as it is moving to the right. But if it’s going right, left, right, left and oscillating, then you cannot integrate it once and for all. That way doesn’t work for friction. Now here’s an even more powerful example of the law of conservation of energy. You have this famous roller coaster and you are here. I say “you are here,” because you’ll never get me on top of one of these guys. So you are here and you got some speed, and I want to know your speed when you are here.

Again, it follows from Newton’s laws. You can calculate from Newton’s laws, because at every point, wherever you may be, there’s the force of gravity acting this way and there is the perpendicular force from the track acting this way, and the perpendicular force is such that it cancels this part of gravity, and this part of gravity will accelerate it and you can find the velocity here. Keep doing this, you can find the velocity at the end. But that’s not how we do this. We use the fact that the gravity force allows us to define an energy, and I’ll come to that in a moment. Then we say kinetic + potential is kinetic + potential in the beginning and at the end, and we can simply find the velocity here if you knew the potential energy here.

All right, so the law of conservation of energy is also a very profound result, which has survived all the revolutions of quantum mechanics. After quantum mechanics, we don’t like to think in terms of force, and we don’t like to think in terms of trajectories. You don’t even know for sure where the particle is. But amazingly, the total energy is always conserved in the problems we study. So the law of conservation of energy is very profound and it’s also very useful for computations. Now the next question I have is, if I go to two dimensions — and it’s the same as in three dimensions. I’m just going to go to two dimensions — what does it take to get the law of conservation of energy? So I’m going to mimic this derivation in 2D and see if I can pull out a law of conservation of energy. So here’s what I will do. So in two dimensions, let me start this time with the kinetic energy. Kinetic energy is ½mv², but v has got 2 components, x and y, therefore it’s ½mvx² + vy², but I’m going to write this ½m v⋅v. Then let me ask, what’s the rate of change of this kinetic energy?

You’ve got to take derivatives. Now you may be a little queasy about taking derivatives from this dot product, but you can check by expanding the component, just like ordinary product, and you can take derivative of this guy times that guy, + derivative of that guy times this guy, and they will both be the same, and you will find it’s equal to m dv/dt, which is the force. I’m sorry. Let me erase this. It’s not m dv/dt; it is mdv/dt⋅v. So if you integrate this, cancel the dt and add all the changes, you will find K₂ - K₁ is m times dr — let me see — I’m sorry. I want to keep the dv/dt here, then I want to write this v as dr/dt. That’s right. So let me do the following, excuse me.

Let me do it again. dK/dt = ma times v and v is dr/dt. Now we cancel the dts and we write the change in K as K₂ - K₁ = integral of F⋅dr from the starting point r₁ to the ending point r₂. In other words, the change in kinetic energy in one dimension is just force times distance for tiny distances. In higher dimensions, like in 2 dimensions, if the particle is moving in the 2 dimensional plane along some path and it moves a distance dr, and at that point, the force is in that direction, you should not simply take the product of the length of the force times the length of the distance, but that times the cosine of the angle.

Why should it take that? Because that’s what gives you the rate of change of kinetic energy. Just go to the kinetic energy and it suggests to you that you take this product and define it to be the work, because if you define it that way, just like in 1D, the work done by the force will be the change in kinetic energy. You guys with me now? In spite of this little mess up here, this ma is my force. The dts got canceled. The dK is added up to K₂ - K₁ and this is the integral. All right, so now it looks pretty good, except for the complication. You have a dot product here. You have this nice result that K₂ - K₁ is the integral F⋅dr from r₁ to r₂. So let’s imagine where r₁ and r₂ are. This is say where you started, and maybe you ended up here at r₂. You’ve just got to integrate this F from r₁ to r₂. Let’s imagine there is no friction. Let’s imagine F depends only on the position. What’s my problem now? What problem do we have? Yes?

Student: _______ going from r₂ to r₁ _______.

Professor Ramamurti Shankar: No, this integral gives me no choice. I must go from r₁ to r₂. Anybody know what’s the problem? Yes?

Student: Your path.

Professor Ramamurti Shankar: Your path. See, in 1 dimension, if I want to go from here to there, I have no choice, okay? Just move along that line. In 2 dimensions, if you want to go from one point to another, I can think of many ways. You can do that, you can take the straight line path, or you can take some other path. That’s an ambiguity we have in higher dimensions. So even though the work energy theorem is valid, and no matter which path you take, the work done by the force will be the change in kinetic energy, this is not in general equal to some function at r₁ - some function at r₂. Because if it had this form, we are done. You agree? If it had this form, then again you write K₂ + U₂ = K₁ + U₁. But no one’s telling you that is true. If it was true, you are in business. But it’s a very unreasonable thing to ask of this force, that its integral from some point to another point, does not depend on what path you integrate it on. In other words, you can take that path, break it up into segments dr, and on each segment, the force is some variable. You take F⋅dr of that little segment, add it up over all the segments and make the segments vanishingly small. That’s the meaning of this integral. And you’re supposed to get the same answer, no matter what you do, including going like this, or anything else. How can they all give the same answer? In fact, I’ll give you a simple example where it’s simply not true, so the answer depends on the path. In fact, just about any force you pick randomly will not have the property, the answer does not depend on the path. Answers typically will depend on what path you take. So let’s take a simple force, F = iy. I can think of more complicated ones, but I don’t have time to do them.

So here’s a simple one: force depends on position. It points in the x direction only and it grows with y. And I’m going to do two calculations. I’m going to find the integral from the origin to the point (1,1) first by going this way and then by going that way. They are two possible paths. And you will see, when I do the calculation, the answer is different. Therefore you cannot write the answer as a difference of something here - something here, because the answer is not a function only of the starting and ending points, but how you got there. And I want to do this just to remind you how to do these integrals. So let’s take this segment here. What is the work done? The work is the force which is i times y times the distance. If you move horizontally, perhaps you guys can see dr for a tiny segment simply i times dx. And x goes from 0 to 1. So how much is that? Can you do this? I want the first person who knows the answer. Yes?

Student: Zero ______________.

Professor Ramamurti Shankar: Right. On this line, y is 0. In fact, if you draw a picture of the force, it looks like this. It gets bigger and bigger and bigger as you go further off. And if you go to negative values, it looks like this. So there is no y, there is no force. That is 0. If you go to this segment, y is definitely non-zero, but notice, the force is like this and the dr is in the y direction and they are perpendicular, so the dot product is 0. Okay? dr and F are perpendicular. One is along i and one is along j dot product to 0. If you take this path, the work done is 0.

Now let’s take the other path, go like this and then like this. On this section, we have the same problem. Force is to the right. You’re moving straight up. dr and F are perpendicular. But on the last segment, from here to here, you get a non-zero contribution, because force is iy, displacement is idx and x goes from 0 to 1, and y is frozen at the value 1, because you’re on this line. Frozen at the value 1. Just put 1 for it. And i⋅i is 1, integral of dx is 1. Therefore the work done by the force going up and then to the right is 1. Going to the right and then up is 0. You can take other paths and keep getting other answers.

So this is a force for which it’s not true, and I bet you that if you just randomly selected forces in the 2 dimensional plane, they will never have the property that the integral is path independent. So it looks like we are onto a hopeless task, that the answer cannot depend on the path. So maybe there is no problem in which you can derive a law of conservation of energy. Of course, you guys know, you can derive a law of conservation of energy for a family of forces. In fact, the answer is, there’s an infinite of numbers of forces you can devise for which this is true, namely, the answer does not depend on the path. Before writing those forces, let me explain to you one other way to write the condition. The condition, namely of a conservative force, is that the integral from here to here is the same as the integral from there to there, from 1 to 2 along path A and 1 to 2 along path B. They must be equal for any starting point and any ending point and any two paths joining them.

So what I’m saying is, F ⋅dr from 1 to 2 along A = integral F ⋅dr from 1 to 2 along B. Let me rewrite this by putting a - sign here and equate it to 0. That’s fine. Shifted it to the left hand side. But then we use the fact that when you go from 1 to 2, with a - sign, this second guy is same as + from 2 to 1 along B of F ⋅dr. In other words, the line integral going this way is - the line integral going backwards, because at every step, dr is opposite but F is the same at the point, F ⋅dr changes sign. Therefore it tells you then that the line integral from here to there and back is 0. We write that by saying the line integral of F ⋅dr in a closed path is 0 for all paths, for all starting points, all ending points, all paths. And that is the force that we want. That’s the conservative force. I’ve just rewritten. I’ve not gotten any closer to finding conservative force, but I’m saying there are two ways to write it. Answer doesn’t depend on the path, or the answer on any closed loop, the line integral on any closed loop, is 0. That means sometimes it’s got to be positive, sometimes it’s got to be negative. So question is, where am I going to get a force with these amazing properties? And I’m just going to give you the answer.

The theorem — if you want, you can call it a theorem, you can call it whatever you like. It’s a fact — number 1, here is a recipe for finding conservative forces. Take any function U(x,y). I’m just going to do it in 2 dimensions. In 3 dimensions, you can make it U(x,y,z). So let’s give an example here. U = x³y². Take that function. You can write your own ticket. Write whatever function you want, time hyperbolic sinh(y) times cosh(x), doesn’t matter. Take a function. Then the force that I want is the property that F_x = -dU/dx and F_y = -dU/dy.

In other words, I’m going to manufacture a force whose x component is - the x derivative of the function I picked and the y component is - the y derivative of the function I picked. So in my example, F_x will be -x derivative partial means keep y as a constant, take the derivative with respect to x. That’s 3x²y and F_y will be -2x³. 3x²y² and -2x³y. Let me see if I got that right, because you guys have been catching me too many times. Okay, I think that’s all right. This force is guaranteed to be conservative. I will show you why this is true.

Why is this magic working? But let me write it in another way that’s more compact. So the force I want to get, written as a vector, is i times F_x, which is i times −du/dx + j times −du/dy. That’s the force I claim is conservative. We’re going to write that as - grad U. This is called the grad and it means gradient, and this symbol is shorthand for this. There’s nothing more to it. Gradient is a machine. Derivative is a machine. You give it a function, sin x, it gives you an output, cosine x. you give it x squared, it gives you 2x. Gradient is a different kind of machine. You feed into it a function U of x and y, and it gives you a vector in all of the xy plane obtained by taking the two derivatives, one along x and one along y, and assembling them into a single vector.

So let’s see for this particular U, the force we have = -i times 3x²y², + -j times 2x³y. Amazingly, it is claimed that this force, when you integrate it from one point to another, answer will not depend on the path; it will depend only on the end points. So we have to understand what makes this work. Before I show that, I will make another statement. I will show you why this works, but the second statement, I will not prove, which is that all conservative forces are of the form - grad U. In other words, not only is this an example of conservative force, that’s it. There are no more examples. Every conservative force you cook up will always be — yes?

Student: Does the j component ___________?

Professor Ramamurti Shankar: j component should be what? Oh, that’s negative. I think I erased it. How’s that? Thank you. Is that what you meant?

Student: Yes.

Professor Ramamurti Shankar: Okay. So how are you guys doing with this? I’m always worried. You are inscrutable, and I don’t know what’s going on in there, whether silence means “I’m with you” or silence means, “I’m so far behind, I don’t even know where to start telling you my problems.” I don’t know. But this is your class. If you don’t speak up, you don’t get service. You have to ask. Don’t assume people next to you know anything. I’ve met them. They don’t. So just speak up. All right. So it’s like one of these psychotherapy sessions. Turn to the person next to you and insult him or her, because that’s what you want to do. Don’t worry about them, okay.

When you go to physics seminar, you never assume the speaker has a problem. I mean, you assume that you don’t have a problem. If you don’t follow it, it’s a problem with the speaker. That’s what makes seminars exciting. It’s the closest thing we have to gladiator fights in the old days. You bring a speaker from somewhere, maybe Harvard, then you put them on the stage and you just roast them for the whole hour. Now you can do that to me just fine, because this is not even my discovery. Somebody else did it hundreds of years ago. You should relish that challenge. If this was a small seminar, you will have time to do it. I don’t have that much time, but I also don’t have that little time that if you don’t follow me, you cannot intervene.

Okay, you have a question? No. Okay. Now we are going to understand why a force manufactured in this fashion is conservative, okay? So let’s calculate integral F ⋅dr for this force from some starting — I’m just going to call it 1 and 2. It means r₁ and r₂. What is F ⋅dr? It is F_xdx + F_ydy. That’s the meaning of the dot product. I hope you know what I mean by this. I’m going from here to here, that vector dr is some amount of dx and some amount of dy. It’s got jdy and idx. And F is some other thing, pointing in that direction, so F ⋅dr is this. Just look at this part. This is what you’re integrating. That becomes, except for this - sign, dU/dx times dx + dU/dy times dy. Now what is that? What is that? What do you think it stands for? In 1 dimension, if I took df/dx times dx, what am I calculating? Come on guys, you know what the derivative means. When you multiply the derivative by some dx, what do you get? Pardon me? Yes, go ahead. Take a shot. Yeah.

Student: You’re getting what you originally ___________.

Professor Ramamurti Shankar: You’re finding the change in the function f, right? The change in the function f is df/dx times dx. There are other changes proportional to dx³ and dx², but for small dx, that’s all it is. That’s the meaning of the rate of change. If you multiply the rate of change by the change you get a change in the function. If you’re in 2 dimensions, if you move from one point to another point, you’ve got a function U that depends on x and on y. Therefore it changes, because you change x and it changes because you change y. It’s changing because of two reasons. This is the change due to change in x; that’s the change due to change in y. For very small, infinitesimal dx and dy, that is the change.

That is just the change in the function U. It follows then that if you add up all these F ⋅dr’s from 1 to 2, this integral will give me U₁ - U₂. Therefore integral F ⋅dr from 1 to 2 is U₁ - U₂. So the magic is the following: you pick a function U, you can think of the function U as measured perpendicular to the blackboard. It’s like a height or something coming out of the blackboard. You cooked up a force which is related to the rate of change of the function U, so that if you add all the F ⋅dr’s, you’re getting the change in the function between two points. And that change in the function is independent of how you got there.

Think of U as a height of a mountain, sitting on top of the xy plane. Then F is really proportional to the rate of change of the height of the mountain. And F ⋅dr is the change when you move a distance dr, a vector distance dr. You add all the changes, what are you adding? You’re adding the height, your elevation. And when you’re done, the total change is the height difference between the final point and the starting point. It doesn’t depend on how you climb that mountain. You can take a long path, you can take a short path.

As long as you’re keeping track of how many feet have I climbed, you’re going to get only one answer. So not every F has this property, but an F derived from U by that trick of taking the gradient has this property. Therefore for that F, you might say this is for conservative force. But we always knew F ⋅dr is K₂ - K₁, therefore we get K₁ + U₁ = K₂ + U₂. In other words, the U that you began with is the potential energy for that problem. So what you’re manufacturing is pairs of potentials and forces. For every U that you dream up, there’s a force, which is obtained by gradient, - gradient of U. And in that problem, where that force is acting on particles, the energy that is conserved will be the kinetic, + the function U at the starting point, will be equal to kinetic plus potential at the end point. And the interesting thing is, there are no other conservative forces, except forces you can get by taking a U and taking its gradient.

So all conservative forces fall under this category, and they all have a conserved quantity where the potential energy is the U from which the F is derived. So we can say it as follows: F = - gradient of U. The gradient is the combination of itimes the x derivative and jtimes the y derivative. And U at 2 - U at 1 is - integral F ⋅dr from 1 to 2. This is the 2 dimensional generalization of something I wrote in 1 dimension somewhere. Oh, way on the top. Here, F = - dU /dx, U(x₂) − U(x₁) is - integral of F. The only difference is the integrand has become this F ⋅dr, rather than the F_x dx. So you can get conservation of energy in higher dimensions for those forces that come from a potential. U is called a potential. U is called a potential energy.

So now we can ask the following question: suppose I give you a force. I know if it’s conservative or not. How are you going to find out? Now there are two options. One is, you are so smart, you can look at the function and say, “Hey, this is the gradient of some other function U.” If you do that, then you are done, because if it’s the gradient of some function, we know it’s conservative. But what if you cannot see that right away? With simple polynomials, you can easily guess. Like the example I gave you, if F_x is -3x²y², F_y is -2x³y, maybe you can guess that U is x³y². But for more complicated functions, you won’t be able to do it. So there’s a process for testing any force to see if it’s conservative. And the process is the following.

It argues as follows: F_x is −dU/dx. F_y is -dU/dy. Then consider dF_x/dy, which is -d²U/dydx. Then consider dF_y/dx, which is -d²U/dxdy. And it’s the property of partial derivatives that the cross derivative, second cross derivative, is independent of whether you take dydx or dxdy. So let’s go to our test case, U = x³y². dU/dx = 3x²y² and dU/dy = 2x³y. Now I’m saying, take the x derivative of this y derivative. That’s written as d²U/dydx. y derivative of the x derivative gives me 6x²y. Then I say take the x derivative of the y derivative. This gives me 3x². That means 6x²y. So cross derivatives are always equal. Yes?

Student: Is this _______ if we’re going ________ taking __________?

Professor Ramamurti Shankar: Yes, that’s correct. So in 2 dimensions, this is the only test you have to apply. The test you have to apply is, are the cross derivatives of the force equal? Namely, is the y derivative of the force F_x = to x derivative of F_y? You want to see if dF_x/dy = dF_y/dx. If that is true, it is conservative. And it comes from the fact that if F is the derivative of some function U, then this mix derivative being equal is a diagnostic for that. In 3 dimensions, there are 2 more equations. We’re not going to use them, but I’ll just mention what they are. You get that by moving every index 1 notch. So this x goes to y, this y goes to z, this y goes to z, this x goes to y.

There’s one more condition. And another condition, z goes back to x, x goes z — let’s see, y goes to z, z goes to x = dF_x/dz. I’m not worried about whether you write this down or not, but I do want you to know in 2 dimensions, you have to remember this condition. So this mathematic extends to all dimensions, but I’m doing it for the case of 2D.

So let’s take stock of what has been done. What has been done is to think from scratch on how to get the law of conservation of energy in high dimensions. You start with the kinetic energy and you ask, why does it change? And you find it changes due to the force, and you find the change in kinetic energy is the line integral of F ⋅dr from start to finish.

But that doesn’t give you a work energy — it gives you a work energy theorem, but not a law of conservation of energy, because this integral can depend on the path. Then you say, maybe there are some functions for which the answer depends only on the end points. Then we get a law of conservation of energy, and we found out that we can manufacture such functions at will, by taking any function of U, taking its gradient, or - the gradient. Therefore that’s a way to generate conservative forces and they are the only conservative forces there are.

There’s no other conservative forces which are not gradients of something. And the relation between the potential and the force, you should know now. It’s what you learned in 1D, suitably generalized to higher dimensions. Change in potential is this and the forces obtained from the potential by taking the gradient, the change in potential is obtained by integrating the force. If you ever get confused about + sign and - sign and so on, go back to the harmonic oscillator, for which the thing is 1 over — is kx²/2.

So let me give you one simple example that we all know, which is the force of gravity near the earth. So here is the earth, so there is some x and y coordinates, and here is z. So I might as well write down the answer, because you guys all know this. U for gravity is mg times z. And the force of gravity, which is -idu/dx - jdu/dy, - kdu/dz, all you got here is -kmg, because it doesn’t depend on x, it doesn’t depend on y. It depends only on z and the z derivative is trivial. Force is -mg pointing down. So the force is the gradient of the potential and you can ask, is the potential difference between two points, U₂ - U₁, is it equal to integral of F ⋅dr from 1 to 2? You can easily check that.

If you go from here to here, straight up, then the integral of -mg times dz from start to finish, with a - sign is just mgz - 0. So indeed, it is true. And if you took two points, not on top of each other, but something like this, you can always do the integral from here to here, where there’s no change in potential, and do the integral there, which will give the same answer. So gravity’s a very trivial example, a very easy case, but you can see that gravitation is a conservative force. And you can see why it’s a conservative force, because if you took a path where you went like this, let’s say, and you went like this, where you get the same answer as here and here. Because in this path and this path, you don’t do any work, because F and dr are perpendicular, right? F is like this, and the displacement is like that, and it’s the same thing, whereas here, if you go like this, F is down and dr is down, F is down, dr is down, and it’s the same F here and here; are the same height so you get the same answer.

Okay, but now, why did we spend quite a bit of time on conservative forces, because that’s not what the primary focus is? It was electricity and electrostatics. So we’ve got to ask ourselves, does the electrostatic force meet the test of a conservative force? Well, we know it will, because otherwise I wouldn’t have spent this much time building up the stuff. But let’s just verify that. It looks like a very formidable task. Let’s see why it’s a formidable task to verify it’s conservative. I’m claiming that you’ve got all kinds of these charges producing a field here and I want to prove that the line integral of this force on any closed loop is 0. Or I want to prove the integral on one path is the integral on any other path, for any 2 paths between any 2 points. I don’t have time to show all that, right? If the theorem is wrong, do you agree it may not be so hard to show that? What will it take to show it is not conservative? Yes?

Student: Just 2 paths for which the integral is not equal.

Professor Ramamurti Shankar: Yes. If you pick any 2 paths, should be the same end points, so always the integral is not equal, that’s it. But the fact that they were equal doesn’t mean that somebody else will not find some other two paths with some other starting points which they are not equal. So how are we going to do this? How are we going to test for every configuration of electric field, every element of charges is conservative? So what do you think a strategy will be? Yes? Any ideas? How am I going to take all possible charges and work it out each time? Yes?

Student: I guess you could find the electric field and use that test to see —

Professor Ramamurti Shankar: Right, but to find the electric field and see if it works, you have to calculate the electric field as a function of x, y and z for every arrangement of charges. Then you can ask, is that true, those derivatives? Yes?

Student: But what works for one charge should work for all charges, because of supersposition.

Professor Ramamurti Shankar: There we go. If it works for one charge, it will work for any arrangement of charges. So you may not have appreciated fully the power of superposition, but you can see now. If you did not realize that, the problem looks insurmountable. Who can handle every arrangement of charges? But if charge by charge it is true that the electric field due to charge 1 is conservative, electric field due to charge 2 is conservative. When I add them up, the net field is E₁ + E₂. The line integral of E₁ + E₂ is the line integral of E₁+ the line integral of E₂. E₁ gives 0 on any closed loop, other will give 0 on any closed loop. The trick is going to be just do it for one charge. That’s it. If you can see it for one charge, you get by superposition that it’s true for any charge, because the field is additive and the integral is additive, okay. That’s the two parts. If you’ve got charges, the net e is E₁ + E₂. The integral of E₁ + E₂ is the integral of E₁ + the integral of E₂.

That’s the way integrals work. If each one vanishes on a closed loop, you’re done. So I’m just going to take one charge. Here’s one charge and let me draw for convenience all the field lines coming out of it. Then I want to see, if I do a line integral, say from here to here, will I get the same answer on some other path? Let’s test that. So the first integral is from 1 to 2 on this path. Then I’m going to take another one, which is different from it, where you can see the answer is the same, then go on making more and more changes, and the answer will not change.

So I’m going to take another path where I go tangent to the lines, along a circle. Then I go here, till I get to that radius, then I turn around and do that. I claim the integral will be the same on this path, this one, that one and that one. So I want you to think about why it’s going to be true, okay? Think of it, why it’s okay to do that. You’ve got a new path but it doesn’t make a difference. Ready? Tell you the answer now? How many people don’t know why it’s true? So other people know why it’s true. I’m going to ask the other people. Okay, why is it true?

Student: I raised my hand.

Professor Ramamurti Shankar: You raised your hand. All right. Yes?

Student: It’s true, because when you’re moving on the paths this way, they’re radial, so the radius is constant, which would mean that there’s no _______.

Professor Ramamurti Shankar: There’s no what?

Student: There’s no change in force because the radius is constant.

Professor Ramamurti Shankar: That’s not a good reason. The force is changing. It’s that way there, it’s that way there. Why is it not?

Student: It’s always perpendicular to the force.

Professor Ramamurti Shankar: The force and the displacement are perpendicular in this section, because force is radial, displacement by definition is tangential along the circle. So dr is perpendicular to F. Oh, by the way, I should tell you something. I usually forget this. We want the electric force to be conservative, right? The electric force in any situation on a charge is q times the electric field.

You will agree that if the electric field is conservative, multiplying by a number q is not going to change the fact. So I’m going to prove the electric field is conservative, then the electric force on any charge is just proportional to it. It will also be conservative. So really taking force on a unit charge and seeing what happens. So this section is for free. This section — we’ll come to that — this section is also for free, because F looks like that and dr looks like that. Here you can pair the sections, so that that section is paired with that section, because you’re moving the same radial distance with a radial force. F ⋅dr, the dot product, is the same here, section by section.

Any work you’re doing here, you’re doing here, therefore the answer is the same. Then you can see that you can go on adding all kinds of changes, radial and angle, radial or angular, and when you’re done, all the radial integrals will give you that part, and the angulars won’t count. So as you move around, it’s only when you’re moving radially you’re even aware that you’re doing work. Going in the angle doesn’t matter. And the net radial change in going from here to here is that distance, no matter how you move. Now you might say, prove to me that that integral is the same as that integral, because this was made up of radial and angular sections.

This doesn’t seem that way, but we can always make a very fine mesh in which every displacement is approximated by radial angle or radial angular. So you can approximate any contour by sequence of radial and angular moves. Angular moves are free. Radial moves keep track of the change in the distance times the force and you can pair them with this reference. Every dr we move there we have a dr here. So that is the pictorial proof of this. So let’s now calculate what is the potential due to that electric field.

So we’re going to do the calculation. So we use the formula that U at r₂ - U at r₁ is - the integral e⋅dr from 1 to 2. You see, we know it is conservative. I’ve shown you with this argument. So if you want to find the potential energy between 1 and 2, you can take any path you like between 1 and 2. So let’s pick the points. Here is point 1 and here is point 2. You can pick any path you like, so here is what I’m going to do. I’m going to go radially out and then I’m going to go along the arc. So this is 1, this is 2, but I’m going to start at some intermediate point 3, then go right.

So work is done here; no work is done there. So let’s just find the work in going from 1 to 3, which is -e_r (is a vector in the radial direction) ¼Πε₀r². That is F . The displacement is e_r times the little length of e_r. Do you understand that? A tiny section here, dr, has a magnitude equal to the change in the radius, and a direction equal to the radius unit rate — the radial direction. And you want to do this from r₁ to r₂. Okay, e_r⋅e_r is 1 so I get — I forgot one thing. You guys know what I forgot here? One thing’s missing. Yes?

Student: Q.

Professor Ramamurti Shankar: When did you notice that? Just now?

Student: Little bit.

Professor Ramamurti Shankar: Don’t wait. Don’t wait even one second. I told you what happens to me, right? When I go home, they tie me to a stool and they force me to watch this tape. And it’s just awful when you think you dropped a symbol, and you’re waiting for somebody to say, “Hey, what about q?” So don’t have any — if you hesitate, it’s only to wait, not to stop me. All right, so look, I’ll get the q because I know the formula from the time I was in a sandbox. I know something is wrong, but I don’t want to get to that last stage, then go back and change everything.

So I forgot the q sitting at the origin here. Okay, so this is q/4Πε₀. Integral of -1/r² is just 1/r, so that’s 1/r₂ − 1/r₁. Again, sorry, I made a mistake now. The change in potential energy is really q times e, but q is the test charge, so I’m going to call this V. So it’s hard for you guys to do it on the notebook, because I want to distinguish between potential energy and potential. In other words, the force — now some other test charge q is the electric field times q. And the force is going to be - the gradient of the potential energy. The electric field is going to be - the gradient of the potential. Yes?

Student: In that last line, should that be a negative?

Professor Ramamurti Shankar: A negative?

Student: Yeah.

Professor Ramamurti Shankar: Let’s see. There’s a negative here, and integral of -1/r² is 1/r.

Student: Oh, okay.

Professor Ramamurti Shankar: So the upper limit, then there’s lower limit.

Chapter 4: Electric Potential [00:59:34]

So let me correct what I said. I’m trying to find the difference in what’s called the potential. It’s not the potential energy. Potential difference between some point and some other point is the work done on a unit charge. If you’ve got some other charge, you should multiply it by the charge. So the energy of some other charge in this field, the potential energy, is q times the potential. So U is q times this thing that I’m calling potential.

So we’re comparing this expression to this expression, we can conclude that V at any point r is q/4Πε₀(1/r) + any constant. Any constant, because you cannot immediately say this guy is equal to this guy and that guy is equal to that guy, because even if you add a constant to both, it will still be true. And we can pick the constant any way we like. It is common to pick the constant so that V goes to 0 as r goes to infinity. That’s what we will do. That’s the convention. Then we can write here, what is the V(r).

This is the V due to a single charge, and the U(r) = to some q₀. I don’t want to call q again, because q is the guy producing the potential. q₀ is the guy experiencing the force. So q₀ times V is the potential energy of charge q, of charge q₀ sitting at that point. Now let’s also verify, if you like, that this V(r) does produce the electric field. In other words, I got the V by integrating electric field. I want to go back, just to get you familiar with this. Getting the electric field from the potential. Let’s try to do that. So E_x — let me just do E_x — it’s −dV/dx, right? -dV /dx = -q/4Πε₀. Derivative of 1/r is -1/r² times dr/dx. What is dr/dx? r is (x² + y² + z²)^1/2. dr/dx is just x divided by this whole thing, is x divided by r. So this becomes q/4Πε₀(1/r²)(x/r). This is E_x.

So what will be the vector e? Can you do that in your head? Multiply this by i, and do similarly for y and z and add them up. What will you find? You will find E will equal q/4Πε₀(1/r²), times r over r. Put r over r, position vector over r, is the unit vector in the radial direction. I’m doing this a little fast, because this is something you can go home and check. I don’t want to spend too much time doing it. I claim that if you took this potential and took its x derivative or y derivative or z derivative, you get the x, y or z components of the electric field.

Okay, if you didn’t follow this, you should go home and check that this is true. All right, so this is not that important for me, this checking, because we know it’s going to work. What is important for me is for you to know that we’ve now found that the potential due to a single charge leads to a conservative force. The field due to a single charge leads to a conservative one, therefore by superposition, if you add any number of charges, if you find the potential due to all of them, then you take its gradient that will give you a conservative electric field.

So what have you got so far? Our gain is the following: we have a law of conservation of energy, which is the following: ½mv² + q (this q is the q of the body you’re interested in) times V at the point r₁ = ½mv₂² + q times v at the point r₂. And what is V at any point r? It’s the sum 1/4Πε₀. Let’s imagine there are many, many charges, q₁, q₂, q₃, each one at r₁, r₂, r₃, etc. This equals q_i which is charge number I divided by the length of the vector from r_i to the r where I’m finding the potential. In other words, this is where I want the potential r. This is r₁and q₁ is sitting there. This is r₂ and q₂is sitting there and so on. And the potential at the point that I’m interested in is obtained by taking each q divided by the distance from that point, and this is the usual 1/4Πε₀.

That is the complete story, and the charges are not discrete but continuous you can write an integral. So every problem, you first find the potential by adding the potential from all the charges. Then its gradient will give it the electric field. But we have now a law of conservation of energy with this as the conserved energy. The second advantage is that it is easier to work with V than with E. So why do you think it’s easier to find V and then find E from it, by taking derivatives than the other way around? Yes?

Student: V is a scalar.

Professor Ramamurti Shankar: Right.

Student: [inaudible]

Professor Ramamurti Shankar: I’m sorry, I didn’t hear the last part.

Student: It’s always easier to take derivatives than to integrate —

Professor Ramamurti Shankar: Very good. Yes. Let me repeat that. If you want to find the potential due to many charges, you simply add the numbers coming from each one. There are no arrows, there are no vectors. Notice the formula for V has no vectors in it. Each one contributes a number. Each charge contributes a number to the point they’re interested in. Add them all up, find the number here, there, everywhere else. Then take derivatives of your answer to get the field. That’s going to be a lot easier than adding the vectors.

So I’m going to illustrate that with one simple problem and that’s the last thing for today. That’s the dipole. So here is our dipole. Let’s take a charge q, -q at -a and +q at +a. And we want to find the field there. So let’s call that distance r₊ and let’s call that distance r_-. These are not vectors; these are just lengths. If you want to find the electric field directly, you know what you have to do. You have to find the vector here and you’ve got to find the vector there due to the second guy, add the two vectors and find the result. Of course, you can do it, but it’s going to be very tedious, because you have to find the direction of the vector here and the direction of the vector here, in terms of all these coordinates.

What we will do instead is to find the potential everywhere, and take the x and y derivatives of the potential to get the E_x and E_y. Well, let’s do that. So let’s find the total potential due to these two guys. The first one is q/4Πε₀ divided by 1/r₊ - 1/r_-. That’s it. Just wanted to simplify the expression a little bit. So this is true for any separation, but I want the limit in which r₊ and r_- are much bigger than the separation between them.

So let’s define an r which is the position vector of the point you’re interested in from the center of the dipole. So we’ve got to find r₊ and r_-. You get that, you’re done. So vector r₊you can see = r - ai/2 and vector r_- = vector r + ai/2. Because the vector ai/2 looks like this, so ai/2 + r₊ should give you r. You can check that. And you can check that ai/2 + r gives you r_-. so these are the two expressions. So what is r₊? r⁺ is the length of the vector r₊. That’s equal to the square root of the length squared. The length squared of r₊ squared = root of (r - ai/2) ⋅ (r - ai/2).

Student: Why is it ai/2?

Professor Ramamurti Shankar: Oh, I’m sorry, it’s not. It is half the distance, but it’s really only a. Thank you. Yeah, in fact, one puts the 2a as a separation to avoid all these factors of 2. So now you take the length squared. It’s equal to r² + a² minus twice r⋅a. The other one would be r² + a² + twice r⋅a. And that we can approximate as r times 1 - twice r⋅a/r²to the 1 half. I’ll tell you what I did here. I’m going to neglect a squared compared to r squared, because r is much bigger than it.

But I’m going to keep this r⋅a term because this has got two powers of a, where a is a small number. It’s got 1 power of a and 1 power of r, so it’s much stronger than this term, so I keep that. Keeping the second term, so I’ve forgotten the a² term. In this one, from the theorem I’ve been giving you guys all the time, it’s 1 - r⋅a/r². Remember 1 + xⁿ is roughly 1 + nx + dot dot dot if x is small, and x is indeed small, because it’s r⋅a over r squared. So now we can see V = q/4Πε₀{1/[r(1-r⋅a/r²)] - 1/[r(1+r⋅a/r²)]} .

This is going to be found in every textbook in the planet, so don’t worry about it. If you simplify this again by taking it upstairs, you’ll get twice a⋅r/r². That’s the potential. Now twice a times q is the dipole moment, so it’s p⋅r/4Πε₀r³. That’s the final piece, the dipole moment. Dipole moment is a charge times the separation between the charges, which is 2a times q. Okay, so what one should do now is, having taken this V, one can take its x and y derivatives very easily to calculate the electric field at each point. It’s very easy to take the derivative of this, just pick the x or y. It’s one of the homework problems I’ve assigned to you, where I show you the dipole field. So the moral of the story is, add the potential due to all the charges, then take the derivative, because derivatives are easier to take. Yes?

Student: The condition for that result, is r much bigger than a?

Professor Ramamurti Shankar: Yes. Up till some point, everything is exact, but then this is long distance property. Any questions? Okay, thank you.

[end of transcript]

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