Our approach is completely opposite to the historical approach in which one develops the subject in terms of the experiments by which the information was obtained. But the subject of physics has been developed over the past 200 years by some very ingenious people, and as we have only a limited time to acquire our knowledge, we cannot possibly cover everything they did. Unfortunately one of the things that we shall have a tendency to lose in these lectures is the historical, experimental development. It is hoped that in the laboratory some of this lack can be corrected. You can also fill in what we must leave out by reading the Encyclopedia Britannica, which has excellent historical articles on electricity and on other parts of physics. You will also find historical information in many textbooks on electricity and magnetism.

Ordinarily, a course like this is given by developing gradually the physical ideas—by starting with simple situations and going on to more and more complicated situations. This requires that you continuously forget things you previously learned—things that are true in certain situations, but which are not true in general. For example, the “law” that the electrical force depends on the square of the distance is not always true. We prefer the opposite approach. We prefer to take first the complete laws, and then to step back and apply them to simple situations, developing the physical ideas as we go along. And that is what we are going to do.

What it means really to understand an equation—that is, in more than a strictly mathematical sense—was described by Dirac. He said: “I understand what an equation means if I have a way of figuring out the characteristics of its solution without actually solving it.” So if we have a way of knowing what should happen in given circumstances without actually solving the equations, then we “understand” the equations, as applied to these circumstances. A physical understanding is a completely unmathematical, imprecise, and inexact thing, but absolutely necessary for a physicist.

It will take you some time to understand what should happen in different circumstances. You will have to solve the equations. Each time you solve the equations, you will learn something about the character of the solutions. To keep these solutions in mind, it will be useful also to study their meaning in terms of field lines and of other concepts. This is the way you will really “understand” the equations. That is the difference between mathematics and physics. Mathematicians, or people who have very mathematical minds, are often led astray when “studying” physics because they lose sight of the physics. They say: “Look, these differential equations—the Maxwell equations—are all there is to electrodynamics; it is admitted by the physicists that there is nothing which is not contained in the equations. The equations are complicated, but after all they are only mathematical equations and if I understand them mathematically inside out, I will understand the physics inside out.” Only it doesn’t work that way. Mathematicians who study physics with that point of view—and there have been many of them—usually make little contribution to physics and, in fact, little to mathematics. They fail because the actual physical situations in the real world are so complicated that it is necessary to have a much broader understanding of the equations.

The physicist needs a facility in looking at problems from several points of view. The exact analysis of real physical problems is usually quite complicated, and any particular physical situation may be too complicated to analyze directly by solving the differential equation. But one can still get a very good idea of the behavior of a system if one has some feel for the character of the solution in different circumstances. Ideas such as the field lines, capacitance, resistance, and inductance are, for such purposes, very useful. So we will spend much of our time analyzing them. In this way we will get a feel as to what should happen in different electromagnetic situations. On the other hand, none of the heuristic models, such as field lines, is really adequate and accurate for all situations. There is only one precise way of presenting the laws, and that is by means of differential equations. They have the advantage of being fundamental and, so far as we know, precise. If you have learned the differential equations you can always go back to them. There is nothing to unlearn.

The vector $\FLPh$ can be defined in another way—in terms of its components. We ask how much heat flows through a small surface at any angle with respect to the flow. In Fig. 2–4 we show a small surface $\Delta a_2$ inclined with respect to $\Delta a_1$, which is perpendicular to the flow. The unit vector $\FLPn$ is normal to the surface $\Delta a_2$. The angle $\theta$ between $\FLPn$ and $\FLPh$ is the same as the angle between the surfaces (since $\FLPh$ is normal to $\Delta a_1$). Now what is the heat flow per unit area through $\Delta a_2$? The flow through $\Delta a_2$ is the same as through $\Delta a_1$; only the areas are different. In fact, $\Delta a_1=\Delta a_2\cos\theta$. The heat flow through $\Delta a_2$ is \begin{equation} \label{Eq:II:2:10} \frac{\Delta J}{\Delta a_2}=\frac{\Delta J}{\Delta a_1}\cos\theta= \FLPh\cdot\FLPn. \end{equation} We interpret this equation: the heat flow (per unit time and per unit area) through any surface element whose unit normal is $\FLPn$, is given by $\FLPh\cdot\FLPn$. Equally, we could say: the component of the heat flow perpendicular to the surface element $\Delta a_2$ is $\FLPh\cdot\FLPn$. We can, if we wish, consider that these statements define $\FLPh$. We will be applying the same ideas to other vector fields.

Let’s make a more precise definition of $\FLPh$: The magnitude of the vector heat flow at a point is the amount of thermal energy that passes, per unit time and per unit area, through an infinitesimal surface element at right angles to the direction of flow. The vector points in the direction of flow (see Fig. 2–3 ). In symbols: If $\Delta J$ is the thermal energy that passes per unit time through the surface element $\Delta a$, then \begin{equation} \label{Eq:II:2:9} \FLPh=\frac{\Delta J}{\Delta a}\,\FLPe_f, \end{equation} where $\FLPe_f$ is a unit vector in the direction of flow.

There are also vector fields. The idea is very simple. A vector is given for each point in space. The vector varies from point to point. As an example, consider a rotating body. The velocity of the material of the body at any point is a vector which is a function of position (Fig. 2–2 ). As a second example, consider the flow of heat in a block of material. If the temperature in the block is high at one place and low at another, there will be a flow of heat from the hotter places to the colder. The heat will be flowing in different directions in different parts of the block. The heat flow is a directional quantity which we call $\FLPh$. Its magnitude is a measure of how much heat is flowing. Examples of the heat flow vector are also shown in Fig. 2–1 .

One way of thinking about scalar fields is to imagine “contours” which are imaginary surfaces drawn through all points for which the field has the same value, just as contour lines on a map connect points with the same height. For a temperature field the contours are called “isothermal surfaces” or isotherms. Figure 2–1 illustrates a temperature field and shows the dependence of $T$ on $x$ and $y$ when $z=0$. Several isotherms are drawn.

The simplest possible physical field is a scalar field. By a field, you remember, we mean a quantity which depends upon position in space. By a scalar field we merely mean a field which is characterized at each point by a single number—a scalar. Of course the number may change in time, but we need not worry about that for the moment. We will talk about what the field looks like at a given instant. As an example of a scalar field, consider a solid block of material which has been heated at some places and cooled at others, so that the temperature of the body varies from point to point in a complicated way. Then the temperature will be a function of $x$, $y$, and $z$, the position in space measured in a rectangular coordinate system. Temperature is a scalar field.

Also we will want to use the two following equalities from the calculus: \begin{gather} \label{Eq:II:2:7} \Delta f(x,y,z)=\ddp{f}{x}\,\Delta x+\ddp{f}{y}\,\Delta y+\ddp{f}{z}\,\Delta z,\\[1ex] \label{Eq:II:2:8} \frac{\partial^2f}{\partial x\,\partial y}= \frac{\partial^2f}{\partial y\,\partial x}. \end{gather} The first equation ( 2.7 ) is, of course, true only in the limit that $\Delta x$, $\Delta y$, and $\Delta z$ go toward zero.

We begin now with the abstract, mathematical view of the theory of electricity and magnetism. The ultimate idea is to explain the meaning of the laws given in Chapter 1 . But to do this we must first explain a new and peculiar notation that we want to use. So let us forget electromagnetism for the moment and discuss the mathematics of vector fields. It is of very great importance, not only for electromagnetism, but for all kinds of physical circumstances. Just as ordinary differential and integral calculus is so important to all branches of physics, so also is the differential calculus of vectors. We turn to that subject.

2–3 Derivatives of fields—the gradient

When fields vary in time, we can describe the variation by giving their derivatives with respect to $t$. We want to describe the variations with position in a similar way, because we are interested in the relationship between, say, the temperature in one place and the temperature at a nearby place. How shall we take the derivative of the temperature with respect to position? Do we differentiate the temperature with respect to $x$? Or with respect to $y$, or $z$?

Useful physical laws do not depend upon the orientation of the coordinate system. They should, therefore, be written in a form in which either both sides are scalars or both sides are vectors. What is the derivative of a scalar field, say $\ddpl{T}{x}$? Is it a scalar, or a vector, or what? It is neither a scalar nor a vector, as you can easily appreciate, because if we took a different $x$-axis, $\ddpl{T}{x}$ would certainly be different. But notice: We have three possible derivatives: $\ddpl{T}{x}$, $\ddpl{T}{y}$, and $\ddpl{T}{z}$. Since there are three kinds of derivatives and we know that it takes three numbers to form a vector, perhaps these three derivatives are the components of a vector: \begin{equation} \label{Eq:II:2:11} \biggl(\ddp{T}{x},\ddp{T}{y},\ddp{T}{z}\biggr)\overset{?}{=}\text{a vector}. \end{equation}

Of course it is not generally true that any three numbers form a vector. It is true only if, when we rotate the coordinate system, the components of the vector transform among themselves in the correct way. So it is necessary to analyze how these derivatives are changed by a rotation of the coordinate system. We shall show that (2.11) is indeed a vector. The derivatives do transform in the correct way when the coordinate system is rotated.

We can see this in several ways. One way is to ask a question whose answer is independent of the coordinate system, and try to express the answer in an “invariant” form. For instance, if $S=\FLPA\cdot\FLPB$, and if $\FLPA$ and $\FLPB$ are vectors, we know—because we proved it in Chapter 11 of Vol. I—that $S$ is a scalar. We know that $S$ is a scalar without investigating whether it changes with changes in coordinate systems. It can’t, because it’s a dot product of two vectors. Similarly, if we have three numbers $B_1$, $B_2$, and $B_3$ and we find out that for every vector $\FLPA$ \begin{equation} \label{Eq:II:2:12} A_xB_1+A_yB_2+A_zB_3=S, \end{equation} where $S$ is the same for any coordinate system, then it must be that the three numbers $B_1$, $B_2$, $B_3$ are the components $B_x$, $B_y$, $B_z$ of some vector $\FLPB$.

Now let’s think of the temperature field. Suppose we take two points $P_1$ and $P_2$, separated by the small interval $\Delta\FLPR$. The temperature at $P_1$ is $T_1$ and at $P_2$ is $T_2$, and the difference $\Delta T=T_2-T_1$. The temperatures at these real, physical points certainly do not depend on what axis we choose for measuring the coordinates. In particular, $\Delta T$ is a number independent of the coordinate system. It is a scalar.

If we choose some convenient set of axes, we could write $T_1=T(x,y,z)$ and $T_2=T(x+\Delta x,y+\Delta y,z+\Delta z)$, where $\Delta x$, $\Delta y$, and $\Delta z$ are the components of the vector $\Delta\FLPR$ (Fig. 2–5). Remembering Eq. (2.7), we can write \begin{equation} \label{Eq:II:2:13} \Delta T=\ddp{T}{x}\,\Delta x+\ddp{T}{y}\,\Delta y+\ddp{T}{z}\,\Delta z. \end{equation} The left side of Eq. (2.13) is a scalar. The right side is the sum of three products with $\Delta x$, $\Delta y$, and $\Delta z$, which are the components of a vector. It follows that the three numbers \begin{equation*} \ddp{T}{x},\ddp{T}{y},\ddp{T}{z} \end{equation*} are also the $x$-, $y$-, and $z$-components of a vector. We write this new vector with the symbol $\FLPgrad{T}$. The symbol $\FLPnabla$ (called “del”) is an upside-down $\Delta$, and is supposed to remind us of differentiation. People read $\FLPgrad{T}$ in various ways: “del-$T$,” or “gradient of $T$,” or “$\grad T$;” \begin{equation} \label{Eq:II:2:14} \grad T=\FLPgrad{T}=\biggl(\ddp{T}{x},\ddp{T}{y},\ddp{T}{z}\biggr). \end{equation}

Using this notation, we can rewrite Eq. (2.13) in the more compact form \begin{equation} \label{Eq:II:2:15} \Delta T=\FLPgrad{T}\cdot\Delta\FLPR. \end{equation} In words, this equation says that the difference in temperature between two nearby points is the dot product of the gradient of $T$ and the vector displacement between the points. The form of Eq. (2.15) also illustrates clearly our proof above that $\FLPgrad{T}$ is indeed a vector.

Perhaps you are still not convinced? Let’s prove it in a different way. (Although if you look carefully, you may be able to see that it’s really the same proof in a longer-winded form!) We shall show that the components of $\FLPgrad{T}$ transform in just the same way that components of $\FLPR$ do. If they do, $\FLPgrad{T}$ is a vector according to our original definition of a vector in Chapter 11 of Vol. I. We take a new coordinate system $x'$, $y'$, $z'$, and in this new system we calculate $\ddpl{T}{x'}$, $\ddpl{T}{y'}$, and $\ddpl{T}{z'}$. To make things a little simpler, we let $z=z'$, so that we can forget about the $z$-coordinate. (You can check out the more general case for yourself.)

We take an $x'y'$-system rotated an angle $\theta$ with respect to the $xy$-system, as in Fig. 2–6(a). For a point $(x,y)$ the coordinates in the prime system are \begin{alignat}{3} \label{Eq:II:2:16} &x'&&=\phantom{-}x\cos\theta&&+y\sin\theta,\\[1ex] \label{Eq:II:2:17} &y'&&=-x\sin\theta&&+y\cos\theta. \end{alignat} Or, solving for $x$ and $y$, \begin{alignat}{3} \label{Eq:II:2:18} &x&&=x'\cos\theta&&-y'\sin\theta,\\[1ex] \label{Eq:II:2:19} &y&&=x'\sin\theta&&+y'\cos\theta. \end{alignat} If any pair of numbers transforms with these equations in the same way that $x$ and $y$ do, they are the components of a vector.

Now let’s look at the difference in temperature between the two nearby points $P_1$ and $P_2$, chosen as in Fig. 2–6(b). If we calculate with the $x$- and $y$-coordinates, we would write \begin{equation} \label{Eq:II:2:20} \Delta T=\ddp{T}{x}\,\Delta x \end{equation} —since $\Delta y$ is zero.