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Feynman Physics Lectures V2 Ch02 1962-10-04 Differential Calculus of Vector Fields

Feynman Physics Lectures V2 Ch02 1962-10-04 Differential Calculus of Vector Fields

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Unformatted text preview: 2 Differential Calculus of Vector Fields 2—] Understanding physics 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 differ- ent circumstances. Ideas such as the field lines, capacitance, resistance, and in- ductance 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 difl‘erent 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 difl‘erential 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. It will take you some time to understand what should happen in difierent 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 equa- tions. That is the diflemnce 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 difl'erential 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 equa- tions. 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. 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 “under- stand” the equations, as applied to these circumstances. A physical understanding is a completely unmathematical, imprecise, and inexact thing, but absolutely neces- sary for a physicist. ' 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 compli- cated 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 situa- tions, developing the physical ideas as we go along. And that is what we are going to do. 2—1 2-1 Understanding physics 2-2 Scalar and vector fields—T and b 2—3 Derivatives of fields—the gradient 2—4 The operator V 2—5 Operations with V 2—6 The difi'erential equation of heat flow 2-7 Second derivatives of vector fields 2—8 Pitfalls Review: Chapter 11, Vol. I, Vectors 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 Ency- clopedia Brittanica, 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. 2—2 Scalar and vector fields—T and h 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 electro- . , magnetism, but for all kinds of physical circumstances. Just as ordinary differential WM MW ('1 AM' and integral calculus is so important to all branches of physics, so also is the differential calculus of vectors. We turn to that subject. Listed below are a few facts from the algebra of vectors. It is assumed that 3“ i I w you already know them. a: —- — E E u‘ n if” E A - 3 = scalar = [1,3, + {4,3, + A3 (2.1) 0““ I"“‘i"" A x 3 = vector (2.2) E. . (A x 3). = 11,3, — A,3, . (A x B), = [4,3, — 3,3,, Walsh.» (AXB)y=Asz—Asz A B (t‘. D E [F G A x A = 0 (2-3) H I J \K u. M N A -(AXB)=0 (2.4) D P Q IR 3 if U A ' (3.x C)= (AXB)-C (2.5) v w x y z A X (B X C) = B(A-C) —- C(A-B) (2.6) 8%“ W w I l I 1 Also we will want to use the two following equalities from the calculus: 8 6 a R. 5' 6 J t f % Af(x,y, z) = £6. Ax + b—J‘éAy + a—{Az’ {s (E .3 U: ’2 1n n 62f = 62f . (2 8) 6x 6y 6y 6x ‘ f M The first equation (2.7) is, of course, true only in the limit that Ax, Ay, and Az 3‘ "3 u go toward zero. . The simplest possible physical field is a scalar field. By a field, you remember, ‘6“ “M W W M“ ' 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. 2—2 Fig. 2—1 . scalar field. 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 tempera- ture 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 2 = 0. Several isotherms are drawn. 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 [1. Its magnitude is a measure of how much heat is flowing. Examples of the heat flow vector are also shown in Fig. 2—1. heat flow 2 Let’s mate a more precise definition of h: 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 AJ is the thermal energy that passes per unit time through the surface element Aa, then M h“—€/, _ M (2.9) where e; is a unit vector in the direction of flow. The vector 11 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 Aaz inclined with respect to Aa 1, which is perpendicular to the flow. The unit vector n is normal to the surface Aag. The 2—3 1 = O) are at the same temperature. are samples of the heat flow vector h. Temperature T is an example of a With each point (X, y, z) in space there is associated a number Tlx, y, z). All points on the surface marked T = 20° (shown as a curve at The arrOWs "fifrmon Fig. 2-2. The velocity of the atoms in a rotating obiect is an example of a vector field. Fig. 2—3. Heat flow is a vector field. The vector It points along the direction of the flow. Its magni- tude is the energy transported per unit time across a surface element oriented perpendicular to the flow, divided by the area of the surface element. Fig. 2—4. The heat flow through A02 is the same as through Acn. angle 0 between n and his the same as the angle between the surfaces (Since h is nor- mal to Aal). Now what is the heat flow per unit area through A02? The flow through A112 is the same as through Aal; only the areas are different. In fact, Aal = A02 cos 0. The heat flow through Aaz is J A 3 cost) = h- n. (2.10) Aaz Aal We interpret this equation: the heat flow (per unit time and per unit area) through any surface element whose unit normal..is n, is given by h - n. Equally, we could say: the component of the heat flow perpendicular to the surface element Aaz is h - n. We can, if we wish, consider that these statements define h. We will be apply- ing the same ideas to other vector fields. 2-3 Derivatives of fields—the gradient When fields vary in time, we can describe the variation by giving their deriva- tives 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 2? 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 aT/ax? 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, aT/ax would cer- tainly be difi'erent. But notice: We have three possible derivatives: aT/ax, aT/ay, and aT/az. 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: 6T 6T 6T _? (a; . 6y — a vector. (2.11) 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 “in- variant” form. For instance, if S = A ' B, and if A and B 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 know that A is a vector, and we have three numbers 3,, B2, and Ba, and we find out that A131 + A1132 + A233 = 59 (2-12) where S is the same for any coordinate system, then it must be that the three numbers Bl, 32, 33 are the components 3,, By, B, of some vector B. Now let’s think of the temperature field. Suppose we take two points P1 and P2, separated by the small interval AR. The temperature at P1 is T1 and at P2 is T 2, and the difference AT = T2 — T1. The temperatures at these real, physical points certainly do not depend on what axis we choose for measuring the coordi- nates. In particular, AT is a number independent of the coordinate system. It is a scalar. 2—4 If we choose some convenient set of axes, we could write T1 = T(x, y, z) and T2 = T(x + Ax, y + Ay, z + A2), where Ax, Ay, and Az are the components of the vector AR (Fig. 2—5). Remembering Eq. (2.7), we can write 6T 0T 6T AT— —Ax+(—9;Ay+~£Az. ax (2.13) The left side of Eq. (2.13) is a scalar. The right side is the sum of three products with Ax, Ay, and Az, which are the components of a vector. It follows that the three numbers 6T 3T 3T _.,.__.,._ 6): 6y dz are also the x-, y-, and z-components of a vector. We write this new vector with the symbol VT. The symbol V (called “del”) is an upside-down A, and is supposed to remind us of diiferentiation. People read VT in various ways: “del-T,” or “gradient of T,” or “grad T;” (2.14) Using this notation, we can rewrite Eq. (2.13) in the more compact form AT = VT-AR. (2.15) 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 VT is indeed a vector. Perhaps you are still not convinced? Let’s prove it in a different way. (Al- though 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 VT transform in just the same way that components of R do. If they do, VTis a vector according to our original definition of a vector in Chapter 11 of Vol. I. We take a new coordi- nate system x’, y’, z’, and in this new system we calculate aT/ax‘, aT/By’, and aT/az’. To make things a little simpler, we let 2 = 2’, 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 0 with respect to the xy-system, as in Fig. 2—6(a). For a point (x, y) the coordinates in the prime system are x’ = xcos 6 + ysin 6, (2.16) y’ = ——x sin 0 + ycos 0. (2.17) Or, solving for x and y, x = x' cos 0 — y’ sin 9, (2.18) y = x’ sin 9 + y’ cos 6. (2.19) 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 P1 and P2, chosen as in Fig. 2—6(b). If we calculate with the x- and y- coordinates, we would write 6T AT _ a; Ax (2.20) ——since Ay is zero. * In our notation, the expreSSion (a, b, c) represents a vector with components a, b, and c. If you like to use the unit vectors 1', j, and k, you may write .6T .6T 6T VT—tax-i-lay-i-kaz. 2—5 Fig. 2—5. The vector AR, whose com- ponents are Ax, Ay, and A2. Fig. 2—6. rotated coordinate system. case of an interval AR parallel to the x-axis. (a) Transformation to 0 lb) Special If we choose some convenient set of axes, we could write T1 = T(x, y, z) and T2 = T(x + Ax, y + Ay, z + Az), where Ax, Ay, and A2 are the components of the vector AR (Fig. 2—5). Remembering Eq. (2.7), we can write AT=gAx+§IAy+ggAz 6x 0y (2'13) The left side of Eq. (2.13) is a scalar. The right side is the sum of three products with Ax, Ay, and A2, which are the components of a vector. It follows that the three numbers 6T 6T 6T __.,—_..,___ 6x 6y 6z are also the x-, y—, and z-components of a vector. We write this new vector with the symbol VT. The symbol V (called “de1”) is an upside-down A, and is supposed to remind us of differentiation. People read VT in various ways: “del-T,” or “gradient of T,” or “grad T;” (2.14) Using this notation, we can rewrite Eq. (2.13) in the more compact form AT = VT-AR. (2.15) 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 VT is indeed a vector. Perhaps you are still not convinced? Let’s prove it in a different way. (Al- though 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 VT transform in just the same way that components of R do. If they do, VTis a vector according to our original definition of a vector in Chapter 11 of Vol. I. We take a new coordi- nate system x’, y’, z’, and in this new system we calculate aT/ax‘, aT/ay’, and aT/az’. To make things a little simpler, we let 2 = 2’, 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 0 with respect to the xy-system, as in Fig. 2—6(a). For a point (x, y) the coordinates in the prime system are x’ = xcos 0 + ysin 6, (2.16) y’ = ——x sin 0 + ycos 0. (2.17) Or, solving for x and y, x = x' cos 0 — y’ sin 9, (2.18) y = x’ sin 0 + y’ cos 6. (2.19) 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 P1 and P2, chosen as in Fig. 2—6(b). If we calculate with the x- and y- coordinates, we would write 6T AT _ 5c— Ax (2.20) —since Ay is zero. “ In our notation, the expression (a, b, 6) represents a vector with components a, b, and c. If you like to use the unit vectors 1', j, and k, you may write .0T .6T 6T VT—Iax-l-lay-I-kaz. 2—5 Fig. 2—5. The vector AR, whose com- ponents are Ax, Ay, and A2. Fig. 2—6. rotated coordinate system. case of an interval AR parallel to the x-axis. (a) Transformation to a (b) Special What would a computation in the prime system give? We would have written aT , 6T , Looking at Fig. 2—6(b), we see that Ax’ = Ax cos 6 (2.22) and Ay’ = ——Ax sin 0, (2.23) since Ay is negative when Ax is positive. Substituting these in Eq. (2.21), we find that AT : Ax cos 0 — 3%: Ax sin 0 (2.24) = cos 0 — 3—; sin 6) Ax. (2.25) Comparing Eq. (2.25) with (2.20), we see that 6T 6T 6T . 5; — by, cos 6 — 3J7 sm 6. (2.26) This equation says that aT/ax is obtained from 6T/6x’ and 6T/6y’, just as x is obtained from x’ and y’ in Eq. (2.18). So aT/ax is the x-component of a vector. The same kind of arguments would show that aT/ay and aT/az are y- and z—com- ponents. So VT is definitely a vector. It is a vector field derived from the scalar field T. 2—4 The operator V Now we can do something that is extremely amusing and ingenious—and characteristic of the things that make mathematics beautiful. The argument that grad T, or VT, is a vector did not depend upon what scalar field we were differ- entiating. All the arguments would go the same if T were replaced by any scalar field. Since the transformation equations are the same no matter what we differ- entiate, we could just as well omit the T and repla...
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