Feynman Physics Lectures V2 Ch03 1962-10-08 Integral Calculus of Vector Fields

Feynman Physics Lectures V2 Ch03 1962-10-08 Integral Calculus of Vector Fields

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Unformatted text preview: 3 Vector Integral Calculus 3-1 Vector integrals; the line integral of VW‘ We found in Chapter 2 that there were various ways of taking derivatives of fields. Some gave vector fields; some gave scalar fields. Although we developed many different formulas, everything in Chapter 2 could be summarized in one rule: the operators a/ax, a/ay, and 6/62 are the three components of a vector operator V. We would now like to get some understanding of the significance of the deriva- tives of fields. We will then have a better feeling for what a vector field equation means. We have already discussed the meaning of the gradient operation (V on a scalar). Now we turn to the meanings of the divergence and curl operations. The interpretation of these quantities is best done in terms of certain vector integrals and equations relating such integrals. These equations cannot, unfor- tunately, be obtained from vector algebra by some easy substitution, so you will just have to learn them as something new. Of these integral formulas, one is practically trivial, but the other two are not. We will derive them and explain their implications. The equations we shall study are really mathematical theorems. They will be useful not only for interpreting the meaning and the content of the divergence and the curl, but also in working out general physical theories. These mathematical theorems are, for the theory of fields, what the theorem of the con- servation of energy is to the mechanics of particles. General theorems like these are important for a deeper understanding of physics. You will find, though, that they are not very useful for solving problems—except in the simplest cases. It is delightful, however, that in the beginning of our subject there will be many simple problems which can be solved with the three integral formulas we are going to treat. We will see, however, as the problems get harder, that we can no longer use these simple methods. We take up first an integral formula involving the gradient. The relation contains a very simple idea: Since the gradient represents the rate of change of a field quantity, if we integrate that rate of change, we should get the total change. Suppose we have the scalar field “x, y, 2). At any two points (1) and (2), the function it will have the values M1) and M2), respectively. [We use a convenient notation, in which (2) represents the point (x2, y2, 22) and M2) means the same thing as :la(x2, y2, 22).] If I‘(gamma) is any curve joining (1) and (2), as in Fig. 3—1, the following relation is true: (2) iI/(Z) - WU) = [m (Vid'ds- (3.1) alongl‘ THEOREM 1. The integral is a line integral, from (I) to (2) along the curve I‘, of the dot product of Wx—a vector—with ds—another vector which is an infinitesimal line element of the curve 1‘ (directed away from (1) and toward (2)). First, we should review what we mean by a line integral. Consider a scalar function f(x, y, z), and the curve I‘ joining two points (1) and (2). We mark off the curve at a number of points and join these points by straight-line segments, as shown in Fig. 3—2. Each segment has the length As“ where iis an index that runs 1,2, 3,... By the line integral (2) f w (1) along I‘ 3-] 3—1 Vector integrals; the line integral of V‘P‘ 3—2 The flux of a vector field 3—3 The flux from a cube; Gauss’ theorem 3—4 Heat conduction; the diffusion equation 3—5 The circulation of a vector field 3—6 The circulation around a square; Stokes’ theorem 3—7 Curl-free and divergence-free fields 3—8 Summary (2) Curve I‘ ds (I) Fig. 3—1. The terms used in Eq. (3.1). The vector Vil/ is evaluated at the line element d3. (2) (I) e b A52 Fig. 3-2. The line integral is the limit of a sum. Closed Surface 5 Fig. 3-3. The closed surface S defines the volume V. The unit vector n is the outward normal to the surface element do, and II is the heat-flow vector at the surface element. we mean the limit of the sum Zfi A35, where f, is the value of the function at the ith segment. The limiting value is what the sum approaches as we add more and more segments (in a sensible way, so that the largest As,- —> 0). The integral in our theorem, Eq. (3.1), means the same thing, although it looks a little different. Instead off, we have another scalar—the component of V1; in the direction of As. If we write (W), for this tangential component, it is clear that (VIII)1AS = (Vip) -As. (3.2) The integral in Eq. (3.1) means the sum of such terms. Now let’s see why Eq. (3.1) is true. In Chapter 1, we showed that the com- ponent of Vzp along a small displacement AR was the rate of change of w in the direction of AR. Consider the line segment As from (1) to point a in Fig. 3—2. According to our definition, 4% = M0) — M1) = (V¢)1'A51- (3-3) Mb) - “61) = (V¢)2 ‘ A52, 0-“) where, of course, (Vinl means the gradient evaluated at the segment Asl, and (V102, the gradient evaluated at A12. If we add Eqs. (3.3) and (3.4), we get Also, we have Mb) " \l/(l) = (Vi/)1 'Asi + (V¢)2'AS2- (35) You can see that if we keep adding such terms, we get the result 1P0) - M1) = 2(V¢)i'Asi. (3-5) The left-hand side doesn’t depend on how we choose our intervals—if (1) and (2) are kept always the same—so we can take the limit of the right-hand side. We have therefore proved Eq. (3.1). You can see from our proof that just as the equality doesn’t depend on how the points a, b, c, . . . . are chosen, similarly it doesn’t depend on what we choose for the curve I‘ to 10m (1) and (2). Our theorem is correct for any curve from (1) to (2). One remark on notation: You will see that there is no confusion if we write, for convenience, (vi/z) - ds = w - ds. (3.7) With this notation, our theorem is (2) M2) - 30(1) = fa) WI - ds. (3-8) any curve from (1) to (2) THEOREM 1. 3—2 The flux of a vector field Before we consider our next integral theorem—a theorem about the divergence —we would like to study a certain idea which has an easily understood physical significance in the case of heat flow. We have defined the vector [1, which represents the heat that flows through a unit area in a unit time. Suppose that inside a block of material we have some closed surface S which encloses the volume V (Fig. 3—3). We would like to find out how much heat is flowing out of this volume. We can, of course, find it by calculating the total heat flow out of the surface S. We write do for the area of an element of the surface. The symbol stands for a two-dimensional differential. If, for instance, the area. happened to be in the xy-plane we would have da = dx dy. 3—2 Later we shall have integrals Over volume and for these it is convenient to con- sider a differential volume that is a little cube. So when we write dV we mean dV = dxdy dz. Some people like to write dza instead of dc to remind themselves that it is kind of a second-order quantity. They would also write d3V instead of dV. We will use the simpler notation, and assume that you can remember that an area has two dimensions and a volume has three. The heat flow out through the surface element da is the area times the com- ponent of h perpendicular to da. We have already defined it as a unit vector pointing outward at right angles to the surface (Fig. 3—3). The component of I: that we want is hn = h - n. (3.9) The heat flow out through da is then h-nda. (3.10) To get the total heat flow through any surface we sum the contributions from all the elements of the surface. In other words, we integrate (3.10) over the whole surface: Total heat flow outward through S = f h - n da. (3.11) s We are also going to call this surface integral “the flux of I: through the sur- face.” Originally the word flux meant flow, so that the surface integral just means the flow of I: through the surface. We may think: It is the “current density” of heat flow and the surface integral of it is the total heat current directed out of the surface; that is, the thermal energy per unit time (joules per second). We would like to generalize this idea to the case where the vector does not represent the flow of anything; for instance, it might be the electric field. We can certainly still integrate the normal component of the electric field over an area if we wish. Although it is not the flow of anything, we still call it the “flux.” We say Flux of E through the surface S = [s E - n da. (3.12) We generalize the word “flux” to mean the “surface integral of the normal com- ponent” of a vector. We will also use the same definition even when the surface considered is not a closed one, as it is here. Returning to the special case of heat flow, let us take a situation in which heat is conserved. For example, imagine some material in which after an initial heating no further heat energy is generated or absorbed. Then, if there is a net heat flow out of a closed surface, the heat content of the volume inside must decrease. So, in circumstances in which heat would be conserved, we say that . _ _9'_Q. [Sh nda _ dt, (3.13) where Q is the heat inside the surface. The heat flux out of S is equal to minus the rate of change with respect to time of the total heat Q inside of S. This interpreta- tion is possible because we are speaking of heat flow and also because we supposed that the heat was conserved. We could not, of course, speak of the total heat inside the volume if heat were being generated there. Now we shall point out an interesting fact about the flux of any vector. You may think of the heat flow vector if you wish, but what we say will be true for any vector field C. Imagine that we have a closed surface S that encloses the volume V. We now separate the volume into two parts by some kind of a “cut,” as in Fig. 3—4. Now we have two closed surfaces and volumes. The volume V1 is enclosed in the surface S1, which is made up of part of the original surface 5,, and of the surface of the cut, Sub. The volume V2 is enclosed by 52, which is made up of the rest of the original surface 5,, and closed off by the cut Sub. Now consider the 3-—3 Fig. 3-4. A volume V contained inside the surface S is divided into two pieces by a ”cut" at the surface Sub. We now have the volume V] enclosed in the surface 51 = 5;. + Sub and the volume V2 enclosed in the surface 52 = $1, + Sub. (x, y. m) s Fig. 3—5. Computation of the flux of C out of a small cube. cut following question: Suppose we calculate the flux out through surface S1 and add to it the flux through surface S2. Does the sum equal the flux through the whole surface that we started with? The answer is yes. The flux through the part of the surfaces Sal, common to both 51 and S 2 just exactly cancels out. For the flux of the vector C out of V1, we can write Flux through SI = f C - n da + C - n1 da, (3.14) Si: Sub and for the flux out of V2, mem$$=mew+sme- (M3 Sb ab Note that in the second integral we have written 111 for the outward normal for 5..., when it belongs to 51, and n2 when it belongs to S2, as shown in Fig. 3—4. Clearly, n1 = —n2, so that C'nl do = — C-ngda. (3J6) Sub Sub If we now add Eqs. (3.14) and (3.15), we see that the sum of the fluxes through S1 and $2 is just the sum of two integrals which, taken together, give the flux through the original surface S = SCl + Sb. We see that the flux through the complete outer surface S can be considered as the sum of the fluxes from the two pieces into which the volume was broken. We can similarly subdivide again—say by cutting V1 into two pieces. You see that the same arguments apply. So for any way of dividing the original volume, it must be generally true that the flux through the outer surface, which is the original integral, is equal to a sum of the fluxes out of all the little interior pieces. 3—3 The flux from a cube; Gauss’ theorem We now take the special case of a small cube* and find an interesting formula for the flux out of it. Consider a cube whose edges are lined up with the axes as in Fig. 3—5. Let us suppose that the coordinates of the corner nearest the origin are x, y, 2. Let Ax be the length of the cube in the x-direction, Ay be the length in the y-direction, and A2 be the length in the z-direction. We wish to find the flux of a vector field C through the surface of the cube. We shall do this by making a sum of the fluxes through each of the six faces. First, consider the face marked 1 in the figure. The flux outward on this face is the negative of the x-component of C, integrated over the area of the face. This flux is —jqwa Since we are considering a small cube, we can approximate this integral by the * The followmg development applies equally well to any rectangular parallelepiped. 3—4 value of CI at the center of the face—which we call the point (l)——multiplied by the area of the face, Ay Az: Flux out of 1 = —CI(1)Ay Az. Similarly, for the flux out of face 2, we write Flux out of 2 = Cz(2) Ay Az. Now C;(1) and Cx(2) are, in general, slightly different. If Ax is small enough, we can write 6C; Cr(2) — Cx(l) + 6x Ax. There are, of course, more terms, but they will involve (A,,)2 and higher powers, and so will be negligible if we consider only the limit of small Ax. So the flux through face 2 is 8C3 6x Flux out of 2 = [Cz(1) + Ax] Ay Az. Summing the fluxes for faces 1 and 2, we get 6C, Flux out of 1 and 2 = 6x Ax Ay A2. The derivative should really be evaluated at the center of face 1; that is, at [x, y + (Ay/2), z + (AZ/2)]. But in the limit of an infinitesimal cube, we make a negligible error if we evaluate it at the corner (x, y, 2). Applying the same reasoning to each of the other pairs of faces, we have Flux out of3 and 4 = 9T? Ax Ay A2 and C Flux out of 5 and 6 = 952—: Ax Ay Az. The total flux through all the faces is the sum of these terms. We find that . _ 6Cz 6—Cy 6C; [C Mia—(ax 6y + aZ>AxAyAz, cube and the sum of the derivatives is just V - C. Also, Ax Ay A2 2 AV, the volume of the cube. So we can say that for an infinitesimal cube / C-nda = (V-C)AV. (3.17) surface We have shown that the outward flux from the surface of an infinitesimal cube is equal to the divergence of the vector multiplied by the volume of the cube. We now see the “meaning” of the divergence of a vector. The divergence of a vector at the point P is the flux—the outgoing “flow” of C—per unit volume, in the neigh- borhood of P. We have connected the divergence of C to the flux of C out of each infinitesimal volume. For any finite volume we can use the fact we proved above—that the total flux from a volume is the sum of the fluxes out of each part. We can, that is, integrate the divergence over the entire volume. This gives us the theorem that the integral of the normal component of any vector over any closed surface can also be written as the integral of the divergence of the vector over the volume enclosed by the surface. This theorem is named after Gauss. GAuss’ THEOREM. [c-nda =/ v-CdV, (3.18) S V where S is any closed surface and V is the volume inside it. 3—5 Fig. 3—6. In the region near a point source of heat, the heat flow is radially outward. 3-4 Heat conduction; the difl'usion equation Let’s consider an example of the use of this theorem, just to get familiar with it. Suppose we take again the case of heat flow in, say, a metal. Suppose we have a simple situation in which all the heat has been previously put in and the body is just cooling off. There are no sources of heat, so that heat is conserved. Then how much heat is there inside some chosen volume at any time? It must be decreasing by just the amount that flows out of the surface of the volume. If our volume is a little cube, we would write, following Eq. (3.17), Heat out = fh'nda = V-ltAV. (3.19) cube But this must equal the rate of loss of the heat inside the cube. If q is the heat per unit volume, the heat in the cube is q AV, and the rate of loss is d _ dq — 3t (q AV) — — 7t AV. (3.20) Comparing (3.19) and (3.20), we see that dq _ _ - 3? — v h. (3.21) Take careful note of the form of this equation; the form appears often in phys- ics. It expresses a conservation law—here the conservation of heat. We have expressed the same physical fact in another way in Eq. (3.13). Here we have the dzfibrential form of a conservation equatiofi, while Eq. (3.13) is the integral form. We have obtained Eq. (3.21) by applying Eq. (3.13) to an infinitesimal cube. We can also go the other way. For a big volume V bounded by S, Gauss’ law says that h'nda = V -th. (3.22) f. Using (3.21), the integral on the right-hand side is found to be just —dQ/dt, and again we have Eq. (3.13). Now let’s consider a different case. Imagine that we have a block of material and that inside it there is a very tiny hole in which some chemical reaction is taking place and generating heat. Or we could imagine that there are some wires running into a tiny res13tor that is being heated by an electric current. We shall suppose that the heat is generated practically at a point, and let W represent the energy liberated per second at that point. We shall suppose that in the rest of the volume heat is conserved, and that the heat generation has been going on for a long time—so that now the temperature is no longer changing anywhere. The problem is: What does the heat vector h look like at various places in the metal? How much heat flow is there at each point? We know that if we integrate the normal component of h over a closed surface that encloses the source, we will always get W. All the heat that is being generated at the point source must flow out through the surface, since we have supposed that the flow is steady. We have the difficult problem of finding a vector field which, when integrated over any surface, always gives W. We can, however, find the field rather easily by taking a somewhat special surface. We take a Sphere of radius R, centered at the source, and assume that the heat flow is radial (Fig. 3—6). Our intuition tells us that h should be radial if the block of material is large and we don’t get too close to the edges, and it should also have the same magnitude at all points on the sphere. You see that we are adding a certain amount of guess- work—usually called “physical intuition”~—to our mathematics in order to find the answer. When h is radial and spherically symmetric, the integral of the normal com- ponent of ’1 over the area is very simple, because the normal component is just 3—6 the magnitude of It and is constant. The area over which we integrate is 41rR2. We have then that [Sh-Ma = 11.471? (3.23) (where h is the magnitude of h). This integral should equal W, the rate at which heat is produced at the source. We get W h _ 47rR2 ’ or W h = 4sz e,, (3.24) where, as usual, er represents a unit vector in the radial direction. Our result says that h is proportional to W and varies inversely as the square of the distance from the source. The result we have just obtained applies to the heat flow in the vicinity of a point source of heat. Let’s now try to find the equations that hold in the most general kind of heat flow, keeping only the condition that heat is conserved. We will be dealing only with what happens at places outside of any sources or absorbers of heat. The differential equation for the conduction of heat was derived in Chapter 2. According to Eq. (2.44), h = ——x VT. (3.25) (Remember that this relationship is an approximate one, but fairly good for some materials like metals.) It is applicable, of course, only in regions of the material where there is no generation or absorption of heat. We derived above another relation, Eq. (3.21), that holds when heat is conserved. If we combine that equation with (3.25), we get dq —7t= v-h= — v-(er), 01' g = KV'VT= xva, (3.26) if x is a constant. You remember that q is the amount of heat in a unit volume and V - V = V2 is the Laplacian operator If we now make one more assumption we can obtain a very interesting equa- tion. We assume that the temperature of the material is proportional to the heat content per unit volume—that is, that the material has a definite specific heat. When this assumption is valid (as it often is), we can write Aq = c, AT or d4 _ LT. E —...
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