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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 21 Understanding physics 22 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 27 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 (23)
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(AC) — C(AB) (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 21
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. 22. 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. 23 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 xaxis, 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 (212) 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 zcomponents of a vector. We write this new vector with
the symbol VT. The symbol V (called “del”) is an upsidedown A, and is supposed
to remind us of diiferentiation. People read VT in various ways: “delT,” or
“gradient of T,” or “grad T;” (2.14) Using this notation, we can rewrite Eq. (2.13) in the more compact form AT = VTAR. (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 longerwinded 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 zcoordinate. (You can check out the more general case for yourself.) We take an x’y’system rotated an angle 0 with respect to the xysystem, 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—taxilayikaz. 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
xaxis. (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 zcomponents of a vector. We write this new vector with
the symbol VT. The symbol V (called “de1”) is an upsidedown A, and is supposed
to remind us of differentiation. People read VT in various ways: “delT,” or
“gradient of T,” or “grad T;” (2.14) Using this notation, we can rewrite Eq. (2.13) in the more compact form AT = VTAR. (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 longerwinded 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 zcoordinate. (You can check out the more general case for yourself.) We take an x’y’system rotated an angle 0 with respect to the xysystem, 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—IaxllayIkaz. 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
xaxis. (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 xcomponent 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 replace Eq. (2.26) by the operator
equation 3?; = cos a — 31:7 sin 0. (2.27)
We leave the operators, as Jeans said, “hungry for something to differentiate.” Since the differential operators themselves transform as the components of a vector should, we can call them components of a vector operator. We can write 6 6 6
= 7— 3 — s — 9 2.2
V (ax 6y 62) ( 8)
which means, of course,
6 6 6
Vx_ﬁ, Vy—Iﬁi Vz—a—z. (2.29) We have abstracted the gradient away from the T—that is the wonderful idea. You must always remember, of course, that V is an operator. Alone, it
means nothing. If V by itself means nothing, what does it mean if we multiply
it by a scalar—say T —to get the product TV? (One can always multiply a vector
by a scalar.) It still does not mean anything. Its xcomponent is a
T a; , (2.30)
which is not at number, but is still some kind of operator. However, according to the algebra of vectors we would still call TV a vector.
2—6 Now let’s multiply V by a scalar on the other side, so that we have the product
(VT). In ordinary algebra
TA = AT, (2.31) but we have to remember that operator algebra is a little different from ordinary
vector algebra. With operators we must always keep the sequence right, so that
the operations make proper sense. You will have no difficulty if you just remember
that the operator V obeys the same convention as the derivative notation. What is
to be differentiated must be placed on the right of the V. The order is important. Keeping in mind this problem of order, we understand that TV is an operator,
but the product VT is no longer a hungry operator; the operator is completely
satisﬁed. It is indeed a physical vector having a meaning. It represents the spatial
rate of change of T. The xcomponent of VT is how fast T changes in the xdirec
tion. What is the direction of the vector VT? We know that the rate of change of
T in any direction is the component of VT in that direction (see Eq. 2.15). It
follows that the direction of VT is that in which it has the largest possible com
ponent—m other words, the direction in which T changes the fastest. The gradient
of T has the direction of the steepest uphill slope (in T). 2—5 Operations with V Can we do any other algebra with the vector operator V? Let us try combining
it with a vector. We can combine two vectors by making a dot product. We could
make the products (a vector)  V, or V  (a vector). The ﬁrst one doesn't mean anything yet, because it is still an operator. What it
might ultimately mean would depend on what it is made to operate on. The
second product is some scalar ﬁeld. (A  Bis always a scalar.) Let’s try the dot product of V with a vector ﬁeld we know, say h. We write
out the components: v  h = v.12. + vyhy + v.21. (2.32)
or
. _ ah, ah, ah.
vh—Fx—+6—y+az (2.33) The sum is invariant under a coordinate transformation. If we were to choose a
different system (indicated by primes), we would have* ah I ah r ah i
I . = :0 _1L 2
V h 6x’ + 6y’ az’ , (2.34) which is the same number as would be gotten from Eq. (2.33), even though it
looks different. That is, V’h = Vh (2.35) {or every point in space. So V  h is a scalar ﬁeld, which must represent some
physical quantity. You should realize that the combination of derivatives in
V ' h is rather special. There are all sorts of other combinations like ahy/ax,
which are neither scalars nor components of vectors. The scalar quantity V  (a vector) is extremely useful in physics. It has been
given the name the divergence. For example, V  h = div h = “divergence of h.” (2.36) As we did for VT, we can ascribe a physrcal signiﬁcance to V t h. We shall, how
ever, postpone that until later. ‘ We think of h as a physical quantity that depends on position in space, and not
strictly as a mathematical function of three variables. When h is “differentiated” with
respect to x, y, and z, or with respect to x', y’, and z’, the mathematical expression for It
must ﬁrst be expressed as a function of the appropriate variables. 2—7 Fig. 2—7. (0) Heat Flow through a
slab. (b) An inﬁnitesimal slob parallel to
an isothermal surface in a large block. First, we wish to see what else we can cook up With the vector operator V.
What about a cross product? We must expect that V X h : a vector. (2.37) It is a vector whose components we can write by the usual rule for cross products
(see Eq. 2.2): ah ah;
(v X h): = Vzhy — Vyhx = 6—; — ‘6‘): (2.38)
Similarly, ah ah
._. _ = ___ﬁ _ _.l
(V X It); — Vultz VJ!” ay 62 (2.39)
and ah ah
(V X It), = V5}?r — Var/1z = *6; — (2.40) The combination V X h is called “the curl of h.” The reason for the name
and the physrcal meaning of the combination will be discussed later.
Summarizing, we have three kinds of combinations with V: VT = gradT = a vector,
V ‘h 2 div h = a scalar,
V X h = curlh = a vector. Using these combinations, we can write about the spatial variations of ﬁelds in a
convenient way—in a way that is general, in that it doesn’t depend on any particular
set of axes. As an example of the use of our vector differential operator V, we write a set
of vector equations which contain the same laws of electromagnetism that we gave
in words in Chapter 1. They are called Maxwell's equations. Maxwell ’3 Equations p
v. =__
(I) E 60
as
(2) VXE——a—t Q41)
(3) vB=0
2 _§£ _J'_
(4)cv><19_at+60 s where p (rho). the “electric charge density.’ is the amount of charge per unit
volume, and j, the “electric current density.“ is the rate at which charge ﬂows
through a unit area per second. These four equations contain the complete
classical theory of the electromagnetic ﬁeld. You see what an elegantly simple
form we can get with our new notation! 2—6 The diﬁ‘erential equation of heat ﬂow Let us give another example of a law of physics written in vector notation.
The law is not a precise one, but for many metals and a number of other sub
stances that conduct heat it is quite accurate. You know that if you take a slab of‘
material and heat one face to temperature T2 and cool the other to a different
temperature T1, the heat will ﬂow through the material from T2 to T1 [Fig. 27(a)].
The heat ﬂow is proportional to the area A of the faces. and to the temperature
difference. It is also inversely proportional to d, the distance between the plates.
(For a given temperature difference, the thinner the slab the greater the heat ﬂow.)
Letting .1 be the thermal energy that passes per unit time through the slab, we write .1 = m, — 71):; (2.42) The constant of proportionality K (kappa) is called the thermal conductivity.
28 What will happen in a more complicated case? Say in an oddshaped block of
material in which the temperature varies in peculiar ways? Suppose we look at a
tiny piece of the block and imagine a slab like that of Fig. 2—7(a) on a miniature
scale. We orient the faces parallel to the isothermal surfaces, as in Fig. 2—7(b), so
that Eq. (2.42) is correct for the small slab. If the area of the small slab is AA, the heat ﬂow per unit time is AA AJ=KATAS , (2.43) where A: is the thickness of the slab. Now AJ/AA we have deﬁned earlier as the ’ magnitude of h, whose direction is the heat ﬂow. The heat ﬂ0w will be from
T1 + AT toward T1, and so it will be perpendicular to the isotherms, as drawn in
Fig. 2—7(b). Also, AT/As is just the rate of change of T with position. And since
the position change is perpendicular to the isotherms, our AT/As is the maximum
rate of change. It is, therefore, just the magnitude of VT. Now since the direction
of VT is opposite to that of h, we can write (2.43) as a vector equation: I: = —x VT. (2.44) (The minus sign is necessary because heat ﬂows “downhill” in temperature.)
Equation (2.44) is the differential equation of heat conduction in bulk materials.
You see that it is a proper vector equation. Each side is a vector if K is just a num
ber. It is the generalization to arbitrary cases of the special relation (2.42) for
rectangular slabs. Later we should learn to write all sorts of elementary physics
relations like (2.42) in the more sophisticated vector notation. This notation is
useful not only because it makes the equations look simpler. It also shows most
clearly the physical content of the equations without reference to any arbitrarily
chosen coordinate system. 2—7 Second derivatives of vector ﬁelds So far we have had only ﬁrst derivatives. Why not second derivatives? We
could have several combinations: (3) V ' (V73 (b) v x (W) (e) v(v  h) (2.45)
(d) V ' (V X h) (e) v x (v x b) You can check that these are all the possible combinations.
Let’s look ﬁrst at the second one, (b). It has the same form as A><(AT)= (AXA)T=0,
since A X A is always zero. So we should have
curl (gradT) = V X (VT) = 0. (2.46) We can see how this equation comes about if we go through once with the com
ponents: [V X (VT)]z ll Va:(VT)y _ V1](VT)1: ,
6 6T 6 6T
5 (a) ‘ a (“7’ ’Which is zero (by Eq. 2.8). It goes the same for the other components. So V X
(VT) = 0, for any temperature distribution—in fact, for any scalar function. 2—9 Now let us take another example. Let us see whether we can ﬁnd another
zero. The dot product of a vector with a cross product which contains that vector is zero:
A  (A X B) = 0. (2.48) because A X B is perpendicular to A, and so has no components in the direction A.
The same combination appears in (d) of (2.45), so we have v  (v x h) = div (curl h) = 0. (2.49) Again, it is easy to show that it is zero by carrying through the operations with components. Now we are going to state two mathematical theorems that we will not prove.
They are very interesting and useful theorems for physicists to know. In a physical problem ,we frequently ﬁnd that the curl of some quantity—say
of the vector ﬁeld A—is zero. Now we have seen (Eq. 2.46) that the‘eurl of a
gradient is zero, which is easy to remember because of the way the vectors work.
It could certainly be. then. that A is the gradient of some quantity, because then
its curl would necessarily be zero. The interesting theorem is that if the curl A is
zero, then A is always the gradient of something—there is some scalar ﬁeld 1% (psi)
such that A is equal to grad 1;. In other words, we have the THEOREM:
If v x A = 0
there is a w
such that A = Vila (2.50) There is a similar theorem if the divergence of A is zero. We have seen in
Eq. (2.49) that the divergence of a curl of something is always zero. If you come
across a vector ﬁeld D for which div D is zero, then you can conclude that D is
the curl of some vector ﬁeld C. T HEOREM:
If V  D = 0
there is a C
such that D = V X C. (2.51) In looking at the posmble combinations of two V operators. we have found
that two of them always give zero. Now we look at the ones that are not zero.
Take the combination V ~ (VT), which was ﬁrst on our list. It is not, in general,
zero. We write out the components: VT = VET + V,,T + VZT.
Then V ' (VT) = VI(VJ'T) + Vy(V1/T) + Vz(VzT)
627” 32T 62T =—#+—+ W W 32—2, (2.52) which would. in general, come out to be some number. lt is a scalar ﬁeld.
You see that we do not need to keep the parentheses, but can write, without any chance of confusion,
v  (VT) 2 v  VT = (v  v)T = v22”. (2.53) We look at V2 as a new operator. It is a scalar operator. Because it appears often
m physics, it has been given a special name—the Laplacran.
. a2 a2 a2
. v = 2 = _._ _.—— — s .
Laplactan V 6x2 + ayz + 622 (2 54)
2~10 Since the Laplacian is a scalar Operator, we may operate with it on a vector—
by which we mean the same operation on each component in rectangular coor
dinates: v2}. = (V271,, V2h,,, V211,). Let’s look at one more possibility: V X (V X h), which was (e) in the list
(2.45). Now the curl of the curl can be written differently if we use the vector
equality (2.6): A x (B x C) = B(A  C) — C(A ~19). (2.55) In order to use this formula, we should replace A and B by the operator V and
put C = In. If we do that, we get VX(VXh)=V(V'h)h(V'V)...??? Wait a minute! Something is wrong. The ﬁrst two terms are vectors all right
(the operators are satisﬁed), but the last term doesn’t come out to anything. It’s
still an operator. The trouble is that we haven’t been careful enough about keeping
the order of our terms straight. If you look again at Eq. (2.55), however, you see
that we could equally well have written it as A X (B X C) = B(A  C) — (A B)C. (2.56)
The order of terms looks better. Now let’s make our substitution in (2.56). We get
V X (V X h) = V(Vh)  (V ' V)h. (2.57) This form looks all right. It is, in fact, correct, as you can verify by computing the
components. The last term is the Laplacian, so we can equally well write V X (V X h) = V(V ' h) — V211. (2.58) We have had something to say about all of the combinations in our list of
double V’s, except for (c), V(V  h). It is a possible vector ﬁeld, but there is nothing
special to say about it. It’s just some vector ﬁeld which may occasionally come up. It will be convenient to have a table of our conclusions: (a) V  (VT) = VZT = a scalar ﬁeld
(b) V X (VT) = 0 (c) V(V h) = a vector ﬁeld (d) V  (V X h) = 0 (c) VX(VXh)=V(V‘h)—V2h
(f) (V  V)h = V2]: = a vector ﬁeld (2.59) You may notice that we haven’t tried to invent a new vector operator (V X V).
Do you see why? 28 Pitfalls We have been applying our knowledge of ordinary vector algebra to the alge
bra of the operator V. We have to be careful, though, because it is possible to go
astray. There are two pitfalls which we will mention, although they will not come
up in this course. What would you say about the following expression, that in
volves the two scalar functions it and «5 (Phi): (W) X (Wt)?
You might want to say: it must be zero because it’s just like (Aa) X (Ab),
2—11 which is zero because the cross product of two equal vectors A X A is always zero.
But in our example the two operators V are not equal! The ﬁrst one operates on
one function, t/x; the other operates on a different function, 4;. So although we rep
resent them by the same symbol V, they must be considered as different operators.
Clearly, the direction of W» depends on the function u», so it is not likely to be
parallel to W). (Vii) X (W) sf 0 (generally). Fortunately, we won’t have to use such expressions. (What we have said doesn’t
change the fact that V X Vw = 0 for any scalar ﬁeld, because here both V’s
operate on the same function.) Pitfall number two (which, again, we need not get into in our course) is the
following: The rules that we have outlined here are simple and nice when we use
rectangular coordinates. For example, if we have V2}: and we want the xcom
ponent, it is 62 a2 a2 _ The same expression would not work if we were to ask for the radial component
of V211. The radial component of V2]; is not equal to V2h,. The reason is that
when we are dealing with the algebra of vectors, the directions of the vectors are
all quite deﬁnite. But when we are dealing with vector ﬁelds, their directions are
different at different places. If we try to describe a vector ﬁeld in, say, polar coordi
nates, what we call the “radial” direction varies from point to point. So we can
get into a lot of trouble when we start to differentiate the components. For ex
ample, even for a constant vector ﬁeld, the radial component changes from point
to point. It is usually safest and simplest just to stick to rectangular coordinates and
avoid trouble. but there is one exception worth mentioning: Since the Laplacian
V2, is a scalar, we can write it in any coordinate system we want to (for example,
in polar coordinates). But since it is a differential operator, we should use it only
on vectors whose components are in a ﬁxed direction—that means rectangular
coordinates. So we shall express all of our vector ﬁelds in terms of their x, y,
and zcomponents when we write our vector differential equations out in com
ponents. 2—I2 ...
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 Spring '09
 LeeKinohara
 Physics

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