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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 repla...
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 Spring '09
 LeeKinohara
 Physics, Vector Calculus, Vector field, Gradient

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