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Unformatted text preview: Chapter 3 Vector Calculus Given a scalar ﬁeld ¢(r) we can deﬁne its gradient VqS, which is a vector ﬁeld, by
6‘25 6 = Dem, (3.1) 83% 1
2‘ V¢>(7‘) E 'L where 7° 2 xi 6), is a spaceﬁxed orthonormal frame, and the mi are rectan—
gular (Cartesian) coordinates. Later in this chapter we will express the gradient
in terms of curvilinear coordinates.
Since dr 2 driéi, the gradient of dB is. related to the differential (one—form)
dgb by 3,415 49“
d¢=W2v¢.dr. (3.2) Thus, at each point in space (speciﬁed by the position vector 1'), Vgt is normal
(perpendicular) to the surface of constant 4’) (equipotential surface) passing
through that point (see Fig. 3.1). Also, V45 is along the direction in which ng
changes the fastest. In classical mechanics the gradient of a scalar field leads to the concept of
a conservative force ﬁeld. In fact, the conservative force ﬁeld F(r) associ—
ated with a given scalar ﬁeld ¢(T), which is called the potential function, is
identiﬁed as F = 47¢ . (3.3) We then have
qu : —F  dr , (3.4) and thus the change in potential energy on going from point 1 to point 2 is given
by a line integral: A¢Z¢2“¢1:—/1#F'd7°, (3.5) 17 r g l 18 CHAPTER 3. VECTOR CALCULUS a“, E:
6‘5? 6,:
where dr is a line element along a given path from 1 to 2 (see Fig. 3.2). We“??? gr 13531 The line integral above is independent of the path chosen, since otherwise A¢,
which only depends on the two end points of the path, would not be a meaningful
quantity.
The divergence of a vector ﬁeld A 2 Ai 6, is deﬁned in terms of Cartesian
coordinates vi by
8A1.
83:"
This operation satisﬁes Gauss’ Theorem (see Fig. 3.3): fDVAdBrzfaDAda , (3.7) where D represents a volume domain of integration, 8D represents the boundary
of D, and do is an inﬁnitesimal outward normal vector area element. The area integral on the RHS of Gauss’ Theorem is called the ﬂux of the
vector ﬁeld A through the area 6D. Thus the divergence of a vector ﬁeld at
a certain pint in space is equal to the flea per unit volume, in the limit of
vanishing volume, at that point. Gauss’ Theorem plays an important role in
Newtonian gravitation and electrostatics. The corresponding laws governing
the gravitational ﬁeld g and the electrostatic ﬁeld E are v  A a = 3,441” . (3.6) V  g 2 ~11va , (3.8) and
V  E = 47Tp, (3.9) respectively, where C is the Newtonian gravitational constant, and p stands for
mass density in the gravitational context and charge density in the electrostatic ‘
context. Gauss’ Theorem immediately leads to the following integral forms for the above laws: %g  da : ~47rG/pd3r = —47TG']\/I, (3.10) jiE  do = 47r/pd3r = 47mg, (3.11) where Ad and Q are the total mass and total charge enclosed by the volume that is integrated over, respectively.
Gauss’ Theorem is also used to give the integral form of the equation of
continuity, which is a statement of conservation obeyed by the ﬂow of electrical 19 currents as well as matter through different regions in space. The differential
form of this equation is given by 6p_
V'Jlb—t—O, (3.12) where J is the current density, measured by the amount of “stuﬁ” (for ex
ample, charge or mass) ﬂowing through a certain area perpendicular to the
direction of ﬂow per unit area per unit time, and p is the volume density of the “stuﬁ”. The integral of the divergence term over a certain volume D is,
according to Gauss’ Theorem, fvJd3r=7i Jda. (3.13)
D 6D Thus the integral form of the equation of continuity appears as, on integrating
(3.12) over a volume D (see Fig. 3.3), if Jda:~——a—/pd3r. (3.14)
6D at D This equation has a very intuitive physical interpretation: the ﬂux of “stuff”7
through a closed area is equal to the negative time rate of change of the amount
of “stuﬁ” contained in the volume that is bounded by that area. The curl of a vector ﬁeld A is deﬁned in terms of Cartesian coordinates by
[compare with (1.37)] 6 8 E) v xA: atriaAt e, = _ ___ 3.15
( " J > 8:221 83:2 8x3 ( )
A1 A2 A3
The curl of a vector ﬁeld satisﬁes Stokes’ Theorem (see Fig. 3.4):
fD(VxA)da=faDAdl , (3.16) where dl is a line element along the boundary (9D of an open surface D, and
has a direction consistent with the “right—handrule”. The integral on the RHS
of the above equation is called the circulation of the vector ﬁeld A around
the loop 6D. Thus Stokes’ Theorem implies that the curl of a vector ﬁeld at
a certain point in space is equal to the circulation per unit area, in the limit of
vanishing area, at that point. Fig. 3.4 20 CHAPTER 3. VECTOR CALCULUS The gradient, divergence and curl satisfy the following two important iden—
tities: V(VXA)=0, (3.17) V x (V45) 2 0 . (3.18) In words, the divergence of the curl of a vector ﬁeld, and the curl of the gradient
of a scalar ﬁeld, always vanish. Prove the above two equations using the tensorial index notation developed in Chapter 1. Eq. (3.18) provides a convenient test for whether a force ﬁeld is conservative. If a force ﬁeld F is conservative, it must be expressible as the gradient of some
scalar ﬁeld qS. Thus F is conservative if and only if V x F = 0. A two—dimensional force ﬁeld is given in terms of its Cartesian components by Fm = y, Fy = —:B (see Fig. 3.5). Show that this force ﬁeld is
not conservative by showing that its curl does not vanish identically. We have the following useful vector calculus identities, which are analogous
to the product rule of ordinary differentiation. For arbitrary scalar ﬁelds (15, 1b
and vector ﬁelds A, B, WW) = ¢>V¢ +¢V¢, (319)
Vx(¢A)=V¢xA+¢VxA, (3.20)
V(¢A):V¢A+¢>V~A, (3.21)
V(AxB):B(V><A)—A.(V><B). (3.22) Prove the above four identities using the index notation. Remark. There is a very powerful mathematical formalism, that of the exterior
calculus of differential forms, that subsumes all the vector calculus opera—
tions we have considered. This formalism is in fact applicable to differentiable
manifolds of arbitrary dimension, whereas our development only applies to Eu
clidean space of dimension three. Under this formalism, it can be seen that the
gradient, divergence and curl are special cases of an operation called the ex—
terior derivative of diﬁ'erential forms, and that Gauss’ Theorem, Stokes’
Theorem, and in addition, Green’s Theorem, as well as the Fundamental The—
orem of Calculus, are all special cases of a general theorem, called Stokes”
Theorem of integrals of differential forms. All the vector calculus iden—
tities are also special cases of a basic property of exterior derivatives: the
second exterior derivative of any differential form vanishes. 21 We will now give the expressions for the gradient, divergence and curl in
terms of arbitrary curvilinear coordinates. Two most commonly occurring sys—
tems of curvilinear coordinates are the spherical and cylindrical coordinates.
A general system is introduced as follows. Referring to Fig. 3.6, we write
the inﬁnitesimal displacement of the position vector in terms of a moving or
thonormal frame ﬁeld {62} as dr 2 0:1 el + a2 62 + a3 63 , (3.23) where the differential 1—forms of are given in terms of a set of curvilinear coor—
dinates 21 as a1 = /\1(r)dz1 , a2 :: /\2(T)d2’2 , a3 = /\3()dz3 . (3.24) In the above equation, the /\,(r) are position—dependent functions determined
by the choice of the coordinate system. Fig. 3.6 Suppose the moving frame {6,} and the space—ﬁxed frame {5,} are related,
at each point in space, by e, = 9,415,. . (3.25)
Then, ’ ‘ I 1
dr 2 clanz 6, = of e, = 0/9,? 6, . (3.26)
Thus ‘ I '
dm‘ 2 £239; , (3.27)
from which it follows that
dml /\ dm2 /\ dz3 : det(g,j) 041 /\ a2 A a3 . (3.28) The “A” in the above equation is called the exterior product of differential
forms, and is a generalization of the cross product for 3dimensional vectors.
The skewsymmetry nature of the exterior product (which is shared'by the cross
product) leads to the presence of the determinant factor. Since ) represents rotations, det(g§) : 1. Then the volume form (19:1 /\ (1:32 /\ dz3 is given by (Z331 /\ (lac? /\ da:3 2 a1 /\ a2 /\ a3 . (3.29) Putting $1 : cc, 3:2 = y, .733 2 z, and using (3.24), we have the following equation relating the volume form in Cartesian coordinates to that in arbitrary
curvilinear coordinates: dz A dy /\ dz = A1A2A3 dzl A dz? A (1.23 . (3.30)
For example, using spherical coordinates, we can put 21:7", 22:0, 23:45. (3.31) 22 CHAPTER 3. VECTOR CALCULUS Then, since
d'r : dr 6, +rdl9 69 +7'sin6dgb e¢ , (3.32)
we have
a1 : d7", a2 = rdﬂ, a3 = rsinﬁdgﬁ; (3.33)
/\1 I , /\2 = T , /\3 Z TSlI'lQ . So
dazAdyAdz :7“2 sinddrAdQAdqﬁ. (3.35) Exercise 3.4 Show that in cylindrical coordinates (zl : 7‘, 22 = 9, z3 =
2), A1 = A3 = 1, A2 = 7‘, and thus the volume element is given by dmAdyAdz=rdrAd0Adz. (3.36) With the above background we now state Without proof the general formulas
for the differential vector operators V (gradient), V (divergence), Vx (curl),
and ﬁnally V2 = V  \7 (Laplacian) in terms of general curvilinear coordinates
(21, 22, 23). (The proofs are most elegantly given using the exterior calculus.) Let qﬁ be a scalar ﬁeld and A : A1 61 + A2 62 + A3 83 be a vector ﬁeld given
in terms of a moving orthonormal frame ﬁeld {6,}. Then 16¢ 18¢ 18¢ VQbZE‘B—gTel‘l—Eazz 62+};ﬁ‘83, _ 1 6 1 6 2 6 3
v A— (1 (2,3 { 6,, (MBA >+ 32,633}: )+ (,2, (A1sz )} , (3.38) A161 /\2 62 /\3 63
1 :2: a 1
A1/\2 As (921 822 82:3 A1111 A2A2 A3443 , _ 1 _6_ may: 6 m3 845 6 m2 345
V ‘15“ A1A2A3{821< ,\1 6z1> +322 ( A2 6z2) + 6,23 < A3 6z3 ' (3.40) Give explicit expressions for (3.37) to (3.40) in terms of i) spher— ical coordinates, and ii) cylindrical coordinates. VXA: , (3.39) ...
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 Spring '08
 LAM
 mechanics

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