Introduction to vector and tensor analysis

line integral in Eq. (2.49), for example, is defined as
C
a · dr = def lim
N→∞
N
a(xp ) · ∆rp,
where all |∆rp| → 0 as N → ∞.
Each of the line integrals in Eq. (2.49) is evaluated over some curve C
that may be either open (A and B being distinct points) or closed (A and B
coincide). In the latter case, one writes
p=
C
to indicate this. The curve C is a
space curve, c.f. section 2.5.1, most often defined in a parametric form. In a
cartesian coordinate system C = (O, R), it becomes
C : r(u) =
xi(u)ei, u0 ≤ u ≤ u1, and [r(u0)] = x(A), [r(u1)] = x(B)
i
(2.50)
The metod of evaluating a line integral is to reduce it to a set of scalar
integrals. It is usual to work in cartesian coordinates, in which case, dr =
i eidxi. The three integrals in Eq. (2.49) then becomes
C
C
φ(r)dr = C φ(r) (i
eidxi) =
i C φ(r)dxi
a(r) · dr = j ejdxj
=
i C ai(r)dxi
C (
i ai(r)ei) ·
ei
(2.51)
and
a(r) × dr C =
C (i
ai(r))
× j ejdxj
=
C a2(r)dx3 −
C a3(r)dx2
e1 +
C a3(r)dx1 −
C a1(r)dx3
+
C a1(r)dx2 −
C a2(r)dx1
e3
(2.52)
Note, that in the above we have used relations of the form
aiejdxj = aidxj ej,
C
which is allowable since the cartesian basis is independent of the coordinates.
If E = E(x) then the basis vectors could not be factorised out in the integral.
The final scalar integrals in Eq. (2.51) and Eq. (2.52) can be solved using the
parametrized form for C, Eq. (2.50). For instance
C
u1
a2(r)dx3 =
u0
C
a2(x1(u), x2(u), x3(u)) dx3
du du.
In general, a line integral between two points will depend on the specific path
C defining the integral. However, for line integrals of the form
C a · dr there
exists a class of vector fields for which the line integral between two points
is independent of the path taken. Such vector fields are called conservative.
A vector field a that has continuous partial derivatives in a simply connected
region R 7 , is conservative if, and only if, any of the following is true
7 A simply connected region R is a region in space for which any closed path can be
continuously shrunk to a point, ie. R has no “holes”
21
e2