 # Conservative fields and potentials

by taratuta

on
Category: Documents
120

views

Report

#### Transcript

Conservative fields and potentials
```11.4 CONSERVATIVE FIELDS AND POTENTIALS
Evaluate the line integral
0
[(ex y + cos x sin y) dx + (ex + sin x cos y) dy] ,
I=
C
around the ellipse x2 /a2 + y 2 /b2 = 1.
Clearly, it is not straightforward to calculate this line integral directly. However, if we let
P = ex y + cos x sin y
and
Q = ex + sin x cos y,
x
then ∂P /∂y = e + cos x cos y = ∂Q/∂x, and so P dx + Q dy is an exact diﬀerential (it
is actually the diﬀerential of the function f(x, y) = ex y + sin x sin y). From the above
discussion, we can conclude immediately that I = 0. 11.4 Conservative ﬁelds and potentials
So far we have made the point that, in general, the value of a line integral
between two points A and B depends on the path C taken from A to B. In the
previous section, however, we saw that, for paths in the xy-plane, line integrals
whose integrands have certain properties are independent of the path taken. We
now extend that discussion to the full three-dimensional case.
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 is conservative if, and only if, any of
the following is true.
B
(i) The integral A a · dr, where A and B lie in/ the region R, is independent of
the path from A to B. Hence the integral C a · dr around any closed loop
in R is zero.
(ii) There exists a single-valued function φ of position such that a = ∇φ.
(iii) ∇ × a = 0.
(iv) a · dr is an exact diﬀerential.
The validity or otherwise of any of these statements implies the same for the
other three, as we will now show.
First, let us assume that (i) above is true. If the line integral from A to B
is independent of the path taken between the points then its value must be a
function only of the positions of A and B. We may therefore write
B
a · dr = φ(B) − φ(A),
(11.6)
A
which deﬁnes a single-valued scalar function of position φ. If the points A and B
are separated by an inﬁnitesimal displacement dr then (11.6) becomes
a · dr = dφ,
387
LINE, SURFACE AND VOLUME INTEGRALS
which shows that we require a · dr to be an exact diﬀerential: condition (iv). From
(10.27) we can write dφ = ∇φ · dr, and so we have
(a − ∇φ) · dr = 0.
Since dr is arbitrary, we ﬁnd that a = ∇φ; this immediately implies ∇ × a = 0,
condition (iii) (see (10.37)).
Alternatively, if we suppose that there exists a single-valued function of position
φ such that a = ∇φ then ∇ × a = 0 follows as before. The line integral around a
closed loop then becomes
0
0
0
a · dr =
∇φ · dr = dφ.
C
C
Since we deﬁned φ to be single-valued, this integral is zero as required.
Now suppose ∇ × a = 0. From
/ Stoke’s theorem, which is discussed in section 11.9, we immediately obtain C a · dr = 0; then a = ∇φ and a · dr = dφ follow
as above.
Finally, let us suppose a · dr = dφ. Then immediately we have a = ∇φ, and the
other results follow as above.
B
Evaluate the line integral I = A a · dr, where a = (xy 2 + z)i + (x2 y + 2)j + xk, A is the
point (c, c, h) and B is the point (2c, c/2, h), along the diﬀerent paths
(i) C1 , given by x = cu, y = c/u, z = h,
(ii) C2 , given by 2y = 3c − x, z = h.
Show that the vector ﬁeld a is in fact conservative, and ﬁnd φ such that a = ∇φ.
Expanding out the integrand, we have
(2c, c/2, h)
2
(xy + z) dx + (x2 y + 2) dy + x dz ,
I=
(11.7)
(c, c, h)
which we must evaluate along each of the paths C1 and C2 .
(i) Along C1 we have dx = c du, dy = −(c/u2 ) du, dz = 0, and on substituting in (11.7)
and ﬁnding the limits on u, we obtain
2 2
I=
c h − 2 du = c(h − 1).
u
1
(ii) Along C2 we have 2 dy = −dx, dz = 0 and, on substituting in (11.7) and using the
limits on x, we obtain
2c
1 3 9 2 9 2
I=
x − 4 cx + 4 c x + h − 1 dx = c(h − 1).
2
c
Hence the line integral has the same value along paths C1 and C2 . Taking the curl of a,
we have
∇ × a = (0 − 0)i + (1 − 1)j + (2xy − 2xy)k = 0,
so a is a conservative vector ﬁeld, and the line integral between two points must be
388
```
Fly UP