Vector Algebra And Calculus Gauss' and Stokes' Theorems

Vector Algebra And Calculus
Professor Michael Brady,
8 Lectures
Michaelmas Term 2000
Gauss' and Stokes' Theorems
...whyweintroduced div; grad and curl!
Gauss' Theorem: enables an integral taken overavolume
! one taken over the surface bounding that volume, or vice
versa
) Computational eciency and/or numerical accuracy
Stokes' Theorem: integral taken around a closed curve !
one taken over any surface bounded by that curve
1

Gauss' Theorem
Suppose a(r) isavector eld.
we want to compute its ux of the eld across the surface S that
bounds a volume V :
Z
S a dS = Z S a ^ndS
where ^n is the unit surface normal on S.
Gauss' Theorem:
ZS a ^ndS = Z
V
div adV
2

Example of Gauss' Theorem
Suppose a = z 3 k and that S is the sphere of radius R centred on
the origin.
On the surface of the sphere, a = R 3 cos 3 k;dS = R 2 sin dd^r
and ^r k = cos . We nd:
ZS adS = Z 2
Z
=0 =0 Z R3 cos 3 R 2 sin dd cos
= 2R 5
0 cos4 sin d
= 2R5 [, cos 5 ] 0
5
= 4R5
5
It was \lucky" that we arranged the angle integral to come out so
nicely!
3

Applying Gauss' Theorem
To apply Gauss' Theorem, we need to gure out div a and decide
how to compute the volume integral. The rst is easy:
diva =3z 2
For the second, we can divide the sphere into discs of constant z
and thickness dz. Then
and
dV = (R 2 , z 2 )dz
ZV div adV = 3 Z R
= 3
= 4R5
5
2
64 R2 z 3
,R z2 (R 2 , z 2 )dz
3
, z5
5
3
75 R ,R
This is a typical example: the surface integral is rather tedious,
whereas the volume integral is straightforward.
imagine how complex the surface integral might become over truly
hairy surfaces. Breaking up the enclosed volume into tiny elements
often makes the computation far easier, faster, and better conditioned
numerically.
4

Extension to Gauss' Theorem
Suppose a = U c but where for present purposes c is a constant
vector, then
div a = Udiv c + gradU c
Since c is assumed constant, divc = 0 and Gauss' Theorem
becomes
Z
S U(c dS) =Z (gradU c)dV
V
taking the constant c out of the integrals
c Z S UdS
= c
Z
V gradUdV
This is still a scalar equation but we now note that the vector c
is arbitrary so that the result must be true for any vector c. This
can be true only if the vector equation
ZS UdS = Z
V gradUdV
5

Example of extension to Gauss' Theorem
Suppose that U = x 2 + y 2 + z 2 and V is the cylinder x 2 + y 2
a 2 ; 0 z h.
6

evaluating the surface integral
Z
S UdS
from symmetry that there is no contribution from the curved surface
of the cylinder since for every vector surface element there
exists an equal and opposite element with the same value of U.
We therefore need consider only the top and bottom faces.
Top face:
Z
UdS =
Z a
Bottom face:
U = x 2 + y 2 + h 2 = r 2 + h 2 and dS =2rdrk
0 (h2 +r 2 )2rdrdk = k
2
4 h 2 r 2 + 1 2 r4 3 5
a
0
= [h 2 a 2 + 1 2 a4 ]k
U = r 2 and dS = ,2rdrk
The contribution from this face is thus , a4 k, and the total integral
is h 2 a 2 2
k.
7

Applying Gauss' Theorem
we have to compute
Z
V
grad UdV
In this case, gradU =2r, so that the i and j components are zero
by symmetry.
The k component gives
Z h
0 2za2 dz = a 2 h 2 k
8

Stokes' Theorem
Stokes' Theorem relates a line integral around a closed path to a
surface integral over what is called a capping surface of the path.
Imagine that the closed path is a bent, originally circular, piece of
wire. There is no requirement for the distorted wire to be planar.
Now let a bubble hang from the wire. The surface of the bubble is
a capping surface. We can choose the capping surface as we wish.
Stokes' Theorem:
IC a dl = Z
S
curl adS
where S is any surface capping the curve C.
9

Example of Stokes' Theorem
Suppose a = x 3 j , y 3 i and C is the circle of radius R centred on
the origin. We want to compute:
On the circle,
I
C a dl
and
a = R 3 (, sin 3 i + cos 3 j)
dl = Rd(, sin i + cos j) :
IC a dl = Z 2
0
R 4 (sin 4 + cos 4 )d = 3 2 R4 ;
since
Z 2
0
Z sin 4 2
d = cos 4 d = 3 0
4
10

Applying Stokes' Theorem
the surface integral requires that curl a be computed. This is
i j k
curl a =
@
@x
@
@y
@
@z
,y 3 x 3 0
=3(x 2 + y 2 )k =3r 2 k
We choose area elements to be circular strips of radius r thickness
dr. Then
dS =2rdrk
and Z S curl a dS =6 Z R
0 r3 dr = 3 2 R4
11

Extension to Stokes' Theorem
suppose a = U c, where c is a constant vector, so that
curl a = Ucurl c + gradU c
Again, curl c is zero. Stokes' Theorem becomes in this case:
I
C U(c dl) =Z S (gradU c dS = Z S
c (dS gradU)
rearranging the triple scalar products and taking the constant c
out of the integrals gives
c I C Udl = ,c Z S
gradU dS
since c is arbitrary, we nd the vector equation:
I
C Udl = , Z S
gradU dS
12

Example of extension to Stokes' Theorem
Consider the case where U = x 2 + y 2 + z 2 and C is the circle
(x , a) 2 + y 2 = a 2 ;z =0. We have to compute:
I
C Udl
Doing this directly, we note the i component is zero by symmetry
and there is no k component. On the circle
U = a 2 (1 + cos ) 2 + a 2 sin 2 =2a 2 (1 + cos )
dl
= ad(cos j , sin i);
so that
I
Udl =2a
3 Z 2
0
(1 + cos )(cos j , sin i)d =2a 3 j
13

applying Stokes' Theorem
On the other hand, over the area of the circle dS = rdrdk, since
r 2 =(x , a) 2 + y 2 ) and
so that
gradU =2r =2(a + r cos )i +2r sin j
dS gradU =[,2r sin i +2(a + r cos )j]rdrd
and
ZS dS gradU =4j Z a
r=0
Z 2
=0 2ardrd =2a3 j
14

Fluid Mechanics - The Continuity Equation
Expression of conservation of mass in a uid ow. The continuity
principle applied to any volume (called a control volume) may
be expressed in words as follows:
" The net rate of mass ow of uid out of the control
volume must equal the rate of decrease of the mass of
uid within the control volume"
denote the velocity of the uid at each point of the ow by q(r),
the density by (r), then rate of ow of mass out of the control
volume is given by the integral over the surface S of the control
volume
Z
S
q dS:
15

The Continuity equation
using Gauss' Theorem, this is
Z
div(q)dV
V
rate of decrease of mass contained in the control volume is:
, @ @t
Z
V dV
Assuming volume of interest is xed, this is equal to
, Z V
@
@t dV:
The two volume integrals can be equal for any control volume V
only if the two integrands are equal at each point of the ow.
the Continuity Equation:
div(q) =, @
@t
for a uid of constant density, div q =0:
16

The Heat Conduction Equation
Flow of heat similar to uid ow, so heat ow satises a similar
continuity equation.
heat ow ischaracterized by its \current density" q(r) (heat
ow per unit area and time)
assuming no mass ow across the boundary of the control volume,
and no source of heat inside it, the rate of ow of heat
out of the control volume by conduction must equal the rate
of decrease of internal energy (constant volume) or enthalpy
(constant pressure) within it
div q = ,c @T
@t ;
where is the density of the conducting medium, c its specic heat
(both are assumed constant) and T is the temperature.
17

the heat conduction equation
to solve for the temperature eld another equation is required,
linking q to the temperature gradient. This is
q = ,kgrad T;
where k is the thermal conductivity of the medium. Combining
the two equations gives
heat conduction equation:
,divq = kdiv grad T = kr 2 T = c @T
@t
where it has been assumed that k is a constant. In steady ow the
temperature eld satises Laplace's Equation r 2 T =0.
18

Conservative elds of force
A conservative eld of force dened as one for which the work
done in moving between two points in the eld is independent of
the path taken.
the work done is:
Z B
A
F dl;
where A and B are the initial and nal points and F(r) is the force
eld.
Since any two dierent paths between A and B together make up
a closed curve C, we see that a conservative eld must satisfy the
condition
using Stokes' theorem:
I
Z
C
S
F dl =0;
curlF dS =0;
where S is any surface bounded by C.
The only way this equation can be satised for any C is for the
integrand to be zero at every point:
So: curlF =0: aconservative eld of force must be irrotational.
the only way is for the force eld to be the gradient of a scalar
eld U, called the potential function, i.e F = ,rU.
19

Inverse Square Law of force
One reason why there are lots of inverse square laws in nature
may be because it is the only central force eld which has zero
divergence:
If F = f(r)r have zero divergence,
Since div F =0we conclude
divF =3f(r) +rf 0 (r):
r df
dr +3f = 0
df
f +3dr = 0
r
Integrating wrt r gives fr 3 = const = A, so that
F = Ar
r 3 ; jFj = A r 2:
note, by symmetry, that any central eld of force is also irrotational.
20

down the plughole...
div F = 0 applies everywhere except at the origin, where the
divergence is innite.
illustration: calculate the outward normal ux out of a sphere of
radius R centered on the origin. This is clearly equal to 4R 2 F =
4A , i.e. a constant, independent of radius. Since this ux must
be equal to
Z R
0 divF 4r2 dr;
so: no contribution to integral from any point except the origin,
and by letting R tend to zero we see that divF must be innite at
the origin.
21

Gravitational eld due to distributed mass
Poisson's Equation
gravitational eld: A = Gm, where m is the mass at the origin
and G the universal gravitational constant.
mass in volume element dV is dV , giving ,4GdV to the ux
integral from the control volume
Z
S F dS = ,4GZ dV: V
Gauss' Theorem gives:
Z
div FdV = ,4GZ dV
V V
true for any V ,so
,div F =4G:
Since gravitational eld is also conservative (i.e. irrotational) it
must have an associated potential function U, so that F = gradU.
So:
Poisson's equation:
r 2 U =4G
22

gravitational eld inside a spherical body
recall that:
Z
S F dS = ,4GZ V dV:
can calculate the gravitational eld inside a spherical body whose
density is a function of radius only.
where F = jFj, or
4R 2 F =4G Z R
0 4r2 dr;
jF j = G R 2 Z R
0 4r2 dr = MG
R 2 ;
where M is the total mass inside radius R. For the case of uniform
density, this is equal to M = 4 3 R3 and jF j = 4 3 GR.
23

Pressure forces in non-uniform ows
a body immersed in a ow experiences a net pressure force
Z
F p = , pdS; S
where S is the surface of the body. If the pressure p is non-uniform,
this integral is not zero.
transforming using Gauss' Theorem:
F p = , Z V
gradp dV;
In the simple hydrostatic case p + gz = constant:
gradp = ,gk
and the net pressure force is simply
F p = gk
Z
V dV
which, in agreement with Archimedes' principle, is equal to the
weight of uid displaced.
24