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Lecture 33
Classical Integration Theorems in the Plane
In this section, we present two very important results on integration over closed curves in the plane,
namely, Green’s Theorem and the Divergence Theorem, as a prelude to their important counterparts
in R3 involving surface integrals.
Green’s Theorem in the Plane
(Relevant section from Stewart, Calculus, Early Transcendentals, Sixth Edition: 16.4)
Let F(x, y) = F1 (x, y) i + F2 (x, y) j be a vector field in R2 . Let C be a simple closed (piecewise C 1 )
curve in R2 that encloses a simply connected (i.e., “no holes”) region D ⊂ R. Also assume that the
∂F2
∂F1
partial derivatives
and
exist at all points (x, y) ∈ D. Then,
∂x
∂y
Z Z I
∂F2 ∂F1
−
dA,
(1)
F · dr =
∂x
∂y
D
C
where the line integration is performed over C in a counterclockwise direction, with D lying to the
left of the path.
Note:
1. The integral on the left is a line integral – the circulation of the vector field F over the closed
curve C.
2. The integral on the right is a double integral over the region D enclosed by C.
Special case: If
∂F1
∂F2
=
,
∂x
∂y
(2)
for all points (x, y) ∈ D,
I then the integrand in the double integral of Eq. (1) is zero, implying that
F · dr is zero.
the circulation integral
C
We’ve seen this situation before: Recall that the above equality condition for the partial derivatives
is the condition for F to be conservative. From this, we would suspect that the circulation integral
over a closed curve C would be zero, because of the Generalized Fundamental Theorem of Calculus
247
(the endpoints are the same point). But also recall that we have to be a bit careful – we must be
concerned about the possibility of “interference” from singularities that might make the circulation
integral nonzero. In this theorem, however, we have assumed that there are no such singularities – the
derivatives are assumed to exist and F is therefore defined over the region. Therefore, we may finally
conclude that, yes, the line integral is zero.
In fact, the integrand in Green’s Theorem, Eq. (1) is the k component of the curl of F. To see
this, let us compute it:
i
j
k
~ × F = ∂/∂x
curl F = ∇
∂/∂y
∂/∂z
F1 (x, y) F2 (x, y)
0
∂F2 ∂F1
= 0i + 0j +
−
k.
∂x
∂y
It is important to keep in mind that that the curl of a vector field in R2 is a vector that points
in the z-direction. This is related to the convention of assigning a velocity vector that points along an
axis of rotation using the right-hand rule. With the above result, we can rewrite Green’s Theorem as
Z Z
I
[curl F]z dA,
(3)
F · dr =
D
C
where [v]z denotes the z-component of vector v. In this case, the z-component of curl F is its only
component. For this reason, we shall omit the subscript z in this section.
Examples:
1. The vector field F = −ωyi + ωxj. You will recall that this is the velocity vector field of a thin
dθ
= ω. Now let CR
plate on the xy-plane that is rotating about the z-axis with angular speed
dt
denote the circle of radius R > 0 centered at the origin. Earlier in the course, we computed the
circulation integral of F over CR to be
I
F · dr = 2πωR2 .
(4)
This was done by a direct calculation of the line integral using the parametrization of CR as
r(t) = (R cos t, R sin t). Let us now compute this result using Green’s Theorem. Here F1 = −ωy
and F2 = ωx so that
∂F2 ∂F1
−
= ω − (−ω) = 2ω.
∂x
∂y
248
(5)
Then the double integral in Green’s Theorem is
Z Z Z Z
∂F2 ∂F1
−
dA = 2ωA(D) = 2ωπR2 ,
dA = 2ω
∂x
∂y
D
D
(6)
where A(D) denotes the area of region D (area of a circular region, radius R).
It is sometimes misleading to present this example because people may associate the curl of F,
which is 2ω, with the origin, about which the rotation is taking place. But the curl of this vector
field is 2ω everywhere. This means that for any simple closed curve in the plane,
I
F · dr = 2ωA(D),
(7)
C
where A(D) denotes the area of the region D enclosed by C.
1. Compute the circulation of the vector field F = −y 2 i + xj around the circle of radius 1 centered
at the origin. Here, F1 = −y 2 and F2 = x. Then
Z Z
Z Z ∂F2 ∂F1
−
(1 + 2y) dA
dA =
∂x
∂y
D
D
Z Z
Z Z
y dA
dA + 2
=
D
(8)
D
= π.
The first integral is the area of the region enclosed by the unit circle. The second integral is
zero. One may confirm this result by computing the integral explicitly, either with Cartesian or
planar polar coordinates. One may also conclude that it is zero because the function y is an odd
function – y > 0 over the region above the x-axis and y < 0 over the region below the x-axis
which is a mirror image of the region above.
Another way of deducing that the integral is zero is to note that it is related to the center of
mass of region D when the density is constant, i.e., the centroid:
Z Z
1
ȳ =
y dA.
A(D)
D
(9)
But by symmetry, ȳ = 0. Therefore the integral is zero.
You can also confirm the above result by explicitly computing the circulation integral using the
parametrization r(t) = (cos t, sin t).
2. The vector field
F=−
x2
y
x
i+ 2
j.
2
+y
x + y2
249
(10)
You may recall that this vector field described, up to a constant, the magnetic field around a
thin current-carrying wire of infinite length lying on the z-axis. Also recall that
~ × F = 0,
curl F = ∇
(x, y) 6= 0.
(11)
Because the curl is not defined at (0,0), we can use Green’s Theorem only for simple closed
curves C that do not enclose the origin (0,0). (Recall that one of the assumptions in Green’s
Theorem was that the partial derivatives existed at all points (x, y) ∈ D, the region enclosed by
C.) In this case, all circulation integrals are zero:
I
F · dr = 0.
(12)
C
We cannot use Green’s Theorem to conclude anything about circulation integrals over closed
curves C that enclose the origin. In this case, we computed earlier in this course that the
circulation integral over a circle CR of radius R centered at the origin is 2π. From this result,
one can actually conclude that the circulation integral over any simple curve C enclosing the
origin is 2π.
3. Some additional special cases: The vector fields
F = −yi,
F = xj,
1
F = (−yi + xj).
2
(13)
In all of these cases,
∂F2 ∂F1
−
= 1.
∂x
∂y
(14)
For these vector fields
I
F · dr =
C
Z Z
dA = A(D), the area of D.
(15)
D
In fact, there are many, many other vector fields F = F1 i + F2 j for which Eq. (14) is satisfied.
Can you find any? Can you find a general family of such fields?
250
Physical interpretation of the curl in terms of the circulation integral
You may recall that when the curl and divergence operators were introduced some time ago, a physical
interpretation of the divergence operator could be provided in terms of the total outward flow of a
vector field from a tiny box of dimensions ∆x and ∆y centered at a point (x, y). A “box-based”
interpretation of the curl operator could not be made at that time because it would have required the
notion of the circulation integral which, of course, was not developed until the previous lecture.
Now that we know about the circulation integral, we can obtain an interpretation of the curl
operator. To start, we consider the rectangular region ABCD centered at (x, y) with dimensions ∆x
and ∆y, as sketched below. We shall estimate the circulation of a planar vector field
F(x, y) = F1 (x, y) i + F2 (x, y) j
(16)
along the closed curve C composed of the four line segments C1 , C2 , C3 and C4 , identified in the
figure.
y + ∆y/2
D
C3
C4
P (x, y)
y
C
C2
y − ∆y/2
A
C1
x − ∆x/2
x
B
x + ∆x/2
From the additivity property of integrals, the circulation integral of F will be given by
Z
Z
Z
Z
I
F · dr.
F · dr +
F · dr +
F · dr +
F · dr =
C
C1
C3
C2
Recall that each of these integrals is computed by means of a parametrization of the form
Z
Z b
F(r(t)) · r′ (t) dt,
F · dr =
Ci
(17)
C4
(18)
a
where the curve Ci is parametrized as r(t) = (x(t), y(t)), a ≤ t ≤ b. The vector r′ (t) = (x′ (t), y ′ (t)) is
the tangent vector to the curve Ci . We now consider the line integral over each of these curves.
251
1. Curve C1 : We may parametrize this curve as
x(t) = x −
∆x
+ t,
2
y(t) = y −
∆y
,
2
0 ≤ t ≤ ∆x,
(19)
implying that the tangent vector is
r′ (t) = (x′ (t), y ′ (t)) = (1, 0).
(20)
As such, the integrand in (18) will be given by
∆y ∼
∆y
′
F(r(t)) · r (t) = F1 x(t), y −
.
= F1 x, y −
2
2
(21)
(The contribution from F2 is zero.) Note that we are approximating this term by setting x(t)
to be constant, namely the value of x at P . Since this approximate integrand is a constant, the
line integral will be the product of this constant with the length of the curve, ∆x. The result is
Z
F · dr ∼
= F1 (x, y −
C1
∆y
)∆x.
2
(22)
∆y
,
2
(23)
2. Curve C3 : We may parametrize this curve as
x(t) = x +
∆x
− t,
2
y(t) = y +
0 ≤ t ≤ ∆x.
implying that the tangent vector is
(x′ (t), y ′ (t)) = (−1, 0).
(24)
The integrand in (18) will be given by
∆y ∼
∆y
F(r(t)) · r′ (t) = −F1 x(t), y +
−F
.
x,
y
+
=
1
2
2
(25)
(The contribution from F2 is again zero.) Once again, since we have approximated the integrand
by a constant, the line integral will simply be the product of this constant with the length of
the curve, ∆x:
Z
∆y
∆x.
F · dr ∼
−F
x,
y
+
=
1
2
C3
Let us now add up the contributions from curves C1 and C3 :
Z
Z
∆y
∆y
∼
F · dr = −F1 x, y +
F · dr +
+ F1 x, y −
∆x.
2
2
C3
C1
252
(26)
(27)
And now perform a few manipulations on the RHS:
[−F1 (x, y + ∆y/2) + F1 (x, y − ∆y/2)] ∆x = − [F1 (x, y + ∆y/2) − F1 (x, y − ∆y/2)] ∆x
F1 (x, y + ∆y/2) − F1 (x, y − ∆y/2)
∆x∆y
= −
∆y
∂F1
∼
∆x∆y.
(28)
= −
∂y
We must now consider the contributions from curves C2 and C4 . The analysis proceeds in a manner
similar to the above. The main difference is that the tangent vectors to C2 and C4 are (0, 1) and
(0, −1), respectively, which implies that the contributions to the line integrals come from the second
component of F, i.e., F2 (x, y). The result is as follows,
Z
F · dr +
Z
F · dr ∼
=
C4
C2
∂F2
∆x∆y.
∂x
(29)
Recalling Eq. (17), we now add up the contributions from curves C1 to C4 to obtain the result,
I
∂F2 ∂F1
∼
F · dr =
−
∆x∆y.
(30)
∂x
∂y
C
In other words, the circulation of the vector field around the rectangular curve C is approximated by
the z-component of the curl of F multiplied by the area ∆A = ∆x∆y of the box. If we divide by this
area and take the limit, then
1
lim
∆A→0 ∆A
I
F · dr =
C
∂F2 ∂F1
~ × F]z .
−
= [∇
∂x
∂y
(31)
The curl at (x, y) is the limiting circulation of the field per unit area. We’ll show this in another way
at the end of this section.
253
The Divergence Theorem in the Plane
(Relevant section from Stewart, Calculus, Early Transcendentals, Sixth Edition: 16.5, p. 1067)
Note: Unfortunately, the discussion of the Divergence Theorem in the Plane in Stewart’s text is
quite minimal, being presented on p. 1067 only as a consequence of Green’s Theorem. It is even more
unfortunate that the physical interpretation of this result, i.e., as measuring the total outward flux
of a vector field through a closed curve C, is missing. For this reason, these lecture notes will have
to serve as the primary source of information. You are invited, of course, to consult other textbooks
on Vector Calculus, e.g., M. Lovric, Vector Calculus, or R.A. Adams, Calculus, Several Variables. (In
past years, these books were used as texts for this course.)
Let F(x, y) = F1 (x, y) i + F2 (x, y) j be a vector field in the plane. Let C be a simple closed
(piecewise C 1 ) curve that encloses a region D. Let N̂ denote the unit outward normal to C assumed
to exist at all points on the curve (except perhaps at a finite set of “corners”). Furthermore, assume
that the divergence of F is defined for all points in D, i.e.,
~ · F(x, y) = ∂F1 (x, y) + ∂F2 (x, y)
div F(x, y) = ∇
∂x
∂y
(32)
is defined for all (x, y) ∈ D.
The Divergence Theorem in the Plane then states that
Z Z
I
div F dA.
F · N̂ ds =
(33)
D
C
The left integral is a line integral around the curve C – it measures the net outward flux of the vector
field F through the closed curve C. The right integral is a double integration over the region D
enclosed by C.
Examples:
1. The vector field F = K(xi + yj). Let CR denote the circle of radius R centered at the origin
(0, 0). During the lectures on line integrals, we computed the net outward flux of this vector
field through CR as a line integral to obtain the result
I
F · N̂ ds = 2πKR2 .
C
254
(34)
Let us now compute this outward flux using the Divergence Theorem. The divergence of this
vector field is simply the value 2K:
∂F1 ∂F2
+
= K + K = 2K.
∂x
∂y
Therefore, by the Divergence Theorem,
Z Z
I
2K dA = 2KA(D) = 2K · πR2 ,
F · N̂ ds =
(35)
(36)
C
C
in agreement with our earlier result. Note that because of the constancy of the divergence in
this case, the circle CR could be placed anywhere in the plane and the net outward flux would
be the same. For a general simple closed curve C,
I
F · N̂ ds = 2KA(D),
(37)
C
where A(D) denotes the area of the region D enclosed by C.
2. The vector field F = x2 i + yj. Let C be the perimeter of the unit square in the first octant with
vertices at (0, 0), (1, 0), (1, 1) and (0, 1). In this case
div F = 2x + 1,
(38)
which is defined at all points in the plane. From the Divergence Theorem,
Z Z
I
(1 + 2x) dA
F · N̂ ds =
C
Z Z
Z ZD
x dA
dA + 2
=
D
(39)
D
1
= 1+2·
2
= 2.
x
y
i+ 2
j, which has been examined a number of times in this
x2 + y 2
x + y2
course. It is, up to a constant, the electrostatic field due to an infinite wire that lies along the
3. The vector field F =
z-axis. You will recall (or can verify for yourself) that
div F = 0
for (x, y) 6= (0, 0).
We may use the Divergence Theorem conclude that
I
F · N̂ ds = 0
C
255
(40)
(41)
for any simple closed curve C that does not contain or enclose the origin (0, 0).
We cannot use the Divergence Theorem to determine the outward flux through curves C that
contain the origin. They must be determined by explicit calculation. Actually, it is sufficient
only to compute the outward flux through a circle of radius R, CR , centered at the origin. We
did this earlier, to find that
I
F · N̂ ds = 2π.
(42)
CR
With this knowledge, one can conclude that the total outward flux for any curve C enclosing
the origin is 2π. We’ll return to this argument in a few lectures.
4. Some additional special cases: The vector fields
F = xi,
F = yj,
1
F = (xi + yj).
2
(43)
In all of these cases,
∂F1 ∂F2
+
= 1.
∂x
∂y
(44)
Therefore, by the Divergence Theorem,
I
F · N̂ ds =
C
Z Z
dA = A(D), the area of D.
(45)
D
In fact, there are many, many other vector fields F = F1 i + F2 j for which Eq. (44) is satisfied.
Once again, can you find any, or perhaps a general family of such fields?
256
Extra reading: Interpretation of “Curl” and “Divergence”
The following material was not covered in the lectures, because of lack of time. You are strongly
recommended to read this material in order to obtain a deeper understanding of the curl and divergence
operations.
Interpretation of the two-dimensional curl via Green’s Theorem
We can use Green’s theorem to determine what the curl of a vector field actually measures – recall
that we had some kind of intuitive picture that it measure the rotation of a vector field. Let F(x, y)
be a vector field in the plane. We shall simply drop the z subscript in Eq. (31) of the previous lecture
and write
curl F(x, y) =
∂F2
∂F1
(x, y) −
(x, y).
∂x
∂y
(46)
In what follows we shall assume that the above partial derivatives, hence the curl of F are continuous
over our region of interest D.
Now let (x0 , y0 ) be a point in D and let CR be a circle of radius R << 1 centered at (x0 , y0 ). Let
DR be the circular region contained in CR . By Green’s Theorem,
I
F · dr =
CR
Z Z
curl F dA.
(47)
DR
By definition, if a function f (x, y) is continuous at (x0 , y0 ), then
f (x, y) → f (x0 , y0 ),
as (x, y) → (x0 , y0 ),
(48)
for any direction of approach. This means that if (x, y) is close to (x0 , y0 ), then f (x, y) is close in
value to f (x0 , y0 ). For R very, very small,
f (x, y) ≈ f (x0 , y0 ) for all points (x, y) ∈ DR .
(49)
Let us now apply this result to f (x, y) = curl F. It means that
Z Z
curl F(x, y) dA ≈ curl F(x0 , y0 )
DR
Z Z
dA = curl F(x0 , y0 ) A(DR ),
(50)
DR
where A(DR ) denotes the area of DR . (It is not important to write the area explicitly as πR2 .) Then
from (47),
I
F · dr ≈ curl F(x0 , y0 ) A(DR ),
CR
257
(51)
which may be rearranged to give
curl F(x0 , y0 ) ≈
H
CR
F · dr
A(DR )
In the limit R → 0, this approximation becomes exact, i.e.
H
CR
curl F(x0 , y0 ) = lim
.
(52)
F · dr
A(DR )
R→0
.
(53)
In other words, the curl of the planar vector field F at a point (x0 , y0 ) is the limit of its circulation
per unit area as the area goes to zero. Therefore, we have established that the curl does measure the
circulation of the field.
Interpretation of the two-dimensional divergence
We can use the Divergence Theorem to determine what the divergence of a vector field actually
measures. You may recall that there was the idea of “net outward flow” of the vector field at a point.
Let F(x, y) be a vector field in the plane. In what follows we shall assume that the divergence of F is
continuous over a region of interest D.
Now let (x0 , y0 ) be a point in D and let CR be a circle of radius R << 1 centered at (x0 , y0 ). Let
DR be the circular region contained in CR . From the Divergence Theorem,
I
F · N̂ ds =
CR
Z Z
~ · F dA.
∇
(54)
DR
Using the same argument as we did for Green’s Theorem and the curl of F, for R very, very small,
~ · F(x, y) ≈ ∇
~ · F(x0 , y0 ) for all points (x, y) ∈ DR .
∇
(55)
Applying this result to the Divergence Theorem,
Z Z
~ · F(x, y) dA ≈ ∇
~ · F(x0 , y0 )
∇
Z Z
~ · F(x0 , y0 ) A(DR ),
dA = ∇
(56)
DR
DR
where A(DR ) denotes the area of DR . (It is not important to write the area explicitly as πR2 .) Then
from (54),
I
~ · F(x0 , y0 ) A(DR ),
F · N̂ ds ≈ ∇
(57)
H
(58)
CR
which may be rearranged to give
~ · F(x0 , y0 ) ≈
∇
258
CR
F · N̂ ds
A(DR )
.
In the limit R → 0, this approximation becomes exact, i.e.
H
F · N̂ ds
~ · F(x0 , y0 ) = lim CR
∇
.
R→0
A(DR )
(59)
In other words, the divergence of the planar vector field F at a point (x0 , y0 ) is the limit of its net
outward flux per unit area as the area goes to zero. Therefore, we have established that the divergence
does measure the limiting net outward flow of the field at a point.
259
Lecture 34
Surface integrals in R3
We come to the final major section of the course. In what follows, we are interested in the integration
of functions over surfaces in R3 . Recall that a surface S ⊂ R3 is a two-dimensional set. As such,
we would expect that the integration of a function over a surface would require two independent
coordinates or parameters, as opposed to one parameter that is needed to integrate over a curve C.
We’ll denote these coordinates as (u, v). The parametrization would then take the following general
form: A point P on the surface S would have the coordinates
r(u, v) = (x(u, v), y(u, v), z(u, v)),
(u, v) ∈ D ⊂ R2 .
(60)
Here, D is the set of parameter values (u, v) that are needed to define the surface S, i.e., to access all
points P on S. We’ll illustrate this idea with some specific examples below.
As in the case of line integrals, there are two major types of integrations that can be performed
over surface integrals.
1. Integration of scalar functions f (x, y, z) over S
Suppose that a scalar-valued function f : R3 → R is
defined at all points of a surface S. Now consider an
infinitesimal element of surface dS centered at a point
(x, y, z) ∈ S and form the product f (x, y, z)dS. Then
sum up, i.e., integrate, over all elements dS to produce
the surface integral,
“
Z
f dS ”.
S
Examples:
1. If f (x, y, z) = 1, then
R
S
f dS =
R
S
dS is the area of S.
2. If f (x, y, z) is the charge density (per unit area), then dq = f dS is the amount of charge in
260
element dS.
total charge Q =
Z
dq =
S
Z
f dS
S
If f is the mass density (per unit area) then dm = f dS, is the amount of charge in element dS,
total mass M =
Z
dm =
S
Z
f dS
S
.
In this very brief section, we shall consider only two types of surfaces, namely, planes and spherical
surfaces. Along with cylindrical surfaces, these are the most commonly employed surfaces in Physics.
(Planes are used to construct cubes.)
Spherical surfaces
A sphere SR of radius R can be defined in terms of the two angular spherical coordinates θ and φ.
The radial variable r is fixed, with r = R. To generate the entire sphere, the parameter space D is
given by
D = {(θ, φ) | θ ∈ [0, 2π], φ ∈ [0, π]}.
(61)
Recall that the infinitesimal element of volume dV in spherical coordinates is given by
dV = r 2 sin φ dr dθ dφ.
(62)
Since r is now fixed, the infinitesimal element of surface area on the sphere is given by
dS = R2 sin φ dr dθ dφ.
(63)
To illustrate, let us compute the surface area of the sphere SR of radius R:
Z
dS =
SR
Z
π
0
= R
2
Z
Z
2π
0
π
R2 sin φ dθ dφ
Z
2π
sin φ dθ dφ
Z 2π Z π
2
dθ
sin φ dφ
= R
0
0
0
0
2
= R (2) (2π)
= 4πR2 .
261
(64)
The above calculation was a simple example of an integration of a scalar function over a surface:
In this case, f = 1. We may easily extend this idea to integrate general functions of the angular
variables θ and φ, i.e.,
Z
f (θ, φ) dS.
(65)
SR
For example, suppose that you wanted to compute the center of mass of a hemispherical and homogeneous shell of radius R. It is, of course, advantageous to position this surface so that its center is
(0, 0, 0) and its circular base sits on the xy-plane. In this case, our surface S is the upper hemisphere of
the surface SR discussed above. And because it is assumed to be homogeneous, i.e., constant density,
the center of mass is the same as the centroid of the surface. Due to the symmetry of the surface with
respect to rotation about the z-axis, the centroid will be situated on the z-axis, i.e., at point (0, 0, z̄).
The coordinate z̄ will be given by
R
z dS
z̄ = RS
.
S dS
(66)
The denominator of this expression is trivially one-half the area of SR , i.e., 2πR2 . In order to compute
the numerator, we shall have to convert the integrand z into spherical coordinates. Since z must lie
on the surface, we have z = R cos φ, so that
1
z̄ =
2πR2
Z
R cos φ dS.
(67)
S
The computation of this integral is left as an exercise. The final result is
z̄ =
R
.
2
(68)
(Just to remind you: in the case of a solid, homogeneous hemisphere, the z coordinate of the cen3
troid/center of mass was found to be z̄ = R.)
8
Because of time limitations, this is all that we can discuss about the integration of scalar-valued
functions on surfaces. That being said, it should be sufficient for your needs. The above discussion gives
a good idea of how to integrate functions on spherical surfaces. And the integration of scalar-valued
functions on planar surfaces is rather straightforward – after all, they are simply double integrals.
We now move on to the very important idea of surface integrals involving vector fields.
262
2. Integration of vector-valued functions F(x, y, z) over surfaces: “Flux integrals”
This is the major topic of the remainder of the course.
At each infinitesimal surface element dS centered at a
point P on S, compute F · N̂ where N̂ is the outward
unit normal to S, i.e., it is perpendicular to the
tangent plane to S at P . Now integrate over all
elements dS comprising the surface S: The result is
“
Z
F · N̂ dS ”,
S
the total outward flux of F through S.
In what follows, we shall develop a method to perform integrations over some simple surfaces by
using rather straightforward parametrizations of surfaces, much as we did for integrations over curves.
By “simple surfaces”, we mean planes and spheres, which are easily parametrized. But we first need
to understand the concept of “flux”.
A closer examination of the idea of “flux” in terms of fluid flow
It is perhaps easiest to visualize the idea of flux with reference to fluid flow. First, consider a region
D that lies in the xy-plane as sketched below. Suppose that a fluid is passing through this region.
For the moment, we assume that motion of the fluid is perpendicular to region D, travelling in the
direction of the positive z-axis. Moreover, we assume that the speed of the fluid particles crossing D
is constant throughout the region. As such, we are assuming that the velocity field of the fluid is given
by
v = vk,
v > 0 (constant) .
(69)
z
F = vk
y
D
x
263
We first ask the question: How much fluid flows through region D during a time interval ∆t?
Consider a tiny rectangular element of area ∆A = ∆x∆y centered at a point (x, y) in D. After a time
∆t, the fluid particles originally situated in this element will have moved a distance v∆t upward. The
volume of fluid that has passed through this element ∆A on D is the volume of the box of base area
∆A and height v∆t:
v∆t∆A.
(70)
This box is sketched below.
z
F = vk
v∆t
x
y
D
∆A
The total volume ∆V of fluid that has passed through region D over the time interval ∆t is
obtained by summing up over all area elements ∆A in D. We let ∆A → dA and integrate over D to
obtain
∆V = v∆t
Z Z
dA = v∆tA(D),
(71)
D
where A(D) denotes the area of D. Of course, this is a rather trivial result: the volume of fluid passing
through D is simply the volume of the solid of base area A(D) and height v∆t. Dividing both sides
by ∆t, we have
∆V
= vA(D).
∆t
(72)
In the limit ∆t → 0, we have the instantaneous rate of volume of fluid passing through region D per
unit time, or simply the rate of fluid flow through region D:
V ′ (t) = vA(D).
(73)
It is useful to express this result in terms of the original vector field v = vk. The result is quite simple
because the vector v points in the same direction as the unit normal vector N̂ = k of the region D,
i.e.,
v = v · k = v · N̂.
264
(74)
As such, we can express Eq. (73) in the form
′
V (t) =
Z Z
v · N̂ dA.
(75)
D
This quantity is the total flux of the vector field v through region D.
Now suppose that the fluid is now moving at a constant speed v through region D but not
necessarily at right angles to it, i.e., not necessarily parallel to its normal vector k. We shall suppose
that
v = v1 i + v2 j + v3 k,
k v k= v
(76)
and let γ denote the angle between v and the normal vector k.
In this case, the fluid particles that pass through the tiny element ∆A after a time interval ∆t
form a parallelopiped of base area ∆A and height v cos γ∆t, as sketched below.
z
F
y
γ
v cos γ
x
∆A
The volume of this box is
v cos γ∆t∆A = v · k∆t∆A = v3 ∆t∆A.
(77)
(Think of this tower of fluid as a deck of playing cards that has been somewhat sheared. When you
slide the cards back to form a rectangular arrangement, the height of the deck is v∆t cos γ.) In other
words, only the vertical component v3 k of the velocity contributes to the flow across the region D.
The total volume ∆V of fluid that has passed through region D over the time interval ∆t is
obtained once again by letting ∆A → dA and integrating over D:
Z Z
dA = v3 ∆tA(D),
∆V = v3 ∆t
(78)
D
Dividing both sides by ∆t, we have
∆V
= v3 A(D).
∆t
265
(79)
In the limit ∆t → 0, we obtain the flux of the vector field v through region D:
V ′ (t) = v3 A(D).
(80)
But recall that v3 = v ·k = N̂, where N̂ once again denotes the unit normal vector to D in the positive
z-direction. We shall rewrite this flux as follows,
′
V (t) = v · N̂ A(D) =
Z Z
v · N̂ dA.
(81)
D
Note that this general case includes the first case, where γ = 0, implying that v1 = v2 = 0 and v3 = v.
Note also that in the case γ = π/2, i.e., v3 = 0, there is no flow through the region D, so the flux is
zero.
A slight generalization of the above – nonconstant velocity field: Of course, the above results
have been rather trivially obtained since (i) the vector fields are constant and (ii) the region D is flat.
Let us now generalize the first case, i.e., the vector field v is assumed to be nonconstant over the
planar region D ⊂ R2 , i.e.,
v(x, y) = v1 (x, y)i + v2 (x, y)j + v3 (x, y)k.
(82)
Once again, the unit normal vector in the positive z-direction is N̂ = k. In this case, the total volume
∆V of fluid that has passed through region D over the time interval ∆t is obtained by summing up
over all area elements ∆A in D:
∆V = ∆t
Z Z
v(x, y) · N̂ dA = ∆t
Z Z
v3 (x, y) dA.
(83)
D
D
Dividing by ∆t and taking the limit ∆t → 0, we obtain
Z Z
′
v(x, y) · N̂ dA,
V (t) =
(84)
D
which is once again the total flux of the (nonconstant) vector field v(x, y) through region D. Note
that the previous two results, Eqs. (75) and (81) are special cases of Eq. (84).
Now suppose that we were concerned with rate of mass flow through region D. The amounts/volumes
of fluid examined earlier would be replaced by amounts of mass flowing through a surface element.
This means replacing the velocity vector field v by the momentum field F = ρ v, where ρ is the mass
density. The rate of transport of mass through region D would then be given by
Z Z
Z Z
′
ρ(x, y) v(x, y) · N̂ dA
F(x, y) · N̂ dA =
M (t) =
D
D
266
(85)
This concludes our discussion of this simple problem involving fluid flow through a flat surface.
Generalization to arbitrary surfaces
We now wish to generalize the above result to general surfaces in R3 . In other words, we do not
require the surface to be flat, as was region D in the plane, but rather a general surface S in R3 –
for example, a portion of a sphere, or perhaps the entire sphere. In the “spirit of calculus,” we divide
the surface S into tiny infinitesimal pieces dS. We then construct a normal vector N̂ to each surface
element dS at a point in dS, as sketched below.
We then form the dot product of the vector field F at that point with the normal vector N̂. This
will represent the local flux of F through the surface element dS. To obtain the total flux through the
surface S, we add up the fluxes of all elements dS – an integration over S that is denoted as
Z
F · N̂ dS.
(86)
S
This is the total “flux” of F through surface S. If F = v the velocity vector field of a fluid moving
in R3 , then the total flux would be the (instantaneous) rate of volume of fluid per unit time passing
through surface S.
Note that in some books, especially Physics books, the vector surface integral is denoted as
Z Z
F · dS.
(87)
S
Here, the infinitesimal surface area element is a vector that is defined as
dS = N̂ dS,
(88)
where dS is the infinitesimal surface element and N̂ is the unit normal vector to the surface element.
267
In other books, the infinitesimal surface element is denoted as dA = N̂dS, so that the flux integral
is denoted as
Z Z
F · dA.
(89)
S
We now compute a very important flux integral – the outward flux of a point charge – that will
provide the basis for many other important results to follow.
An important flux integral that can be computed in a rather simple manner:
We consider a stationary charge Q situated at the origin (0, 0, 0) of a coordinate system. Also consider
an arbitrary point P with coordinates (x, y, z) and position vector r = xi + yj + zk. As you well know,
the electrostatic field vector E(r) at P due to the presence of Q is given by
E(r) =
Q
r,
4πǫ0 r 3
r = krk.
(90)
The situation is pictured below.
If Q > 0, then the vector field E points outward; If Q < 0, then it points inward. In the above sketch,
without loss of generality, we consider the case Q > 0.
We now wish to compute the total outward flux of E through the spherical surface SR of radius
R > 0, i.e., the surface integral
Z Z
E · N̂ dS,
(91)
SR
where N̂ denotes the unit outward normal vector to SR at a point. The vector field E and sphere SR
are sketched below.
It is necessary to compute the integrand E · N̂ in the above surface integral. At each point P on
SR , we have r = krk = R, so that the field E(r) is given by
E(r) =
Q
r.
4πǫ0 R3
268
(92)
And at point P , the outward unit normal N̂ is given by
N̂ = r̂ =
rr
.
rR
(93)
Therefore, at point P , the integrand of the surface integral becomes
E · N̂ =
=
=
Q 1
r
r·
3
4πǫ0 R
R
Q 1
· R2
4πǫ0 R4
Q
.
4πǫ0 R2
This is the “flux” through the infinitesimal element dS at P . Note that it is a constant over the sphere
SR . To obtain the total flux over S, we must integrate over the entire surface SR .
Z Z
E · N̂ dS =
Z Z
SR
SR
=
Q
dS
4πǫ0 R2
Z Z
Q
4πǫ0 R2
|
dS
{z }
SR
surf ace area of SR =4πR2
=
=
Q
· 4πR2
4πǫ0 R2
Q
.
ǫ0
(The fact that the surface area of SR is 4πR2 was established in the previous lecture by integration
using spherical polar coordinates.)
Of course, this result is no suprise to you – you have seen it in your Electricity and Magnetism
course. It’s Gauss’ Law – the total flux of the electric field through a surface is equal to the amount
of charge Q contained inside the surface. But the point is that you have been simply told this
269
result in your E&M course - it is the goal of this course to derive this important result.
For the moment, the above result is all that we can assume in this course.
The above result is somewhat general, since the outward flux is INDEPENDENT of the radius R
of the sphere SR . But this is still a far cry from stating that Gauss’ Law holds for arbitrary surfaces.
Beofre considering another example, we’ll state that the above result – as limited as it may seem
1
1
– is due to the fact that the electric field vector is given by 3 r and not 3+a r, where a is some small
r
r
(or even large!) nonzero number.
Exercise: Compute the total flux of the electric field vector through a sphere of radius SR in the case
that the field is given by
E(r) =
Q
r.
4πǫ0 krk3+a
270
(94)
Lecture 35
Surface integrals in R3 (cont’d)
“Surface flux integrals” of vector fields (cont’d)
OK, so we showed Gauss’ Law for spheres in the previous lecture. Before going to the general case,
let’s check if it applies to another class of “easy” surfaces that we can integrate, i.e., cubes. Consider
the following construction. We shall put six planes together to form a cubic surface with unit side
lengths and centered at (0, 0, 0).
Once again, we wish to compute the total flux
E(r) =
Z Z
E · N̂ dS for the vector field
S
K
r,
r3
K=
Q
.
4πǫ0
(95)
We begin with the top surface, which we shall call S1 :
Surface S1 is comprised of the points (x, y, 12 ), − 12 ≤ y ≤ 21 . At each of these points, the electrostatic field vector due to point charge Q at (0, 0, 0) is (up to constant K)
E(r) =
(x2
+
y2
K
1
1 2 3/2 (xi + yj + 2 k)
+ (2) )
(96)
It should be clear that the normal vectors N̂ to this surface are the unit vectors ±k. We choose the
unit outward normal N̂ = k.
271
The integrand in the surface vector integral will therefore be
E · N̂ =
=
1
K
(x, y, 1/2) · (0, 0, 1)
2
2
2 (x + y + 14 )3/2
1
K
.
2
2
2 (x + y + 14 )3/2
(97)
To obtain the total outward flux through S1 , we integrate over S1 :
Z Z
E · N̂ dS =
S1
K
2
Z
1
2
− 12
Z
1
2
− 12
1
(x2
+
y2
+ 14 )3/2
dx dy
.
(use trig substitution)
= ..
K 4
=
π
2 3
2
Kπ.
=
3
Recall that K =
Q
, we have the final result that
4πǫ0
Z Z
2
Q
Q
· π=
.
E · N̂ dS =
4πǫ
3
6ǫ
0
0
S1
(98)
The contributions from each of the other five surfaces S2 · · · S6 will be identical to this result by
symmetry. The final result is that the total flux of the vector field through the cubic surface will be
Z Z
E · N̂ dS = 6 ·
S
Q
Q
= ,
6ǫ0
ǫ0
(99)
which agrees with Gauss’ Law.
That was indeed a good deal of work, and still for a not-so-complicated surface. In order to obtain
Gauss’ Law for arbitrary surfaces, we’ll have to make use of another powerful result attributed to –
guess who? – Gauss.
Gauss Divergence Theorem in R3
In this section, we are concerned with the outward flux of a vector field (through a smooth/piece-wise
smooth) surface S that encloses a region V ⊂ R3 . Typically, in physics such surfaces are spheres,
boxes, parallelpipeds or cylinders. This is the subject of the celebrated Gauss Divergence Theorem,
the three-dimensional version of the Divergence Theorem in the Plane of a previous lecture.
272
The Gauss Divergence Theorem is one of the most important results of vector calculus. It is not
as important for computational purposes as for conceptual developments. It provides the basis for the
important equations in electromagnetism (Maxwell’s equations), fluid mechanics (continuity equation)
and continuum mechanics in general (heat equation, diffusion equation).
It is sufficient to consider a somewhat simplified version of the general Divergence Theorem.
A simplified version of the Divergence Theorem:
Let S be a “nice” (i.e., piecewise smooth) closed and nonintersecting surface that encloses a region
D ⊂ R3 , such that an outward unit normal vector N̂ exists at all points on S. Also assume that a
vector field F and its derivatives are defined over region D and its boundary S.
The Divergence Theorem states that:
Z Z
|
S
Z Z Z
F · N̂ dS =
{z
} |
surf ace integral
div F dV .
{z
}
(100)
D
volume integral
Once again, we have assumed that div F exists at all points in V .
You have already seen a version of this theorem – the two-dimensional Divergence Theorem in
the plane. It expressed the total outward flux of a 2D vector field F through a closed curve C in the
plane as an integral of the divergence of F over the region D enclosed by C:
I
F · N̂ ds =
Z Z
div F dA.
(101)
D
C
Examples: In what follows, unless otherwise indicated, the surface S is an arbitrary surface in R3
satisfying the conditions of the Divergence Theorem.
1. The vector field F = k = (0, 0, 1). This vector field could be viewed as the velocity field of a
fluid that is travelling with constant speed in the positive z-direction:
The divergence of this vector field is zero:
div F =
∂
∂
∂
(0) +
(0) +
(1) = 0.
∂x
∂y
∂z
273
(102)
y
z
x
More importantly, it exists at all points in R3 , i.e., there are no singularities, so that we may
employ the Divergence Theorem. Therefore, for any surface S enclosing a region D, we have
Z Z Z
Z Z Z
Z Z
0 dV = 0.
(103)
div F dV =
F · N̂ dS =
D
D
S
In other words, the total outward flux of F over the surface S is zero. In terms of the fluid
analogy, fluid is entering the region through surface S from the bottom at the same rate that it
is leaving it at the top. There is no creation of extra fluid anywhere inside region D that would
cause a nonzero flux.
2. The vector field F = zk = (0, 0, z). A sketch of the vector field is given below.
y
z
x
This field could be viewed as the velocity field of a liquid that originates from the xy-plane and
travels upward and downward away from it. As it moves away, it accelerates, since the velocity
is proportional to the distance from the xy-plane.
The divergence of this field is
div F =
∂
∂
∂
(0) +
(0) +
(z) = 1.
∂x
∂y
∂z
(104)
Once again, the divergence exists at all points in R3 . Therefore, by the Divergence Theorem
Z Z Z
Z Z Z
Z Z
1 dV = V (D),
(105)
div F dV =
F · N̂ dS =
SR
D
D
274
the volume of region D.
Note that the same result for the flux, i.e., Eq. (105), would be obtained for the following vector
fields:
(i)
F = xi,
(ii)
F = yj,
(106)
since the divergence of each of these vector fields is 1. And the list does not stop here. Consider
the set of all vector fields of the form
F = c1 xi + c2 yj + c3 zk,
where c1 + c2 + c3 = 1.
(107)
In all cases, we have div F = 1, so that the total outward flux of each of these fields through the
surface S is V (D), the volume of region D.
3. The vector field F = z 2 k = (0, 0, z 2 ). Here, all arrows of F point upward, as sketched below.
y
z
x
This could be visualized as fluid that emanates from the xy-plane to travel upward, accelerating
as it moves away from the plane, along with fluid that approaches the xy-plane from below,
decelerating as it gets closer.
The divergence of F is
div F =
∂
∂ 2
∂
(0) +
(0) +
(z ) = 2z
∂x
∂y
∂z
(108)
Therefore, by the Divergence Theorem
Z Z
S
F · N̂ dS =
Z Z Z
div F dV = 2
D
Z Z Z
z dV.
(109)
D
The value of this integral will depend on the region D. In principle, if we knew the region, we
could integrate over it, using the techniques for integration in R3 developed earlier in the course.
275
There is one interesting point regarding this integral: It is related to the z coordinate of the
centroid of region D. Recall that
RRR
RRR
z dV
D z dV
D
,
.=
z̄ = R R R
V (D)
D dV
implying that
Z Z Z
(110)
z dV = V (D)z̄.
(111)
F · N̂ dS = 2z̄V (D).
(112)
D
Therefore, Eq. (109) becomes
Z Z
S
Note that if the surface S is located in the upper half-plane, i.e., z > 0, then z̄ > 0, implying that
the total outward flux is positive. However, if the surface S is located in the lower half-plane,
i.e., z < 0, the total outward flux is negative. Why is this so? And why would the total outward
flux be directly proportional to the volume V (D) of the region D enclosed by the surface S?
The Divergence Theorem and Point Charges
We now return to the class of vector fields having the form E =
K
r. In particular, we focus on the
r3
field
E(r) =
Q
r,
4πǫ0 r 3
(113)
which is the electrostatic field at r due to the presence of a point charge Q at the origin. Recall that
for this class of vector fields
div E(r) = 0,
for all (x, y, z) ∈ R, (x, y, z) 6= (0, 0, 0).
(114)
The divergence is undefined at the point where the point charge Q is situated. From these facts we
can conclude from the Divergence Theorem that
Z Z
E · N̂ dS = 0
(115)
S
for any surface S that does not contain or enclose the point charge Q that it situated at the origin
(0, 0, 0). As you well know, this result is consistent with Gauss’ Law from your Electricity and
Magnetism course.
276
Note, however, that we cannot use the Divergence Theorem to make any conclusions
about the total flux of E through a surface S that encloses the point charge Q. This is
because div E does not exist at the location of the charge, which violates the assumptions of the
Divergence Theorem.
Now recall that we did compute the total flux of E for a special class of surfaces that enclosed the
charge, namely the spherical surfaces SR centered at the origin with radius R. We found that
Z Z
E · N̂ dS =
SR
Q
,
ǫ0
(116)
independent of the radius R. We also found this to be the result for a unit cube centered at the origin.
Let’s now qualify that there is nothing special about having the charge Q at the origin. The above
result applies to any sphere SR with radius R that is centered at the point where charge Q is located.
Gauss’ Law for arbitrary surfaces
The question now remains, “What is the total flux of E through an arbitrary surface S that encloses
point charge Q?”. The situation is sketched in the figure below on the left. We suspect that the
Q
– in fact, this is what you’ve been told in your Electricity and Magnetism course - but
answer is
ǫ0
we must somehow derive this result mathematically! In order to do so, we shall have to extend our
earlier statement of the Divergence Theorem and then use a very clever trick.
In what follows, we shall adopt a convenient mathematical shorthand notation: We shall denote
the boundary of a region D ⊂ R3 , as ∂D. In the situations we have encountered so far, the boundary
of the region has been a single surface S, for which we write
∂D = S.
We shall extend this notion very shortly.
First, let us construct a surface SR centered at Q with radius R > 0 but sufficiently small so
that SR lies inside surface S, as pictured below on the right. Now let D′ denote the region that lies
inside surface S and outside surface SR , including points from both surfaces. Region D′ is a kind of
three-dimensional “donut” - it has an inner hole. More importantly, its boundary is composed of two
277
N̂1
N̂
Region D′
Region D
N̂2
Q
Q
SR
Arbitrary surface S enclosing Q
Arbitrary surface S enclosing Q
∂V ′ = S ∪ SR
∂V = S
surfaces:
∂D′ = S ∪ SR .
(117)
Note that region D ′ does not contain the troublesome point charge Q. It follows that
for all x, y, z ∈ D ′ ,
div E(x, y, z) = 0,
(118)
which will be a nice fact to be used below.
We now state a slightly generalized version of the earlier Divergence Theorem:
Let R ⊂ R3 be a region (or a “domain”, as the textbook calls it) with smooth boundary ∂R.
The boundary ∂R may be composed of several surfaces Si , 1 ≤ i ≤ M . Let N̂ denote the outer unit
normal, defined for all points on ∂R. Also assume that F(r) is a vector field for which div F(r) is
defined at all points r ∈ R. Then
Z Z
F · N̂ dS =
S
M Z Z
X
i=1
F · N̂ dS =
Si
Z Z Z
div F dV.
(119)
R
The above result allows us to employ the Divergence Theorem to the vector field E over region
D ′ , where div E = 0, noting that the boundary of D ′ is the union of both S and SR :
Z Z Z
Z Z
div E dV = 0.
E · N̂ dS =
D′
∂D ′
But we must be careful in defining the total outward flux of E from region D′ : It is the sum of
278
(120)
1. the total flux of E through the outer surface S pointing outward from D′ , in the direction of the
normal vectors N̂1 shown in the above figure and
2. the total flux of E through the inner surface SR pointing outward from D′ , hence toward the
point charge Q, in the direction of the normal vectors N̂2 shown in the above figure.
From the above, and from Eqs. (119) and (120),
Z Z
E · N̂ dS =
∂D ′
Z Z
E · N̂1 dS +
S
Z Z
E · N̂2 dS = 0.
(121)
SR
In all of these integrals, the unit normals N̂1 and N̂2 on surfaces S and SR , respectively point outward
from region D ′ .
But we know what the second integral on the RHS of the above equation is:
Z Z
E · N̂2 dS = −
SR
Q
.
ǫ0
(122)
This is because the unit normal vector N̂2 points in a direction opposite to that of the unit outward
normal vector to the spherical region DR which contains Q. Therefore, from Eq. (121), we have
Z Z
implying that
E · N̂1 dS −
S
Z Z
Q
= 0,
ǫ0
(123)
Q
.
ǫ0
(124)
E · N̂1 dS =
S
This is our desired result since N̂1 , the “normal” unit outward normal to S! This is true for an
arbitrary surface S enclosing (but not containing) the point charge Q. No matter how close Q may
be to S, we may always find an R > 0 sufficiently small so that the surface SR lies inside of S. We
have finally proved Gauss’ Law! And we have used the Divergence Theorem to do so!
279
Lecture 36
Gauss Divergence Theorem in R3 (cont’d)
Extension to many point charges
The procedure of the previous lecture can be extended to handle the case where a surface S encloses
n point charges Qk , k = 1, 2, · · · n located at points rk . The electrostatic field E(r) at any point in R3
due to these n charges is given by
E(r) =
n
Qk
1 X
[r − rk ].
4πǫ0
k r − rk k3
(125)
k=1
Also note that
div E(r) = 0,
except at r = rk , k = 1, 2, · · · , n,
(126)
where it is undefined.
We now place tiny spherical surfaces SRk around each charge Qk , where the radii Rk are sufficiently
small so that they do not intersect with each other or with the outer surface S. Now define region
V as the region enclosed by surface S but not including the interiors of the surfaces SRk , as shown
below.
Qn
Region V
Q1
Arbitrary surface S enclosing charges Qk
SR1
SRn
Q4
Q2
Q3
Note that the boundary of V is comprised not only of S but also of all the SRk :
∂V = S ∪ SR1 ∪ · · · ∪ SRn .
From the (Generalized) Divergence Theorem,
Z Z Z
E · N̂ dS = 0.
∂V
280
(127)
(128)
But this outward flux is the sum of the outward fluxes from region V through surface S and the
surfaces SRk . For the same reason as before, this total outward flux becomes
Z Z
E · N̂ dS −
S
n Z Z
X
SRk
k=1
E · N̂SRk dS = 0.
(129)
where N̂SRk denotes the unit outward normal through surface SRk that points away from the charge
Qk it encloses. The total flux through each of these tiny spherical surfaces due to the charge Qk they
Qk
. Thus
enclose is known – it’s simply
ǫ0
Z Z
n
X
Qk
E · N̂ dS −
S
or
Z Z
ǫ0
k=1
E · N̂ dS =
S
n
X
Qk
k=1
where
Q=
n
X
ǫ0
= 0,
=
Q
,
ǫ0
Qk
(130)
(131)
(132)
k=1
is the total charge enclosed by surface S. This is Gauss’ Law for electrostatic charges.
Note that any charges outside surface S would not contribute to this flux: the divergence of the
electostatic field due to these charges vanishes at all points inside surface S.
Extension to continuous distribution of charge
In the previous lecture, we proved Gauss law for the case of a single charge Q situated at the origin
(it didn’t have to be), followed by the case for a finite number of charges Qi situated at positions ri .
There is a final and important extension of the above result – the case of a continuous distribution
of charges over a region as defined by a charge density function ρ(r). There are no longer any point
charges – all charge has been “smeared out” into a continuous distribution.
Suppose that a surface S encloses a region D over which there is defined such a continuous charge
density function ρ(r). Then at each point r′ = (x′ , y ′ , z ′ ) in D, there is an infinitesimal element dV
that contains an infinitesimal element of charge
dq = ρ(r′ )dV.
281
(133)
This element of charge at r′ will contribute to a total electrostatic field vector E(r) according to
Coulomb’s law. We can now “add up” or integrate over the contributions of all such charge elements
dq over the region to produce the field vector E(r). (Such an integration procedure to compute
total electrostatic or gravitational potentials was described in an Appendix in an earlier lecture. The
extension to computation of the field vector is fairly straightforward.) We can also consider each
element of charge dq as a infinitesimal point charge that contributes to the total outward flux of the
electric field vector. Summing over all charges yields the result,
Z Z
Q
E · N̂ dS = ,
ǫ
0
S
where
Q=
Z
(134)
ρ(r) dV
(135)
D
is the total charge contained in region D enclosed by surface S. This is the integral form of Gauss’
Law.
Consequence of Gauss’ Law for a continuous charge distribution – Maxwell’s equation
Let us now carry the above results one step further. We now assume that we may apply the Divergence
Theorem to the outward flux integral since the electrostatic field vector E is produced by a continuous
distribution of charge and not an ensemble of point charges (in which case the charge densities would
be infinite at the location of the charges). Thus
Z Z Z
Z Z
∇ · E dV.
E · N̂ dS =
(136)
D
S
From Gauss’ Law above, we then have
Z Z Z
1
∇ · E dV =
ǫ0
D
Z Z Z
ρ dV,
(137)
1
∇ · E(r) − ρ(r) dV = 0.
ǫ0
(138)
V
which we shall now rewrite as
Z Z Z D
Of course, just because the integral of a function is zero, it does not mean that the function itself is
zero. BUT, we can make use of two additional points here:
1. We are going to assume that the integrand, i.e., the function ∇ · E(r) −
all r in D.
282
1
ρ(r), is continuous for
ǫ0
2. The result in (138) is valid for all smooth surfaces S and corresponding enclosed regions D. In
other words, for any surface S, the electrostatic field vector E produced by the charge distribution
enclosed by S must satisfy this relation.
If these two conditions are satisfied, then we may conclude, using the “duBois - Reymond Lemma”
– see below, that the integrand in (138) is zero at all points r, i.e.,
∇ · E(r) =
1
ρ(r).
ǫ0
(139)
This is known as the differential form of Gauss’ Law. In some books, it is called Maxwell’s first
equation for electrostatics.
The “du Bois - Reymond Lemma:” Before we go on with this fundamental equation,
let us return to the conclusion made regarding the integrand of (138). The justification
of this conclusion is called the “du Bois - Reymond lemma”. A simple one-dimensional
version might help:
Suppose we are given that f (x) is continuous on [a, b] and that
Z
b
f (x) dx = 0.
(140)
a
Of course, we can not conclude that f (x) = 0 on [a, b]. But if we are given that
Z d
f (x) dx = 0 for all c, d such that a ≤ c < d ≤ b,
(141)
c
then we may prove that f (x) = 0 identically on [a, b]. This is a one-dimensional version of
the duBois-Reymond lemma.
The proof of this lemma is rather straightforward. In a nutshell, assume that f satisfies
the above conditions and that there is a point x0 ∈ [a, b] such that f (x0 ) 6= 0. Without
loss of generality, assume that f (x0 ) > 0. Since f is assumed to be continuous (first
assumption), there must be a δ > 0 defining an interval I = [x0 − δ, x0 + δ] over which f
Rd
is positive. This implies that for any interval [c, d] ⊂ I, c f (x) dx > 0, contradicting the
second assumption. Therefore, no such x0 can exist.
283
Now back to the Maxwell equation. Notice that where there are no charges, there is no divergence
of E, i.e.,
If ρ(r) = 0, then ∇ · E(r) = 0.
In many problems in electricity and magnetism, one is required to find the electrostatic field vector
E(r) that corresponds to a given charge distribution. Often it is more convenient to solve the problem
using the potential function V associated with E, where E(r) = −∇V (r). Then the Maxwell equation
becomes
∇ · E(r) = ∇ · (−∇V (r)) =
1
ρ(r),
ǫ0
(142)
or
∇2 V (r) = −
1
ρ(r),
ǫ0
(143)
which is known as Poisson’s equation. Note that the left-hand side of this equation involves the
Laplacian operator introduced earlier in the course. In the absence of charge, i.e., ρ(r) = 0 for r in
some region D, then Poisson’s equation becomes Laplace’s equation
∇2 V (r) = 0.
(144)
Because of its importance in physics and engineering, a great deal of effort was spent by mathematicians over the past three centuries on these equations, including how to solve them and the
properties of solutions. You will encounter these equations in your future courses on electricity and
magnetism as well as fluid mechanics.
END OF COURSE
284
Appendix 1: Electrostatic field vector for a homogeneous spherical
charge distribution
The following example was not discussed in the lecture but is provided below for the
interested reader.
Here is a quite simple example to illustrate the meaning of the above equation. Consider a solid
sphere of radius R with a constant charge density ρ0 . This, of course, implies that the total charge
4
on the sphere is Q = πR3 ρ0 . Let us determine the electrostatic field vector E(r) produced by this
3
charge distribution.
1. Case 1: For k r k> R, the charged sphere may be considered as a point charge of Q situated at
the center of the sphere, which we shall assume to be the origin of our coordinate system. This
conclusion follows from the results of Problem Set No. 9, where you determined the gravitational
field produced by a spherically symmetric earth. Thus,
E(r) =
Q
ρ0 R3
r
=
r.
4πǫ0 r 3
3ǫ0 r 3
(145)
In this case, we know the divergence of the vector field, having computed it many times during
this course:
∇ · E(r) = 0,
(146)
which is consistent with the Maxwell equation, since there is no charge for r > R.
2. Case 2: For k r k≤ R. We may determine E(r) in the same way that we determined the
gravitational force F(r) at a point P inside the earth. In that case, the nonzero contribution to
the force comes from all points that lie inside the spherical surface that passes through P – in
other words, all points inside a sphere of radius r.
The same holds for the electrostatic vector – it will be given by
E(r) =
where Qr =
Qr
r,
4πǫ0 r 3
(147)
4 3
πr is the amount of charge contained in a sphere of radius r. Thus we arrive at
3
the result,
E(r) =
ρ0
r.
3ǫ0
(148)
Note that k E(r) k grows linearly as we move from the center of the sphere until we reach the
outer surface at r = R.
285
The divergence of this vector field is easily computed to be (since ∇r = 3)
∇ · E(r) =
ρ0
,
ǫ0
(149)
which is consistent with the Maxwell equation, since we stipulated that the charge density in
the sphere was ρ0 .
Note that the electrostatic field vector E(r) is continuous at r = R.
Finally, we could have solved this problem by solving for the potential function V (r). This method
of solution will be added as an Appendix to this lecture.
286
Appendix 2: Another consequence of the Divergence Theorem – The
“Continuity Equation”
This material was also not covered in class but provided for the interested reader. You
will encounter these ideas in a future course in Continuum Mechanics.
We now proceed to derive an important equation that has applications in theoretical physics and
applied mathematics, including fluid mechanics.
Let us consider the case of fluid flow. Let v(x, y, z, t) = v(r, t) represent the velocity field of a
fluid moving in R3 . (We acknowledge that the field – in particular, the components of v – can change
in time.) And let ρ(r, t) be a scalar field representing the mass density at a point r and time t. Then
the vector field
F(r, t) = ρ(r, t)v(r, t),
(150)
which is the momentum field of the fluid, describes the rate of mass transfer, or “transport”, at a
point r at time t. The rate of mass transfer is k F(r, t) k and the direction of flow is v(r, t).
Now consider a fixed surface S that encloses a bounded region D in R3 . (In other words, neither
the surface S nor the region D change in time.) At a given time t, the total amount of mass in region
D is given by
M (t) =
Z Z Z
ρ(r, t) dV,
(151)
D
where the integration is performed over the variables r ∈ D. The instantaneous rate of change of mass
is given by
d
dM
=
dt
dt
Z Z Z
ρ(r, t) dV =
Z Z Z
D
D
∂ρ
(r, t) dV.
∂t
(152)
We were able to take the partial derivative into the integral since the region D is presumed to be fixed,
hence independent of time.
Note: If you are worried about the above result, i.e., taking the derivative operator inside the integral,
consider the former definition of the derivative,
M (t + h) − M (t)
,
h→0
h
M ′ (t) = lim
(153)
and apply it to (151). You will obtain (152).
Since we assume that matter is neither created nor destroyed in region D, the rate of change
of mass in D is determined only by the rate of entry/escape through the boundary surface S. By
287
definition, the total outward flux of F, given by
Z Z
F · N̂ dS,
(154)
S
where N̂ is the unit outward normal to S, measures the outward flux of F through surface S, hence
the rate of escape of mass through S. By conservation of mass, it follows that
Z Z
dM
F · N̂ dS,
=−
dt
S
or
Z Z Z
S
∂ρ
(r, t) dV = −
∂t
Z Z
F · N̂ dS.
(155)
(156)
S
We now assume that the Divergence Theorem can apply to the right-hand side, (i.e., div F exists at
all points in D) so that the above equation becomes
Z Z Z
Z Z Z
∂ρ
div F(r, t) dV.
(r, t) dV = −
S ∂t
D
We now rewrite this equation as follows, also substituting F = ρv,
Z Z Z ∂ρ
(r, t) + div [ρ(r, t)v(r, t)] dV = 0.
D ∂t
(157)
(158)
Since this result is assumed to apply to arbitrary surfaces S with enclosed regions D, we once
again invoke the duBois-Reymond lemma to conclude that
∂ρ
(r, t) + div [ρ(r, t)v(r, t)] = 0,
∂t
(159)
∂ρ
+ ∇ · (ρv) = 0.
∂t
(160)
or in more simple form
This important equation is known as the Continuity Equation. It is a conservation relation that
represents the first step in the analysis of fluid mechanics, continuum mechanics and field theory.
If ρ is constant, then the continuity equation implies that
∇ · v = 0,
(161)
i.e., v is incompressible.
The conservation approach with which the continuity equation was derived can also be used for
a number of other physical phenomena including electric currents and heat transfer. For example,
288
regarding electric current, we interpret the vector field ρv as the current density of charge transfer –
the rate of transfer of charge. The scalar ρ represents charge density and v the velocity of the charge.
It is customary to let J = ρv denote the charge density field so that the continuity equation for charge
transfer becomes
∂ρ
+ ∇ · J = 0.
∂t
(162)
And finally, we mention that one can derive the heat equation
σρ
∂T
= k∇2 T,
∂t
(163)
where T (r, t) denotes the temperature of a solid object at point r at time t, k is the thermal conductivity, σ the specific heat and ρ the mass density.
289
Appendix 3: Appendix 2 revisited – solving Poisson’s equation for
spherical charge distribution
We now return to the problem studied earlier of a sphere of radius R with constant charge density ρ0 .
The problem was to find the electrostatic field vector E(r) produced by this charge distribution. We
now show how one can solve for the potential V (r) and then use this result to produce E using the
relation E = −∇V .
First of all, the physical situation is spherically symmetric – the charged body is a sphere and
the charge density function is constant, hence spherically symmetric. Thus we expect E and V to be
spherically symmetric as well. So we shall assume that the potential function V is a function only of
the radial coordinate r, i.e., V = V (r). And, of course, it is convenient to express the Laplacian in
spherical polar coordinates. For the case of the function V (r) which is a function only of r,
∇2 V (r) =
2 dV
d2 V
.
+
2
dr
r dr
(164)
You will note that we have changed the partial derivatives to ordinary derivatives w.r.t. r since V is
a function only of r. Poisson’s equation then becomes
∇2 V (r) =
d2 V
2 dV
1
+
= ρ(r).
2
dr
r dr
ǫ
(165)
1. Case 1: r > R. For the same reasons as in our derivation in the main text, we may treat
the charged body as a point charge Q concentrated at the origin. Technically, we need to solve
Laplace’s equation here, i.e.,
∇2 V (r) = 0,
(166)
since ρ(r) = 0 for r > 0. But we already know the solution – the potential will have the general
form
V (r) =
Q
+ C,
4πǫ0 r
(167)
where we include the arbitrary constant. The convention is to set C = 0 so that the potential
V (r) → 0 as r → ∞. The constant is actually irrelevant for the problem at hand – we wanted
to find the electrostatic field E, which will be given by
Q
r.
4πǫ0 r 3
4
(We’ve done this calculation many times.) Since Q = πR3 , we may also write E as
3
ρ0 R3
E(r) =
r,
3ǫ0 r 3
E(r) = −∇V (r) =
290
(168)
(169)
in agreement with the result obtained in the main section. After having rewritten Q in this way,
the potential function becomes
V (r) =
ρ0 R 3
.
3ǫ0 r
(170)
2. Case 2: 0 ≤ r ≤ R. We must solve Poisson’s equation for the charge density in the sphere, i.e.,
∇2 V (r) =
d2 V
2 dV
ρ0
+
= .
2
dr
r dr
ǫ
(171)
This is a second order linear differential equation in V (r) with nonconstant coefficients, normally
the subject of a third-year course on differential equations or mathematical methods in physics.
Apart from seeing whether such an equation can be written in terms of one of the standard second
order linear DEs (e.g. Bessel’s DE, Laguerre DE, Hermite DE, etc.), one can try assuming a
power series for the solution V (r). With an eye to the “answer at the back of the book,” we shall
first try something simpler and see if it works, namely, assuming that V (r) is a simple power of
r, i.e.,
V (r) = Ar α ,
(172)
where A and α can hopefully be determined. (If they can’t, then our method fails and we have
to try something else.)
Substitution of (172) into (171) yields
Aα(α − 1)r α−2 + A2αr α−2 = −
ρ0
,
ǫ0
(173)
or
Aα(α + 1)r α−2 = −
ρ0
.
ǫ0
(174)
You will note that the right-hand side is constant, but the left-hand side has a power of r. The
only way for this equation to hold for all values of r ≥ 0 is that the exponent α − 2 is zero, i.e.,
α = 2. In this case,
A=−
ρ0
,
6ǫ0
(175)
so that the potential function is
V (r) = −
ρ0 2
r + D.
6ǫ0
(176)
Note that V (r) is determined only to a constant since the derivatives in the Laplacian remove the
constant. For the moment, we ignore the constant, since our primary interest is the electrostatic
291
field vector E(r), given by
E(r) = −∇V (r) =
ρ0
r,
3ǫ0
(177)
where we have used the fact that ∇r 2 = 2rr. The result for E(r) agrees with the result obtained
in the main lecture text (as it should).
Finally, let us summarize the results of our potential function determination from above:

 ρR3 ,
r≥R
3ǫ0 r
V (r) =
 − ρ0 r 2 + D, 0 ≤ r ≤ R.
6ǫ0
(178)
In the above, we have set C = 0 so that V (r) → 0 as r → ∞. In order that V (r) be continuous at
ρ0 2
r = R, it is necessary that D =
R . Thus the final result is:
2ǫ0

 ρR3 ,
r≥R
3ǫ0 r
(179)
V (r) =
 ρ0 (3R2 − r 2 ), 0 ≤ r ≤ R.
6ǫ0
We note the behaviour of V (r) at three important points:
ρ0
but V ′ (0) = 0, which implies that E(0) = 0.
2ǫ0
ρ0
.
2. r = R: V (R) =
3ǫ0
1. r = 0: V (0) =
3. r → ∞: V (r) → 0.
A qualitative sketch of V (r) vs. r is presented below.
ρ0
2ǫ0
V (r) vs. r
2ρ0
3ǫ0
r
0
R
Potential function V (r) for a sphere of radius R with constant charge density ρ0 .
292
“Lecture 37”
(This lecture, which would have been the next lecture in the course, was not given this year because
of lack of time. Nevertheless, it is included for your own information. You are not responsible for this
material for the final examination.)
Surface integrals of vector-valued functions (cont’d): Stokes’ Theorem
Relevant section of textbook by Stewart:
16.8
We now arrive at the final important result of surface integration, namely, Stokes’ Theorem, which
is a three-dimensional version of Green’s Theorem in the Plane, covered earlier in the course. Recall
that Green’s Theorem in the Plane concerns line integrals of vector fields over closed curves, i.e.,
circulation integrals, in the plane: If S is a closed curve in R2 enclosing a region D, and F = F1 i + F2 j
is a vector field for which the partial derivatives ∂F2 /x and ∂F1 /y exist at all points in D, then
Z Z I
∂F2 ∂F1
F · dr =
−
dx dy.
(180)
∂x
∂y
D
C
The circulation integral over C on the left is equal to an integral over region D on the right.
Recall that the integrand on the right side is the k component of the curl of F – in fact, it is the
only component of curl F:
∂F
∂F
1
2
~ × [F1 (x, y)i + F2 (x, y)j] =
−
k.
curl F = ∇
∂x
∂y
(181)
Note that k is a unit normal vector to the xy-plane: By convention, in fact, because of the “righthand rule”, it is the unit normal vector associated with the counterclockwise motion of the circulation
integral around C. So we may rewrite Green’s Theorem in Eq. (180) as follows:
Z Z
I
curl F · N̂ dS.
F · dr =
C
(182)
D
Here, dS = dA, an infinitesimal element of area on the flat surface D enclosed by curve C.
Stokes’ Theorem is a three-dimensional generalization of this result involving nonplanar surfaces
S in R3 .
Let S be a piecewise smooth surface in R3 , with unit normal vectors N defined at all
points of S. Furthermore, suppose that S has a boundary C which is a piecewise smooth
293
closed curve. If F is a “smooth” (i.e., its derivatives exist and are continuous functions)
over S, then
I
F · dr =
Z Z
curl F · N̂ dS.
(183)
S
As in Green’s Theorem, the left-hand-side integral is a line integral and the right-hand-side integral
is a surface integral. A generic picture is sketched below.
Surface S
Boundary curve C
The remarkable point of this result is that the surface S is not necessarily “contained” inside the
curve C – it can be like a deformed soap bubble that has C as its boundary. Thanks to the continuity
of the derivatives involved in the curl of F, the circulation around C is transmitted over the surface,
just as was the case for Green’s Theorem in the Plane.
It is not the intention of this course to compute a great variety of integrals using Stokes’ Theorem.
We’ll consider one simple example, if only to illustrate the point.
Example: Let CR be the circle x2 + y 2 = R2 in the plane, and let S be the upper hemispherical
H
surface x2 + y 2 + z 2 = R2 , z ≥ 0, with CR as its boundary. Compute CR F · dr using Stokes’ Theorem
where F = −yi + xj + zk.
The i and j components of F correspond to the vector field of the rotating turntable encountered
~ × F = 2k, which is the curl
earlier in the course. The curl of this vector is easily computed to be ∇
of the turntable vector field – the k-component does not contribute to the curl.
In fact, we could in principle write down the solution to the problem from this information. The
294
curve CR lies in the plane, so we can use Green’s Theorem in the plane:
I
F · dr =
Z Z
curl F · NdA =
2 dA = 2πR2 .
(184)
DR
DR
CR
Z Z
Here DR is the circular region x2 + y 2 ≤ R2 in the plane enclosed by CR .
But let us compute the result using Stokes’ Theorem as applied to the hemispherical surface S.
Using the parametrization for the sphere from earlier, the outward normal vector to the surface S is
N = R sin vr(u, v).
(185)
We now compute the surface integral in Stokes’ Theorem as follows:
Z Z
curl F · N̂ dS =
Z Z
curl F · N du dv.
(186)
D
S
The integrand in the integral on the right is
curl F · N = (R sin vr(u, v)) · (2k) = 2R2 sin v cos v.
(187)
This follows from the fact that z(u, v) = R cos v on the sphere. The surface integral then becomes
Z Z
curl F · N dS = 2R2
SR
Z
0
π/2 Z π
in agreement with our earlier result.
295
0
sin v cos v du dv = 2πR2 ,
(188)
The Biot-Savart effect for distribution of moving charges: Ampère’s Circuital Law
In an earlier lecture, we examined the Biot-Savart effect: a current of charge flowing in a conducting
wire creates a magnetic field B surrounding the wire. We studied the case in which the conductor
is a straight, thin wire such that the direction of current is given by the unit vector û. The current
vector is then given by I = I û. The physical situation is once again sketched below: P is a point of
observation with position vector r.
û direction vector
of current
B(r)
magnetic field B(r)
created by current I = I û
P
θ
O
r
wire carrying electric current I
From electrodynamics, the magnetic field vector at P is given by
B(r) =
µ0 I û × r
.
2π k û × r k2
(189)
Here, µ0 denotes the permeability of the vacuum.
In the special case that the wire lies on the z-axis, i.e. û = k̂, the the current vector is I = I k̂
and the magnetic field vector becomes (see Lecture 27 for details)
µ0 I
−y
x
B(r) =
i+ 2
i + 0k .
2π x2 + y 2
x + y2
(190)
Note that B is undefined for (x, y) = (0, 0), i.e. the z-axis, which is the location of the current-carrying
wire. This is a consequence of the infinite density of charge on the wire.
In Lecture 27, we also computed the following circulation integral,
I
B · dr = µ0 I,
(191)
CR
where CR denotes the circle x2 + y 2 = R2 . This is the circulation of the magnetic field over the circle
CR . It is noteworthy that the circulation is independent of R.
We also noted that
~ × B(r) = 0,
∇
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(x, y) 6= (0, 0).
(192)
If we restrict our attention to the plane, we cannot derive the circulation result in Eq. (191) using
Green’s Theorem in the Plane, because of the singularity of curl B at (0, 0). Nor could we use Stokes’
Theorem, since the curl B is undefined over the entire z-axis – any surface with CR as boundary would
have to include at least one point from the z-axis.
However, we can use the result in Eq. (191) to derive the result that
I
B · dr = µ0 I,
(193)
C
for any closed curve C in the plane that encloses (0, 0), but does not contain it. The way to prove this
result is analogous to our proof of Gauss’ Law for arbitrary surfaces S containing the point Q: We
introduce a circle CR with R > 0 sufficiently small so that CR lies entirely inside the region enclosed
by curve R, as sketched below.
arbitrary curve
C
CR
We then apply Green’s Theorem to the region lying outside CR and inside C, etc., to eventually
derive Eq. (193). The details are left as an exercise for the reader.
Extension to several thin wires
The above result may now be extended to cover the case of several thin wires. For simplicity, we that
there are n such wires, all parallel to the z-axis. Each wire Wk carries a current Ik , 1 ≤ k ≤ n, as
sketched schematically below.
The current from each wire Wk will produce a magnetic field Bk (r). For any closed curve Ck
enclosing wire Wk , the circulation of the vector field Bk will be µ0 Ik . The net magnetic field B(r) will
be the vector sum of these fields,
B(r) =
n
X
k=1
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Bk (r).
(194)
Wn
Arbitrary closed curve C enclosing infinitely thin
parallel wires Wk with currents Ik
W1
W4
W2
W3
Moreover, for any closed curve C that encloses all of these wires, the circulation integral of B(r) will
be
I
B · dr = µ0
C
n
X
Ik = µ0 I,
(195)
k=1
where I is the sum of the currents of the individual wires.
Generalization to charge distributions/current densities
Instead of infinitely thin wires with point currents, we now consider continuous charge distributions
that are moving in space. Let ρ(r, t) denote the charge density at a point r at time t. And let v(r, t)
denote the instantaneous velocity at point r at time t. We then define the current density vector
J(r, t) as
J(r, t) = ρ(r, t)v(r, t).
(196)
The element of current crossing a surface area element dS centered at r with unit normal N̂(r) is then
given by
J(r, t) · N̂(r) dS.
(197)
Now let C be a smooth, closed curve in R3 that serves the boundary of a smooth surface S. Then
the total amount of current crossing this surface will be given by the surface integral
Z Z
J · N̂ dS,
(198)
S
where N̂ denotes the unit normal vector on S. Each element of curent J(r, t) passing through S
will contribute to the total magnetic field B in particular, to the net circulation of the field over the
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boundary curve C as follows:
I
B · dr =
C
Z Z
J · N̂ dS.
(199)
S
We now assume that B and its derivatives exist – thanks to the continuous distribution of electric
charge ρ and therefore the current density J, so that Stokes’ theorem can be applied to the left integral:
Z Z
curl B · N̂ dS =
S
We now rewrite this result as
Z Z
Z Z
J · N̂ dS.
(200)
S
[curl B − J] · N̂ dS = 0.
(201)
S
Assuming that the integrand is a continuous function over some region D ∈ R3 , and the above result
holds for arbitrary surfaces S with boundary curves C, it follows, from the du Bois-Reymond Lemma,
that
curl B(r) = J(r).
(202)
This is known as Ampère’s circuital law. It is a kind of magnetic analogy to the differential form of
Gauss’ Law: Where there is current, there is a magnetic field.
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