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A lecture note from math s21a: multivariable calculus, covering the topics of curl and divergence in three dimensions. It explains the definition, properties, and significance of curl and divergence, and how they relate to vector fields and nabla calculus.
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Math S21a: Multivariable calculus
Oliver Knill, Summer 2011
We have seen the curl in two dimensions: curl(
x
y
. By Greens theorem, it had been
the average work of the field done along a small circle of radius
r
around the point in the limit
when the radius of the circle goes to zero. Greens theorem so has explained what the curl is. Inthree dimensions, the curl is a vector:
The
curl
of a vector field
is defined as the vector field
curl(
y
z
z
x
x
y
Invoking nabla calculus, we can write curl(
. Note that the third component of the
curl is for fixed
z
just the two dimensional vector field
is
x
y
While the curl
in 2 dimensions is a scalar field, it is a vector in 3 dimensions.
In
n
dimensions, it would have
dimension
n
n
n
dimensions.
The curl measures the ”vorticity” of the field.
If a field has zero curl everywhere, the field is called
irrotational
The curl is often visualized using a ”paddle wheel”. If you place such a wheel into the field intothe direction
v
, its rotation speed of the wheel measures the quantity
~v
Consequently, the
direction in which the wheel turns fastest, is the direction of curl(
). Its angular velocity is the
length of the curl. The wheel could actually be used to measure the curl of the vector field at anypoint. In situations with large vorticity like in a tornado, one can ”see” the direction of the curlnear the vortex center. In two dimensions, we had two derivatives, the gradient and curl. In three dimensions, there arethree fundamental derivatives, the
gradient
, the
curl
and the
divergence
The
divergence
of
is the scalar field div(
x
y
z
The divergence can also be defined in two dimensions, but it is not fundamental.
The
divergence
of
is div(
x
y
In two dimensions, the divergence is just the curl of a
90 degrees rotated field
because div(
x
y
= curl(
). The divergence measures the ”expansion” of a field. If a
field has zero divergence everywhere, the field is called
incompressible
With the ”vector”
x
y
z
, we can write curl(
and div(
. Formulating
formulas using the ”Nabla vector” and using rules from geometry is called
Nabla calculus
. This
works both in 2 and 3 dimensions even so the
vector is not an actual vector but an operator.
The following combination of divergence and gradient often appears in physics:
f
= div(grad(
f
f
xx
f
yy
f
zz
It is called the Laplacian of
f
We can write ∆
f
2
f
because
f
div(grad(
f
We can extend the Laplacian also to vector fields with
) and write
2
Here are some identities:
div(curl(
curlgrad(
curl(curl(
)) = grad(div(
Proof.
Question:
Is there a vector field
such that
x
y, z, y
2
= curl(
Answer:
No, because div(
) = 1 is incompatible with div(curl(
Show that in simply connected region, every irrotational and incompressible field can bewritten as a vector field
= grad(
f
) with ∆
f
= 0. Proof. Since
is irrotational, there exists
a function
f
satisfying
= grad(
f
). Now, div(
) = 0 implies divgrad(
f
f
Find an example of a field which is both incompressible and irrotational.
Solution.
Find
f
which satisfies the Laplace equation ∆
f
= 0, like
f
x, y
x
3
xy
2
, then look at its
gradient field
f
. In that case, this gives
x, y
x
2
y
2
xy
If we rotate the vector field
by 90 degrees =
π/
2, we get a new vector field
. The integral
∫
C
ds
becomes a
flux
∫
γ
dn
of
through the boundary
of
, where
dn
is a normal vector with length
r
′
dt
. With div(
x
y
), we see that
curl(
) = div(
Green’s theorem now becomes
∫
∫
R
div(
dxdy
∫
C
dn ,
where
dn
x, y
) is a normal vector at (
x, y
) orthogonal to the velocity vector
r
′
x, y
) at (
x, y
This new theorem has a generalization to three dimensions, where it is called Gauss theoremor divergence theorem. Don’t treat this however as a different theorem in two dimensions.It is just Green’s theorem in disguise.This result shows:
The divergence at a point (
x, y
) is the average flux of the field through a small circle
of radius
r
around the point in the limit when the radius of the circle goes to zero
We have now all the derivatives together. In dimension
d
, there are
d
fundamental derivatives.
grad
grad
curl −→
grad
curl −→
div −→
They are incarnations of the same derivative, the so called
exterior derivative
To the end, let me stress that it is important you keep the dimensions. Many books treat two di-mensional situations using terminology from three dimensions which leads to confusion. Geometryin two dimensions should be treated as a ”flatlander”
1
in two dimensions only. Integral theorems
become more transparent if you look at them in the right dimension. In one dimension, we hadone theorem, the fundamental theorem of calculus. In two dimensions, there is the fundamentaltheorem of line integrals and Greens theorem. In three dimensions there are three theorems: thefundamental theorem of line integrals, Stokes theorem and the divergence theorem. We will lookat the remaining two theorems next time.
1
A. Abbott, Flatland, A romance in many dimensions,
Find a nonzero vector field
x, y
x, y
x, y
in each of the following cases:
a)
is irrotational but not incompressible.
b)
is incompressible but not irrotational.
c)
is irrotational and incompressible.
d)
is not irrotational and not incompressible.
The terminology in this problem comes from fluid dynamics where fluids can be incompressible, irrotational.
The vector field
x, y, z
x, y,
z
satisfies div(
Can you find a vector field
x, y, z
) such that curl(
? Such a field
is called a
vector potential
Hint.
Write
as a sum
x,
z
, y,
z
and find vector potentials for each of the
summand using a vector field you have seen in class.
Evaluate the flux integral
∫ ∫
S
, yz
dS
, where
is the surface with parametric equation
x
uv, y
u
v, z
u
v
on
u
2
v
2
Evaluate the flux integral
∫
∫
S
curl(
dS
for
x, y, z
xy, yz, zx
, where
is the part of
the paraboloid
z
x
2
y
2
that lies above the square [
1] and has an upward
orientation.
a) What is the relation between the flux of the vector field
g/
g
through the surface
g
with
g
x, y, z
x
6
y
4
z
8
and the surface area of
b) Find the flux of the vector field
g
through the surface
Remark
This problem, both part a) and part do not need any computation. You can answer
each question with one sentence. In part a) compare
dS
with
dS
in that case.