Download The most basic introduction to mechanics requires vector concepts

VECTOR CALCULUS
The most basic introduction to mechanics requires vector concepts in addition to scalar ones.
Electric and magnetic vector fields are fundamental concepts for understanding the interactions of
charged particles and the behavior of light. Our calculus techniques must be generalized to include
differential and integral operations on vector quantities.
Concepts of primary interest:
Vector and scalar fields
Gradient
‘grad’, ‘del’ or , is more formally known as: nabla.
Divergence and flux
Curl and circulation
Helmholtz theorem
Divergence of the gradient: Laplacian  2 F
The integral theorems: Why do we study them? FIVE STAR TOPIC
Search for ‘The Integral Theorems’ and study thoroughly.
Sample calculations:
Gravitational potential due to a point mass
Curl of a curl
Application examples:
Scaled directional derivative: Force on an Electric Dipole
Tools of the trade:
Parameterized line integrals
Finding the Potential Function for a Vector Field
Permutation symbol product identity
Operator notation
Curl of Curl  Laplacian of a vector field
Product Rules for Vector Calculus
Note:  is being replaced by  as the cylindrical angular coordinate. Be alert for incomplete changes!!!

Hints regarding vector integration appear first followed by developments of vector differential
operations. Natural definitions for the divergence and curl make the integral theorems of Gauss and
Stokes obvious. Your goal should be to master the differential operators on fields, scalar and vector
valued functions of position, in Cartesian, cylindrical and spherical coordinates. More general
Contact: [email protected]
results appear for enrichment; you may skip them on first reading. The first order operators: the
gradient, divergence and curl are defined and illustrated. The capstone element of this section is
Helmholtz theorem that provides an expression for computing the position dependence of a vector
field if its divergence and curl are known. Study the theorem carefully as the reward is a prescription
for the forms of Maxwell's equations. The Tools of the Trade sections illustrate techniques required
to master the application of the concepts presented.
Vector Integration: Conceptually vector integrations are no more difficult than the standard single
variable (1D) integrals studied in freshman Calculus. However they are quite ominous in appearance
and often require more steps to complete their evaluation. In fact, most vector integrals are evaluated
by a careful reduction to a set of 1-D integrals. As an example, the work done on a particle of mass m
is computed as it moves along a path from an initial position ri to a final position r f subject to the
gravitational attraction of a particle of mass M located at the origin.
z
F 
GMm
rˆ
r2
rf
ri
y
M
x
The work done on the particle by the gravitational force is given by:
f
W   F  dr  GMm 
i
f
i
rˆ  dr
r2
[VCA.1]
The line element dr in spherical coordinates is dr rˆ  r d ˆ  r sin  dˆ . The coordinate
directions are orthogonal and normalized, thus: rˆ  rˆ  1, rˆ ˆ  0 and rˆ ˆ  0 . The initial and final


positions are labeled by their coordinates as  ri ,i ,i  and rf , f , f . Substituting, we find:
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
GMm 
GMm
rˆ  dr rˆ  r d ˆ  r sin  d ˆ    2 dr
2


r
r
f
W   F  dr  GMm 
i
f
i
dr
 GMm  r1  r1 
2
i
 f

r
The integral collapses to a simple integration w.r.t. r because the force field only has a radial
component and the increment of the work done on the particle is the magnitude of the force times the
component of the differential displacement in the direction of the force.
A Guide for Vector Integrations
1. Choose a coordinate system that is appropriate for the problem. For the problem above, the
spherical symmetry of the force field makes a spherical coordinate system centered on the mass M a
good candidate.
2. Express all vectors in terms of the coordinate directions for that coordinate system, and compute
all inner (dot) and cross products.
3. If after step 2, there is a unit vector ê in the integrand that is not constant w.r.t. the integration
variables, replace that vector by its representation in terms of the constant directions: iˆ , ĵ and kˆ .
That representation makes the dependence of the direction ê on the integration variables explicit.
4. By this step, the integration has been reduced to either a scalar integration or to a set of scalar
integrals multiplying constant directions. If they are present, the constant directions should be taken
outside the integral leaving a sum of terms with each term being an integral multiplying a constant
direction. This form is a generalization of the component-wise addition of vectors.
5. Multiple integrals are to be computed as a nested set of single integrations. The techniques to try
are:
a. Change of variable: A first guess is the argument of the most complicated function in the
integrand. Be sure to change your limits appropriately and simultaneously!
t
t
0
f ( x)
x (t )
dx
dt ' 
f ( x) dx
x (t0 )
dt '
Note that a dummy label t ' is used to represent the integration variable as the
integration variable must be distinct from the limits.
The arguments of functions are dimensionless. Adopting dimensionless variables is a wise and
common practice. Try it!
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b. Trig substitutions: Indicated if a. has failed and if the square root of the sum or difference of
squares is present.
a 2  x 2  tan   x a as 1  tan 2   sec2  and d (tan  )  sec2  d
a 2  x 2  sin   x a as 1  sin 2   cos 2  and d (sin  )  cos  d
and sometimes forms like 1  ( x b) can use sin 2   x b
Be aware that the angle  chosen as the new variable is meaningful. Identify  on your figure.
Interpret your results in terms of this angle if possible. Square roots are often used to represent
distances in physics. If this is the case, only the positive root is meaningful. Be alert and examine
cases. (See page 4 of the EFieldRing Handout for a trig identities trick.)
The hyperbolic functions should be tried in these cases if trig
substitution fails (See the Basic Integration handout.).
c. Integration by parts:
d (u v)
dv du
dv d (u v) du
u

v or u


v
dx
dx dx
dx
dx
dx
d. Any tricks presented by your instructor in the course.
e. Integral tables, calculators or Mathematica, but only as a last resort.
6. Reflect on your efforts. Review the techniques that were successful and attempt to identify clues
that would lead you to select them more quickly.
7. Attempt to re-express your results in the language of the problem statement and without any
direct reference to the particular coordinate system that was used. For example: The electric field due
to a long straight uniformly charged wire varies as the inverse of the distance from the wire, and it is
directed perpendicularly away from the wire at any point.
Several sample line integral calculations appear as the first unit in the Tools of the Trade section.
Review them for practical hints.
The Vector Differential Operations:
Introduction: There are three first order operations involving vectors and partial derivatives that
play a central role in the representation of physical laws. These operations are the gradient, the
divergence, and the curl (circulation density). Second order differential operators can be formed by
repeated operations with the three first order operators. For example, the divergence of the gradient
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is called the Laplacian, an operator of cosmic significance. Less impressive is the curl of the gradient
which vanishes for 'well-behaved' functions. All of these operators act on fields. A field is a function
(physical quantity) that has a value assigned at each point in a region of space. If that value is a scalar
such as the temperature, then the field is a scalar field. If the value is a vector such as the flow
velocity in the Severn River, then the field is a vector field. The detailed definitions and properties of
scalars and vectors are discussed in intermediate mechanics. You should reread this handout after
completing the first six weeks of that course.
Line Elements and Metrics: As undergraduates, you will use Cartesian, cylindrical and spherical
coordinates almost exclusively. You should review the coordinate systems handout describing these
systems before proceeding. A short digression presents more generic systems called generalized
orthonormal coordinates in the next few paragraphs. A basic requirement for a system of
orthonormal curvilinear coordinates is that they represent the differential displacement dr and the
associated distance dr in the standard forms below. A general set of such coordinates in three
dimensions is (q1, q2, q3). The three directions, ê1 , ê2 , and ê3 , may vary from point to point, but at
any point are mutually perpendicular. The line element becomes:
dr  h1  q1, q2 , q3  dq1 eˆ1  h2  q1, q2 , q3  dq2 eˆ2  h3  q1, q2 , q3  dq3 eˆ3
[VCA.2]
where the unit vector eˆi associated with qi is the direction in which the coordinate point moves
when the coordinate qi is given a small positive increment while the other coordinates are held fixed.
The functions hi(q1,q2,q3 ) are the (metric) scale factors for the coordinates and, in particular, the
length of dr squared is:
ds 2  dr  dr  dr 2 
 h1  q1 , q2 , q3  
2
 dq1    h2  q1 , q2 , q3   dq2    h3  q1 , q2 , q3   dq3 
2
2
2
2
[VCA.3]
2
The line elements in our three standard coordinate systems are:
drCart  dx iˆ  dy ˆj  dz kˆ
drcyl  dr rˆ  r d ˆ  dz kˆ
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[VCA.4]
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drsph  dr rˆ  r d ˆ  r sin dˆ
rsph  x2  y 2  z 2
Beware: The symbol r is the distance from the axis in cylindrical coordinates and the distance from
the origin in spherical coordinates. This dual usage can cause tremendous difficulties unless you
think about what you are doing. Think about what you are doing at all times. Some authors use
the symbol  or s to represent rcylindrical.
Note: gradient symbol, ‘del’ or , is more formally called nabla.
The Gradient: The gradient of a scalar function of position is a vector-valued function of position.
[ex: E (r )  V (r ) ]. If one chose to study the collection of the three partial derivatives for a
coordinate system, the various derivatives could have different dimensions. The gradient is an
improvement on partial derivatives in which the components in the several directions have the same
dimensions. A partial derivative is computed by taking the ratio of the change in the value of the
function to the change in one argument when that argument is varied while the other arguments are
held fixed.
F
F ( x, y  y, z )  F ( x, y, z )
 Limit

y

0
y
y
[VCA.5]
The gradient is an associated generalization of the derivative for functions of position in a two, three,
or n dimensional space. The gradient of a scalar function G is a vector function G (r ) that has, as
its component in a direction, the rate of change of G with respect to distance, not with respect to the
change in the corresponding coordinate. This distinction between distance and coordinate change is
important, for example, when an angular coordinate is varied. In the spherical case, a coordinate
change d becomes a distance r d. Also, all the components of the gradient have the same
dimensions, the dimensions of the function divided by length.
The point is that physics can depend on rates of change with respect to distance, a physical concept.
The physics does not depend on a particular set of coordinates. Physics therefore is more naturally
described by gradients that have as their components the rates of change with respect to distance for
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each of the independent directions in the coordinate system. Further the gradient of a scalar field is a
vector field while the collection of the three partial derivatives in spherical coordinates is a mongrel
lacking the good transformation properties as it is composed of items that do not even share the same
dimensions (units). [More about this point appears in the linear algebra section.]
More formally, the component of the gradient in the direction ê is the rate of change of G with
respect to distance for an infinitesimal displacement in the direction of ê , the directional
derivative. [That is: G

for d in the direction of interest.] Thus, as x is the displacement in
the x direction when x is varied to x + x, the x component of the gradient is the limit of (G/x) as
x approaches zero (= G
x
), the same as the partial with respect to x). As r  is the displacement
in the  direction when  is varied to  + , the  component of the gradient is the limit of
(G/r) as  approaches zero which is r 1 G
which is: G


. This result differs from the partial derivative
.
A prescription for computing the components of the gradient is the directional derivative given
below.
 G (r  s eˆ)  G (r ) 

s


Def n : eˆ G (r )  G (r )   Lim 

e
s 0
[VCA.6]
 G (r   dr  eˆ  eˆ)  G (r ) 

dr  eˆ


 Lim 
dr 0
For example, the x component of the gradient is the x directional derivative:
 G (r   dr  iˆ  iˆ)  G (r ) 
 G (r  s iˆ)  G (r ) 



iˆ G (r )  G (r )   Lim 

 Lim 
x
s
dr

iˆ

s 0 
 dr 0 
and the  component in cylindrical coordinates is:
 G (r   dr  ˆ  ˆ)  G (r ) 
 G (r  r d ˆ )  G (r ) 


ˆ

 G (r )  G (r )   Lim 
  Lim 

ˆ


r
d

dr  
d 0 
 dr 0 

Review these examples and then derive the forms for all three components of the gradient in each of
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the three coordinate systems.
Normal Derivative: The derivative of a function in the direction of the normal at a surface or
f
 nˆ f , the direction derivative of f (r ) in the
n
interface appears in a variety of problems. It is
normal direction.
Directional Derivative: The directional derivative of a function f (r ) is the rate of change of that
function with respect to distance moved in a direction. For the direction ê , it is computed as: ê f
Note that an alternative definition of the gradient is implicit in the equation
dG  G(r  dr )  G(r )  G(r )  dr
[VCA.7]
that must be true for arbitrary infinitesimal dr . The equation states that the inner product of the
gradient of G and the line element must be equal to the total differential of G with respect to its
spatial arguments. This definition also makes explicit the fact that G (r ) points in the direction in
which the function G is increasing most rapidly with respect to distance and that the magnitude of
G (r ) is the rate of change with respect to distance in that direction. The area patch formed by a
full set of differential displacements perpendicular to is an equi-G surface patch. As an example, in
cylindrical coordinates we would have:
dG  Gr dr  G d  Gz dz   G r rˆ   G  ˆ   G  z kˆ   dr rˆ  r d ˆ  dz kˆ 

 
 

 
 G dr  G r d  G dz
r

z
As the coordinates can be varied independently, we must equate the coefficients of dr, d and dz
individually with the results that:
 
  r d ; Gz dz  G  dz
G dr  G dr ; G d  G
r

r

[VCA.8]
z
Exercise: Use Cartesian coordinates to compute  r  rˆ where r is the distance from the
origin and r̂ is the direction away from the origin. Repeat the calculation using spherical
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coordinates. Motivate the result that r̂ is direction of most rapid change and that the gradient has
magnitude 1. Note that  r  rˆ is a vector statement; given the definitions above for r and r̂ ,
it is valid in all coordinate systems.
The Gradient Operators
Cart  iˆ
cyl  rˆ
sph  rˆ
 ˆ  ˆ 
j
k
x
y
z
 ˆ 1  ˆ 
  
k
r
z
 r  
[VCA.9]
 ˆ 1  ˆ  1  
  
 

r
 r  
 r sin   
The gradient operators have been written with the differential operators rightmost to emphasize that
the derivatives do not operate on the unit vectors or coordinate forms involved in the representation
of the gradient operator itself. These operators are occasionally written with their coordinate
directions on the right. This variant is just a notation, and it is not intended that their evaluation be
altered. As represented above, the differential operators are to act on everything to their right.
Exact Forms: A vector field F ( r ) is exact if it is the gradient of a scalar field ( F (r )  U (r ) ).
The scalar field U (r ) is the potential function for the vector field.
rf
rf
rf
ri
ri
ri
U (rf )  U (ri )   dU  U (ri )   U  dr  U (ri )   F  dr
The form dU  F  dr is then an exact differential. A requirement for exactness is that
 F  dr  0 for all paths.
(This requirement is met if  F  0 .)
Gradient Summary:
i.) The direction of the gradient of a scalar-valued field (function of position) f ( r ) is the direction of
change of the argument for which f ( r ) increases most rapidly.
ii.) The magnitude of the gradient is the rate of change of f ( r ) with respect to the distance by which
the argument is changed in the direction for which f ( r ) increases most rapidly.
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In other words: The component of the gradient of a scalar functions in any direction is the rate of
change of that function with respect to the distance by which the argument point is moved in that
direction.
Fundamental Theorem of Vector Integral Calculus: df  f (r  dr )  f (r )  f (r )  dr
leading to the fundamental theorem of integral calculus for scalar functions of position in higher
dimension.
rf
rf
ri
ri
f (rf )  f (ri )   df   f  dr
[VCA.10]
Divergence: The divergence is a first order vector differential operation on vectors (vector fields)
that yields scalar fields [ex:  E  
0
]. The divergence of a vector field is defined by:
Defn: div F  Limit
V 0
 F  nˆ dA
V
[VCA.11]
That is: the divergence of a vector field at a point r is defined as the limit as the enclosed volume
approaches zero of the integral of the outward directed normal component of the vector over a closed
surface that encloses the point r divided by the volume enclosed, or, in other words, as the limit of
the ratio of the flux of F out of a closed surface to the volume enclosed by the surface. [The limit is
taken for well-behaved surface choices for which the greatest distance between two points on the
surface, d, and the area of the surface tend smoothly to zero as the enclosed volume approaches zero.
Further, the closed surface is to be composed of a finite number of smooth (differentiable) surface
elements. Our choices, the surfaces of coordinate 'cubes', meet these requirements.] The flux of a
vector field though a surface is the integral of the normal component of that vector field over the
surface. The divergence is the net flux out of the surface bounding a volume per volume. If one
identifies F with v the flow velocity of an incompressible fluid, like water, then the flux integral
over the closed surface gives the net volume flow rate of water out of the volume. If it is non-zero,
then there must be a water source (or sink) in the volume.
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(x+x,y+y,z+z)
z - up
F ( x, y , z )
n̂  iˆ
n̂  iˆ
k̂
y - into page
F ( x  x, y , z )
ĵ
Ax= y z
(x,y,z)
iˆ
x - to right
Cartesian-Coordinate Cube: Flux evaluation for the faces with normal directions iˆ and - iˆ .
An expression for the divergence in Cartesian coordinates can be developed using a small coordinate
'cube' with edges that are coordinate line elements. The cube has corners at (x,y,z) and at
(x+x,y+y,z+z). The flux of F out of the volume is:
Faverage value  nˆoutward Aside
 F  nˆ dA six
sides
[VCA.12]
The sides naturally form pairs for each coordinate. One such pair per example provides a model for
the complete calculation.
 F  nˆ dA  F  nˆ A  F  x  x, y, z   iˆ  y z  F  x, y, z     iˆ  y z  ...
Only terms for two of the six sides are displayed, the sides perpendicular to the x-axis with iˆ and - iˆ
being the respective outward-directed normals. The symbol y represents a mean value of y where
(y< y <y+y). Substituting into the definition,
Defn: div F  Limit
V 0
 F  nˆ dA 
V


 F ( x  x, y , z )  iˆ  F ( x, y , z )  iˆ y z

Limit 
 ...  
x , y , z 0 
x y z



Noting that F  iˆ  Fx and using the definition of the partial derivative, this expression reduces to:
F ( x  x, y , z )  Fx ( x  x, y , z )
 F
div F    F  Limit  x
 ...  ...  x  ...  ...
x
x
x , y , z 0
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div F    F 
Fx Fy Fz


(Cartesian)
x y
z
[VCA.13]
Note that in the Cartesian representation there are contributions to the divergence of a vector field
from any component that varies when its associated coordinate varies. A non-zero contribution
results if Fx depends on x (
Fx
x
 
 0 ). The Cartesian form is very suggestive and leads to the
 
notation: div F   F . Both div F and  F are read as the divergence of F where, by
the inflection of your voice, the quantity F is clearly specified to be a vector field.
One must understand the meaning of 'gradient dot vector' before using it to develop expressions for
divergence. It is correct in general if the differential operators act on the coordinate directions in F
as well as the components of F .
Fy
 iˆ 
 
Fx


div F    F   iˆ  ˆj  kˆ   Fxiˆ  Fy ˆj  Fz kˆ  iˆ  iˆ
 Fx  iˆ    iˆ  ˆj
 ....
y
z 
x
x
 x
 x 

  
 
with the expansion continuing to 18 terms. Six terms vanish immediately because the coordinate
directions are orthonormal ( iˆ  iˆ  1; iˆ  ˆj  0,... ). Nine more terms vanish because the coordinate
directions are constants. The Cartesian coordinate direction set ( iˆ, ˆj, kˆ ) are global constants, but
several of the partials of coordinate directions are non-zero in cylindrical and spherical coordinates.
As a consequence, the expressions for the divergence have more than three terms in these systems.
Even for these cases, the extra terms are usually combined into just three terms. These combined
terms require product rule differentiation and thereby reveal the additional terms during evaluation.
That is, the extra terms are hidden.
Lesson Learned: Use the expressions for divergence from the vector identities handout.
Do not attempt to interpret gradient dot vector as the divergence of the vector field.
Pitfall Alert: The surface integral
 F  nˆ dA  F ( x , y , z )  nˆ A can be approximated by a constant
value F ( x , y , z )  nˆ times the area only as long as the area is infinitesimal so that the continuously
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varying and differentiable F ( x , y , z )  nˆ is essentially constant over that area. There are problems in
which one is asked to verify Gauss’s theorem {  F  nˆ dA     F dV } for a specific case. The areas
V
V
are not infinitesimal patches in these cases, and the surface integrals must be computed using the
standard, tedious methods. Some guidance is provided in the Tools of the Trade section in Vector
Calculus B. An analogous comment applies to problems that require verifying that Stokes’ theorem
{  F  d   ( F )  nˆ dA } for a specific case.
A
A
The vector field increases in magnitude for displacements in the direction of the field. This behavior
indicates a positive divergence (   F  
Fx Fy Fz
). In the figure below, the field only varies


x y
z
for changes in position perpendicular to the field. The divergence vanishes in this case.
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Non-constant field with zero divergence
The change is perpendicular to the field rather than parallel
A third (and extremely important example) is the vector field
r
x iˆ  y ˆj  z kˆ

that is
r 3  x 2  y 2  z 2  3/ 2


represented by the drawing below in the field line picture of the vector field. In this picture, the field
is tangent to the lines, and its magnitude is represented by the density of lines, the number per area
perpendicular to the line direction. In this picture the field has a positive (negative) divergence in a
region of space in which lines are starting (ending). A straightforward application of the Cartesian
form for divergence shows that  
r
 0 . The field decreases in magnitude for radial displacements,
r3
but the transverse components grow for displacements perpendicular to the radial direction yielding a
net zero divergence. The lines are spreading out, not starting or stopping. Alternatively, the flux
involves the component of the field normal to the area. The area of the concentric spherical surface
grows like r 2 which balances variation of the field magnitude that decreases like r -2.
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This field pattern has a point at which lines start, r = 0. The expression for the field is singular at
that point rendering normal techniques for computing derivatives inadequate.
A more detailed analysis shows that  
r
 4  3 (r ) where the delta function vanishes except for
3
r
the point at which its argument vanishes where it explodes huge positive. This field pattern models
the electrostatic field due to a point charge.
Exercise: Compute  
Compute

r
explicitly using Cartesian coordinate representations for r  0 .
r3
r
 n̂ dA , the flux out of a closed surface of radius R centered on the origin. Comment
r3
on the dependence of the result on R.
Next, the definition of the divergence is exercised in cylindrical coordinates.
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z
dz k̂
rd ˆ
dr r̂
zkˆ

r rˆ
y
x
Note that for cylindrical coordinates, the two faces of the cube with radial normals have slightly
different areas Ar. Consider: F  Fr rˆ  F ˆ  Fz kˆ . If Fr is constant, the flux into the inner radial
face will not equal the flux out of the outer radial face. Terms like
  Fr Ar 
final representation of the divergence, but they 'reduce' to (z 
  Fr r 
r
r
must appear in the
as r is the only factor in
Ar = z r  that varies for the radial displacement. The area Ar has the value z r on the
inner surface, and it is equal to z (r +r) on the outer surface.
 F  nˆ dA  F  nˆ A  F  r  r,  , z   rˆ  r  r   z  F  r ,  , z    rˆ  r  z  four more terms
 F  nˆ dA  F  nˆ A  F  r  r , , z   r  r   z  F  r , , z  r  z  four more terms
r
r
 F  nˆ dA  F  nˆ A  G  r  r, , z   z  G  r , , z 
 z  four more terms
 F  nˆ dA  F  nˆ A   1r  G r  r r  z   four more terms
where G(r) = r Fr(r).
That is: the differential of the product r Fr(r) appears leading to the derivative of that product rather than just
the derivative of Fr.
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d[r Fr(r)] = [Fr(r+r,,z) (r+r)] - [Fr(r,,z) r]  [(r Fr)/r ] r
Multiplication of the differentials by r completes the volume element inside the last set of brackets. The
division by r before the derivative compensates for that multiplication. Recall that the definition of
divergence requires a division by the volume element V   r r  z  so that factor should appear
explicitly in each term representing the flux out of the volume to facilitate the derivation.

The opposite face areas are equal for the other two directions and thus:
 1    r Fr 
 F  nˆ dA   r  

  Fz  
 1     F  
r r  z    
r r  z  
r r  z


r 
 
z 
 r  


Note that the first two terms have a factor of r-1 times the volume element.

 1   r Fr    1   F     Fz   

ˆ
F

n
dA


 r 

  r
  r r  z

r



z

 



 

 
 
Hence after division by V   r r  z  :
 1    r Fr   1  F Fz
divF   F   
 

 r  r
 r   z
(cylindrical)
[VCA.14]
The definition of the divergence directs that the limit in which the size of the volume and of the
lengths of its sides approach zero is to be taken. The mean value theorem for integrals of continuous
b
functions declares that
f ( x )  f ( x) so
a
x x
x
b
f ( x) dx  f ( x )  dx for some x [a, b] . As the limit progresses,
a
f ( x) dx  f ( x) x .
CAUTION: NEVER take a non-constant factor outside an integral unless the integration interval is
infinitesimal. Problems VC: 8, 9 and 33 are examples in which the infinitesimal requirement is not
met.
The divergence has a flux term for each pair of surfaces of the coordinate cube which is the
difference in the vector component times the corresponding area element evaluated at the larger
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value of the coordinate minus the quantity evaluated at the smaller value. This net flux out is divided
by the volume to give the divergence.
 F r z 
r r
r
 F r 
r
r r


  Fr r  z r    F r z     F r z    Fz r r  z z   Fz r r  z 
r r  z
  Fr r r 
r r
 F 




  F 
 Fz

r 
 
z   Fz z z 
z

1    r Fr   1


r  r  r
 F  Fz


   z
Finally, we will discuss the definition of the divergence in spherical coordinates. The area elements
are dAr = r sin d r d, dA = r sin d dr, and dA = r d dr. In this case, the dAr and dA surfaces
pairs have different areas. As a result, every term contains factors related to the metric of the line
element in spherical coordinates.
 F A 
r
r r r

  Fr Ar r    F A
    F A      F A     F A  
r 2 sin  r  
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
1
{  Fr r 2 sin     r r   Fr r 2 sin     r
r sin  r  
2


  F r sin   r     F r sin   r 


  F r r      F r r  




2
 1   sin  F
 1  F
 1   r Fr
  F  2 




r

r 
 r sin  
 r sin   
}
[VCA.15]
Gauss's Theorem: One application of the divergence is as a computational step in the
transformation of surface integrals to volume integrals using Gauss's Theorem.
 V F  nˆ dA  V  F dV
 F  nˆ dA 
  div F
V 0  V 


Defn: Limit  

 i  div F i Vi
 F  nˆ dA  volumes
This development was just a little cavalier. One should demonstrate that the contributions from the
elements of surface that are internal to the final assembled volumes cancel such that only the flux out
of the net bounding surface survives. It works for all reasonable volumes - even those with a few
enclosed voids as long as one keeps track of the contributions for the surfaces of those voids.
Exercise:
Argue that the net flux out of the
surface that bounds both boxes is just the sum
of the fluxes out of the individual boxes.
HINT: Consider the outward directed normal
for each surface segment.
Spherical Divergence Example:
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 g   4  G M
Application: Gauss’s Law in integral and differential form.
(There is an electrostatic analog to this example:   E 
charge
 0 .)
The differential form of Gauss’s law for gravitation:  g   4  G M .
The divergence of the gravitational field is minus 4 times the Newton’s gravitational constant times
the volumetric mass density. The negative divergence is related to the attractive nature of the gravity.
Develop integral and differential forms; use both to compute the gravitational field for a uniform mass
sphere. Integrate to find the gravitational potential. Use both integral and differential approaches.
 0 r  R
0 rR
M  
As a vector must know which way to point, the only choice that can be made for problems with
spherical symmetry is away from (or toward) the center of symmetry. Hence a vector defined by a
spherically symmetric problem will be radially directed. As the problem is spherically symmetric, there
can be no  or  dependence. The conclusion is that:
g (r )  gr (r ) rˆ
2
1  d  r gr  
   4  G M .
The divergence for g (r )  g r (r ) rˆ takes the form, 2 
r  dr 


For this example, M  0 for r  R and  0 for r  R .
d  r 2 gr 
dr
d  r 2 gr 
dr
 0 for r  R  r 2 g r  C or g r  C r 2
 4  G o r 2 for r  R
 
 
r 2 g r   4 3 G o r 3  D  g r   4 3 G o r  D r 2
The mass distribution is smooth and non-singular for r < R so the field should be well-behaved at r =
0  D  0 . (No r -2divergence is expected at r = 0.) In fact, at the center (r = 0) there is no preferred
direction so the vector must vanish as no direction can be chosen.
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4  R3 o
The expressions for gr should match at r = R  C  
.
3
 
 4 G o r rˆ
3

g (r )   4  R3 G 
o
rˆ
 
2
3
r

 for
r  R
 for r  R 
Using the integral forms, a spherical surface of radius r concentric with the origin is chosen.
 g  nˆ dA     g  dV
 g  rˆ dA  g
r
4 r 2 
    g  dV    4  G   dV
M
inside
g r 4  r 2   4  G M inside  g (r )  
M inside
G M inside
rˆ
r2
  4  r 3  M total r 3
rR
 0 

R3
  3 

3
   4 R   M
rR
total
 0 3 



 
 4 G o r rˆ
3

g (r )   4  R3 G 
o
rˆ
 
2
3
r

 for
r  R
 for r  R 
The same final answer follows from the differential and integral approaches.
Review of Coordinate Systems: You must be familiar with the representations of the position
vector, line element and a general vector in each coordinate system. Take the time to represent a
vector field in the form native for the coordinate system chosen. In spherical coordinates for
example, that requires that you identify the explicit forms of Fr, F and F before proceeding with
the vector calculus operations.
TABLE I: Position Vectors, Line Elements, and General Vectors in the Big Three
SYSTEM
POSITION r
LINE ELEMENT d  dr
VECTOR F
Cartesian
xiˆ  y ˆj  z kˆ
dxiˆ  dy ˆj  dz kˆ
Fx iˆ  Fy ˆj  Fz kˆ
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cylindrical
r rˆ  z kˆ
dr rˆ  r d ˆ  dz kˆ
Fr rˆ  F ˆ  Fz kˆ
spherical
r rˆ
dr rˆ  r d ˆ  r sin  d ˆ
Fr rˆ  F ˆ  F ˆ
Exercise: Compute the divergence of the position vector in our three coordinate systems. Begin by
finding explicit forms of the components of the position vector in each system.
WARNING: LITTLE HAS BEEN GAINED BY READING THE PRECEEDING PAGES. YOU MUST
WORK THROUGH THE DEVELOPMENTS DISPLAYING, IN PARTCULAR, THE TERMS REPRESENTED
BY '...' IN SO MANY PLACES.
FINALLY, YOU MUST LEARN DEFINITIONS AND COMPLETE A
REPRESENTATIVE SET OF THE PROBLEMS FOR THIS SECTION.
Circulation, curl and Stokes' Theorem:
The circulation of a vector around a closed path is defined as:
The circulation integral:
F  dr

the path
[VCA.16]
If one identifies F as v , the flow velocity of water, the circulation integral is non-zero for a
circulating flow pattern like the one around the drain of a bathtub as it empties. Just to run through
the arguments backwards, we deduce the definition of curl F from its desired integral property,
Stokes' Theorem.

( F )  nˆ dA 
any surface
bounded byC

F  dr
[VCA.17]
pathC
This relation only specifies the value of the integral of the curl over a surface. In order to deduce the
value of the curl at a point, we must follow a limiting procedure under which the path shrinks down
around the point. Thus, we define the component of the curl in the direction of the normal to the area
bounded by the curve as:
Def n: (curlF )  nˆ 
A F  dr
Limit A
enclosed
A,Aenc o
[VCA.18]
Note that the normal direction to the area is uniquely determined as A  0 as long as the surface
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is smooth (differentiable). Hence the component of the curl normal to the area in the sense of the
right-hand rule is the circulation per area. Be alert: the direction matters. The definition is more
clearly written once you understand the terms as:


curlF   Limit 
 A o
 enc
Aenc
A
o
nˆ dA   Limit


Aenc
F  dr
enc
The curl of a vector field is essentially another vector field.
The curl of a vector field behaves differently under a coordinate inversion than does a vector field. It is close
enough to a vector that it is to be treated as one. Review this point when you enter graduate school.
Green's Theorem is a special case in which Stokes' theorem is restricted to two dimensions.
Consider a problem with a closed path in the x-y plane. That path is translated upward (z direction)
one unit to sweep out a ‘vertical’ band. Surfaces parallel to the x-y plane are added to yield a closed
surface. Next we imagine a vector field with only x and y components that depends only on x and y,
and that have no z component; Green’s theorem follows. Prepare a representative sketch.
Reduce GUASS to theorem to Stokes 2D:  F  nˆ dA     F dV Consider
V

V
F  nˆ dA 
V
  F  nˆ drdz  (1)  F  nˆ dr   F dy  F dx
pathC
pathC
pathC
x
y
The outward directed normal to the surface described above is parallel to dr  kˆ . As
dr  dxiˆ  dy ˆj so n̂dr  dyiˆ  dxjˆ Comparing the leftmost integral to the rightmost above
leads some to refer to

Fx dy  Fy dx as the flux of F out of the curve. Recalling Gauss’s theorem
pathC
 Fx Fy 
 Fx Fy 
 Fx Fy 

dV


(1)
dA





V
V  x y  V  x y 
V  x  y dA
 F F 
Fx dy  Fy dx  A  x  y dA
Let Fx = M and Fy = -L
y 
 x
pathC  A
F  nˆ dA 

Green’s Theorem:

A

L dx  M dy   M
x
A
 Ly
dA
Let us apply the definition (VCA.18) to find the representation of the x component of curl F in
Cartesian coordinates.
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The direction of integration around the path was chosen such that the RHR normal to the surface is
iˆ . Check this by applying the right hand rule. For this path, we find:
 F  dr  F  x, y, z   y ˆj  F  x, y  y, z   z kˆ  F  x, y, z  z    y  ˆj  F  x, y, z    z  kˆ
 F  dr  F  x, y, z  y  F  x, y  y, z  z  F  x, y, z  z  y  F  x, y, z  z
y
z
y
z
After computing the dot products and dividing by the area y z, we find:
 Fy ( x, y , z )  Fy ( x, y , z  z) 

y


 F ( x, y  y, z )  F ( x, y, z ) 
 curl F   iˆ  Limit
  Limit
y 0 
z 0 
y
z
z


curl F iˆ  curl F 
 Fz
x
 
 y

Fy 

z 
[VCA.19]
This result can be quickly extended to the y and z components by cyclically permuting the
coordinates.
(x y  z  x  ...)
 curl F   ˆj   curl F 
y
 F F 
 x  z 
x 
 z
 curl F   kˆ   curl F 
z
 F F 
 y  x 
 x y 
These equations can be summarized in a determinant form for the case of Cartesian components.
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(DO NOT USE THE CROSS PRODUCT METHOD TO FIND THE CURL IN ANY
COORDINATE SYSTEM OTHER THAN CARTESIAN. REFER TO THE IDENTITIES AS
LISTED IN THE SECTION SUMMARY JUST BEFORE THE TOOLS OF THE TRADE. )
curl F 
iˆ
ˆj

x

y
kˆ

  F
z
[VCA.20]
Fx Fy Fz
This result motivates the other common notation for the curl (curl F   F ). The determinant
representation is to be avoided in cylindrical and spherical coordinates because, if not carefully
interpreted, it omits the terms that arise from the action of the differential operators on the coordinate
directions in the representation of F .
Alternative representation using the permutation symbol:
As notation, represent



by 1 ,
by  2 ,
by  3 and Fx by F1 , Fy by F2 , Fz by F3 . After
x
y
z
invoking the permutation symbol, the expression for the ith component of the curl of F becomes:
 F  
i

j , k 12,3
 i j k  j Fk
The result for cylindrical coordinates is:
 


 1  F F 
 Fr Fz  ˆ  1    r F
Fr  ˆ
z
 F   




k
 rˆ  
z 
r   r   r
 
 z
 r  


[VCA.21]
Summary of the method: Select one face of a coordinate cube that has its normal in the direction of
the component of the curl to be computed. Pick a starting corner and integrate around the path that is
the boundary of the face using the RHR to determine the positive sense for traversing the path.
Identify the differential line element that is each side of the path (such as d

1
 r d ˆ ). Evaluate

F  d for each side. (Example F  r d ˆ  F r d .) Compute the difference of the contributions
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VCA-25
from the 'parallel' sides to form the numerators in the definitions of partial derivatives. Divide by the
area of the face. Collect terms, complete the identification of each partial derivative and identify the
leftovers factors in each denominator.
The radial component of the curl in spherical coordinates is to be developed from the definition next.
Identify each step in the summary of the method.
Starting at  r, ,  and following the path in the sense indicated (which by the RHR selects the
normal  r̂ ),
 F  dr 








F r ,  ,  r   F r ,    ,  r sin       F r ,  ,    r   F r ,  ,  r sin  








 F r,   ,  sin      F r, ,  sin   r    F r , ,    F r , ,  r 




After dividing by the area (r 2 sin  ) and using the definition of a partial derivative, we find:

5/1/2017
 1     sin  F  F 
 F  rˆ  




 
 r sin   

Vector Calculus
[VCA.22]
VCA-26
The other two components are: (Derive the forms and record them in the space below. Check your
answers by comparing with the forms in the summary just before Tools of the Trade.)
[VCA.23]
[VCA.24]
The curl is the net circulation of F normal to the area per area.
The component of the curl for each coordinate direction is the net circulation around a coordinate
'square' with that normal divided by the area of the 'square'. Each component has two terms of
opposite sign that are the products of the lengths of the opposite sides times the corresponding
component of the vector. These four terms sum to the net circulation that is divided by the area to
give the RHR normal component of the curl. Taking cylindrical coordinates as an example:
 F  
r
 Fz z     Fz z 
  F  

r  z
 F r  
  F r   z 1  Fz
z z

 
r  z
r  
 Fr r z z   Fr r z  Fz z r r   Fz z r
r z

r z

Fr
z

 F

 z
Fz
r
 F r  
  F r   r  Fr r     Fr r  1    r F  Fr 
r r
 F   

 


z
r  z
r  z
r  r
 
Observe that a circulation requires that the vector field vary for position shifts perpendicular to the
direction of the field. The left-side field below has a curl that is out of the page corresponding to a
counter-clockwise circulation in the plane of the page. Imagine a small paddlewheel mounted on a
axis perpendicular to the page. Dip it into the field; what rotation would result? The field on the
right is non-constant, but is irrotational (it has no circulation). Dip the paddlewheel into the field on
the right-hand side. If the field represented flowing water, would it turn the wheel?
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As a final example consider
r
x iˆ  y ˆj  z kˆ

which is proportional to the electrostatic field
3/ 2
3
r
 x 2  y 2  z 2 
pattern of a point charge. Try the paddle wheel test. The field is divergence-free (solenoidal) except
at r = 0 and curl-free (irrotational).
Exercise: Compute the curl of
r
x iˆ  y ˆj  z kˆ

in the Cartesian coordinate representation.
r 3  x 2  y 2  z 2  3/ 2


A more fruitful example for curl and circulation is the field v  ( kˆ  r ) r 2  (  y iˆ  x ˆj ) r 2 where
k̂
is the z-direction. The field is illustrated in the x-y plane.
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y
x
y
x
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The first and third figures are in the itty-bitty arrow picture and the second in the field line picture.
Clearly
 v d
 0 for a circular path about the origin, but patient calculation shows that
 v  0 for r  0 . The field is irrotational everywhere except at x = y = 0. On that line, the curl
is singular. That is:  v  2
 ( x)  ( y) kˆ where x) is the Dirac delta function.
Exercise: Show that:  v  0 , where v  ( kˆ  r ) r 2  (  y iˆ  x ˆj ) r 2 , for non-zero r.
Exercise: Compute
 v d
for a circular path of radius R centered on the origin in the x-y plane.
The Integral Theorems
The following theorems relate integrals over volume to integrals over the surface that bounds the
volume and integrals over surfaces to integrals around the path that bounds that surface. These
results are incredibly important. Do not sleep until you have completely assimilated them into your
knowledge web.
ˆ where n̂ is
The flux of a vector field F through a surface element dA is defined as  F  F  ndA
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VCA-30
the chosen normal direction for the surface element. The net flux out of a volume is
 V F  nˆ dA
where the outward directed normal is chosen.
The circulation integral of a vector field is defined as the integral of the component of the field
tangential to the path integrated with respect to path length around a closed path,
 A F  dr .
Gauss's (Divergence) Theorem Where V is the closed surface that bounds the volume V.

V
 F dV   F  nˆ dA
[VCA.25]
V
The net flux of a vector field out of a volume can be computed directly or as the integral of the
divergence of the field integrated with respect to volume over the volume enclosed by the surface.
Where A is the closed path that bounds the surface A.
Stokes' (Circulation) Theorem
 A ( F )  nˆ dA   A F  dr
[VCA.26]
The circulation of a vector field around a closed path can be computed directly or as the integral of
the normal component the curl of the field integrated with respect to area over the surface enclosed
by the path. The normal is chosen in accord with the right-hand rule and the sense of the path
integral.
Fundamental Theorem of Vector Integral Calculus
rf
r
i
g   dr
 g (r )  g (r )
f
i
  g   dr
Given that the scalar valued function exists, it follows that
 g  dr vanishes for all closed paths.
equivalently
g (r )  g (r )  
f
i
rf
ri
g   dr
[VCA.27]
is path independent or
If either of these were not true, the equation
would yield multiple or indeterminate values for g (rf ) . These
issues also arise when relations such as: V (rf )  V (ri ) 
rf
r
i
 E   dr
are employed to define
potential functions V given an electric field E . The procedure yields a single-valued potential only
if  E  0 . The path integral of a vector field can be used to define a scalar (potential) field if the
vector field is curl-free as this condition guarantees that the integral is path-independent.
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Helmholtz Theorem (the Fundamental Theorem of Vector Analysis): The Helmholtz Theorem
states that a vector field is determined up to an additive constant vector by its divergence and
circulation. This result accounts for, among other things, the differential forms of Maxwell's
equations as presented in undergraduate texts. Each equation specifies the divergence or the curl of
one of the fields. Once the divergence and curl are known for a vector field, the field is determined
everywhere to within an additive constant. For electromagnetism, we will need four equations; div
and curl for E and for B . The E and B fields usually may be assumed to approach zero as
r   . In this case, the undetermined constant vector must be zero. Even better, the Helmholtz
Theorem provides a prescription for computing a vector field from its divergence and curl.
 1
 4

F (r )   
  F (r ) 3 

d r   C
r  r

all r 


F (r )  (r )   A(r )  C
or
( r ) 

  F (r ) 3 

 1
d r     
r  r

 4
all r 



1
4
  F (r ) 3
1
d r  and A(r ) 
r  r
4
all r 

[VCA.28]
where
  F (r ) 3
d r
r  r
all r 

[VCA.29]
Each unprimed  acts only on the unprimed field coordinates while each primed  acts only on the
primed source coordinates. The field coordinate r specifies the point at which the electric, magnetic
or other field is to be computed while the source coordinate r ranges over all space where sources
such as charge and current are distributed. The function (r ) is called the scalar potential for F ,
and the function A(r ) is called the vector potential for F . The undetermined constant vector is C .
The power of this theorem is that it states that only two attributes, the divergence and curl, of a
vector field need be determined to essentially fix a vector field. Further the relations above cleanly


separate the two attributes of the vector field as (r )  0 and   A  0 . There are
cases for which only one of the potentials is required to represent a vector field. If the field is curl
free, only  is needed, and the field is said to irrotational. If the field is divergence-free, only the
vector potential A is required, and the vector field is said to be solenoidal.
Suggestion: Whenever you encounter a new vector field, attempt to find and memorize
expressions for the divergence and curl (circulation density) of that field.
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NOTES FOR FUTURE ADDITIONS -- YOU MAY SKIP THEM!
The theorem above is the 'instantaneous' Helmholtz theorem. It is valid in the case that all fields
are evaluated over all space at a single instant in time according to an inertial observer. This
means that the theorem applies to electrostatics and magnetostatics in addition to the general case
in which all fields are evaluated at a single instant. There is a time dependent form that is more
appropriate for the study of time dependent problems such as radiating electromagnetic waves.
See Am. J Phys 72, page 412 March 2004; The Validity of the Helmholtz Theorem, F. Rohrlich.
A statement of the time dependent form appears in an appendix following Tools of the Trade in
section B of the vector calculus handout.
More than Helmholtz promises: One potential pair for E&B
Article in Jul, 2007 AJP Continuity implies Maxwell
Coulomb’s Law and the Law of Biot and Savart:
The Helmholtz theorem with time-dependence is the basis for two familiar laws for the electric and
magnetic field. First, note that the differential operators external to the integrals act on the field-point
variables, not the source-point variables. The field point values only appear in | r  r  | .
 1
 4

F (r )   

  F (r ) 3 

 1
d r     
r  r

 4
all r 



  F (r ) 3 

d r   C
r  r

all r 


[VCA.30]
After applying a few of the identities developed in this handout,
F (r ) 
1
4
  F (r )   r  r  
1
d 3r  
3

4
r  r
all r 
  F (r )    r  r  
d 3r   C
3


r r
all r 
[VCA.31]
In electrostatics, the electric field has a divergence source, charge density, and no circulation (curl)
source. All such irrotational, time-independent fields obey a Coulomb’s Law:
F (r ) 
1
4
 divF (r )   r  r  
d 3r 
3

r  r
all r 
[VCA.32]
Coulomb's Law Form
For the electric field, divE   01  and curlE  0 so:
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VCA-33
E (r ) 
1
4  0

 (r )  r  r  
r  r
all r 
3
d 3r 
Coulomb's Law
The constant vector has been suppressed as the fields usually vanish at infinity. In magnetostatics,
the magnetic induction has no divergence, but circulates around current. All such solenoidal, timeindependent fields obey a Biot-Savart Law:
F (r ) 
1
4
curlF (r )    r  r  
d 3r 
3


r r
all r 
[VCA.33]
Biot-Savart Law Form
For the magnetic field, divB  0 and curlB  0 J so:
B(r ) 
0
4

all r 
J (r )   r  r  
r  r
3
d 3r 
Biot-Savart Law
Exact Forms: A vector field is exact if it is the gradient of a scalar field. The scalar field is the
potential function for the vector field. An exact vector field has no curl and hence requires no vector
potential for its representation according to Helmholtz. His theorem provides a representation of the
field as the gradient of a scalar potential: F (r )   (r )  C  r  . The choice of plus or minus


the gradient of a scalar field is of no particular consequence. The negative sign is used when the
vector field is to be directed from regions with higher potential toward regions with lower potential.
The line integral of an exact vector field around a closed path vanishes.
Vector Differential Operators in Generalized Orthonormal Coordinates
A requirement for a system of orthonormal curvilinear coordinates is that they represent the
differential displacement dr and the associated distance dr as shown below. A general set of such


coordinates will be represented as q1, q2 , q3 . The three directions ê1 , ê2 , and ê3 may vary from
point to point, but, at any point, they are mutually (orthogonal) perpendicular. The line element
becomes (SEE PROBLEM #11.):
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dr  h1  q1, q2 , q3  dq1 eˆ1  h2  q1, q2 , q3  dq2 eˆ2  h3  q1, q2 , q3  dq3 eˆ3
[VCA.34]
where the unit vector eˆi associated with qi is the direction in which the coordinate point moves
when the coordinate qi is given a small positive increment while the other coordinates are held fixed.
The scale functions hi(q1,q2,q3) squared are the elements of the metric and, in particular, the length
of dr squared is:
ds 2  dr  dr   h1  q1 , q2 , q3   dq1    h2  q1 , q2 , q3   dq2    h3  q1 , q2 , q3   dq3 
2
2
2
2
2
2
[VCA.35]
The elements of the metric may be computed from the definitions of the Cartesian coordinates (x,y,z)
= (x1, x2, x3) in terms of the q's. (SEE PROBLEM #11)
2
 x 
g11  h    i  ,
i 1  q1 
3
2
1
 x 
g 22  h    i 
i 1  q2 
3
2
2
2
 x 
g33  h    i 
i 1  q3 
3
2
2
3
,
[VCA.36]
For example, the line element in spherical coordinates is:
drsph  dr rˆ  r d ˆ  r sin  dˆ
[VCA.37]
For the systems studied above:
TABLE II: Metric Factors and Coordinate Directions for the Big Three
Coordinates
h1
ê1
h2
ê2
h3
ê3
Cartesian
x,y,z
1
iˆ
1
ĵ
1
k̂
Cylindrical
r,,z
1
r̂
r
ˆ
1
Spherical
r,,
1
r̂
r
ˆ
k̂
ˆ
System
r sin
Gradient: The gradient is particularly easy to find as it has as its component in a direction the rate
of change with respect to distance in that direction.
G   limit  hGq   h1  qG 
i
qi 0
i
i
i
[VCA.38]
i
Use this relation and the table above to generate the components of the gradient in cylindrical and
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Vector Calculus
VCA-35
Cartesian coordinates.
Divergence: The divergence in general orthonormal-curvilinear coordinates is a little more difficult
to develop.
(q1+q1,q2+q2,q3+q3)
nˆ  eˆ3
nˆ  eˆ1
h3 dq3 ê3
h2 dq2 ê2
(q1,q2,q3)
nˆ  eˆ2
nˆ  eˆ1
h1 dq1 ê1
We consider a coordinate 'cube' with one corner at
 q1  q1, q2  q2 , q3  q3  .
 q1, q2 , q3  and the opposite corner at
In the limit that the q ' s  0 , each ê1 approaches the other ê1 's.
[Note that ê2 drawn in the center of the figure is 'into the page'.] Following the definition, we wish
to compute the flux of F out of the volume.
 F  nˆ dA   F (q , q , q )  (eˆ )
1
2
3
1
h2 (q1 , q2 , q3 ) h3 (q1 , q2 , q3 ) q2 q3
  F (q1  q1 , q2 , q3 )  (eˆ1 )  h2 (q1  q1 , q2 , q3 ) h3 (q1  q1 , q2 , q3 ) q2 q3

(terms for sides with outward normals  eˆ2 and  eˆ3 )
Using F   eˆ1   F1 and so forth, we continue by dividing by the volume of the 'cube'.
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VCA-36

F  nˆ dA [ F1 h2 h3  ( q q ,q ,q )   F1 h2 h3  ( q ,q ,q ) ] q2 q3
1
1 2 3
1 2 3

 ..........
V
h1 q1 h2 q2 h3 q3
As the q ' s  0 ,
 F1 h2 h3  ( q q ,q ,q )   F1 h2 h3  ( q ,q ,q )
1
1
2
3
1
V
1
h1 h2 h3
3
q1
h1 h2 h3
 F  nˆ dA 
2

  F1 h2 h3 
1
h1 h2 h3
q1
   F1 h2 h3    F2 h1 h3    F3 h1 h2  



   F
q1
q2
q3


The last two terms follow by permuting the indices 1  2  3  1 .... although one can compute
them directly if so inclined.
In cylindrical coordinates, we have hr = 1, h = r , hz = 1 so:
1  ( Fr r ) ( F ) ( Fz r )  1 ( Fr r ) 1 ( F ) ( Fz )
 F  





r  r

z  r r
r 
z
Curl: The curl in general orthonormal curvilinear coordinates follows a little more easily. Following
the definition, the circulation around a coordinate 'square' is computed.
(curl F )  eˆ1  Limit
A0

 ( A )
F  dr
A
 Limit
q ' s 0

 ( A )
F  dr
h2 q2 h3 q3
The notation (A) specifies the path that is the boundary of the area A.
(q1,q2+q2,q3+q3)
nˆ   eˆ1
h3 dq3 ê3
h2 dq2 ê2
(q1,q2,q3)
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VCA-37

 ( A )
F  dr  F (q1 , q2 , q3 )  eˆ2 h2 (q1 , q2 , q3 ) q2  F (q1 , q2  q2 , q3 )  eˆ3 h3 (q1 , q2  q2 , q3 ) q3
 F (q1 , q2 , q3  q3 )  (eˆ2 ) h2 (q1 , q2 , q3  q3 ) q2  F (q1 , q2 , q3 )  (eˆ3 ) h3 (q1 , q2 , q3 ) q3

 ( A )
F  dr   F2 h2  ( q ,q ,q )   F2 h2  ( q , q , q q )  q2   F3 h3  ( q , q q ,q )   F3 h3  ( q , q , q )  q3
1 2 3
1 2 3
3 
1 2
2 3
1 2 3 


[ F3 h3 ]
[ F2 h2 ]


q2 q3 
q2 q3
q3
q2
(curl F )  eˆ1  Limit

 ( A )
F  dr
A
A0


(curl F )1   F  
1
[ F3 h3 ]
[ F2 h2 ]
q2 q3 
q2 q3
q3
q2
h2 q2 h3 q3
1  (h3 F3 ) (h2 F2 ) 



h2 h3  q2
q3 
The remaining components can be found by permuting the indices.
   (hk Fk )  (h j Fj ) 
  F   ijk 



 i h h  q
qk 
j k 
j
[VCA.39]
Applying this relation to the radial component of the curl in spherical coordinates, we find:
(curl F )r 
 (r sin  F ) (r F ) 
1  (h F ) (h F ) 
1






h h  
  r r sin  

 
(curl F )r 
1  (sin  F ) ( F ) 



r sin  

 
After you have mastered the derivation of the expressions for the various differential operators in
cylindrical and spherical coordinates, it is permissible to just plug into the general curvilinear results
if you need the operators in some other locally orthonormal coordinate system.
The appearance of the ijk in [VCA.39] suggest that a determinant is lurking in the area. In fact,
h1 eˆ1 h2 eˆ2
1


 F 
q2
h1 h2 h3 q1
h1 F1 h2 F2
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h3 eˆ3

q3
.
h3 F3
Vector Calculus
VCA-38
In practice, students make as errors using this form. It is recommended that you start with:
(curl F )1   F  
1
1  (h3 F3 ) (h2 F2 ) 



h2 h3  q2
q3 
Next, permute the indices to write out the expressions for the 2 and 3 components. Then execute a
careful plug and chug.

The Laplacian of a scalar field is generated by computing the divergence of the gradient of that scalar
field.
Laplacian:  2 g 
1
h1 h2 h3
   h2 h3 g    h1 h3 g    h1 h2 g  








q
h

q

q
h

q
1

1
1

2

2
2
 q3  h3 q3  




[VCA.40]

The Laplacian of a vector field is found using the identity 2 F     F    F .
Summary: Differential Operators in Generalized Orthonormal Coordinates
dr  h1  q1, q2 , q3  dq1 eˆ1  h2  q1, q2 , q3  dq2 eˆ2  h3  q1, q2 , q3  dq3 eˆ3
 1 g 
 1 g 
 1 g 
1 g
 or g i 
  eˆ2 
  eˆ3 
hi qi
 h1 q1 
 h2 q2 
 h3 q3 
g  eˆ1 
 F 
h1 eˆ1
1

 F 

h1h2 h3 q1
h1 F1
2 g 
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1
h1 h2 h3
1    F1 h2 h3    F2 h1 h3    F3 h1 h2  



h1 h2 h3  q1
q2
q3

h2 eˆ2
h3 eˆ3

q2

q3
h2 F2
h3 F3
or
   (hk Fk )  (h j Fj ) 
  F   ijk 



 i h h  q
qk 
j k 
j
   h2 h3 g    h1 h3 g    h1 h2 g  







 q1  h1 q1  q2  h2 q2  q3  h3 q3  
Vector Calculus
VCA-39
Second Order Vector Differential Operations
 f   2 f

(Defines the Laplacian.)
[VCA.41]
   F  0

[VCA.42]
  f  0
[VCA.43]




  F     F   2 F


[VCA.44]


Note that the expression 2 F     F    F defines 2 F , the Laplacian of a vector field,
in terms of operations that have been previously defined. The expression 2 F is equivalent to
eˆ1  2 F1  eˆ2  2 F2  eˆ3  2 F3 only in Cartesian coordinates. In all other systems, the curvilinear nature
and non-constant coordinate directions lead to additional terms.






2 F     F    F


[VCA.45]
The identity also has the reordering,    F   2 F    F . The gradient of the divergence is
a possible second order operation. It is included implicitly in the list above. 
The Extended Integral Theorems
Note: V is the surface that bounds the volume V and A the path bounding A.
  f  g  f g  dV  
2
  f  g  g 
2
2
f  dV 
V
 f  dV  
V
f g  nˆ dA
f nˆ dA 
V
  f g  gf   nˆ dA
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[VCA.47]
V

f dA
[VCA.48]
V
  F  dV    F  nˆ dA
V
[VCA.46]
V
V
[VCA.49]
V
Vector Calculus
VCA-40
 f  nˆ dA    f
[VCA.50]
d
A
A
The equation [VCA.46] is designated as the first form of Green's theorem. It plays a prominent role
in the Laplace equation handout
To prove the results above:
When you use a constant vector, rewrite the answer for c  iˆ, ˆj , and kˆ.
1.) For [VCA.46], apply the divergence theorem to the vector: f g .
2.) For [VCA.47], use the extended theorem [VCA.46] twice.
3.) For [VCA.48], use the vector f C in the divergence theorem where C is constant and f is a
scalar function.
4.) For [VCA.49], apply the divergence theorem to F  C where C is constant.
5.) For [VCA.50], apply Stokes' theorem to the vector f C where C is constant and f is a scalar
function.
Surface Integrals – areas
Some Properties of the Surface Area Integral
Definition: vector area:
A 
surface
nˆ dA
[VCA.51]
1.) The vector area of any closed surface is zero.
2.) Any two smooth surfaces with the same boundary have the same vector area.
3.) A  1 2
A r  d
where A is the path bounding A.
4.) For any constant vector C , A  C 
A C  r  d
where A is the area enclosed by the path.
To prove the area results:
1.) Use extended integral theorem [VCA.48].
2.) Use area property (1.) from the list just above.
3.) Imagine a line from the origin to a point on the curve. As that point moves along the path, the
line sweeps out the surface area of a complicated cone. For a small movement along the path, the
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VCA-41
line sweeps out an area
1
2
r  d  .
4.) Use extended integral theorem [VCA.50] with f  C  r where C is a constant vector and r is
the position vector.
Sample Calculation: Surface area integral of a right circular cone
zmax = R cot

n̂
R cot
R
Sketch a right circular cone concentric with the z-axis with its base of radius R in the x-y plane and
its vertex a distance h above that plane. The cone half-angle is  where tan() = R/h. Compute the
area of the curved side surface of the cone. The edge of the cone is defined by z = (R – r) cot.
Adopting cylindrical coordinates with the naming conventions (r, , z), a patch of area on the surface
is illustrated. It has edge vectors (dr rˆ  dz kˆ) and r d ˆ . The area of the patch has the form:
dA  r d dz rˆ  dr dz ˆ  r d dr kˆ . A problem remains. A surface patch has two normal directions.
One or the other must be chosen. The default dA  r d dz rˆ  dr dz ˆ  r d dr kˆ convention might
not be appropriate. Examining the figure, one concludes that the normal with positive radial and z
components is needed.
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Choose a set of parameters: It is a surface, so it can be described by two parameters. A popular
choice is to select a subset of the coordinates. At any point on the surface, varying  transports the
point to another point on the surface without changing r or z. The coordinate  must be chosen as one
of the parameters as it is independent of r and z. There is a firm functional relation between r and z
so either one can be chosen as the second parameter. With no further thought or justification, r and 
are chosen as the parameters.
tan() = R/h;
r(z) = R – z tan ;
z = (R - r) cot();
dz = - dr cot
Our integration limits are to be chosen to run in the positive sense so that dr and dz are positive.
With these choices, the signs are adjusted to generate a normal with positive radial and z
components, dA   cot( ) dr r d rˆ  dr ( cot  dr ) ˆ  r d dr kˆ .
The strange form of the phi component suggests that there is no phi component. See Further
discussion below.
That can be checked as the area element is the cross product of two edge elements for the patch. The
less ad hoc cross product procedure is presented next.
FORMAL PROCEDURE:
i.) Set the normal: The area element has a normal direction which is outward for a closed surface.
For this problem, the normal is chosen to have a positive z component. That direction is the same as
that of the cross product of the two edge vectors in the drawing. dA nˆ  (dr rˆ  dz kˆ)  r d ˆ
ii.) Choose a set of parameters: It is an area, so it can be described by two parameters. A popular
choice is to select a subset of the coordinates. At any point on the surface, varying  transports the
point to another point on the surface without changing r or z. The coordinate  must be chosen as one
of the parameters as it is independent of r and z. There is a firm functional relation between r and z
so either one can be chosen as the second parameter. With no further thought or justification, r and 
are chosen as the parameters.
iii.) Identify the limits: The limits required to sweep out the surface are: 0  r  R and 0   < 2.
iv.) Identify the edges: Find a corner of the area patch from which you can draw edges
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corresponding to positive increments in the chosen parameters. Begin with the line element
dr rˆ  r d ˆ  dz kˆ and generate the vector representations for the two edges, dr rˆ  dz kˆ and r d ˆ .
Choose the order that such that edge 1 cross edge 2 is in the chosen normal direction for positive
increments of the parameters.
v.) Generate dA using the Cross Product Method:
dA  (dr rˆ  dz kˆ)  r d ˆ  r dr d kˆ  r d dz rˆ .
Do not fret; dz is negative so the direction has a positive z component as specified.
vi.) Reduce the variable set to the parameter set: dz = - dr cot
dA  (dr rˆ  dz kˆ)  r d ˆ  r dr d kˆ  cot( ) r d dr rˆ
vi.) Assemble and integrate:
R
2
o
0
 dA   
R
 cot( ) r d dr rˆ  r dr d kˆ  


 o

2
0
 cot( ) r d dr (cos  iˆ  sin  ˆj )  r dr d kˆ 


Note that the non-constant coordinate direction was immediately replaced by its representation in
terms of the Cartesian directions. After a couple of quick integrations,
 dA  
R
o
r dr 2 kˆ   R2 kˆ
[VCA.52]
You should visualize adding little vectors with magnitude the area patch size and with the normal
direction all over the surface. Clearly, only a z component will survive, and it will have a magnitude
equal to the projection of the surface on a constant z plane.
Note: After the choice of normal was set, the process proceeded automatically according to the rules
on calculus and algebra without the need for further intervention. The more formal cross product
method is preferred.
ˆ
A slightly bolder approach uses: rˆ   d d .
 dA  
R
o

2
0
 cot( ) r d dr  dˆ   r dr d kˆ   R cot( ) r d dr ˆ(2 )  ˆ(0)   r dr 2 kˆ
d 



 o

Exercise: Complete the argument to show that the evaluation above yields:  R 2 kˆ .
One might need to know the volume of paint necessary to cover the surface of the cone with a paint
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Vector Calculus
VCA-44
layer of uniform thickness. In this case one needs:  | dA | .
|dA| = r dr d [ (cot)2 + 1]½ = r dr d csc.
 | dA |  
R
o

2
0
cot 2 ( )  1 r dr d  csc( ) 
R
o

2
0
r dr d  csc( )  R2
Clearly, the net vector area was not the desired quantity when you are planning to purchase paint.
The vector area is desired to calculate projected areas.
Picture the various small patches of area and mentally compute their vector sum. The result is
clearly correct; it does not give the total surface area of the side of the cone. Rather, the magnitudes
of each area patch are to be computed and summed:

dA  
R
o

2
0
R
ˆ
cot( ) r d dr d d  r d dr kˆ  
o

2
0
r d dr (cot[ ]) 2  1   R 2 csc( )
The magnitude of the vector area patch follows as the square root of the sum of the squares of its
components – it follows in the normal fashion.
To verify that dA  r d dr csc( ) , invoke the cross product. The magnitude of the cross product is
the parallelogram area patch with the vectors as adjacent sides. Pick area patches with a side r d ˆ
directed around the cone and a side dr rˆ  dr cot( ) kˆ directed down the side of the cone away from
its vertex. Make a sketch! Believe that the surface could be covered with little patches of this


description. The cross product r d ˆ  dr rˆ  dr cot( ) kˆ  r dr d  kˆ  cot( ) rˆ  has magnitude


csc( ) r dr d as promised.
A right circular cone can be constructed by
rolling up a pie-shaped wedge cut from a circle.
The vertex to base plane distance is
2R
R 2  h2
included
and so is the radius of the circle is chosen to be
R 2  h2
that value. The included angle in the wedge must
be the circumference of the base divided by the
radius of the wedge. included =
2 R
R2 h2
.
Roll so as to join the two dashed edges.
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Vector Calculus
VCA-45
The area of the wedge is
1

included  wedge radius   1 2
2
2
2 R
R 2  h2


R 2  h2
2
or  R R 2  h2 .
Exercise: Show that  R R 2  h2 =  R 2 csc( ) for the cone described above.
Further discussion: In the less formal method, there was another foggy issue which never even
arose when the more formal procedure was followed..
dA   cot( ) dr r d rˆ  dr ( cot  dr ) ˆ  r d dr kˆ
The center term dr cot dr needs to be addressed. Imagine integrating it with respect to r using the
Riemann sum approach. Slicing the range 0 to R into N equal slices, dr = R/N.

R
0
N
  NR   0
R
dr dr  Limit  N
N 
k 1
The term vanishes.
*** Add the variable density paint problem from the BVCG handouts.
Alternative Definitions: The extended integral theorems and the area integral can be used to show
that the following definitions are equivalent to the ones used in the current development.

grad ( )    Limit
 nˆ dA
V

V ,V 0
V

curlF    F  Limit
V
V ,V 0
V

 dA
V
dA nˆ  F

V
V ,V 0
dV
dV

 Limit

 Limit
V ,V 0
V
dV
[VCA.53]
dA  F

V
dV
[VCA.54]
which when taken with
divF    F  Limit

V
V ,V 0
dA nˆ  F

V
dV
 Limit
V ,V 0

V

V
dA  F
dV
[VCA.55]
define the three operations in analogous fashions.
Summary of the Differential Operators
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Vector Calculus
VCA-46
Cartesian Vector Calculus Operations
G ˆ G ˆ G
G  iˆ
j
k
x
y
z
gradient
F
divergence
 F F
 F  iˆ  z  y
z
 y
curl
 2G 
Laplacian
G  rˆ
gradient
divergence
ˆj  Fx  Fz
x
 z
 ˆ  Fy Fx 


k 

 x y 
r  x2  y 2
G ˆ  1  G ˆ G
  
k
r
z
 r  
 1    r Fr   1  F Fz
 F   
 

 r  r
 r   z
 1  F F 
 F F 
 1     r F  Fr  ˆ
 F    z    rˆ   r  z  ˆ    

k
r 
 
 r   r
 z
 r   z 
curl
Laplacian
2
2
 1    G   1   G  G
 2G     r


  2 2
z 2
 r  r  r   r  
Spherical Vector Calculus Operations
G  rˆ
gradient
divergence
5/1/2017



 2G  2G  2G

 2
x 2 y 2
z
Cylindrical Vector Calculus Operations
curl
Fx Fy Fz


x y
z
r  x2  y 2  z 2
G ˆ  1  G ˆ  1  G
  
 

r
 r  
 r sin   
2
 1    r Fr   1    sin  F   1  F
  F  2 




r
sin



 r  r


 r sin   
    sin  F  F 
 1   1

 rˆ    


 
 r   sin 
 
 1     r F  Fr  ˆ
 

 
 
 r   r
 1
 F  
 r sin 
Vector Calculus
 Fr   r F   ˆ



r 
 
VCA-47
Laplacian
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  2G
G  
1
 1    G   1   
 2G   2   r 2

sin


 

  2
 
   r 2 sin 2    2
 r  r  r   r sin    
Vector Calculus
VCA-48