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Chapter 16
Curvilinear VectorAnalysis
All the vector analytical quantities discussed in the previous chapters canalso be calculated in other coordinate systems. The general procedure is tostart with definitions of quantities in a coordinate-free way and substitute theknown quantities in terms of the particular coordinates we are interested inand “read off” the vector analytic quantity. Instead of treating cylindrical andspherical coordinate systems separately, we lump them together and derive re-lations that hold not only in the three familiar coordinate systems, but also inall coordinate systems whose unit vectors form a set of right-handed mutuallyperpendicular vectors. Since the geometric definitions of all vector-analyticquantities involve elements of length, we start with the length elements.
16.1 Elements of Length
Consider curvilinear coordinates1 (q1, q2, q3) in which the primary line curvilinearcoordinateselements are given by
dl1 = h1(q1, q2, q3) dq1, dl2 = h2(q1, q2, q3) dq2, dl3 = h3(q1, q2, q3) dq3,
where h1, h2, and h3 are some functions of coordinates. By examining theprimary line elements in Cartesian, spherical, and cylindrical coordinates, wecan come up with Table 16.1.
Denoting the unit vectors in curvilinear coordinate systems by e1, e2, ande3, we can combine all the equations for the elements of length and writethem as a single vector equation:
dr = d�l = e1dl1 + e2dl2 + e3dl3 = e1h1dq1 + e2h2dq2 + e3h3dq3. (16.1)1As will be seen shortly, Cartesian coordinates are also included in such curvilinear
coordinates. The former have lines (and planes) as their primary lengths and surfaces, thusthe word “linear” in the name of the latter.
424 Curvilinear Vector Analysis
Curvilinear Cartesian Spherical Cylindricalq1 x r ρq2 y θ ϕq3 z ϕ zh1 1 1 1h2 1 r ρh3 1 r sin θ 1
Table 16.1: The specifications of the three coordinate systems in terms of curvilinear
coordinates.
This equation is useful in its own right. For example, we can obtain the curvi-linear unit vectors as follows. Rewrite Equation (16.1) in terms of increments:
Δr ≈ e1h1Δq1 + e2h2Δq2 + e3h3Δq3.
Keeping q2 and q3 constant (so that Δq2 = 0 = Δq3), divide both sides byΔq1 to obtain
ΔrΔq1
≈ e1h1.
In the limit, the LHS becomes a partial derivative and we get
e1 =1h1
∂r∂q1
. (16.2)
The other two unit vectors can be obtained similarly. We thus have
Box 16.1.1. The ith unit vector of a curvilinear coordinate system isgiven by
ei =1hi
∂r∂qi
, i = 1, 2, 3. (16.3)
This is a useful formula for obtaining the Cartesian components of curvilinearunit vectors, when the Cartesian components of the position vector are givenin terms of curvilinear coordinates.
Example 16.1.1. As an illustration of the above procedure, we calculate the unitvectors in spherical coordinates. First we write
r = xex + yey + zez = exr sin θ cos ϕ + eyr sin θ sin ϕ + ezr cos θ.
Now we differentiate with respect to r to get
e1 ≡ er =∂r
∂r= ex sin θ cos ϕ + ey sin θ sin ϕ + ez cos θ.
16.2 The Gradient 425
Similarly,
e2 ≡ eθ =1
r
∂r
∂θ= ex cos θ cos ϕ + ey cos θ sin ϕ − ez sin θ,
e3 ≡ eϕ =1
r sin θ
∂r
∂ϕ= −ex sin ϕ + ey cos ϕ,
where we have used Table 16.1. These are the results we obtained in Chapter 1 frompurely geometric arguments. �
We are now in a position to find the gradient, divergence, and curl ofa vector field in general curvilinear coordinates. Once these are obtained,finding their specific forms in cylindrical and spherical coordinates entailssimply substituting the appropriate expressions for q1, q2, and q3 and h1, h2,and h3.
16.2 The Gradient
The gradient is found by equating
df =∂f
∂q1dq1 +
∂f
∂q2dq2 +
∂f
∂q3dq3
to the differential of f in terms of the gradient:
df = ∇f · dr = (∇f)1h1 dq1 + (∇f)2h2 dq2 + (∇f)3h3 dq3.
The last two equations yield
(∇f)1h1 =∂f
∂q1, (∇f)2h2 =
∂f
∂q2, (∇f)3h3 =
∂f
∂q3,
which gives gradient incurvilinearcoordinates
Box 16.2.1. The gradient of a function f in a curvilinear coordinatesystem is given by
∇f = e11h1
∂f
∂q1+ e2
1h2
∂f
∂q2+ e3
1h3
∂f
∂q3. (16.4)
This result, in conjunction with Table 16.1, agrees with the expression ob-tained for the gradient in the Cartesian coordinate system. In cylindricalcoordinates, we obtain
∇f = eρ∂f
∂ρ+ eϕ
1ρ
∂f
∂ϕ+ ez
∂f
∂z, (16.5)
426 Curvilinear Vector Analysis
so that the operator ∇ in cylindrical coordinates is given by
∇ = eρ∂
∂ρ+ eϕ
1ρ
∂
∂ϕ+ ez
∂
∂z. (16.6)
Similarly, in spherical coordinates, we getgradient of afunction insphericalcoordinates
∇f = er∂f
∂r+ eθ
1r
∂f
∂θ+ eϕ
1r sin θ
∂f
∂ϕ(16.7)
with the operator ∇ given by
∇ = er∂
∂r+ eθ
1r
∂
∂θ+ eϕ
1r sin θ
∂
∂ϕ. (16.8)
Example 16.2.1. The electrostatic potential of an electric dipole was given inExample 10.5.1 in spherical coordinates. With the expression for the gradient givenabove, we can find the electric field E = −∇Φ of a dipole in spherical coordinates:
Er = −∂Φdip
∂r= − ∂
∂r
(kep cos θ
r2
)
=2kep cos θ
r3,
Eθ = −1
r
∂Φdip
∂θ= −1
r
∂
∂θ
(kep cos θ
r2
)
=kep sin θ
r3,
Eϕ = − 1
r sin θ
∂Φdip
∂ϕ= − 1
r sin θ
∂
∂ϕ
(kep cos θ
r2
)
= 0.
Summarizing, we haveelectric field of anelectric dipole
Edip =kep
r3(2er cos θ + eθ sin θ). (16.9)
This is the characteristic field of a dipole. �
Example 16.2.2. Just as electric charges can produce electric dipoles, electriccurrents can produce magnetic dipoles. We saw this in Subsection 15.2. In thisexample, we will calculate the magnetic field of a dipole directly. Consider themagnetic field of a circular loop of current as given in Equations (4.24) and (4.26).We change the coordinates of the field point P to spherical and assume that P is faraway from the loop, i.e., that a is very small compared to r. Writing r2 for ρ2 + z2
and r sin θ for ρ, we expand the integrands of (4.24) and (4.26) in powers of a/rkeeping only the first nonzero power. Thus,
1
(r2+a2−2ra sin θ cos t)3/2=
1
r3
[
1 +(a
r
)2
− 2(a
r
)sin θ cos t
]−3/2
=1
r3
[1 + 3
(a
r
)sin θ cos t
]+ · · · ,
r sin θ cos t−a
(r2+a2−2ra sin θ cos t)3/2=
1
r2
(sin θ cos t − a
r
) [
1 +(a
r
)2
− 2(a
r
)sin θ cos t
]−3/2
=1
r2
(sin θ cos t − a
r
) [1 + 3
(a
r
)sin θ cos t
]+ · · ·
=1
r2
(
sin θ cos t − a
r+
3a
rsin2 θ cos2 t
)
.
16.3 The Divergence 427
Substituting these in the integrals of (4.24) and (4.26) yields
Bρ =kmIaz
r3
∫ 2π
0
cos t
(
1 +3a
rsin θ cos t
)
dt =3kmIπa2 cos θ sin θ
r3,
where we substituted r cos θ for z. In an analogous way, we also obtain
Bz = −kmIa
r2
∫ 2π
0
(
sin θ cos t − a
r+
3a
rsin2 θ cos2 t
)
dt
= −kmIa
r2
(
−2πa
r+
3aπ
rsin2 θ
)
.
We are interested in the spherical components of the magnetic field. To findthese components, we first write
B = Bρeρ + Bz ez
and take the dot product with appropriate unit vectors:
Br = B · er = Bρeρ · er + Bz ez · er = Bρ sin θ + Bz cos θ
=3kmIπa2 cos θ sin θ
r3sin θ +
kmIa
r2
(2πa
r− 3aπ
rsin2 θ
)
cos θ
=2kmIπa2
r3cos θ.
Similarly,
Bθ = B · eθ = Bρeρ · eθ + Bz ez · eθ = Bρ cos θ − Bz sin θ
=3kmIπa2 cos θ sin θ
r3cos θ − kmIa
r2
(2πa
r− 3aπ
rsin2 θ
)
sin θ
=kmIπa2
r3sin θ.
Summarizing, we write magnetic field of amagnetic dipole
B =kmIπa2
r3(2er cos θ + eθ sin θ). (16.10)
This has a striking resemblance to Equation (16.9). In fact once we identify Iπa2
as the magnetic dipole of the loop, and change all magnetic labels to electric ones,we recover Equation (16.9). �
16.3 The Divergence
To find the divergence of a vector A, we consider the volume element ofFigure 16.1 and find the outward flux through the sides of the volume. Forthe front face we have
Δφf = Af · e1Δaf ,
428 Curvilinear Vector Analysis
A
A1 ˆ e 1
A2 e 2
A3 e 3
P(q1, q2, q3)
Δ
Δ
Δ
l1
l3
l2
Figure 16.1: Point P and the surrounding volume element in curvilinear coordinates.
Note that the midpoints of the front and back faces are Δq1/2 away from P in the
positive and negative e1 directions, respectively. Similarly for the other four faces.
where Af means the value of A at the center of the front face and Δaf is thearea of the front face. Following the arguments presented for the Cartesiancase, we write
Δφf ≈ Af · e1Δaf = A1fΔl2fΔl3f
= A1f (h2Δq2)f (h3Δq3)f = A1fh2fh3fΔq2Δq3
The subscript 1 in A1f , for example, means component of A in the directionof the first coordinate. The subscript f implies evaluation—at the midpoint—on the front side whose second and third coordinates are the same as P , andwhose first coordinate is q1 + Δq1/2. Thus, we have
Δφf ≈ A1
(
q1 +Δq1
2, q2, q3
)
h2
(
q1 +Δq1
2, q2, q3
)
× h3
(
q1 +Δq1
2, q2, q3
)
Δq2Δq3
because, unlike the Cartesian case, h1, h2, and h3 are functions of the co-ordinates. Using Taylor series expansion for the functions A1, h2, and h3
yields
Δφf ≈{
A1(q1, q2, q3) +Δq1
2∂A1
∂q1
}{
h2(q1, q2, q3) +Δq1
2∂h2
∂q1
}
×{
h3(q1, q2, q3) +Δq1
2∂h3
∂q1
}
Δq2Δq3.
16.3 The Divergence 429
Multiplying out and keeping terms up to the third order (corresponding tothe order of a volume element by which we shall divide shortly), we obtain
Δφf ≈{
A1h2h3 + A1h2∂h3
∂q1+ A1h3
∂h2
∂q1+ h2h3
∂A1
∂q1
}Δq1
2Δq2Δq3
=
{
A1h2h3 +∂
∂q1(h2h3A1)
}Δq1
2Δq2Δq3,
where we left out the explicit dependence of the functions on their independentcoordinate variables. For the back face we have
Δφb ≈ Ab · (−e1Δab) = −A1bΔl2bΔl3b = −A1b(h2Δq2)b(h3Δq3)b
= −A1
(
q1 −Δq1
2, q2, q3
)
h2
(
q1 −Δq1
2, q2, q3
)
× h3
(
q1 −Δq1
2, q2, q3
)
Δq2Δq3.
Taylor expanding the three functions A1, h2, and h3 as above, and multiplyingout yields
Δφb ≈ −{
A1h2h3 −∂
∂q1(h2h3A1)
}Δq1
2Δq2Δq3.
Adding the front and back contributions, we obtain
Δφ1 ≡ Δφf + Δφb ≈ ∂
∂q1(h2h3A1) Δq1Δq2Δq3.
Similarly, the fluxes through the faces perpendicular to e2 and e3 are
Δφ2 ≈ ∂
∂q2(h1h3A2)Δq1Δq2Δq3,
Δφ3 ≈ ∂
∂q3(h1h2A3)Δq1Δq2Δq3. (16.11)
Adding the three contributions and dividing by the volume
ΔV = Δl1Δl2Δl3 = h1h2h3Δq1Δq2Δq3
and finally taking the limit of smaller and smaller volumes—which turns allapproximations into equalities—we get divergence in
curvilinearcoordinatesTheorem 16.3.1. The divergence of a vector field A in a curvilinear coordi-
nate system is given by
∇ ·A =1
h1h2h3
{∂
∂q1(h2h3A1) +
∂
∂q2(h1h3A2) +
∂
∂q3(h1h2A3)
}
.
430 Curvilinear Vector Analysis
Now that we have a general formula for the divergence, we can use Table16.1 to write the divergence in a specific coordinate system. For instance,substituting the entries of the second column gives the formula in Theorem13.2.1, and the third column yieldsdivergence of a
vector field insphericalcoordinates
∇ · A =1
r2 sin θ
{∂
∂r
(r2 sin θAr
)+
∂
∂θ(r sin θAθ) +
∂
∂ϕ(rAϕ)
}
=1r2
∂
∂r
(r2Ar
)+
1r sin θ
{∂
∂θ(sin θAθ) +
∂Aϕ
∂ϕ
}
. (16.12)
To obtain the divergence in cylindrical coordinates, we use the last columnand obtain
∇ · A =1ρ
{∂
∂ρ(ρAρ) +
∂
∂ϕ(Aϕ) +
∂
∂z(ρAz)
}
=1ρ
∂
∂ρ(ρAρ) +
1ρ
∂Aϕ
∂ϕ+
∂Az
∂z. (16.13)
Example 16.3.2. Consider the vector field defined by
A = krαer,
where k and α are constants. Let us verify the divergence theorem for a sphericalsurface of radius R (see Figure 16.2). The total flux is obtained by integrating overthe surface of the sphere:
φ =
∫ ∫
S
A · da =
∫ ∫
S
kRαer · enR2 sin θ dθ dϕ
= kRα+2
∫ ∫
S
sin θ dθ dϕ = 4πkRα+2.
dθ dϕ
e n = e r
R
x
y
z
Figure 16.2: The element of area and its unit normal for a sphere.
16.4 The Curl 431
On the other hand, using the expression for divergence in the spherical coordinatesystem and noting that Aθ = 0 = Aϕ, we obtain
∇ · A =1
r2
∂
∂r(r2Ar) =
1
r2
d
dr
(krα+2
)= (α + 2)krα−1,
where we have assumed that α �= −2. Therefore,∫ ∫
V
∫∇ · A dV =
∫ R
0
(α + 2)krα−1r2 dr
∫ π
0
sin θ dθ
∫ 2π
0
dϕ = 4πkRα+2
which agrees with the surface integration.For α = −2 the divergence appears to vanish everywhere. However, a closer
examination reveals that the statement is true only if r �= 0. In fact, as we discussedbefore, the divergence of A is proportional to the Dirac delta function, δ(r) in thiscase [see Equation (15.2)]. �
16.4 The Curl
To calculate the curl, we choose a closed path perpendicular to one of the unitvectors, say e1 and calculate the line integral of A around it. The situation isdepicted in Figure 16.3. We calculate the contribution to the line integral frompath (1) in detail and leave calculation of contributions from the remainingthree paths to the reader. In all calculations, terms of higher order than thesecond will be omitted
∫
(1)
A · dr ≈ Al · Δrl = Al · (−e3Δll) = −A3lΔll = −A3lh3lΔq3
= −A3
(
q1, q2 −Δq2
2, q3
)
h3
(
q1, q2 −Δq2
2, q3
)
Δq3
= −{
A3 −Δq2
2∂A3
∂q2
}{
h3 −Δq2
2∂h3
∂q2
}
Δq3
≈ −A3h3Δq3 +∂
∂q2(h3A3)
Δq2
2Δq3.
P(q1, q2, q3)
(1) (2)
(3)
(4)
h2 b dq2
h2 t dq2
h3l dq3 h3r dq3
e2
e3
e1
Figure 16.3: Path of integration for the first component of the curl of A in curvilinear
coordinates.
432 Curvilinear Vector Analysis
Following similar steps, the reader may check that∫
(2)
A · dr ≈ A3h3Δq3 +∂
∂q2(h3A3)
Δq2
2Δq3,
∫
(3)
A · dr ≈ A2h2Δq2 −∂
∂q3(h2A2)
Δq3
2Δq2, (16.14)
∫
(4)
A · dr ≈ −A2h2Δq2 −∂
∂q3(h2A2)
Δq3
2Δq2.
Summing up all these contributions, we obtain∮
A · dr ≈{
∂
∂q2(h3A3) −
∂
∂q3(h2A2)
}
Δq2Δq3.
Dividing this by the area enclosed by the path
Δa = Δl2Δl3 = h2h3Δq2Δq3
we obtain the first component, the component along the unit normal to thearea:
(∇ × A)1 =1
h2h3
{∂
∂q2(h3A3) −
∂
∂q3(h2A2)
}
.
Corresponding expressions for the other two components of the curl canbe found by proceeding as above. We can put all of the components togetherin a mnemonic determinant form:curl in curvilinear
coordinatesTheorem 16.4.1. The curl of a vector field A in a curvilinear coordinatesystem is given by
∇ × A =1
h1h2h3
∣∣∣∣∣∣∣∣∣∣∣
e1h1 e2h2 e3h3
∂∂q1
∂∂q2
∂∂q3
h1A1 h2A2 h3A3
∣∣∣∣∣∣∣∣∣∣∣
. (16.15)
Note that ∇×A is not a cross product (except in Cartesian coordinates),warning! ∇ × A isnot a crossproduct in generalcurvilinearcoordinates!
but a vector defined by the determinant on the RHS of (16.15).If we substitute the appropriate values for h’s and q’s in spherical coordi-
nates, we obtain
∇ × A =1
r2 sin θ
∣∣∣∣∣∣∣∣∣∣∣
er eθr eϕr sin θ
∂∂r
∂∂θ
∂∂ϕ
Ar rAθ r sin θAϕ
∣∣∣∣∣∣∣∣∣∣∣
. (16.16)
16.4 The Curl 433
In cylindrical coordinates we get
∇ × A =1ρ
∣∣∣∣∣∣∣∣∣∣∣
eρ eϕρ ez
∂∂ρ
∂∂ϕ
∂∂z
Aρ ρAϕ Az
∣∣∣∣∣∣∣∣∣∣∣
. (16.17)
Example 16.4.2. We have already calculated the magnetic field of a dipole inExample 16.2.2. Here we want to obtain the same result using the vector potentialof a dipole given in Equation (15.12). We take μ to be along the z-axis. Then
μ = μez = μ(er cos θ − eθ sin θ)
andμ × er = μ(− sin θeθ × er) = μ sin θeϕ.
Therefore,
B = ∇ × A = ∇ ×(
kmμ × er
r2
)
= ∇ ×(
kmμ sin θeϕ
r2
)
=kmμ
r2 sin θ
∣∣∣∣∣∣∣∣∣∣∣∣
er eθr eϕr sin θ
∂∂r
∂∂θ
∂∂ϕ
0 0 r sin θsin θ
r2
∣∣∣∣∣∣∣∣∣∣∣∣
=kmμ
r2 sin θ
∣∣∣∣∣∣∣∣∣∣∣∣
er eθr eϕr sin θ
∂∂r
∂∂θ
∂∂ϕ
0 0sin2 θ
r
∣∣∣∣∣∣∣∣∣∣∣∣
=kmμ
r2 sin θ
[
er
(2 sin θ cos θ
r
)
− reθ
(
− sin2 θ
r2
)]
=kmμ
r3(2 cos θer + sin θeθ),
which is the expression obtained in Example 16.2.2. �
Example 16.4.3. Consider the vector field B described in cylindrical coordinatesas
B =k
ρeϕ,
where k is a constant. The curl of B is easily found to be zero:
∇ × B =1
ρ
∣∣∣∣∣∣∣∣∣∣∣
eρ eϕρ ez
∂∂ρ
∂∂ϕ
∂∂z
0 ρ(k/ρ) 0
∣∣∣∣∣∣∣∣∣∣∣
= 0.
However, for any circle (of radius a, for example) centered at the origin and locatedin the xy-plane, we get2
∮
C
B · dr =
∫ 2π
0
k
aeϕ · (eϕa dϕ) = 2πk �= 0.
2See also Example 14.3.3 which discusses this same vector field in Cartesian coordinates.
434 Curvilinear Vector Analysis
The reason for this result is that the circle is not contractible to zero: At theorigin—which is inside the circle and at which ρ = 0—B is not defined.
This vector field should look familiar. It is the magnetic field due to a longstraight wire carrying a current along the z-axis. According to Ampere’s circuitallaw, the line integral of B along any closed curve encircling the wire, such as theabove circle, gives, up to a multiplicative constant, the current in the wire, and thiscurrent is not zero. �
Example 16.4.4. A vector field that can be written ascentral force fieldsare conservative
F = f(r)r,
where r is the displacement vector from the origin, is conservative. It is instructiveto show this using both Cartesian and spherical coordinate systems.
First, in Cartesian coordinates
F = xf(r)ex + yf(r)ey + zf(r)ez
and the curl is
∇ × F =
∣∣∣∣∣∣∣∣∣∣∣
ex ey ez
∂∂x
∂∂y
∂∂z
xf yf zf
∣∣∣∣∣∣∣∣∣∣∣
= ex
{∂
∂y(zf) − ∂
∂z(yf)
}
+ ey
{∂
∂z(xf) − ∂
∂x(zf)
}
+ ez
{∂
∂x(yf) − ∂
∂y(xf)
}
.
Concentrating on the x-component first and using the chain rule, we have
∂
∂y(zf) = z
∂f
∂y= z
df
dr
∂r
∂y= zf ′ ∂r
∂y.
But∂r
∂y=
∂
∂y
√x2 + y2 + z2 =
y
r.
Thus,∂
∂y(zf) = yzf ′.
Similarly,∂
∂z(yf) = yzf ′.
Therefore, the x-component of ∇×F is zero. The y- and z-components can also beshown to be zero, and we get ∇ × F = 0.
On the other hand, using spherical coordinates, we easily obtain
∇ × F =1
r2 sin θ
∣∣∣∣∣∣∣∣∣∣∣
er eθr eϕr sin θ
∂∂r
∂∂θ
∂∂ϕ
rf(r) 0 0
∣∣∣∣∣∣∣∣∣∣∣
= 0.
Obviously, the use of spherical coordinates simplifies the calculation consi-derably. �
16.4 The Curl 435
The preceding example shows that
Box 16.4.1. Any well-behaved vector field whose magnitude is only afunction of radial distance, r, and whose direction is along r is conserva-tive. Such vector fields are generally known as central vector fields.
16.4.1 The Laplacian
Combining divergence and the gradient gives the Laplacian. Using Equation(16.4) in Theorem 16.3.1, we get Laplacian in
curvilinearcoordinatesTheorem 16.4.5. The Laplacian of a function f is the divergence of gradient
of f and—in a curvilinear coordinate system—is given by
∇2f =1
h1h2h3
{∂
∂q1
(h2h3
h1
∂f
∂q1
)
+∂
∂q2
(h1h3
h2
∂f
∂q2
)
+∂
∂q3
(h1h2
h3
∂f
∂q3
)}
.
For cylindrical coordinates the Laplacian is
∇2f =1ρ
∂
∂ρ
(
ρ∂f
∂ρ
)
+1ρ2
∂2f
∂ϕ2+
∂2f
∂z2(16.18)
and for spherical coordinates it is
∇2f =1r2
∂
∂r
(
r2 ∂f
∂r
)
+1
r2 sin θ
{∂
∂θ
(
sin θ∂f
∂θ
)
+1
sin θ
∂2f
∂ϕ2
}
. (16.19)
Equations (16.7) and (16.19) allow us to write the angular momentumdifferential operator derived in Example 15.3.1 in spherical coordinates, whichis the most common way of writing it. We note that
∂
∂r
(
r2 ∂f
∂r
)
= 2r∂f
∂r+ r2 ∂2f
∂r2 ,
andr · (∇f) = r
∂f
∂r,
and
(r · ∇)2f = r∂
∂r
(∂f
∂r
)
= r∂f
∂r+ r2 ∂2f
∂r2 .
Substituting these plus (16.19) in (15.22) yields
L2f = − 1sin θ
{∂
∂θ
(
sin θ∂f
∂θ
)
+1
sin θ
∂2f
∂ϕ2
}
. (16.20)
Therefore, the angular momentum operator depends only on angles in spher-ical coordinates.
436 Curvilinear Vector Analysis
16.5 Problems
16.1. The divergence of a vector can be obtained in any coordinate systemby brute force calculation. In this problem you are asked to find ∇ · A incylindrical coordinates.(a) Express Ax in terms of cylindrical coordinates and components. Hint:Write A in cylindrical ccordinates and take the dot product with ex expressingeverything in terms of cylindrical ccordinates.(b) Use the chain rule
∂Ax
∂x=
∂Ax
∂ρ
∂ρ
∂x+
∂Ax
∂ϕ
∂ϕ
∂x+
∂Ax
∂z
∂z
∂x
where Ax is what you found in (a).(c) Do the same with Ay and Az , and add the three terms to obtain thedivergence in cylindrical coordinates.
16.2. Find the divergence of a vector in spherical coordinates following theprocedure outlined in Problem 16.1.
16.3. Find the gradient of a function in cylindrical and spherical coordinatesfollowing a procedure similar to the one outlined in Problem 16.1.
16.4. Find the curl of a vector in cylindrical and spherical coordinates fol-lowing a procedure similar to the one outlined in Problem 16.1.
16.5. Start with the Laplacian in Cartesian coordinates.(a) By using the chain rule and expressing the second derivatives in cylindricalcoordinates, find the Laplacian in cylindrical coordinates.(b) Do the same for spherical coordinates.
16.6. The elliptic cylindrical coordinates (u, θ, z)are given by
x = a coshu cos θ
y = a sinh u sin θ
z = z
where a is a constant.(a) What is the expression for the gradient of a function f in elliptic cylindri-cal coordinates?(b) What is the expression for the divergence of a vector A in elliptic cylin-drical coordinates?(c) What is the expression for the curl of a vector A in elliptic cylindricalcoordinates?(d) What is the expression for the Laplacian of a function f in elliptic cylin-drical coordinates?
![Vector Calculus & General Coordinate Systems Orthogonal curvilinear coordinates For orthogonal curvilinear coordinates, recall, Vector Calculus & General Coordinate Systems [, ]](https://img.dokumen.tips/doc/110x75/5b0d24927f8b9a8b038d43de/vector-calculus-general-coordinate-systems-orthogonal-curvilinear-coordinates-for.jpg)
