Electromagnetic Theory Prof. Ruiz, UNC Asheville ...

Electromagnetic Theory
Prof. Ruiz, UNC Asheville, doctorphys on YouTube
Chapter L Notes. Curvilinear Coordinates

L1. Curvilinear Coordinates

Cartesian (x, y, z) Cylindrical (ρ,φ, z) Spherical (r,θ ,φ )

Cylindrical to Cartesian

x = ρ cosφ
y = ρ sinφ

z=z

Spherical to Cartesian

x = r sinθ cosφ
y = r sinθ sinφ
z = r cosθ

Cartesian to Cylindrical: ρ = x2 + y2 tan φ = y z=z
x

Cartesian to Spherical: r = x2 + y 2 + z 2 tan φ = y cosθ = z
x r

∧ ∧ ∧ ∧ ∧∧ ∧∧
Unit Vectors: ρ = cosφ i + sin φ j φ = k × ρ = cosφ j − sin φ i

∧ ∧∧

r = sinθ ρ + cosθ k

∧ ∧∧ ∧ ∧∧
PL1 (Practice Problem). Use θ = φ × r to show that θ = cosθ ρ − sin θ k .

∧ ∧∧

PL2 (Practice Problem). Arrive at the result θ = cosθ ρ − sin θ k by inspecting

the figure on the next page. Do not take any cross products. Simply write the answer

∧

down by clever inspection of the figure, where we have moved θ down carefully

without changing its direction or length.

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Unit-Vector Summary:

∧ ∧∧

ρ = cosφ i + sin φ j

∧ ∧∧

φ = − sin φ i + cosφ j

∧ ∧∧

r = sinθ ρ + cosθ k

∧ ∧∧

θ = cosθ ρ − sinθ k

Cylindrical Coordinates Differential Line and Volume Elements. Figure from Tony Saad.

For the volume element we write

dV = dAdz , where the dA is the

area of the base in the figure. This is

simply dA = ρ dφ d ρ . But if you

want to be super careful, use the
average of the shorter and longer arc
lengths (see left). As we take the limit
to the infinitesimal, we a product of
three differentials will vanish faster

than the product of two. Therefore, we can toss the d ρ dφ d ρ term compared to the
dφ d ρ term. The volume element is then

dV = ρdφd ρdz .

The line element for the diagonal of the "cube-like" region is found by using the

Pythagorean theorem twice: once for the floor d ρ 2 + ρ 2dφ 2 and then the diagonal
of the floor with the height dz .

ds2 = d ρ 2 + ρ 2dφ 2 + dz2

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PL3 (Practice Problem with Solution). To verify that we can use length times width
times height for curved boundaries let's integrate out the volume of a can this way.

Figure from Tony Saad.

Let the can have radius R and height h.
We want to check out

dV = ρdφd ρdz

to see if it integrates out to the right
volume.

dV = ∫ ∫R ρd ρ 2π dφ ∫ h dz
0 00

dV = ρ2 R 2π z h = R2 2π h = π R2h

φ
2 00 2
0

This is the volume of a cylinder. So we are good.

Spherical Coordinates Differential Line and Volume Elements. Figure from Tony Saad.
The volume element is found using
length times width times height for the
tilted volume element.

dV = (r sinθ dφ)(rdθ )dr

We don't even take the average of
the widths since to first order, the
differential

(r sinθ + dθ )dφ
becomes r sinθ dφ .

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The volume element is usually written as

dV = r2 sinθ drdθ dφ .

The differential line element is the square of the diagonal as before. So we use simple

square each of the differential sides. ds2 = dr 2 + r 2dθ 2 + r 2 sin 2 θ dφ 2

PL4 (Practice Problem with Solution).
Let's see if we get the volume of a
sphere with radius R, using the volume
formula for a rectangular solid even
though our boundary lines are curved.

dV = r2 sinθ drdθ dφ

∫ ∫ ∫dV = R r2dr π sinθ dθ 2π dφ
00 0

r3 R 2π R3 (−1) [ −1 − 1] 2π R3 4 π R3

dV = [− cosθ ] π φ = = 2 ⋅ 2π =
3 00 3 33
0

This checks out. So we are okay with how we are setting things up.

Here is the summary for the differential elements for all three coordinate systems.

Cartesian: ds2 = dx2 + dy2 + dz2
ds2 = d ρ 2 + ρ 2dφ 2 + dz2
dV = dxdydz ds2 = dr2 + r2dθ 2 + r 2 sin2 θ dφ 2

Cylindrical:

dV = ρ d ρdφdz

Spherical:

dV = r2 sinθ drdθ dφ

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We next consider an infinitesimal solid in general curvilinear coordinates. Note that each
variable differential has a factor in front. These "h" factors are called scale factors.
Familiar examples are the arc lengths we just investigated for cylindrical and spherical

coordinates: ds = rdθ or ds = r sin θ dφ .

Differential Line Element. We use the
Pythagorean theorem twice for the
diagonal of the solid. First, for the base, we
have for the square of the hypotenuse

h12dq12 + h22dq22 .

Second, adding to this the square of the
height gives

ds2 = h12dq12 + h22dq22 + h32dq32 .

Differential Volume Element. Here we
use the length times width times height

idea to arrive at the volume element.

dV = (h1dq1)(h2dq2 )(h3dq3 )

dV = h1h2h3dq1dq2dq3

General Case: dV = h1h2h3dq1dq2dq3 and ds2 = h12dq12 + h22dq22 + h32dq32 .

Using the coordinates and scale factors for each of our coordinate systems, we can

find dV and ds2 from the above master formulas.

Cartesian: q1 = x , q2 = y , q3 = z with h1 = 1, h2 = 1 , h3 = 1
Cylindrical: q1 = ρ , q2 = φ , q3 = z with h1 = 1, h2 = ρ , h3 = 1
Spherical: q1 = r , q2 = θ , q3 = φ with h1 = 1, h2 = r , h3 = r sin θ

Scale Factors are in general functions. That makes coordinates curvilinear.

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L2. The Gradient. The Gradient in Cartesian coordinates is given by

∇f = ∂f ∧ ∂f ∧ ∂f ∧
∂x ∂z
i+ ∂y j+ k.

We can obtain the gradient in curvilinear coordinates by comparing the differentials. In
Cartesian coordinates we have these three differentials:

dx , dy , and dz .

In cylindrical coordinates we have: d ρ , ρ dφ , and dz . Note that each has

dimension of length. Our gradient in cylindrical coordinates is then

∇f = ∂f ∧ 1 ∂f ∧ ∂f ∧
∂ρ ρ
ρ+ φ+ k .
∂φ ∂z

In spherical coordinates we have: dr , rdθ , and r sin θ dφ . Note that each has

dimension of length. Our gradient in spherical coordinates is then

∇f = ∂f ∧ + 1 ∂f ∧ 1 ∂f ∧
∂r r sinθ ∂φ
r r ∂θ θ+ φ .

In curvilinear coordinates we have: h1dq 1 , h2dq 2 , and h3dq 3 . Note that each has

dimension of length. Our gradient in curvilinear coordinates is then

∇f = 1 ∂f ∧ 1 ∂f ∧ 1 ∂f ∧
h1 ∂q2 ∂q3 e3 .
e1 + e2 +
∂q1 h2 h3

Caution: If you write the operator by itself, put the unit vectors on the left since
they are in general functions of the coordinates.

∇ ∧ 1 ∂ ∧ 1 ∂ ∧ 1 ∂
h1 ∂q1 h2 ∂q2 h3 ∂q3
= e1 + e2 + e3

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L3. The Divergence Theorem. Here is a quick review of the derivation of the

divergence theorem in Cartesian coordinates.

We are interested in calculating the flux
through the enclosed surface:

∫∫ E ⋅ dA .

Ebottom = Ez ( x, y, z) and its
counterpart Etop = Ez ( x, y, z + ∆z) .

The net flux out of the surface of our
cube is given as follows.

∫∫ E ⋅ dA => Ez (x, y, z + ∆z)∆x∆y − Ez (x, y, z)∆x∆y

∫∫ E ⋅ dA => Ez (x, y, z + ∆z) − Ez (x, y, z) ∆x∆y∆z

∆z

∫∫ E ⋅ dA = ∫∫∫ ∂Ez dxdydz
∂z

∫∫ E ⋅ dA = ∫∫∫  ∂Ex + ∂E y + ∂Ez 
 ∂x ∂y ∂z  dxdydz
 

∫∫ E ⋅ dA = ∫∫∫ ∇ ⋅ E dxdydz

PL5 (Practice Problem). Show that starting we Ebottom = Ez (x + ∆x , y + ∆y , z) and
2 2

its appropriate counterpart that you get the same result. You will find that taking limits as

the extra deltas go to zero do not lead to any derivatives. They just go away.

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L3. The Divergence Theorem in Curvilinear Coordinates

We are going to redo our derivation of the divergence theorem in general now for

curvilinear coordinates. We consider (q 1 , q 2 , q 3) = (u, v, w) with scale factors

(h u , h v , h w ). It is easier to work without subscripts for the coordinates.

The important thing to notice is
that the scale factors are in
general functions of the
coordinates, i.e.,

hw = hw (u, v, w) .

As a specific example, in
spherical coordinates

hφ = hφ (r,θ ,φ ) = r sinθ .

As before, consider a simplified
vector field such that

∧

E(u, v, w) = Ew ew .

We then subtract the flux out at the top from the flux in at the bottom along this

∫∫dimension. Our enclosed integral E ⋅ dA will include different areas at the bottom

and top.

dAbottom = [hu (u, v, w)∆u][hv (u, v, w)∆v]
dAtop = [hu (u, v, w + ∆w)∆u][hv (u, v, w + ∆w)∆v]

∫∫Therefore, E ⋅ dA =>

Ew (u, v, w + ∆w)[hu (u, v, w + ∆w)∆u][hv (u,v, w + ∆w)∆v]
−Ew (u, v, w)[hu (u, v, w)∆u][hv (u, v, w)∆v]

∫∫ E ⋅ dA => (Ewhuhv ) w+∆w − (Ewhuhv ) w  ∆u∆v

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∫∫ E ⋅ dA =>  ( Ew hu hv ) w+∆w − ( Ew hu hv ) w  ∆u∆v∆w
 
 ∆w 

∫∫ E ⋅ dA => ∂(Ewhuhv ) ∆u∆v∆w
∂w

Now we want the differential volume element

dV = (hu ∆u)(hv∆v)(hw∆w) = huhvhw∆u∆v∆w in there.

∫∫ E ⋅ dA => 1 ∂(Ewhuhv ) hu hv hw ∆u∆v∆w
hu hvhw ∂w

Now it's time to put the q-variables back in.

∫∫ E ⋅ dA => 1 ∂(E3h1h2 ) h1h2h3∆q1∆q2∆q3
h1h2h3 ∂w

∫∫ E ⋅ dA = ∫∫∫ 1 ∂(E3h1h2 ) dV

h1h2h3 ∂q3

From this we can generalize to the general form of the divergence.

∇⋅ E = 1  ∂( E1h2 h3 ) + ∂(E2h1h3 ) + ∂(E3h1h2 ) 
h1h2 h3  ∂q1 ∂q2 ∂q3 
 

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PL6 (Practice Problem). Show that the divergence in curvilinear coordinates, i.e.,

∇⋅ A = 1  ∂( A1h2 h3 ) + ∂( A2h1h3 ) + ∂( A3h1h2 ) 
h1h2h3  ∂q1 ∂q2 ∂q3 
  ,

reduces to the following in Cartesian, cylindrical, and spherical coordinates.

Cartesian: ∇ ⋅ A = ∂Ax + ∂Ay + ∂Az
∂x ∂y ∂z

Cylindrical: ∇ ⋅ A = 1 ∂(ρ Aρ ) + 1 ∂Aφ + ∂Az
ρ ∂ρ ρ ∂φ ∂z

Spherical: ∇ ⋅ A = 1 ∂(r2 Ar ) + 1 ∂(sinθ Aθ ) + 1 ∂Aφ
r2 ∂r r sinθ ∂θ r sinθ ∂φ

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L4. Stoke's Theorem

In a previous chapter we considered a vector field B and proceed to do a closed line
integral of this field.

∫ B ⋅ dl

You recognize this as the left side of one of our Maxwell equations. The vector field can
be any vector field. To simplify, we will pick the field to be in the x-y plane.

∫We obtained B ⋅ dl =>

By (x + ∆x, y, z) − By (x, y, z) ∆x∆y − Bx (x, y + ∆y, z) − Bx (x, y, z) ∆x∆y
∆x ∆y

PL7 (Practice Problem). Show where each piece above comes from.

We can extend this to curvilinear coordinates (u,v,w) with scale factors (hu,hv,hw).

∫ B ⋅ dl => [ Bv (u, v, w)hv (u , v, w)] + ∆u − [ Bv (u , v, w)hv (u, v, w)] ∆u∆v
u u

∆u

− [ Bu (u, v, w)hu (u, v, w)] +∆v − [Bu (u, v, w)hu (u, v, w)] ∆u∆v
v v

∆v

∫ ∫∫B ⋅ dl = 1  ∂( Bv hv ) − ∂( Bu hu )  dAw
hu hv  ∂u ∂v 
A

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Introducing the q-variables (q 1 , q 2 , q 3) = (u, v, w) with (h1 , h 2 , h 3).

∫ ∫∫B ⋅ dl = 1  ∂ ( B2 h2 ) − ∂ ( B1h1 )  dA3
 ∂q1 ∂q2 
h1h2 A  

The third component of the cross product defined as

(∇ × B)3 = 1  ∂ ( B2 h2 ) − ∂ ( B1h1 ) 
h1h2  ∂q1 ∂q2 
  .

leads to Stoke's Theorem in curvilinear coordinates.

∫ B ⋅ dl = ∫∫ (∇ × B ) ⋅ dA

A

∇×B = 1  ∂ ( B2 h2 ) − ∂ ( B1h1 )  ∧
h1h2  ∂q1 ∂q2 
Our cross product   e3 is better written as

∇× B = 1  ∂ ( B2 h2 ) − ∂ ( B1h1 )  h3 ∧
h1h2h3  ∂q1 ∂q2 
  e3

Then we can use the following determinant to express this.

∧ ∧ ∧

h1 e1 h2 e2 h3 e3
∂ ∂
∇×B = 1 ∂
h1h2h3 ∂q1 ∂q2 ∂q3

h1B1 h2 B2 h3 B3

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L5. The Laplacian

We find the Laplacian of a function, i.e., ∇2 f , by applying the divergence to the

gradient of a function: ∇2 f = ∇ ⋅ (∇f ) . We start with our previous result for the

gradient

∇f = 1 ∂f ∧ 1 ∂f ∧ 1 ∂f ∧
h1 ∂q1 h2 ∂q2 ∂q3
e1 + e2 + h3 e3 .

and then use our previous result for the divergence

∇⋅ A = 1  ∂( A1h2 h3 ) + ∂( A2h1h3 ) + ∂ ( A3h1h2 ) 
h1h2h3  ∂q1 ∂q2 ∂q3 
  ,

where A = ∇f . This means substituting

A1 = 1 ∂f , A2 = 1 ∂f , and A3 = 1 ∂f .
h1 ∂q1 h2 ∂q2 h3 ∂q3

We obtain for ∇2 f the following.

1 ∂ 1 ∂f h2h3 ) + ∂ 1 ∂f h1h3 ) + ∂ 1 ∂f h1h2 ) 
h1h2h3  ( ∂q1 ∂q2 ( ∂q2 ∂q3 ( ∂q3 
 ∂q1 h1 h2 h3 

This simplifies to

∇2 f = 1  ∂ ( h2h3 ∂f ) + ∂ ( h1h3 ∂f ) + ∂ ( h1h2 ∂f 
h1h2h3  ∂q1 h1 ∂q1 ∂q2 h2 ∂q2 ∂q3 h3 ∂q3 )
 

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PL8 (Practice Problem). Show our Laplacian in curvilinear coordinates, which is

∇2 = 1 ∂ ( h2h3 ∂ )+ ∂ ( h1h3 ∂ )+ ∂ ( h1h2 ∂
h1h2h3  h1 ∂q1 ∂q2 h2 ∂q2 ∂q3 h3 ) 
 ∂q1 ∂q3  ,

reduces to the following in Cartesian, cylindrical, and spherical coordinates.

Cartesian: ∇2 = ∂2 + ∂2 + ∂2
∂x2 ∂y 2 ∂z 2

Cylindrical: ∇2 = 1 ∂ (ρ ∂ )+ 1 ∂2 + ∂2
ρ ∂ρ ∂ρ ρ2 ∂φ 2 ∂z 2

Spherical: ∇2 = 1 ∂ (r2 ∂ )+ 1 ∂ (sin θ ∂ )+ r2 1 ∂2
r2 ∂r ∂r r 2 sinθ ∂θ ∂θ sin2 θ ∂φ 2

Many years ago Mr. Samuel S. Ensor, my Calculus teacher at St. Joseph's College in
Philadelphia (now university) gave us a project in Calculus III that was long, but very
useful and productive (Spring 1969). It is given below. Everyone aspiring to be a
physicist or engineer should do this calculation once in a lifetime. It will correct any
rough edges you have in taking partial derivatives and using the chain rule.

PL9 (Practice Problem for the Summer). Derive the Laplacian in spherical
coordinates the long way! Start with

x = r sinθ cosφ , y = r sinθ sin φ , z = r cosθ and

∇2 = ∂2 + ∂2 + ∂2 .
∂x2 ∂y 2 ∂z 2

Then you start cranking: ∂ = ∂r ∂ + ∂θ ∂ + ∂φ ∂ and so on. Have fun!
∂x ∂x ∂r ∂x ∂θ ∂x ∂φ

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