CYLINDRICAL COORDINATES
1.The expressions of the components of a vector:
In this system, we have:
P = (Px , Py, Pz), or:
P = Pxex + Pyey + Pzez
Where:
eρ, eφ and ez are the vector
unit in this system; that is they are perpendicular to each
other and their magnitude is equal to 1.
Remark that that ez stays the same over the z axis.
But for eρ and eφ, we do the following
calculations:
They are in the same plan (yx). Their projections over y
and x axis give the vector unit over y and x:
- Over y : eρ sin φ + eφ cos φ = ey (1) - Over x : eρ cos φ - eφ sin φ = ex (2) - Over z : ez = ez (3) P can be written as: P = Px[eρ cos φ - eφ sin φ] + Py[eρ sin φ + eφ cos φ] + Pz[ez] P = [Pxcos φ + Py sin φ]eρ + [- Px sin φ + Pycos φ]eφ + Pz[ez] = Pρeρ + Pφeφ + Pzez
2. Unit vectors transformations:
The related transformation (eρ, eφ, ez) → (eρ, eφ, ez) can be obtained by writing: [Pρ, Pφ, Pz] = MATRIX [Px, Py, Pz] and therfore inverse the matrix MATRIX to get the coordinates of P in (x, y, z) system from(ρ, φ, z) system. Where:
We can obtain these transformations just by:
- Multiplying (1) by cosφ
eρ sin φcosφ + eφcos φ = cosφey (4)
- Multiplying (2) by sinφ
eρcos φsinφ - eφ sin φ2 = sinφex (5)
(4) - (5) gives:
eφ = cosφey - sinφex
- Multiplying (2) by cosφ
eρ cos φ2 - eφ sin φcosφ = excosφ (6)
- Multiplying (1) by sinφ
eρ sin φ2 + eφcos φsinφ = eysinφ (7)
(6) + (7) gives:
eρ = excosφ + eysinφ
Then: eρ = cosφex + sinφey eφ = cosφey - sinφex ez = ez
3.The Gradiant:
V(x,y,z) : any vector. x = ρ cosφ y = ρ sinφ z = z The inverse transformation is: ρ = [x2 + y2]1/2 φ = tg-1(x/y) = arctg (x/y) We use the relationships: tg φ = y/x 1 + tg2φ = 1/cos2φ The variable x depends on ρ and φ. It follows that: ∂/∂x = (∂/∂ρ)(∂ρ/∂x) + (∂/∂φ)(∂φ/∂x) Then: ∂ρ/∂x = (1/2) [x2 + y2] - 1/2 . 2.x = x/ρ = cos φ ∂φ/∂x = ? First: ∂(tgφ)/∂x = [∂(tgφ)/∂φ] φ/∂x = [1 + tg2φ] ∂φ/∂x = ∂(y/x)/∂x = - y/x2 = - tg φ/ρcos φ Then: φ/∂x = cos2φ[- tg φ/ρcos φ]= - sin φ/ρ Then: ∂/∂x = (∂/ρ)(cos φ) + (∂/φ)(- sin φ/ρ) ∂/∂x = cos φ ∂/∂ρ - sinφ/ρ ∂/∂φ ∂/∂y = (∂/∂ρ)(∂ρ/∂y) + (∂/φ)(∂φ/∂y) ∂ρ/∂y = (1/2) [x2 + y2] - 1/2 . 2.y = y/ρ = sin φ ∂φ/∂y = cos2φ[∂(tg φ)/∂y ]= (1/x) cos2φ = cos φ/ρ Then: ∂/∂y = (∂/∂ρ)( sin φ) + (∂/∂φ)(cos φ/ρ) ∂/∂y = sinφ ∂/∂ρ + cos φ/ρ ∂/∂φ To recap:
∂/∂x = cos φ ∂/∂ρ - sinφ/ρ ∂/∂φ ∂/∂y = sinφ ∂/ρ + cos φ/∂ρ ∂/∂φ ∂/∂z = ∂/∂z
From the above figure, we have: ex = eρ cos φ - eφ sin φ ey = eρ sin φ + eφ cos φ ez = ez Then: ∇ = ∂/∂x ex + ∂/∂y ey + ∂/∂z ez= [cos φ ∂/∂ρ - sinφ/ρ ∂/∂φ][eρ cos φ - eφ sin φ] + [sinφ ∂/∂ρ + cos φ/ρ ∂/∂φ][eρ sin φ + eφ cos φ ] + [ ∂/∂z][ez] = cos φ ∂/∂ρeρ cos φ - sinφ/ρ ∂/∂φeρ cos φ - eφ sin φcos φ ∂/ρ + eφ sin φsinφ/ρ ∂/∂φ + sinφ ∂/ρeρ sin φ + cos φ/ρ ∂/∂φeρ sin φ + sinφ ∂/ρ eφ cos φ + cos φ/ρ ∂/∂φ eφ cos φ + ∂/∂z ez = ∂/∂ρ eρ + (1/ρ) ∂/∂φ eφ + ∂/∂z ez ∇ = ∂/∂ρ eρ + (1/ρ) ∂/∂φ eφ + ∂/∂z ez
4.The Laplacien:
∂2/∂x2 = [(∂ρ/∂x)(∂/∂ρ) + (∂φ/∂x)(∂/∂φ)] (∂/∂x)(∂/∂ρ) (∂/∂x) = cosφ ∂2/∂ρ2 + (sinφ/ρ2) ∂/∂φ (∂/∂φ) (∂/∂x) = - (sinφ) ∂/∂ρ - (cosφ/ρ)∂/∂φ - (sinφ/ρ) ∂2/∂φ2 Then: ∂2/∂x2 = (sin2φ/ρ) ∂/∂ρ + 2( cosφsinφ/ρ2) ∂/∂φ + (cos2φ) ∂2/∂ρ2 + (sin2φ/ρ2) ∂2/∂φ2 ∂2/∂y2 = [(∂ρ/∂y)(∂/∂ρ) + (∂φ/∂y)(∂/∂φ)] (∂/∂y) (∂/∂ρ) (∂/∂y) = sinφ ∂2/∂ρ2 - (cosφ/ρ2) ∂/∂φ (∂/∂φ) (∂/∂y) = (cosφ) ∂/∂ρ - (sinφ/ρ)∂/∂φ + (cosφ/ρ) ∂2/∂φ2 Then: ∂2/∂y2 = (cos2φ/ρ) ∂/∂ρ - 2( cosφsinφ/ρ2) ∂/∂φ + (sin2φ) ∂2/∂ρ2 + (cos2φ/ρ2) ∂2/∂φ2 ∂2/∂z2 = ∂2/∂z2 It follows that: Δ = (1/ρ) ∂/∂ρ + ∂2/∂ρ2 + (1/ρ2) ∂2/∂φ2 + ∂2/∂z2
5.The volume element:
The volume element is:
dV = ρdρdφdz
The surface elements are:
ρdφdz eρ
dρdφdz eφ
ρdρdφ ez
6. The divergence of a vector: ∇ . A
We know that: ∇ = ∂/∂ρ eρ + (1/ρ) ∂/∂φ eφ + ∂/∂z ez And: eρ = cosφ ex + sinφ ey eφ = - sinφ ex + cosφ ey ez = ez Let A = (Aρ, Aφ, Az) ∇ . A = [∂/∂ρ eρ + (1/ρ) ∂/∂φ eφ + ∂/∂z ez] . [Aρ eρ + Aφ eφ + Az ez] = ∂Aρ/∂ρ eρ . eρ + (1/ρ) ∂Aρ/∂φ eφ . eρ + ∂Aρ/∂z ez . eρ + Aρ eρ . ∂eρ/∂ρ + (1/ρ)Aρ eφ . ∂eρ/∂φ + Aρ ez . ∂eρ/∂z + ∂Aφ/∂ρ eρ . eφ + (1/ρ) ∂Aφ/∂φ eφ . eφ + ∂Aφ/∂z ez . eφ + Aφ eρ . ∂eφ/∂ρ + (1/ρ) Aφ eφ . ∂eφ/∂φ + Aφ ez . ∂eφ/∂z ] + ∂Az/∂ρ eρ . ez + (1/ρ) ∂Az/∂φ eφ . ez + ∂Az/∂z ez . ez + Aρ eρ . ∂ez/∂ρ + (1/ρ)Az eφ . ∂ez/∂φ + Az ez . ∂ez/∂z We will use: ∂eρ/∂ρ = 0 ∂eρ/∂φ = eφ ∂eρ/∂z = 0 ∂eφ/∂ρ = 0 ∂eφ/∂φ = - eρ ∂eφ/∂z = 0 ∂ez/∂ρ = 0 ∂ez/∂φ = 0 ∂ez/∂z = 0 Then: ∇ . A = ∂Aρ/∂ρ + 0 + 0 + 0 + (1/ρ)Aρ + 0 + 0 + (1/ρ) ∂Aφ/∂φ + 0 + 0 + 0 + 0] + 0 + 0 + ∂Az/∂z + 0 + 0 + 0 We find then: ∇ . A = ∂Aρ/∂ρ + (1/ρ)Aρ + (1/ρ) ∂Aφ/∂φ + ∂Az/∂z Or: ∇ . A = (1/ρ)∂(ρ Aρ)/∂ρ + (1/ρ) ∂Aφ/∂φ + ∂Az/∂z
7. The curl of a vector: ∇ x A
We know that: ∇ = ∂/∂ρ eρ + (1/ρ) ∂/∂φ eφ + ∂/∂z ez And: eρ = cosφ ex + sinφ ey eφ = - sinφ ex + cosφ ey ez = ez Let A = (Aρ, Aφ, Az) ∇ x A = [∂/∂ρ eρ + (1/ρ) ∂/∂φ eφ + ∂/∂z ez] x [Aρ eρ + Aφ eφ + Az ez] = ∂Aρ/∂ρ eρ x eρ + (1/ρ) ∂Aρ/∂φ eφ x eρ + ∂Aρ/∂z ez x eρ + Aρ eρ x ∂eρ/∂ρ + (1/ρ)Aρ eφ x ∂eρ/∂φ + Aρ ez x ∂eρ/∂z + ∂Aφ/∂ρ eρ x eφ + (1/ρ) ∂Aφ/∂φ eφ x eφ + ∂Aφ/∂z ez x eφ + Aφ eρ x ∂eφ/∂ρ + (1/ρ) Aφ eφ x ∂eφ/∂φ + Aφ ez x ∂eφ/∂z + ∂Az/∂ρ eρ x ez + (1/ρ) ∂Az/∂φ eφ x ez + ∂Az/∂z ez x ez + Aρ eρ x ∂ez/∂ρ + (1/ρ)Az eφ x ∂ez/∂φ + Az ez x ∂ez/∂z We will use:Then: ∇ x A = 0 - (1/ρ) ∂Aρ/∂φ ez + ∂Aρ/∂z eφ + 0 + 0 + 0 + ∂Aφ/∂ρ ez + 0 - ∂Aφ/∂z eρ + 0 + (1/ρ) Aφ ez + 0 + - ∂Az/∂ρ eφ + (1/ρ) ∂Az/∂φ eρ + 0 + 0 + 0 + 0 = - (1/ρ) ∂Aρ/∂φ ez + ∂Aρ/∂z eφ + ∂Aφ/∂ρ ez - ∂Aφ/∂z eρ + (1/ρ) Aφ ez - ∂Az/∂ρ eφ + (1/ρ) ∂Az/∂φ eρ = [(1/ρ) ∂Az/∂φ - ∂Aφ/∂z]eρ + [∂Aρ/∂z - ∂Az/∂ρ ]eφ + [(1/ρ) Aφ + ∂Aφ/∂ρ - (1/ρ) ∂Aρ/∂φ] ez We get then: ∇ x A = [(1/ρ) ∂Az/∂φ - ∂Aφ/∂z] eρ + [∂Aρ/∂z - ∂Az/∂ρ ] eφ + (1/ρ)[ Aφ + ρ ∂Aφ/∂ρ - ∂Aρ/∂φ] ez Or: ∇ x A = [(1/ρ) ∂Az/∂φ - ∂Aφ/∂z] eρ + [∂Aρ/∂z - ∂Az/∂ρ ] eφ + (1/ρ)[ ∂(ρ Aφ)/∂ρ - ∂Aρ/∂φ] ez
eρ x eρ = 0 eφ x eφ = 0 ez x ez = 0 eρ x eφ = ez eφ x ez = eρ ez x eρ = eφ ∂eρ/∂ρ = 0 ∂eρ/∂φ = eφ ∂eρ/∂z = 0 ∂eφ/∂ρ = 0 ∂eφ/∂φ = - eρ ∂eφ/∂z = 0 ∂ez/∂ρ = 0 ∂ez/∂φ = 0 ∂ez/∂z = 0