Download Curvilinear Coordinates - Lecture Notes | PH 106 and more Exams Physics in PDF only on Docsity!
University of Alabama
Department of Physics and Astronomy
PH 106-4 / LeClair Fall 2008
Curvilinear Coordinates
Note that we use the convention that the cartesian unit vectors are ˆx, ˆy, and ˆz, rather than ˆı, ˆ,
and
k, using the substitutions ˆx = ˆı, ˆy = ˆ, and ˆz =
k.
Definition of coordinates
Table 1: Relationships between coordinates in different systems.
†
Cartesian ( x, y, z ) Cylindrical ( R, ϕ, z ) Spherical ( r, θ, ϕ )
R =
x
2
2 x = R cos ϕ x = r sin θ cos ϕ
ϕ = tan
− 1
y
x
y = R sin ϕ y = r sin θ sin ϕ
z = z z = z z = r cos θ
r =
x
2
2
2 r =
R
2
2 R = r sin θ
θ = tan
− 1
x
2 +y
2
z
θ = tan
− 1
R
z
ϕ = ϕ
ϕ = tan
− 1
y
x
ϕ = ϕ z = r cos θ
† See also http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates and references therein.
Definition of unit vectors
Cartesian ( x, y, z ) Cylindrical ( R, ϕ, z ) Spherical ( r, θ, ϕ )
R =
x
R
ˆx +
y
R
ˆy xˆ = cos ϕ
R − sin ϕ ϕˆ ˆx = sin θ cos ϕ ˆr + cos θ cos ϕ
θ − sin ϕ ϕˆ
ϕ ˆ = −
y
R
ˆx +
x
R
yˆ yˆ = sin ϕ
R + cos ϕ ϕˆ yˆ = sin θ sin ϕ ˆr + cos θ sin ϕ
θ + cos ϕ ϕˆ
ˆz = ˆz ˆz = ˆz ˆz = cos θ ˆr − sin θ
θ
ˆr =
1
r
(x xˆ + y ˆy + z ˆz) ˆr =
1
r
R
R + z ˆz
R = sin θ ˆr + cos θ
θ
θ =
1
rR
xz ˆx + yz ˆy − R
2 ˆz
θ =
1
r
z
R − Rˆz
ϕ ˆ = ϕˆ
ϕ ˆ =
1
R
(−y ˆx + x ˆy) ϕˆ = ϕˆ ˆz = cos θ ˆr − sin θ
θ
Line, surface, and volume elements in different coordinate systems.
Cartesian
d
l = ˆx dx + ˆy dy + ˆz dz
d
S =
ˆx dy dz yz-plane
ˆy dx dz xz-plane
ˆz dx dy xy-plane
dV = dx dy dz
Cylindrical
d
l =
R dR + ϕˆ R dϕ + ˆz dz
d
S =
R R dϕ dz curved-surface
ϕ ˆ dR dz meridional-plane
ˆz R dR dϕ top/bottom-plane
dV = R dR dϕ dz
Spherical
d
l = ˆr dr +
θ r sin θ dθ + ϕˆ r dϕ
d
S =
ˆr r
2 sin θ dθ dϕ curved-surface
θ r sin θ dr dϕ meridional-plane
ϕ ˆ r dr dθ xy-plane
dV = r
2 sin θ dr dθ dϕ
~a ·
b = |~a||
b| cos θ
Put another way, given two vectors, the angle between them can be found readily:
θ = cos
− 1
~a ·
b
|~a||
b|
Of course, this implies that if ~a and
b are orthogonal (at right angles), their dot product is zero:
if ~a ⊥
b, then ~a ·
b = 0
Moreover, two vectors are orthogonal (perpendicular) if and only if their dot product is zero, and
they have non-zero length, providing a simple way to test for orthogonality. A few other properties
are tabulated below.
Table 2: Algebraic properties of the scalar product
formula relationship
~a ·
b =
b · ~a commutative
~a · (
b + ~c) = ~a ·
b + ~a · ~c distributive
~a · (r
b + ~c) = r(~a ·
b) + r(~a · ~c) bilinear
(c 1
~a) · (c 2
b) = (c 1
c 2
)(~a ·
b) multiplication by scalars
if ~a ⊥
b, then ~a ·
b = 0 orthogonality
Vector products
The ‘cross’ or vector product between these two vectors results in a pseudovector , also known as an
‘axial vector.’
i An easy way to remember how to calculate the cross product of these two vectors,
~c =~a ×
b, is to take the determinant of the following matrix:
ˆx ˆy ˆz
a x
a y
a z
b x
b y
b z
i Pseudovectors act just like real vectors, except they gain a sign change under improper rotation. See for example,
the Wikipedia page “Pseudovector.” An improper rotation is an inversion followed by a normal (proper) rotation,
just what we are doing when we switch between right- and left-handed coordinate systems. A proper rotation has no
inversion step, just rotation.
Or, explicitly:
~c = det
ˆx ˆy ˆz
a x
a y
a z
b x
b y
b z
= (aybz − az by) xˆ + (az bx − axbz ) ˆy + (axby − aybx) ˆz
Note then that the magnitude of the cross product is
|~a ×
b| = |~a||
b| sin θ
where θ is the smallest angle between ~a and
b. Geometrically, the cross product is the (signed)
volume of a parallelepiped defined by the three vectors given. The pseudovector ~c resulting from a
cross product of ~a and
b is perpendicular to the plane formed by ~a and
b, with a direction given
by the right-hand rule:
~a ×
b = ab sin θ ˆn
where ˆn is a unit vector perpendicular to the plane containing ~a and
b. Note of course that if ~a
and
b are collinear ( i.e., the angle between them is either 0
◦ or 180
◦ ), the cross product is zero.
Right-hand rule
- Point the fingers of your right hand along the direction of ~a.
- Point your thumb in the direction of
b.
- The pseudovector ~c = ~a ×
b points out from the back of your hand.
The cross product is also anticommutative , distributive over addition, and has numerous other
algebraic properties:
Finally, note that the unit vectors in a orthogonal coordinate system follow a cyclical permutation:
ˆx × ˆy = ˆz
yˆ × ˆz = ˆx
ˆz × xˆ = ˆy
I
2
I
3
R
3
− V
2
R
2
... and plug it in to the second one:
V
1
= R
1
I
2
+ (R
1
+ R
3
) I
3
= R
1
I 3 R 3 − V 2
R
2
+ (R
1
+ R
3
) I
3
V
1
V 2 R 1
R 2
R
1
+ R
3
R 1 R 3
R 2
I
3
I
3
V
1
V 2 R 1
R 2
R
1
+ R
3
R 1 R 3
R 2
I
3
V 1 R 2 + V 2 R 1
R
1
R
2
+ R
2
R
3
+ R
1
R
3
Now that you know I 3 , you can plug it in the expression for I 2 above.
What is the second way to solve this? We can start with our original equations, but in a different
order:
I
1
− I
2
− I
3
R
2
I
2
− R
3
I
3
= −V
2
R 1 I 1 + R 3 I 3 = V 1
The trick we want to use is formally known as ‘Gaussian elimination,’ but it just involves adding
these three equations together in different ways to eliminate terms. First, take the first equation
above, multiply it by −R 1 , and add it to the third:
[−R
1
I
1
+ R
1
I
2
+ R
1
I
3
] = 0
+ R
1
I
1
+ R
3
I
3
= V
1
=⇒ R 1 I 2 + (R 1 + R 3 ) I 3 = V 1
Now take the second equation, multiply it by −R 1
/R
2
, and add it to the new equation above:
R
1
R
2
[R
2
I
2
− R
3
I
3
] = −
R
1
R
2
[−V
2
]
+ R
1
I
2
+ (R
1
+ R
3
) I
3
= V
1
R
1
R
3
R
2
+ R
1
+ R
3
I
3
R
1
R
2
V
2
+ V
1
Now the resulting equation has only I 3 in it. Solve this for I 3 , and proceed as above.
Optional: There is one more way to solve this set of equations using matrices and Cramer’s rule,
ii
if you are familiar with this technique. If you are not familiar with matrices, you can skip to the
next problem - you are not required or necessarily expected to know how to do this. First, write
the three equations in matrix form:
R 1 0 R 3
0 R
2
−R
3
I 1
I
2
I
3
V 1
−V
2
aI = V
The matrix a times the column vector I gives the column vector V, and we can use the determinant
of the matrix a with Cramer’s rule to find the currents. For each current, we construct a new matrix,
which is the same as the matrix a except that the the corresponding column is replaced the column
vector V. Thus, for I 1 , we replace column 1 in a with V, and for I 2 , we replace column 2 in
a with V. We find the current then by taking the new matrix, calculating its determinant, and
dividing that by the determinant of a. Below, we have highlighted the columns in a which have
been replaced to make this more clear:
I 1 =
V
1
0 R
3
−V
2
R
2
−R
3
det a
I 2 =
R
1
V
1
R
3
0 −V
2
−R
3
det a
I 3 =
R
1
0 V
1
0 R
2
−V
2
det a
Now we need to calculate the determinant of each new matrix, and divide that by the determinant
of a.
iii First, the determinant of a.
ii See ‘Cramer’s rule’ in the Wikipedia to see how this works.
iii Again, the Wikipedia entry for ‘determinant’ is quite instructive.
