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Curvilinear Coordinates - Lecture Notes | PH 106, Exams of Physics

Material Type: Exam; Class: Generl Physics W/Calc II; Subject: PH-Physics; University: University of Alabama; Term: Fall 2008;

Typology: Exams

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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 ˆ
k, using the substitutions ˆx =ˆı,ˆy =ˆ, and ˆz =ˆ
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=px2+y2x=Rcos ϕ x =rsin θcos ϕ
ϕ= tan1y
xy=Rsin ϕ y =rsin θsin ϕ
z=z z =z z =rcos θ
r=px2+y2+z2r=R2+z2R=rsin θ
θ= tan1x2+y2
zθ= tan1R
zϕ=ϕ
ϕ= tan1y
xϕ=ϕ z =rcos 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)
ˆ
R=x
Rˆx +y
Rˆy ˆx = cosϕˆ
Rsin ϕˆϕˆx = sin θcos ϕˆr + cos θcos ϕˆ
θsin ϕˆϕ
ˆϕ=y
Rˆx +x
Rˆy ˆy = sinϕˆ
R+ cos ϕˆϕˆy = sinθsin ϕˆr + cos θsin ϕˆ
θ+ cos ϕˆϕ
ˆz =ˆz ˆz =ˆz ˆz = cos θˆr sin θˆ
θ
ˆr =1
r(xˆx +yˆy +zˆz)ˆr =1
rRˆ
R+zˆzˆ
R= sin θˆr + cos θˆ
θ
ˆ
θ=1
rR xz ˆx +yz ˆy R2ˆzˆ
θ=1
rzˆ
RRˆzˆϕ=ˆϕ
ˆϕ=1
R(yˆx +xˆy)ˆϕ=ˆϕˆz = cos θˆr sin θˆ
θ
pf3
pf4
pf5
pf8
pf9

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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

  • y

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

  • y

2

  • z

2 r =

R

2

  • z

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

  1. Point the fingers of your right hand along the direction of ~a.
  2. Point your thumb in the direction of

b.

  1. 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.