Vector Calculus & General Coordinate Systems

1 2 3 1 2 3 1 2 2 21 2 3

1cos , sin , ( ) ( )2

x x xθ θ θ θ θ θ θ θ⎡ ⎤= = = −⎣ ⎦

Homework #4, Prob 2:

Curvilinear Systems: Spherical CoordinatesThe curvilinear spherical coordinate system is probably familiar to all of you. In engineering and physics this coordinate system is used to take advantage of spherical symmetry. Let’s examine this coordinate system in detail.The curvilinear transformations and inverse transformations that define the spherical system are given by,

1 2 2 2

q r x y z

zqx y z

= = + +

⎛ ⎞= = ⎜ ⎟

⎜ ⎟+ +⎝ ⎠⎛ ⎞= = ⎜ ⎟⎝ ⎠

sin cossin sincos

x ry rz r

θ φθ φθ

and scale factors and fundamental metric components are

1/ 22 2 2

(sin cos ) (sin cos ) cos 1

rx y zh hr r r

h h rh h r

θ φ θ φ θ

⎡ ⎤∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤= + + =⎣ ⎦= == =

2 1111 1 2

2 2 2222 2 2 2

2 2 2 3333 3 2 2 2

1 1sinsin

g h gh

g h r gh r

= = = =

Since this is an orthogonal curvilinear system,

gij = gij = 0, i ≠ j

rr = const

= const

= constφ

z ˆ re

Spherical coordinates basis and dual basis:

We can now write the basis and dual basis,

ˆ (no summation)i i i

ˆ ˆˆˆ

ˆˆsin

rr r r

θ θθ θ

φφφ φθ

e e e eee e e

ee e e

Basis vectors in terms of the Cartesian basis:

In matrix format, these three equations are,

So this matrix equation gives spherical basis in terms of the Cartesian basis.

1 ˆˆ (no summation in )

1 1 1ˆ ˆ ˆˆ ˆ ˆ, ,

r j j jrr

x ih q

x x xh q h q h qθ φθ φ

∂ ∂ ∂= = =

∂ ∂ ∂

e i e i e i

ˆˆ sin cos sin sin cosˆˆ cos cos cos sin sin

ˆ ˆsin cos 0

θ φ θ φ θθ φ θ φ θ

⎛ ⎞⎛ ⎞ ⎡ ⎤ ⎜ ⎟⎜ ⎟ ⎢ ⎥= − ⎜ ⎟⎜ ⎟ ⎢ ⎥ ⎜ ⎟⎜ ⎟ −⎢ ⎥⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠

iee ie i

Now for the inverse transformation that gives the Cartesian basis in terms of the spherical basis. We can start with the now familiar relation,

Then, for example,

In matrix format,

ˆ ˆ ˆ ˆ( ) ( )ii i i j j= ⋅ → = ⋅a a e e i i e e

ˆ ˆsin cos cos cos sinˆ ˆsin sin cos sin cosˆ ˆcos sin 0

θ φ θ φ φθ φ θ φ φθ θ

⎛ ⎞ ⎛ ⎞−⎡ ⎤⎜ ⎟ ⎜ ⎟⎢ ⎥=⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟−⎢ ⎥⎜ ⎟ ⎣ ⎦ ⎝ ⎠⎝ ⎠

i ei e

ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( )x x r r x xθ θ φ φ= ⋅ + ⋅ + ⋅i i e e i e e i e e

Note that if we designate the coefficient matrix of the transformation as R, then the inverse transformation coefficient matrix is R−1 = RT Thus, R is orthogonal (AEM Sec. 3.5, 3.8). It can be shown that any orthogonal transformation represents a rotation (possibly combined with a reflection). Physical components of an arbitrary vector a:

1ˆ ˆ (no summation),

ˆ ˆ ,

ˆ ˆ sin .sin

i ii i i

r rr r

a a h a ah

a a a aaa a rar

aa a r a

θ θ θθ

φφ φφ θ

A convenient consequence of using spherical coordinates is that the position arrow has a single component, i.e.,

ˆ .rr=r e

Curvilinear Systems: Cylindrical Coordinates

Now we examine this familiar curvilinear coordinate system in detail.

Transformation and inverse transformation:

Scale factors:

q R x y

φ −

⎛ ⎞= = ⎜ ⎟⎝ ⎠

cossin

x Ry Rz Z

1/ 22 2 2

x y zh hR R R

h h R h hφ

⎡ ⎤∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + + =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦= = = =

Fundamental metric components:

Basis and dual:

2 1111 1 2

2 2 2222 2 2 2

2 3333 3 2

g h gh

g h R gh R

g h gh

= = = =

Again, since this is an orthogonal curvilinear system,

gij = gij = 0, i ≠ j

ˆ (no summation)i i ij ij

= ⎫→⎬

ˆ ˆˆ

RR R R

ZZ z z

RRφφ

e e e ee

e e e e

In terms of the Cartesian basis:

In matrix format, these three equations are:

and the inverse transformation is:

1 ˆˆ (no summation in )j

x ih q∂

ˆˆ cos sin 0ˆˆ sin cos 0

ˆ ˆ0 0 1

φ φφ φ

⎛ ⎞⎛ ⎞ ⎡ ⎤ ⎜ ⎟⎜ ⎟ ⎢ ⎥= − ⎜ ⎟⎜ ⎟ ⎢ ⎥ ⎜ ⎟⎜ ⎟ ⎢ ⎥ ⎜ ⎟⎝ ⎠ ⎣ ⎦ ⎝ ⎠

iee ie i

ˆ ˆcos sin 0ˆ ˆsin cos 0ˆ ˆ0 0 1

φ φφ φ

⎛ ⎞ −⎡ ⎤⎛ ⎞⎜ ⎟ ⎜ ⎟⎢ ⎥=⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎣ ⎦ ⎝ ⎠⎝ ⎠

i ei e

You can again verify that the coefficient matrix R is orthogonal, i.e., R−1 = RT.

The physical components of an arbitrary vector a:

Finally, the position arrow is

1ˆ ˆ (no summation),

ˆ ˆ ,

ˆ ˆ .

i ii i i

Z ZZ Z

a a h a ah

a a a aa

a a RaR

a a a a

φφ φφ

ˆ ˆ .R ZR Z= +r e e

General Transformation between Two Curvilinear SystemsUp to this point we have explored how to transform to and from the Cartesian system to a curvilinear system and, in particular, the spherical and cylindrical systems. But what of the situation where we need a transformation between curvilinear systems, say, cylindrical to spherical or vice versa?

We now present the rules for doing these types of transformations. Consider two curvilinear systems, and as before denote them as the “unbarred” and “barred” systems. The transformations and inverse transformations are written (for i = 1, 2, 3) as

Since the vector dr must be the same, regardless of the coordinate system,

Similarly, if we dot with the dual of the barred system,

( ),{ , , },

r e1 2 3

( ),{ , , },

s i ji j

i s j si j

s s jj

s i ji j

s i si

Recall from multivariable differential calculus, the chain rule for a differential,

By comparison with the covariant and contravarianttransformation laws in the Vector Algebra section, we see that,

We can now write the

and .s s

s j s ij j

q qdq dq dq dqq q∂ ∂

= =∂ ∂

and .s s

s s s sj j i ij i

q qq q

α β∂ ∂⋅ = ≡ ⋅ = ≡

∂ ∂e e e e

Covariant Transformation Law:

Contravariant Transformation Law:

An example of the summations:

s j s js s

q qa aq q∂ ∂

= =∂ ∂

s i s ii i

q qa aq q

∂ ∂= =∂ ∂

1 1 2 31 1 1

1 1 11 1 2 3

q q qq q qq q qq q q

∂ ∂ ∂= + +∂ ∂ ∂

e e e e

Note these transformation laws allow one to determine the basis and dual basis of one system in terms of the other by using the given transformation relations,

An additional relation can also be shown,

1 2 3 1 2 3( , , ) and ( , , )i i j jq q q q q q q q q q= =

ji is js

j iis s j

qqq qq q

⎛ ⎞∂= ⋅⎜ ⎟∂⎝ ⎠∂

= ⋅∂

∂ ∂=∂ ∂

Look familiar? In fact, everything we’ve done here is similar to the transformations we derived between the curvilinear system and Cartesian. In fact, we could write the present relations in matrix format just as we did in the previous section.

Actually, none of this should surprise you since the Cartesian system is just a particular (albeit special for we humans) curvilinear system.

Analytical Definition of a Vector If the ordered triples,

satisfy

1 2 3 1 2 3

( , , ) ( , , )

a a a a a a

q q q q q q⇓ ⇓

qa aq∂

then and are the covariant components of vector a.

Note: If and are rectangular Cartesian coordinates, the covariant and contravariant are identical.

A vector quantity is independent of any coordinate system, thus is invariant to a coordinate transformation

Vector Calculus & General Coordinate Systemsja ia

Example: Vector ain two Cartesian systems.

The barred coordinate system is translated from the unbarred,

3xExample: Non-invariance of the position “vector” P

0= +r r r

Both are rectangular Cartesian thus,

The contravariant transformation law gives, for some a,

Similarly, for these Cartesian systems, the covariant components are,

Thus, the components of vector a are unchanged by the coordinate transformation. But!

0ˆ ˆ and .i i ii i x x x= = +i i

i j j i j ijj j

x xa a a a ax x

δ∂ ∂= = = =∂ ∂

.i ia a=

0 since .j j j j jx x x x x≠ = +

Because the tail of the position “vector” is, by definition, located at the origin of the coordinate system, it is tied to that origin. Therefore, the position “vector” is not invariant to a coordinate translation. Consequently, the position “vector” r is not a vector under a translation transformation.

PSome ordered triples are vectors for certain types of transformations, but not others. For instance, r transforms as a vector for a rotation transformation.

Derivatives of an Orthonormal BasisAn orthonormal basis vector triad can be viewed as a rotating rigid body, i.e., the orientation of the triad may change, but the vectors remain fixed with respect to each other.

The have constant unit magnitude but variable orientation.

Note we use δΦ instead of δΦ since Φ is a finite rotation, thus is not a vector.

dφ dφ

ω δΦ

Recall, for rigid-body rotation,

Since |r| = const for rigid-body rotation, we can separately look at each of the using,

= ×v ω r

d ddt dt

= × → = ×r δΦ r r δΦ r

ˆ ˆ ˆ , 1,2,3i i id i= → = × =r e e δΦ e

Example: Cylindrical coordinates In the cylindrical system, a change in position causes a rigid-body rotation of the basis-vector triad if it involves the angle φ(a rotation about the z-axis). Based on this, we develop expressions for the differential changes.

ˆRe ˆφe

For this simple rotation about the z-axis, the differential change due to the rotation is δΦ. In the cylindrical coordinate system, we have the following for the differentials of the basisvectors:

Vector Calculus & General Coordinate SystemsˆZe

ˆRe ˆφedφ

arc-length formula:s = rφ for constant r, ds = r dφ

ˆZdφ=δΦ e

ˆ ˆ( )ˆ

dφ φ

ˆ ˆR

ds Rd dd d φ

φ φφ

= ==e e

Example: Spherical coordinates

ˆ ˆ ,ˆ ˆ ˆ ˆ ˆ( ) ,ˆ ˆ ˆ ˆ ˆ( ) ,ˆ ˆ ˆ ˆ( ) .

Z Z Z Z

dd d d

φ φ φ

= ×= × = × =

= × = × = −

= × = × =

e δΦ ee δΦ e e e e

e δΦ e e e e

e δΦ e e e 0

This case is a bit more complex, since two angles (θ,φ) are involved. A change in position results in a superposition of two angular rotations.

φˆθe

ˆ reˆφe

ˆθe1r

ˆφe ˆ zdφe

ˆd φθe

dφ + =δΦ

ˆ ˆzd d φφ θ= +δΦ e e

The vector is not a member of the spherical triad, so we write it in terms of the spherical basis,

The superposition of differential rotations, Eq. (9), becomes,

Employing Eq. (8), we write the differentials of the orthonormal spherical basis vectors,

Vector Calculus & General Coordinate Systemsˆ ze

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) cos sin .z z r r z z rθ θ φ φ θθ θ⊥→=

= ⋅ + ⋅ + ⋅ = −e e e e e e e e e e e e

ˆ ˆ ˆcos sin .rd d dθ φφ θ φ θ θ= − +δΦ e e e

ˆ ˆ ,ˆ ˆ ˆ ˆsin ,ˆ ˆ ˆ ˆcos ,ˆ ˆ ˆ ˆcos sin .

dd d dd d d

θ θ φ

φ φ θ

φ θ θ

φ θ φ θ

= ×= × = +

= × = −

= × = − −

e δΦ ee δΦ e e ee δΦ e e e

e δΦ e e e

We will define the curl vector-differential operator soon, however we will introduce it now, a bit prematurely perhaps, to state that for a general orthogonal system,

This expression is obtained by taking the curl of the equation(how this is done will become clear later). Here we

have introduced the gradient or del operator “∇” and the curl operation.

1 1curl( ) ( ).2 2

d d= = ∇×δΦ r r

= ×v ω r

Curl of a VectorThe curl of an arbitrary vector a is written as

where ∇ (del, nabla) is a vector differential operator, the general form of which we will define. For now, in an orthogonal curvilinear system,

1 1 2 2 3 3

1 2 31 2 3

1 2 31 1 2 2 3 3

ˆ ˆ ˆ1curl( )

( ) ( ) ( )

d dh h h q q q

h h dq h h dq h h dq

∂ ∂ ∂≡ ∇× =

∂ ∂ ∂

curl ≡ ∇×a a

Now, for example:

Vector Calculus & General Coordinate Systems3 2 2 21

3 22 32 3

1 2 3 221 33 1

2 2 1 232 11 2

ˆcurl( ) ( ) ( )

ˆ( ) ( )

d dq h dq hh h q q

dq h dq hh h q q

⎡ ⎤∂ ∂= −⎢ ⎥∂ ∂⎣ ⎦

⎡ ⎤∂ ∂+ −⎢ ⎥∂ ∂⎣ ⎦

2 2 1 222 11 2

3 2 1 233 11 3

1ˆ ˆ ˆ( )2

ˆ( ) ( )

dq h dq hh h q q

= × = ∇× ×

⎡ ⎤∂ ∂= −⎢ ⎥∂ ∂⎣ ⎦

⎡ ⎤∂ ∂+ −⎢ ⎥∂ ∂⎣ ⎦

e δΦ e r e

The differentials and are computed similarly. Comparing these results to the total differential

We can identify the partial derivatives of the orthonormal base vectors

Vector Calculus & General Coordinate Systems2ˆde 3ˆde

ˆˆ ,jii jd dq

31 1 1 11 2 3

3 31 2 2 12 1 3 1

ˆˆ ˆ,

ˆˆ ˆ ˆ, .

h hq h q h q

hhq h q q h q

∂ ∂ ∂= − −

∂ ∂ ∂∂∂ ∂ ∂

= − =∂ ∂ ∂ ∂

ee e e

3 31 1 2 21 3 2 3

3 3 31 23 1 2

ˆ ˆˆ ˆ, ,

ˆ ˆ ˆ.

h hq h q q h q

h hq h q h q

∂ ∂∂ ∂= =

∂ ∂ ∂ ∂∂ ∂ ∂

= − −∂ ∂ ∂

e ee e

32 1 1 2 1 2 21 2 2 1 3

3 323 2

ˆˆ ˆ ˆ ˆ, ,

h h hq h q q h q h q

hq h q

∂ ∂ ∂ ∂ ∂= = − −

∂ ∂ ∂ ∂ ∂∂∂

=∂ ∂

ee e e e

Rotating Reference Frames A primary application for the following analysis is in classicalmechanics (dynamics). For this problem of relative motion, the unbarred frame represents a fixed (inertial) reference frame. The rotating “barred” frame may also be translating with respect to the fixed frame.

We will show how the rotation of an observer in the barred frame affects the measurement of the time rate of change of an arbitrary vector quantity b and how this relates to an observer in the fixed frame. For simplicity, assume the are a constant (magnitude and direction) basis in the nonrotating frame. Since the basis is constant, the time derivative of a vector b is

In the rotating frame, the basis vectors have constant magnitude also, but variable orientation. So, for an observer in the rotating frame,

1 2 3{ , , }e e e

id dbdt dt

dd db bdt dt dt

= +eb e

Since the have constant magnitude, the rate of change is dueonly from the rigid-body rotation of the frame, i.e.,

Note that because the rotating observer is rotating with the barred coordinates, the observer does not detect a change in orientation (to this observer, the inertial frame appears to be rotating). We then define the time rate of change of b, as observed by the observer in the rotating frame,

Thus we have,

idb ddt dt

= ×e ω e

What does this mean? An observer in the inertial frame is not rotating so sees the absolute derivative db/dt. The rotating observer, however, sees the derivative (db/dt)rot and an additional part due to the fact that the observer is rotating. Now define a vector differential operator:

.d ddt dt

= + ×ω

d d bdt dt

= + ×

b b ω e

b ω b

Velocity and Acceleration in Rotating FramesVelocityWe now inspect the determination of velocity and acceleration at point P with respect to a rotating coordinate frame. We apply the vector differential operator (10) to ,

d ddt dt

⎛ ⎞= × +⎜ ⎟⎝ ⎠

= + + ×

r ω r r

r r ω r

0= +r r r

where,

Note that when dr0/dt = 0 and , the point P is fixed with respect to the rotating frame and we have a rigid-body rotation with respect to the inertial frame.

absolute velocity,

absolute velocity of the origin of the rotating frame

velocity of object at with respect to rotating frame

velocity of point in the rotating frame.

ddtddtd Pdt

rot( / )d dt =r 0

AccelerationApplying the operator (10) to the velocity, we determine the acceleration,

2rot rot

2 2rot rotrot

2 2rotrot

dd d ddt dt dt dt

d d d ddt dt dt dt

⎛ ⎞⎛ ⎞= × + + ×⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= + + × + × + × ×

rr rω ω r

r r rω r ω ω ω r

r r ω rr ω ω ω r

where,

absolute acceleration

absolute acceleration of the origin of the rotating frame

tangential component of acceleration in the plane

of and ( / ) and perpendicular to

ddtddtd

d dtdt

⎧× = ⎨

ω rr r r

Coriolis acceleration

( ) centripetal acceleration

× × =

ω ω r

Example:

D8″ 30°

Given: At the instant shown,

The velocity and acceleration of D, relative to the rod, are 50 in/s and 600 in/s2 upward, respectively.

Find: The velocity and acceleration of the collar D.

ˆ20 rad/s

ˆ200 rad/s

OABOAB

= = −

ω jωα j

Solution: Note that for this problem, . First find position and velocity at D,

The two parts of the velocity are

ˆ ˆ ˆ ˆ (8 in)(sin30 + cos30 ) (4 in) + (6.93 in) ( )D D OAB

= ° ° == + ×

r i j i jv v ω r

ˆ ˆ( ) (50 in/s)(sin30 + cos30 )ˆ ˆ(25 in/s) (43.3in/s)

ˆ ˆ ˆ( 20 rad/s) [(4 in) (6.93 in) ]ˆ80 in/s

ˆ ˆ ˆ(25 in/s) (43.3 in/s) 80 in/s

= ° °

× = − × +

= + + ⇐

ω r j i j

v i j k

Now the acceleration,

The individual terms are

ˆ ˆ(600 in/s )(sin30 + cos30 )

ˆ ˆ(300 in/s ) (520 in/s )

ˆ ˆ ˆ( 200 rad/s ) [(4 in) (6.93 in) ]

ˆ(800 in/s )

= ° °

× = − × +

i jω r j i j

2 2 ( ) ( )D OABOAB

d ddt dt

+ + × + × + × ×r ω r ω v ω ω r

ˆ ˆ ˆ2 ( ) 2( 20 rad/s) [(25 in/s (43.3 in/s) ]ˆ(1000 in/s )

ˆ ˆ( ) ( 20 rad/s) (80 in/s) ˆ( 1600 in/s )

ˆ ˆ ˆ( 1300 in/s ) (520 in/s ) (1800 in/s )

× = − × +

× × = − ×

= − + × ⇐

ω v j i j

ω ω r j k

a i j k

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