Vector Calculus & General Coordinate Systems
Homework #4, Prob 2:
1
x1 = θ 1θ 2 cosθ 3 , x2 = θ 1θ 2 sin θ 3 , x3 = ⎡⎣(θ 1 ) 2 − (θ 2 ) 2 ⎤⎦
2
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Vector Calculus & General Coordinate Systems
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Vector Calculus & General Coordinate Systems
Curvilinear Systems: Spherical Coordinates
The 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,
q1 = r = x 2 + y 2 + z 2
⎛
z
−1
2
q = θ = cos ⎜
⎜ x2 + y 2 + z 2
⎝
⎛ y⎞
q 3 = φ = tan −1 ⎜ ⎟
⎝ x⎠
⎞
⎟
⎟
⎠
x = r sin θ cos φ
y = r sin θ sin φ
z = r cosθ
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Vector Calculus & General Coordinate Systems
and scale factors and fundamental metric components are
⎡⎛ ∂x ⎞ ⎛ ∂y ⎞ ⎛ ∂z ⎞
h1 = hr = ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟
⎢⎣⎝ ∂r ⎠ ⎝ ∂r ⎠ ⎝ ∂r ⎠
2
2
2 1/ 2
⎤
⎥
⎥⎦
= ⎡⎣(sin θ cos φ ) + (sin θ cos φ ) + cos θ ⎤⎦
h2 = hθ = r
2
2
2
1/ 2
=1
h3 = hφ = r sin θ
g11 = h12 = 1
g 22 = h = r
2
2
g 11 =
2
g
g33 = h = r sin θ
2
3
2
2
22
1
=1
2
h1
1
1
= 2 = 2
h2 r
1
1
g = 2 = 2 2
h3 r sin θ
33
Since this is an
orthogonal
curvilinear
system,
gij = gij = 0, i ≠ j
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Vector Calculus & General Coordinate Systems
θ
eˆ φ
θ
θ = const
r = const
eˆ r
z
r
r
φ
φ = const
eˆθ
y
x
φ
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Vector Calculus & General Coordinate Systems
Spherical coordinates basis and dual basis:
ei = hi eˆ i (no summation)
e j = g ij ei
We can now write the basis and dual basis,
e r = eˆ r
e r = eˆ r
eˆθ
θ
ˆ
eθ = reθ
e =
r
eˆ φ
φ
eφ = r sin θ eˆ φ e =
r sin θ
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Vector Calculus & General Coordinate Systems
Basis vectors in terms of the Cartesian basis:
1 ∂x j ˆ
eˆ i =
i (no summation in i )
i j
hi ∂q
1 ∂x j ˆ
1 ∂x j ˆ
1 ∂x j ˆ
eˆ r =
i , eˆθ =
i , eˆ φ =
i
θ j
φ j
r j
hr ∂q
hθ ∂q
hφ ∂q
In matrix format, these three equations are,
cosθ ⎤ ⎛ ˆi x ⎞
⎜ ⎟
⎥
− sin θ ⎜ ˆi y ⎟
⎥
0 ⎥⎦ ⎜⎜ ˆi z ⎟⎟
⎝ ⎠
So this matrix equation gives spherical basis in terms of the
Cartesian basis.
⎛ eˆ r ⎞ ⎡ sin θ cos φ
⎜ ⎟ ⎢
⎜ eˆθ ⎟ = ⎢cosθ cos φ
⎜ eˆ φ ⎟ ⎢ − sin φ
⎝ ⎠ ⎣
sin θ sin φ
cosθ sin φ
cos φ
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Vector Calculus & General Coordinate Systems
Now for the inverse transformation that gives the Cartesian
basis in terms of the spherical basis. We can start with the now
familiar relation,
a = (a ⋅ ei )ei
→ ˆii = (ˆii ⋅ eˆ j )eˆ j
Then, for example,
ˆi = (ˆi ⋅ eˆ )eˆ + (ˆi ⋅ eˆ )eˆ + (ˆi ⋅ eˆ )eˆ
x
x
r
r
x
θ
θ
x
φ
φ
In matrix format,
⎛ ˆi x ⎞ ⎡sin θ cos φ
⎜ ⎟ ⎢
⎜ ˆi y ⎟ = ⎢ sin θ sin φ
⎜ ⎟
⎜ ˆi z ⎟ ⎢⎣ cosθ
⎝ ⎠
cosθ cos φ
cosθ sin φ
− sin θ
− sin φ ⎤ ⎛ eˆ r ⎞
⎜ ⎟
⎥
cos φ ⎜ eˆθ ⎟
⎥
0 ⎥⎦ ⎜⎝ eˆ φ ⎟⎠
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Vector Calculus & General Coordinate Systems
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:
1
ai (no summation),
hi
A convenient
consequence of using
= aˆr = a r = ar ,
spherical coordinates
aθ
θ
= aˆθ = ra = ,
is that the position
r
arrow has a single
aφ
φ
component, i.e.,
.
= aˆφ = r sin θ a =
r sin θ
r = reˆ r .
aˆ i = aˆi = hi a i =
aˆ r
aˆθ
aˆ φ
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Vector Calculus & General Coordinate Systems
Curvilinear Systems: Cylindrical Coordinates
eˆ Z
Now we examine this
familiar curvilinear
coordinate system in
detail.
eˆ φ
z
R
eˆ R
r
Z
φ
y
x
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Vector Calculus & General Coordinate Systems
Transformation and inverse transformation:
q1 = R = x 2 + y 2
x = R cos φ
y = R sin φ
z=Z
⎛ y⎞
q = φ = tan ⎜ ⎟
⎝ x⎠
q3 = Z = z
2
−1
Scale factors:
⎡⎛ ∂x ⎞ ⎛ ∂y ⎞ ⎛ ∂z ⎞
h1 = hR = ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟
⎢⎣⎝ ∂R ⎠ ⎝ ∂R ⎠ ⎝ ∂R ⎠
h2 = hφ = R
h3 = hZ = 1
2
2
2 1/ 2
⎤
⎥
⎥⎦
=1
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Vector Calculus & General Coordinate Systems
Fundamental metric components:
1
2
11
g11 = h1 = 1
g = 2 =1
h1
g 22 = h = R
2
2
2
g33 = h = 1
2
3
g
22
1
1
= 2 = 2
h2 R
1
g = 2 =1
h3
33
Again, since this is an
orthogonal curvilinear
system,
gij = gij = 0, i ≠ j
Basis and dual:
e R = eˆ R
ei = hi eˆ i (no summation) ⎫
⎬→
j
ij
e = g ei
⎭
eφ = Reˆ φ
e Z = eˆ z
e R = eˆ R
eˆ φ
φ
e =
R
e Z = eˆ z
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Vector Calculus & General Coordinate Systems
In terms of the Cartesian basis:
1 ∂x j ˆ
eˆ i =
i (no summation in i )
i j
hi ∂q
In matrix format, these three equations are:
ˆ
⎛ eˆ R ⎞ ⎡ cos φ sin φ 0 ⎤ ⎛ i x ⎞
⎜ eˆ ⎟ = ⎢ − sin φ cos φ 0 ⎥ ⎜ ˆi ⎟
⎜ φ⎟ ⎢
⎥⎜ y ⎟
⎜ˆ ⎟
⎜ eˆ ⎟ ⎢ 0
0
1
⎥
⎝ Z⎠ ⎣
⎦ ⎜⎝ i z ⎟⎠
and the inverse transformation is:
⎛ ˆi x ⎞ ⎡cos φ
⎜ ⎟ ⎢
⎜ ˆi y ⎟ = ⎢ sin φ
⎜ˆ ⎟ ⎢ 0
⎜ iz ⎟ ⎣
⎝ ⎠
− sin φ
cos φ
0
0 ⎤ ⎛ eˆ R ⎞
0 ⎥ ⎜ eˆ φ ⎟
⎥⎜ ⎟
1 ⎥⎦ ⎝⎜ eˆ Z ⎟⎠
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Vector Calculus & General Coordinate Systems
You can again verify that the coefficient matrix R is
orthogonal, i.e., R−1 = RT.
The physical components of an arbitrary vector a:
1
i
i
aˆ = aˆi = hi a = ai (no summation),
hi
aˆ R = aˆ R = a R = aR ,
φ
φ
aˆ = aˆφ = Ra =
aφ
R
aˆ Z = aˆZ = a Z = aZ .
,
Finally, the position arrow is r = Reˆ R + Zeˆ Z .
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Vector Calculus & General Coordinate Systems
General Transformation between Two Curvilinear Systems
Up 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
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Vector Calculus & General Coordinate Systems
q i = q i (q j ),
{e1 , e 2 , e3 },
q i = q i (q j ),
{e1 , e2 , e3 },
dr = dq i ei
dr = dq i ei
Since the vector dr must be the same, regardless of the
coordinate system,
e s ⋅ (dq i ei = dq j e j )
dq iδ is = dq j (e s ⋅ e j )
dq s = (e s ⋅ e j )dq j .
Similarly, if we dot with the dual of the barred system,
e s ⋅ (dq i ei = dq j e j )
( e s ⋅ ei )dq i = dq s .
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Vector Calculus & General Coordinate Systems
Recall from multivariable differential calculus, the chain rule
for a differential,
s
s
∂
q
∂
q
j
s
i
and
.
dq s =
dq
dq
dq
=
j
j
∂q
∂q
By comparison with the covariant and contravariant
transformation laws in the Vector Algebra section, we see that,
s
s
∂
q
∂
q
s
s
s
α
and
β
es ⋅ e j =
≡
e
⋅
e
=
≡
j
i
i .
j
i
∂q
∂q
We can now write the
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Vector Calculus & General Coordinate Systems
Covariant Transformation Law:
∂q j
es = s e j
∂q
∂q j
as = s a j .
∂q
Contravariant Transformation Law:
s
∂
q
e s = i ei
∂q
s
∂
q
a s = i ai .
∂q
An example of the summations:
∂q1
∂q 2
∂q 3
e1 = 1 e1 + 1 e 2 + 1 e3 ,
∂q
∂q
∂q
1
1
1
q
q
q
∂
∂
∂
e 1 = 1 e1 + 2 e 2 + 3 e3 .
∂q
∂q
∂q
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Vector Calculus & General Coordinate Systems
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,
q i = q i (q 1 , q 2 , q 3 ) and q j = q j (q1 , q 2 , q 3 )
An additional relation can also be shown,
⎛
∂q j ⎞ i
⎜ es = s e j ⎟ ⋅ e
∂q
⎝
⎠
j
∂
q
δ si = s (e j ⋅ e i )
∂q
j
i
∂
∂
q
q
δ si = s j
∂q ∂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.
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Vector Calculus & General Coordinate Systems
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,
(a1 , a2 , a3 )
(a1 , a2 , a3 )
⇓
⇓
(q1 , q 2 , q 3 ) (q 1 , q 2 , q 3 )
satisfy
∂q i
aj =
a,
j i
∂q
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Vector Calculus & General Coordinate Systems
then a j and a i are the covariant components of vector a.
i
j
Note: If q and q 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
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Vector Calculus & General Coordinate Systems
x3
x
x2
1
a1
x
Example: Vector a
in two Cartesian
systems.
3
a3
a3
a
a2
a2
a1
x2
x1
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Vector Calculus & General Coordinate Systems
Example: Noninvariance of the
position “vector”
x3
P
r
x3
r
x2
r0
x1
x2
x1
The barred coordinate system is translated from the unbarred,
r = r0 + r
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Vector Calculus & General Coordinate Systems
Both are rectangular Cartesian thus,
ˆi = ˆi and x i = x i + x i .
0
i
i
The contravariant transformation law gives, for some a,
∂x i j ∂x i j
i
a = j a = j a = δ ij a j = a i .
∂x
∂x
Similarly, for these Cartesian systems, the covariant
components are,
ai = ai .
Thus, the components of vector a are unchanged by the
coordinate transformation. But!
x j ≠ x j since x j = x j + x0j .
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Vector Calculus & General Coordinate Systems
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.
Some ordered triples are vectors
for certain types of
transformations, but not others.
For instance, r transforms as a
vector for a rotation
transformation.
P
r
r
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Vector Calculus & General Coordinate Systems
Derivatives of an Orthonormal Basis
An 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 eˆ i have constant unit magnitude
but variable orientation.
ê3
ω
ê1
dφ
δΦ
Note we use δΦ instead of δΦ since
Φ is a finite rotation, thus is not a
vector.
dφ
ê 2
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Vector Calculus & General Coordinate Systems
Recall, for rigid-body rotation,
v = ω×r
then,
dr δΦ
=
× r → dr = δΦ × r
dt
dt
Since |r| = const for rigid-body rotation, we can separately look
at each of the eˆ i using,
r = eˆ i
→ deˆ i = δΦ × eˆ i , i = 1, 2,3
(8)
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Vector Calculus & General Coordinate Systems
Example: Cylindrical coordinates
In the cylindrical system, a change in position causes a rigidbody 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.
eˆ Z
eˆ Z
eˆ φ
eˆ R
eˆ φ
r1
eˆ R
r2
φ
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Vector Calculus & General Coordinate Systems
eˆ Z
δΦ = dφ eˆ Z
arc-length formula:
s = rφ for constant r,
ds = r dφ
dφ
eˆ R
dφ
eˆ φ
ds = Rdφ = dφ
deˆ R = dφ eˆ φ
deˆ φ = dφ (−eˆ R )
= − dφ eˆ R
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 basis
vectors:
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Vector Calculus & General Coordinate Systems
deˆ i = δΦ × eˆ i ,
deˆ R = δΦ × eˆ R = dφ (eˆ φ × eˆ R ) = dφ eˆ φ ,
deˆ φ = δΦ × eˆ φ = dφ (eˆ Z × eˆ φ ) = − dφ eˆ R ,
deˆ Z = δΦ × eˆ Z = dφ (eˆ Z × eˆ Z ) = 0.
eˆ r
Example: Spherical coordinates
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.
eˆ φ
θ
r1
eˆ r
eˆθ
r2
eˆθ
eˆ φ
φ
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Vector Calculus & General Coordinate Systems
eˆ z
dθ eˆ φ
eˆ r
dφ
dθ
dφ
eˆ φ
eˆθ
eˆ φ
+
=
dφ eˆ z
δΦ
dθ
dφ
δΦ = dφ eˆ z + dθ eˆ φ
(9)
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Vector Calculus & General Coordinate Systems
The vector eˆ z is not a member of the spherical triad, so we
write it in terms of the spherical basis,
eˆ z = (eˆ z ⋅ eˆ r )eˆ r + (eˆ z ⋅ eˆθ )eˆθ + (eˆ z ⋅ eˆ φ ) eˆ φ = cosθ eˆ r − sin θ eˆθ .
⊥ → =0
The superposition of differential rotations, Eq. (9), becomes,
δΦ = dφ cosθ eˆ r − dφ sin θ eˆθ + dθ eˆ φ .
Employing Eq. (8), we write the differentials of the
orthonormal spherical basis vectors,
deˆ i = δΦ × eˆ i ,
deˆ r = δΦ × eˆ r = dφ sin θ eˆ φ + dθ eˆθ ,
deˆθ = δΦ × eˆθ = dφ cosθ eˆ φ − dθ eˆ r ,
deˆ φ = δΦ × eˆ φ = − dφ cosθ eˆθ − dφ sin θ eˆ r .
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Vector Calculus & General Coordinate Systems
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,
1
1
δΦ = curl(dr ) = (∇ × dr ).
2
2
This expression is obtained by taking the curl of the equation
v = ω × r (how this is done will become clear later). Here we
have introduced the gradient or del operator “∇” and the curl
operation.
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Vector Calculus & General Coordinate Systems
Curl of a Vector
The curl of an arbitrary vector a is written as
curl a ≡ ∇ × a
where ∇ (del, nabla) is a vector differential operator, the
general form of which we will define. For now, in an
orthogonal curvilinear system,
h1eˆ1
h2eˆ 2
h3eˆ 3
1
∂
∂
∂
curl(dr ) ≡ ∇ × dr =
h1h2 h3
∂q1
∂q 2
∂q 3
h1 (h1dq1 ) h2 (h2 dq 2 ) h3 (h3 dq 3 )
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Vector Calculus & General Coordinate Systems
eˆ1 ⎡ 3 ∂
2
2 ∂
2 ⎤
curl(dr ) =
dq
(h3 ) − dq
(h2 ) ⎥
⎢
2
3
h2 h3 ⎣
∂q
∂q
⎦
eˆ 2 ⎡ 1 ∂
2
3 ∂
2 ⎤
dq
(h1 ) − dq
(h3 ) ⎥
+
⎢
3
1
h1h3 ⎣
∂q
∂q
⎦
⎡ 2 ∂
2
1 ∂
2 ⎤
dq
(
h
)
dq
(
h
−
2
1 )⎥
⎢
1
2
∂q
∂q
⎣
⎦
1
Now, for example: deˆ1 = δΦ × eˆ1 = (∇ × dr ) × eˆ1
2
eˆ 2 ⎡ 2 ∂
2
1 ∂
2 ⎤
dq
(h2 ) − dq
(h1 ) ⎥
=
⎢
1
2
2h1h2 ⎣
∂q
∂q
⎦
eˆ 3
+
h1h2
eˆ 3 ⎡ 3 ∂
2
1 ∂
2 ⎤
dq
(h3 ) − dq
(h1 ) ⎥
+
⎢
1
3
h1h3 ⎣
∂q
∂q
⎦
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Vector Calculus & General Coordinate Systems
The differentials deˆ 2 and deˆ 3 are computed similarly.
Comparing these results to the total differential
∂eˆ i
deˆ i = j dq j ,
∂q
We can identify the partial derivatives of the orthonormal base
vectors
∂eˆ1
eˆ1 ∂h1 eˆ 3 ∂h1
=−
−
,
1
2
3
∂q
h2 ∂q h3 ∂q
∂eˆ1
eˆ 2 ∂h2
∂eˆ1 eˆ 3 ∂h3
,
.
=−
=
2
1
3
1
h1 ∂q
h1 ∂q
∂q
∂q
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Vector Calculus & General Coordinate Systems
eˆ1 ∂h2 eˆ 3 ∂h2
∂eˆ 2 eˆ1 ∂h1
∂eˆ 2
=
=−
−
,
,
1
2
2
1
3
∂q
∂q
h2 ∂q
h1 ∂q h3 ∂q
∂eˆ 2 eˆ 3 ∂h3
=
.
3
2
∂q
h2 ∂q
∂eˆ 3 eˆ1 ∂h1
∂eˆ 3 eˆ 2 ∂h2
=
=
,
,
1
3
2
3
∂q
∂q
h3 ∂q
h3 ∂q
∂eˆ 3
eˆ1 ∂h3 eˆ 2 ∂h3
=−
−
.
3
1
2
∂q
h1 ∂q h2 ∂q
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Vector Calculus & General Coordinate Systems
Rotating Reference Frames
A primary application for the following analysis is in classical
mechanics (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.
x3
x3
b
ω
e
P
3
e3
e1
x1
x2
r
r0
e1
e2
x1
e2
x2
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Vector Calculus & General Coordinate Systems
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 {e1 , e 2 , e3 } are a constant (magnitude
and direction) basis in the nonrotating frame. Since the basis is
constant, the time derivative of a vector b is
db dbi
=
ei .
dt
dt
In the rotating frame, the basis vectors {e1 , e2 , e3 } have constant
magnitude also, but variable orientation. So, for an observer in
the rotating frame,
de
db db i
=
ei + b i i .
dt
dt
dt
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Vector Calculus & General Coordinate Systems
Since the ei have constant magnitude, the rate of change is due
only from the rigid-body rotation of the frame, i.e.,
d ei
= ω × ei .
dt
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,
db i
db
.
ei ≡
dt
dt rot
Thus we have,
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Vector Calculus & General Coordinate Systems
db db
=
+ b i (ω × ei )
dt dt rot
db
=
+ (ω × b).
dt rot
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
d
=
dt dt
+ ω× .
rot
(10)
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Vector Calculus & General Coordinate Systems
Velocity and Acceleration in Rotating Frames
Velocity
We 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 r = r0 + r ,
⎞
dr ⎛ d
ω × ⎟ (r0 + r )
=⎜
dt ⎝ dt rot
⎠
dr0 d r
=
+
+ ω × r.
dt dt rot
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Vector Calculus & General Coordinate Systems
where,
dr
= absolute velocity,
dt
dr0
= absolute velocity of the origin of the rotating frame
dt
dr
= velocity of object at P with respect to rotating frame
dt rot
ω × r = velocity of point P in the rotating frame.
Note that when dr0/dt = 0 and (d r / dt ) rot = 0 , the point P is
fixed with respect to the rotating frame and we have a rigidbody rotation with respect to the inertial frame.
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Vector Calculus & General Coordinate Systems
Acceleration
Applying the operator (10) to the velocity, we determine the
acceleration,
d 2r ⎛ d
=⎜
2
dt
⎝ dt
⎞⎛ dr0 d r
ω × ⎟⎜
+
rot
⎠⎝ dt dt
d 2r0 d 2 r
= 2 + 2
dt
dt
d
+
dt
rot
d 2r0 d 2 r
= 2 + 2
dt
dt
+
rot
⎞
+ ω× r ⎟
rot
⎠
dr
ω× r + ω×
dt
rot
dω
dr
× r + 2ω ×
dt
dt
+ ω × (ω × r )
rot
+ ω × (ω × r )
rot
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Vector Calculus & General Coordinate Systems
where,
d 2r
= absolute acceleration
2
dt
d 2r0
= absolute acceleration of the origin of the rotating frame
2
dt
⎧ tangential component of acceleration in the plane
dω
×r = ⎨
dt
⎩of r and (dr / dt ) rot and perpendicular to r
dr
2ω ×
dt
= Coriolis acceleration
rot
ω × (ω × r ) = centripetal acceleration
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Vector Calculus & General Coordinate Systems
Example:
Given: At the instant shown,
ω = −20 rad/s ˆj
Y
B
OAB
A
8″
dωOAB
= −200 rad/s 2 ˆj
α OAB =
dt
The velocity and acceleration of D,
relative to the rod, are 50 in/s and 600
in/s2 upward, respectively.
30°
D
O
Z
X
Find: The velocity and acceleration of
the collar D.
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Vector Calculus & General Coordinate Systems
Solution:
Note that for this problem, r = r . First find position and
velocity at D,
r = (8 in)(sin30° ˆi + cos30° ˆj) = (4 in)ˆi + (6.93 in)ˆj
v D = ( v D )OAB + ω × r
The two parts of the velocity are
( v D )OAB = (50 in/s)(sin30° ˆi + cos30° ˆj)
= (25 in/s) ˆi + (43.3in/s) ˆj
ω × r = (−20 rad/s) ˆj × [(4 in) ˆi + (6.93 in) ˆj]
= 80 in/s kˆ
v D = (25 in/s) ˆi + (43.3 in/s)ˆj + 80 in/s kˆ ⇐
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Vector Calculus & General Coordinate Systems
Now the acceleration,
d 2r0 d 2 r
+ 2
aD =
2
dt
dt
+
OAB
dω
× r + 2ω × ( v D )OAB + ω × (ω × r )
dt
The individual terms are
d 2r0
dt 2
= (600 in/s 2 )(sin30° ˆi + cos30° ˆj)
OAB
= (300 in/s 2 ) ˆi + (520 in/s 2 ) ˆj
dω
× r = (−200 rad/s 2 ) ˆj × [(4 in) ˆi + (6.93 in) ˆj]
dt
= (800 in/s 2 ) kˆ
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Vector Calculus & General Coordinate Systems
2ω × ( v D )OAB = 2(−20 rad/s) ˆj × [(25 in/s ˆi + (43.3 in/s) ˆj]
= (1000 in/s 2 ) kˆ
ω × (ω × r ) = (−20 rad/s) ˆj × (80 in/s) kˆ
= (−1600 in/s 2 ) ˆi
a D = (−1300 in/s 2 ) ˆi + (520 in/s 2 ) ˆj × (1800 in/s 2 ) kˆ ⇐
174