Kinematics in 3-D
1. Position
a) Rectangular Coordinates: x(t), y(t), and z(t)
r = xx + yy + zz, where x, y and z are constant unit vectors.
b) Spherical Coordinates: r(t), q(t) and f(t)
Here r is the radial coordinate, a distance measured from the origin to the location of the point. The angle q is the angle r makes with the z-axis. The angle f is the angle the projection of r on the x-y plane makes with the x-axis.
We need to determine the unit vectors for this system, and we need to determine the transformation equations to the rectangular system:
r = sin(q)cos(f)x + sin(q)sin(f)y + cos(q)z = r(q,f)
q = cos(q)cos(f)x + cos(q)sin(f)y - sin(q)z = q (q,f)
f = -sin(f)x + cos(f)y = f(f)
Notice that the q direction has a direction 90o different from the r direction (replace q with q+90o). Notice that the f direction has a direction 90o different from the projection direction (replace f with f+90o in the normal projection direction of cos(f)x + sin(f)y .
In spherical coordinates, like polar, the position vector is simply: r = rr . While the r dependence is explicit, both the q and the f dependence are "hidden" in the unit vector, r.
2. Velocity
a) Rectangular Coordinates:
v = dr/dt = d[xx + yy + zz]/dt = (dx/dt)x + (dy/dt)y + (dz/dt)z = vxx + vyy + vzz.
b) Spherical Coordinates:
v = dr/dt = d[rr]/dt = (dr/dt)r + r(dr/dt) .
We note that the unit vector, r, is not a constant. As we did with polar, we must determine what the derivatives of the unit vectors in spherical coordinates are. Note that dr/dt = (¶r/¶q)(dq/dt) + (¶r/¶f)(df/dt).
¶r/¶q = ¶[sin(q)cos(f)x + sin(q)sin(f)y + cos(q)z]/¶q =
cos(q)cos(f)x + cos(q)sin(f)y - sin(q)z = q .
In similar fashion, we can find the derivatives of all three unit vectors with respect to both q and f . We summarize these below:
¶r/¶q = q
¶r/¶f = sin(q)f
¶q/¶q = -r
¶q/¶f = cos(q)f
¶f/¶q = 0
¶f/¶f = -sin(q)r - cos(q)q .
We now have for the velocity in spherical coordinates:
v = dr/dt = d[rr]/dt = (dr/dt)r + r(dr/dt) =
r'r + r[(¶r/¶q)(dq/dt) + (¶r/¶f)(df/dt)] =
r'r + rq'q + rf'sin(q)f = v .
We recognize the first term as the regular radial velocity, the second term is the circular velocity around an axis perpendicular to the z axis (wr), and the third term is the circular velocity around the z axis - but this has a radius not of r, but of [r sin(q)]. On the earth these three velocities would correspond to: 1) velocity up or down (away from or toward the earth; 2) velocity North or South; and 3) velocity East or West.
Notice that each term has one r (distance) and one ' (per time) as required for a velocity.
3. Acceleration
a) Rectangular Coordinates:
a = dv/dt = d[vxx + vyy+ vzz]/dt = (dvx/dt)x + (dvy/dt)y+ (dvz/dt)z = axx + ayy+ azz .
b) Spherical Coordinates:
a = dv/dt = d[r'r + rq'q + rf'sin(q)f ]/dt
= (dr'/dt)r + r'(dr/dt) + (dr/dt)q'q + r(dq'/dt)q + rq'(dq/dt) + (dr/dt)f'sin(q)f + r(df'/dt)sin(q)f + rf'(d[sin(q)]/dt)f + rf'sin(q)(df/dt)
= r''r + r'[q'q + f'sin(q)f] + r'q'q + rq''q + rq'[q'(-r) + f'cos(q)f] + r'f'sin(q)f + rf''sin(q)f + rf'[cos(q)q']f + rf'sin(q)f'[-sin(q)r - cos(q)q ]
= [r'' - rq'2 - rsin2(q)f'2]r
+ [2r'q' + rq'' - rsin(q)cos(q)f'2]q
+ [2r'f'sin(q) + 2rq'f'cos(q) + rsin(q)f'']f
Analysis of each term:
For the r component: 1) the r'' is just the regular straight line acceleration in the radial direction (up-down); 2) the -rq'2 term is the centripetal acceleration due to rotation around an axis perpendicular to the z-axis (due to a North-South motion on the earth); 3) the -rsin2(q)f'2 term is the r-component of the centripetal acceleration due to rotation around the z-axis, with r sin(q) the radius of the circular motion (due to an East-West motion on the earth).
For the q component: 1) the 2r'q' is the Coriolis term due to the radius changing affecting the q rotational motion (N-S); 2) the rq'' term is the regular tangential acceleration causing the tangential (N-S) speed to change; 3) the -rsin(q)cos(q)f'2 term is the q-component of the centripetal acceleration due to rotation around the z-axis (E-W), with r sin(q) the radius of the circular motion.
For the f component: 1) the 2r'f'sin(q) is a Coriolis term due to the radius changing affecting the f rotational motion (E-W); 2) the 2rq'f'cos(q ) is a Coriolis term due to the f-radius (r sin(q) changing due to the rq' motion; 3) the rsin(q)f'' is the regular tangential acceleration causing the tangential (E-W) speed to change.
Note that each term has one r (distance) and two '' (per time squared) are required by an acceleration.
Homework Problem #16: Given that A is a vector function, find d2A/dt2 in cylindrical coordinates (a 3-D vector problem).
4. The del operator
a) Rectangular Coordinates:
Ñ
= (¶ /¶x) x + (¶
/¶y) y + (¶ /¶z)
z
dr = dx x + dy y
+ dz z
df = (¶f/¶x)dx + (¶f/¶y) dy + (¶f/¶z) dz
Ñf · dr = df = dr · Ñf
b) Spherical Coordinates:
dr = dr r + rdq q + rsin(q)df f
(since a small distance in the radial direction is simply dr, a small distance in the q direction is rdq , and a small distance in the f direction is r sinq df )
df = (¶f/¶r)dr + (¶f/¶q )dq + (¶f/¶f )df
dr · Ñf = df
To make this last statement work, we need for the del operator:
Ñ = (¶ /¶r)r + (1/r)(¶ /¶q )q + (1/r sin(q))(¶ /¶f )f .
5. Divergence (Ñ · A)
a) Rectangular Coordinates
Ñ
= (¶ /¶x) x + (¶
/¶y) y + (¶ /¶z)
z
A = Axx
+ Ayy + Azz
Ñ · A = (¶Ax/¶x) + (¶Ay/¶y) + (¶Az/¶z)
b) Spherical Coordinates
Ñ = (¶ /¶r)r + (1/r)(¶ /¶q)q + [1/r sin(q)](¶ /¶f)f
A = Arr + Aqq + Aff
Ñ · A = r · ¶A/¶r + q/r · ¶A/¶q + f /[r sin(q)] · ¶A/¶f =
r · [(¶Ar/¶r)r + Ar(¶r/¶r) + (¶Aq/¶r)q + Aq (¶q /¶r) + (¶Af/¶r)f + Af(¶f /¶r)] +
q /r · [(¶Ar/¶q)r + Ar(¶r/¶q) + (¶Aq/¶q)q + Aq(¶q/¶q) + (¶Af/¶q)f + Af(¶f/¶q)] +
f /[r sin(q )] · [(¶Ar/¶f)r + Ar(¶r/¶f) + (¶Aq/¶f)q + Aq(¶q/¶f) + (¶Af/¶f)f + Af(¶f/¶f)]
We can eliminate all terms that have (¶r/¶r), (¶q /¶r) and (¶f/¶r) since all these unit vectors do not depend on r; we can also eliminate the term that has (¶f /¶q) since f does not depend on q . Recall that (¶r/¶q)=q ; (¶r/¶f)=sin(q)f ; (¶q/¶q)=-r; (¶q/¶f)=cos(q)f ; and (¶f /¶f )= [-sin(q)r - cos(q)q ].
Ñ · A =
r · [(¶Ar/¶r)r + 0 + (¶Aq/¶r)q + 0 + (¶Af/¶r)f + 0] +
q /r · [(¶Ar/¶q)r + Arq + (¶Aq/¶q)q + Aq(-r) + (¶Af/¶q)f + 0] +
f
/[r sin(q)] · [(¶Ar/¶f )r + Ar sin(q)f + (¶Aq/¶f)q + Aq cos(q)f + (¶Af/¶f )f + Af (-sin(q)r - cos(q)q )]
Now we can take the dot products which will further reduce the number of terms:
Ñ · A = (¶Ar/¶r) + Ar/r + (¶Aq/¶q)/r + Ar/r + Aq cos(q)/[r sin(q)] + (¶Af/¶f)/[r sin(q)] =
(¶Ar/¶r) + 2Ar/r + (¶[Aq
sin(q)]/¶q)/[r
sin(q)] + (¶Af/¶f)/[r sin(q)]
= Ñ ·
A
where we have combined the second and fourth terms into the 2Ar/r term, and we have combined the third and fifth terms into the (¶[Aq sin(q)]/¶q)/[r sin(q)] term.