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 (labeled r in the above diagram) 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. To see this more
clearly, let’s use an intermediate distance, r. Note that the z component of r is r cos(q), and the horizontal (xy plane) component of
r is
r where r = r sin(q). The x component of r is r cos(f), and the y component of r is r sin(f). Therefore, the x component of r is r cos(f) = [r sin(q)] cos(f), and the y component of r is r
sin(f) = [r sin(q)] sin(f). The r
unit vector is simply r/r, so we get:
r = sin(q)cos(f)x + sin(q)sin(f)y + cos(q)z = r(q,f) .
The direction for q
is simply that of r but with q increased by 90o, so the sin(q+90o)
becomes cos(q)
and the cos(q+90o)
becomes –sin(q)
, and we get:
q
= cos(q)cos(f)x + cos(q)sin(f)y -
sin(q)z = q (q,f) .
The direction for f is going to be 90o from that of r and that means its x component is going to be cos(f+90o) = -sin(f), and its y
component is going to be sin(f+90o) = cos(f), and we get:
f =
-sin(f)x + cos(f)y = f(f)
Note that the magnitudes of all three unit
vectors are equal to 1 as they should be:
{[sin(q)cos(f)]2 + [sin(q)sin(f)]2 + cos(q)2 }1/2 = 1, and
{[cos(q)cos(f)]2 + [cos(q)sin(f)]2 + [- sin(q)]2
}1/2 = 1, and
{[-sin(f)]2
+ [cos(f)]2}1/2
= 1.
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
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
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.
Example: Let’s check the Divergence Theorem (Gauss’ Law) for a gravitational field due to a uniform distribution of mass in a sphere of radius, R.
òòò
Ñ · g dV =
òò closed surface
g·
dS = -4pGMenclosed . We
saw in the previous section on Gradient, Divergence and Curl that òò
closed surface g· dS = -4pGMenclosed . We now try to calculate òòò Ñ · g dV to
see if we get the same result.
Using the òò closed surface g· dS =
-4pGMenclosed result, we see that there is a net
gravitational field at a point (point A in the figure below) on any sphere that
we might specify (dotted sphere in the diagram below) that is inside the total
sphere (full, big circle in the figure below), and that field is due to the
mass inside the dotted sphere, and not to any mass that is outside of the
dotted sphere. This comes from the
observation that the gravitational field inside a hollow sphere must be zero. We can show this by using òò closed surface g· dS =
-4pGMenclosed for a hollow sphere and noting that Menclosed for a hollow sphere is zero! To “see” this, consider point A on the chosen
(dotted) sphere inside the actual sphere of mass. All the mass to the left of the line drawn creates a gravitational field
directed left, while all the mass to the right of the line creates a gravitational field directed
right. The mass to the left is nearer,
and so each bit of mass has a bigger effect, but the mass on the right is
greater. The net effect is that these
two fields cancel. Thus the
gravitational field at point A is just that due to the mass inside the dotted
sphere, so it has a magnitude of g = G*menclosed/r2
directed down toward the center of the sphere, where menclosed
= r*V = r*[(4/3)pr3], so the magnitude of g becomes: g = G* r*[(4/3)pr3]/r2 = (4/3)pGrr directed down
(towards the center). This depends on r
to the first power and this should make sense because the gravitational field
at the center of the sphere (r=0) should be zero by symmetry.
Now we need to take Ñ · g where g
= -(4/3)pGrr r [note that radial component
of g, gr
= -(4/3)pGrr , and the theta
and phi components of g are zero].
Ñ · g =
(¶gr/¶r) +
2gr/r + (¶[gq sin(q)]/¶q)/[r sin(q)]
+ (¶gf/¶f)/[r sin(q)] = -(4/3)pGr + 2(-(4/3)pGr) + 0 + 0 = -4pGr .
When we do the integration: òòò
Ñ · g dV =
-4pGr (V) = -4pGMenclosed
(since
M = r*V) . This
is the same result we obtained for òò
closed surface g· dS .
