General Central Force

1.  Definition and Notations:

Here vectors will be displayed in bold blue, unit vectors in bold plum, and scalars (and components) in regular black.

Since force is a vector, F should be a vector, F.

If the force depends on position only, we can express force as a function of position as F(r), which in three dimensions means F(x,y,z) or F(r,q,f) .

A central force is one where the force is directed along the radial direction, r, and not in the other two directions, q or f  .  It also depends only on the radial distance, r, and not on q or f.  This can be expressed as F = F(r) r .  Be aware, however, that the radial direction, r, does depend on q and f  (see 3-D Kinetics for review):

r(x,y,z)  =  xx + yy + zz   =   r sin(q) cos(f) x  +  r sin(q) sin(f) y  +  r cos(q) =   rr

where r = r(q,f) =  sin(q)cos(f) x  +  sin(q) sin(f) y  +  cos(q) .

2.  Is a central force (always, sometimes, never) a conservative force?

To answer this, recall Stokes Theorem (see Gradient, Divergence & Curl for review):

òò (Ñ´F) · n dS  =  òclosed loop F · dl              (where S is an area and n is the unit vector perpendicular to the surface)

so that if Ñ´F = 0 everywhere, then òclosed loop F · dl  =  0, and so liòlf F · dl  is independent of path, and so a potential energy can be defined:

DV  =  - riòrf F · dl  .

Let’s now calculate Ñ´F for a central force (using spherical coordinates):

Ñ´[r /r  + q (1/r) /¶q  +  f (1/r sin(q)) /¶f´  F(r) r   =

(F(r)/r) (r ´ r)        +       F(r) (r ´ r/r)                   +   (1/r) (F(r)/¶q) (q ´ r

+   (F(r)/r) (q ´ (r/¶q))   +  (1/r sin(q)) (F(r)/¶f) (f´ r)   +   (1/r sin(q)) F(r) (f ´ r/f) .

There are six terms, so let’s look at these one by one.

The first term is zero since (r ´ r) = 0  (since any vector crossed itself is zero).

The second term is zero since r/r = 0  (since r depends only on q and f).

The third term is zero since F(r)/¶q = 0  (since F(r) does not depend on q).

The fourth term is zero since r/¶q  =  q,  and (q ´ q) = 0.

The fifth term is zero since F(r)/¶f = 0 (since F(r) does not depend on f).

The sixth term is zero since r/¶f = sin(q) f, and (f ´ f) = 0.

Therefore,  Ñ´F = 0 for F = F(r)r, so a central force is always a conservative force!  We can then define a potential energy function:

V(r)  =  - r-standardòr F(r) dr .

3.  Torque and Angular Momentum for a Central Force

Starting with the definition of torque as  t  =  r ´ F,  for a central force, F = F(r)r

we have (recalling that r = rr):   t  =  r ´ F(r)r  =  r F(r) r ´ r  =  0.  Central forces can create no torque.

Now we recall the relation between torque and angular momentum (L):  t  =  L/t .  Since torque under a central force is zero, the angular momentum must remain constant.  Let’s now get an expression for angular momentum in polar coordinates.  We can use    2-D polar coordinates (instead of 3-D) since a central force has no component of either motion or force perpendicular to the (r ´ v) plane.  By definition,  L  =  r ´ r ´ m

In polar form,  =  dr/dt  =  d(rr)/dt  =  (dr/dt)r + r(dr/dt)  =  r’r + rqq

(where r’ = dr/dt, and q’ = dq/dt) .

Therefore,

L  =   rr ´ m(r’r + rqq)   =   mrr’(r ´ r)  +  mr2q’(r ´ q)    =   0 + mr2qz  =  constant.

Note that if we know the initial conditions, ro and qo, we can calculate the value of L.
If we just consider the magnitude, since the direction will not change,

L = mr2q’ = constant.

This equation provides a relation between the radius, r and the angular speed, q’ (which we sometimes have called w).  We can write this as:

q’ = L/mr2  .

4.  Newton’s Second Law for a Central Force

S F = ma  is a vector equation, which means it is one equation for each dimension.  For a central force, F = F(r) r, in polar coordinates we have  Fr = F(r) ,  and  Fq = 0 .  The acceleration expressed in polar form is:  ar = r’’ –  rq2   (where r’’ = d2r/dt2 is the usual linear acceleration in the radial direction, and rq2 = w2r is the circular acceleration);  and  aq = rq’’ + 2r’q’  (where rq’’ is r(d2q/dt2), the acceleration in the circular direction, and 2r’q’ is that strange Coriolis acceleration).   Therefore, in component form, Newton’s Law becomes:

r component:  F(r)  =  m(r’’ – rq2)    ,   or    F(r) + mrq2  =  mr’’

[where the term mrq2 = mw2r  can be considered the “centrifugal” force]

q component:    Fq = 0,     so        0    =  m(rq’’ + 2r’q’)  .

If we use the angular momentum equation, q’ = L/mr2 in the radial equation, we get:

F(r) + mr(L/mr2)2 = mr’’,   or    F(r) + L2/mr3 = mr’’

which is an expression that is equivalent to a one-dimensional problem since there is no mention of q.

5.  Conservation of Energy

Since a central force is a conservative force, we can define a potential energy.  If there are no other forces, then energy is conserved and we can write this in an equation:

KE + PE  =  E  =  constant

v =  vrr + vqq ,   and   v2  =  vr2 + vq2   =   (r’)2 + (rq’)2    so we get:

½ m r’2 + ½ m r2 q2 + V(r)  =  E .

If we use the angular momentum expression, q’ = L/mr2,  this becomes:

½ m r’2 + ½ m r2 (L/mr2)2 + V(r)    =    E   =    ½ m r’2 + ½ L2 /mr2 + V(r) .

If we treat the term ½ L2 /mras an additional potential energy (since it only depends on r), we can define an effective potential energyVeff(r)   =   V(r) + L2 /2mr2 .  We now have an equation that is essentially one-dimensional:

½ m r’2 + Veff(r)  =  E   =   constant .

Note that if we treat the mass times centripetal acceleration, m*(– rq2r), on the acceleration side of Newton’s Second Law as a “centrifugal force”, + mrq2 r = mr(L/mr2)2 r = L2/(mr3) r, on the force side of Newton’s Second Law, then this acts like another central force (in the  r direction with the magnitude depending only on r), we should be able to get a potential energy associated with this “centrifugal force”:  DVcf  =  - riòrf Fcf · dl  =  - riòrf L2/(mr3) dr  =  + L2 /2mr2 - 0 (where we let ri = rstandard = infinity).  This is just the “extra” term in the effective potential energy we had above!

6.  Straightforward solutions

a)  Use the Conservation of Energy (1-D) equation with the equivalent potential energy, and use the technique we used earlier in the course to first get r(t) and then to get q(t):

½ mr’2  +  Veff(r)  =  E               (where Veff(r)  =  V(r) + L2 /2mr2 )

dr/dt  =  r’  =  [2(E-Veff(r))/m ]½

0òt dt  =  roòr {1 / [2(E-Veff(r))/m ]½ } dr

Assuming we can actually do the integration over r, we get an expression for t(r).

Then we must find the inverse to get r(t).  This is sometimes very difficult.

If we can get r(t), we can then use the angular momentum expression:  L = mr2q’  to get an expression for q’:

q’  =  L/mr(t)2 .

We can then use  q’ = dq/dt  to get:

q0òq dq   =   0òt q’ dt   =   0òt [L/mr(t)2] dt  .

If we can do the integration, we then have an expression for q(t) and the problem is solved.

b)  Use Newton’s Second Law to try to get r(q)  [instead of r(t) and q(t)].  While this will lose some information (time information), it has the advantage of giving us the orbits directly.  We start with the equivalent 1-D relation (see part 4 above):

F(r) + L2/mr3  =  mr’’  .

Since we want r(q), we note that (recalling that q’ = L/mr2) :

r’  =  dr/dt  =  (dr/dq)*(dq/dt)  =  (dr/dq)*q’  =  (dr/dq)*(L/mr2) .

r’’ = d(r’)/dt  =   (dr’/dq)*(dq/dt)  =   d[(dr/dq)*(L/mr2)]/dq * q’

=  (d2r/dq2)*(L/mr2)*q’  +  (dr/dq)*(L/m)*(d[1/r2]/dq)*q

note:  d[1/r2]/dq  =  (d[1/r2]/dr)*(dr/dq)  =  (-2/r3)*(dr/dq)

r’’  =  (d2r/dq2)*(L/mr2)*q’  +  (dr/dq)*(L/m)* (-2/r3)*(dr/dq)*q

(and again using q’ = L/mr2 )

r’’  =   (d2r/dq2)*(L2/m2r4)  - (2L2/m2r5)*(dr/dq)2 .

Using this, we can now write Newton’s 2nd Law (r component equation) as:

F(r) + L2/mr3    =    (d2r/dq2)*(L2/mr4)  - (2L2/mr5)*(dr/dq)2 .

This doesn’t look all that easy to solve.

7.   Tricky way to solve for r(q)

Sometimes things look nicer in one form than in another.  We will try a trick that involves substitution:  let’s define u = 1/r  (or r = 1/u).

L  =  mr2q’  =  mq’/u2,      or      q’  =  Lu2/m

r’  =  dr/dt  =  (dr/du)*(du/dt)  = (dr/du)*(du/dq)*(dq/dt)  = (d(1/u)/du)*(du/dq)*q

=   (-1/u2)*(du/dq)*q’   =   (-1/u2)*(du/dq)*(Lu2/m)   =   (-L/m)*(du/dq).

r’’  =  dr’/dt  =  (dr’/dq)((dq/dt)   =   d[(-L/m)*(du/dq)]/dq * q’

=   (-L/m)*(d2u/dq2)*(Lu2/m)    =    (-L2u2/m2)*d2u/dq2

Newton’s Second Law (r component equation)

F(r) + L2/mr3  =  mr’’

thus becomes:

F(1/u) + L2u3/m   =    m*(-L2u2/m2)*d2u/dq2  ,  or

(-m/L2u2)*F(1/u)  +  (-m/L2u2)*(L2u3/m)   =   d2u/dq2   ,  or

(-m/L2u2)*F(1/u)  +  -u     =    d2u/dq2  .

This looks better than the straightforward way above because the r’’ term gives only one term instead of two terms when we use this particular substitution.  This is because the r’ term does not have any u’s in it – a result of our particular choice for substitution.

8.  Special Cases

a)   One of Kepler’s Laws of planetary motion

Consider an area of the orbit, dS, that is swept out in a small unit of time, dt:

The distance covered by vrdt does not cover an area since it’s direction is parallel to r.  The distance covered by vqdt does cover an area of a triangle, so dS = ½r*(rq’dt) = ½r2q’dt.   We now use the relation   q’ = L/mr2    to get

dS/dt = ½r2(L/mr2) = L/2m = constant.

This is one of Kepler’s laws for planetary motion:  a planet sweeps out equal areas in equal time intervals.  Note that this law applies for ANY central force, not just for the inverse square gravity force.

We can go even further to get:

ò dS = (L/2m) òdt,

or if we integrate over the complete orbit, we get:

S = (L/2m)T                where T is the period of orbit and S is the area of the orbit..

b)  Oscillations near equilibrium

If we use the effective potential (see part 5 above):

Veff(r)  =  V(r) + L2 /2mr2 ,

we now have an equation that is essentially one-dimensional:

½ m r’2 + Veff(r)  =  E   =   constant .

If there is a stable equilibrium position (where dVeff/dr = 0, and Veff is a minimum at this value of r), then we should be able to get circular motion (where r’=0), since the central force (if it is attractive) can cause the centripetal acceleration.   If we provide a little more energy than necessary for this stable circular motion, we should get oscillations (in the value of r) about this stable value of r.  (See the first part of the course, under Potential Wells and Oscillations.)  The frequency, wosc , in which r oscillates is given by

wosc = [k/m]½   ,  where k = (d2Veff/dr2)r-equilibrium  .

Note:  the above angular frequency, wosc, is the frequency that r oscillates around the circular radius, requilibrium
This
wosc is not the same thing as q, which is the angular frequency that the motion has going around the center of the force.

Example:  Given a spherically symmetric harmonic oscillator with V = ½ kr2.
(To visualize this, picture a mass on a spring in outer space – no gravity – where the spring is free to rotate about one end with the mass on the other end.)

a)  What is the frequency for circular motion at a radius of ro ?

b)  What is the frequency for small oscillations of r about ro ?

c)  Describe the orbits for small oscillations of r about ro .

a-1)  For circular motion, r = constant so r’ = 0 and r’’ = 0, so the r component equation of Newton’s 2nd Law: F(r)  =  m(r’’ – rq2)  becomes, with  q’ = wc :

Fr = mar = m(-wc2r).   Fr = -dV/dr = -kr .  Putting these two expressions together gives:  -kr = -mwc2r ,  or  wc = [k/m]1/2  .

a-2)  For stable orbit, want dVeff /dr = 0  where Veff  =  ½ kr2 + L2/2mr2 so we get the condition:  kr – L2/mr3  =  0,  or  ro4  =  L2/mk  .  We have L  =  mvqr = m(wcr)r = mwcr2 , or
wc   =    L/mro2   =   L/m(L/[mk]1/2   =   [k/m]1/2  .
Note that wc does not depend on the radius!  As the radius gets bigger, wc stays the same and the vq = wc*r gets bigger.

b)  For small oscillations about the bottom of the potential well,  wosc = [k’/m]1/2

where  k’ = (d2Veff/dr2)ro   =   (d2[½ kr2 + L2/2mr2] /dr2)ro   =   k + 3L2 /mro4  ,  so

wosc  =  [k’/m]1/2   =  [ {k + 3L2 /mro4}/m ]1/2   =   [(k/m) + 3L2/m2ro4]1/2 =  [wc2 + 3wc2]1/2 , or       wosc = 2wc .

c)  For one cycle around the circle, the radius grows bigger and gets smaller and grows bigger again twice.  The origin is still right in the center.

Homework Problem #21:  For an inverse square law force with effective potential energy,    Veff(r) = K/r + L2/2mr2 ,

b)  show that this frequency of small radial oscillations is equal to the frequency of the circular orbit.  (This makes the circular orbit into an elliptical one).