Precession In A Circle Suppose a particle is moving counter-clockwise around the perimeter of a circle, and the circumferential speed of the particle varies as a function of its position according to the formula where U and e are constants with e < 1.  Thus the maximum speed is U(1+e), which occurs at q = 0, and the minimum speed is U(1-e), which occurs at q = p.  If r is the radius of the circle, we have v = rw = r dq/dt, and by a suitable choice of units we can set U/r = 1 without loss of generality, so the above equation can be written in the form Integrating both sides gives (with suitable constant of integration) Solving for the angular position q as a function of the time t gives Of course, if  e = 0, we have v = 1, and the above expression reduces to q = t/r.  On the other hand, if  e = 1/2, the position of the particle as a function of time is as shown in the plot below. Now consider a second particle moving counter-clockwise around the same circle, but with a velocity profile that precesses over time.  In other words, the velocity profile is of the form for some small positive constant  k.  This implies that the angular locations of maximum and minimum speed precess in the clockwise direction.  In fact, the entire velocity profile precesses slowly clockwise.  Still, at any given instant the two particles have precisely the same overall speed profile, except for the orientation.  Sometimes the first particle is moving more rapidly than the second, whereas at other times the second is moving more rapidly than the first.  It's interesting to determine how this precession affects the long-term position of the particle, relative to the position of the particle whose speed profile is not precessing. We have so if we define the auxiliary variable f = q + kt  then we have df = dq + k.  Substituting into the above equation gives Hence we have This is the same form of differential equation as we saw above, except that the leading coefficient 1 is replaced with the constant 1 + k.  So, integrating as before, solving for f, and converting back to the variable q gives Expanding this into a series in terms of e, we get The zeroth-order term is independent of k, and the first-order term as well as the first part of the second-order term contain the time parameter t only in arguments of sine functions, so they have no long-term cumulative effect.  However, the last part of the second-order term is directly proportional to the time t, so it increases without limit.  This is called a secular term, because it progresses cumulatively with time.  Now, the speed profile of the first particle had no precession, so with k = 0 the second-order secular component of its position is simply -(e2/2)t, whereas the speed profile of the second particle precesses due to a non-zero value of k, and hence the second-order secular term is -((e2/2)/(1+k))t.  Consequently, if we consider the position of the second particle relative to the position of the first particle, the highest-order secular contribution is the difference This shows that the particle with the (backwards) precessing speed profile will move ahead of the other particle in the long run.  To illustrate, for e = 1/4 and k = 1/10, the difference in positions of the two particles is as shown below. Initially the two particles pass each other alternately, but the net secular tend is for the particle with the precessing speed profile to pull ahead.  The solid sloping line is the predicted secular difference given by the preceding formula. Return to MathPages Main Menu