48Lecture 6 Thursday Feb 5 2015Particle motion for forces in 3-d that are both Conservative and Central,(for example, gravity in the solar system)- Reduction to a 1-d problemWe will consider forces of the form rF(rr ) = f (r)ˆrthat are both central and conservative.We will describe the particle position using three dynamicalcoordinates, equal to the familiar spherical coordinates. r, θ, φdefined in the figure. (sometimes the angle θ is defineddifferently: relative to the equator or south pole instead of thenorth pole) . In general three degree of freedom dynamical systems resist analytic solution, butfor forces that are central and conservative, it turns out, we can reduce the problem to one degreeof freedom and say lots about the motions.This is a central force so the angular momentum rLO≡rrO→ particle× mrv; |rLO| = m|rr ||rv | sinαtakes on a fixed (vector) value. This is because drLO/ dt ≡rv × mrv +rrxrF = 0 + 0. Throughout the trajectory of theparticle, as its speed and direction and distance vary, this vector isconstant, in both direction and magnitude.The vector L is perpendicular to the plane defined by the velocity v and the position vector r.Because L is a constant, this implies that the particle motion remains in that plane - defined by theinitial velocity and the initial position vector. Further analysis can be simplified by choosing theequatorial plane of our spherical coordinate system to be that plane. Thus we see that θ = π/2, andit does not change dθ/dt = 0. The problem of finding the particle's trajectory is thus reduced to 2-d; we may use planar polar coordinates. We need further consider only how r and φ might bechanging.Velocity in spherical coordinates is rv =&rˆr + r&θˆθ+ r sinθ&φˆφ( If you picture the above diagram as the earth, you can recognize that v is dr/dt upwards + r dθ/dtsouthwards + r sinθ dφ/dt eastwards )49On taking θ = π/2 = constant, this becomes rv =&rˆr + r&φˆφ - like polar coordinates (~p27).Then rL ≡rr × mrv = mrˆr × [&rˆr + r&φˆφ] = mr2&φˆθ= −mr2sinθ&φˆk = mr2&φˆk.We know this is constant, therefore we state r2 dφ/dt = constant during the motion (value set by initial conditions).Whatever the trajectory of the particle, we conclude that the particle is moving faster around thecenter when it is closer to the center. (This sounds like conservation of energy, but it appliesfor any central force, conservative and attractive or not.) Do not make the mistake of setting,L = m r v, as there is also a factor of the sine of some angle in the magnitude of a cross product.L = m r v only at those points of the trajectory in which the velocity is perpendicular to theradius vector.-----------------Our form for the force rF(rr ) = f (r)ˆr is also conservative. Grad in spherical coordinates is(see Appendix F.3 of Thornton and Marion, or the inside back cover of Taylor, or the physicsformulary) r∇ =ˆr∂∂r+ˆθ∂r∂θ+ˆφ1r sinθ∂∂φThus central forces rF(rr ) = f (r)ˆr can be represented as −r∇U(rr ) by choosing U=U(r) as thenegative anti-derivative of f(r). For example, if rF(rr ) = −Kr2ˆr(as we will have for thegravitational field around a point mass) thenU = -K/r + constantso that f(r) = -K/r2 = -dU/dr. Typically we choose the constant to be zero, thus setting U to vanishat ∞. This is not necessary, but it is conventional and convenient. It has the peculiar effect ofallowing total energy E = T + U to be negative.For example, given the pictured trajectory,we know that |L|/m = r1v1 = r3 v3 . Wealso know that |L|/m is equal to r2v2 timesthe sine of the angle between the vectors r2and v2, so it is not equal to r2v2.Given the distances r, you cannot tell thecorresponding speeds unless you knowthese angles.50Kinetic energy is m/2 times the sum of the squares of the orthogonal components of v: T=12mrv ⋅rv =12m{&r2+ r2&θ2+ r2sin2θ&φ2}So we conclude that total energy, takes the form E = U + T =12m{&r2+ r2&θ2+ r2sin2θ&φ2} −Kr,and is conserved. It is a constant of the motion. Its value may be determined by consideringinitial conditions. Note that E may be negative or positive depending on the particle's speed andposition.--------We now combine the information from knowing E=constant and L = constant. Recalling that wehave planar motion θ = π/2, &θ= 0 (because L had constant direction and we chose the mostconvenient orientation of our coordinate system), we have E = T + U=12m{&r2+ r2&φ2} −KrAnd recalling |rL |= mr2&φ, we can substitute for dφ/dt: E = U + T =12m{&r2+ (Lmr)2} −KrOn rearrangement, this may be put in a suggestive form: E =12m&r2+ (L22mr2−Kr)= Tradial+Ueff(r)which looks like the constant energyassociated with the motion of a particle in1-dimension (r) with an effectivepotential that includes the true potentialenergy and a term that depends on L51Ueff(r) =L22mr2−KrThe term in L2 looks like Potential Energy – because it depends on position r and not on thespeeds. It has its origin however in the KE associated with the motion &φ. Thus for central forceswhose strength depends only on r, we have reduced the 3-d problem to a 1-d problem for r(t).We can solve for dr/dt as a function of r: &r =2m(E −L22mr2+Kr)we could now if we wished write a time – r relation in the formdt =dr / √ 2(E / m − L2/ 2mr2+ K / r)∫∫but this is usually not so useful as getting a qualitative understanding of the motion,see below.Examine the above sketch of the effective potential.At small r, Ueff is dominated by the term in L2 and goes to +∞ like L2 /2mr2 for any L ≠ 0. Thisis called the centrifugal barrier. As far as radial motion looks, there seems to be a strong repulsionfrom the origin! Of course it is not a real barrier; when observed in 2-d, we merely see a particlemoving by the origin, with some closest approach. At that closest approach dr/dt = 0 where weidentify a turning point in r(t), the particle still has a tangential velocity r&φ.At large r, where the term in L2 / r2 is relatively negligible, Ueff is dominated by the term –K/r:At infinite distance there is no effective force. At large, but not infinite, distance the centrifugalbarrier is negligible compared to the weak attractive potential – K/r.At intermediate r , Ueff is negative.Thus there must be a potential well at intermediate r.IF E < 0 then there is a region at intermediate r in which the particlecan