PHY 5246: Theoretical Dynamics, Fall 2015 Assignment # 5, Solutions 1 Graded Problems Problem 1 (1.a) Using the equation of the orbit or force law d2 1 1 mr2 2 + = 2 F (r) , (1) dθ r r − l with r(θ)= keαθ one finds α2 1 mr2 + = F (r) , (2) r r − l2 from which (1 + α2)l2 1 F (r)= . (3) − m r3 (1.b) For a central force motion we have that l l 2αθ θ˙ = = e− , (4) mr2 mk2 where l is the magnitude of the conserved angular momentum. We can easily integrate this equation by separation of variables, i.e. l 1 l e2αθdθ = dt e2αθ + C = t , (5) mk2 −→ 2α mk2 where C a constant of integration. Isolating the exponential term and taking the logarithm of both l.h.s and r.h.s. one gets 1 2αl θ(t) θ0 = ln t + C′ , (6) − 2α " mk2 # where C′ = 2αC is determined by the initial conditions on θ . − 0 Substituting θ(t) into the expression of r(θ) one gets 1/2 1/2 2αl 2αl 2 r(t)= K t + C′ = t + k C′ , (7) "mk2 # " m # where K = keαθ0 . (1.c) The total energy of the orbit is 1 E = m(r ˙2 + r2θ˙2)+ V (r) , (8) 2 where, using Eqs. (6)-(7), we can calculate the kinetic energy as 1 (1 + α2)l2 1 T = m(r ˙2 + r2θ˙2)= , (9) 2 2m r2 while the potential energy (modulus a constant of integration) is (1 + α2)l2 1 (1 + α2)l2 1 V (r)= F (r)dr = 3 = 2 , (10) − Z − Z "− m r # − 2m r and E = T + V = 0. Problem 2 (2.a) The problem is easily discussed in terms of the effective potential 2 2 1 l 1 l k V ′(r)= + V (r)= + βr . (11) 2 mr2 2 mr2 In order for a circular orbit to exist the effective potential has to have a minimum for some finite value of r. The minimum condition is 2 ∂V ′(r) l k 1 =0 + βkr − =0 , (12) ∂r −→ −mr3 which admits a real solution only if β and k are either both positive or both negative. In which case the radius of the circular orbit is 1 l2 k+2 r0 = . (13) mkβ ! (2.b) Since about the equilibrium position r = r0 the system behaves as a linear harmonic oscillator 1 2 subject to a restoring force F (r)= α(r r ), with potential energy V ′(r)= V ′(r )+ α(r r ) , − − 0 0 2 − 0 we can find α by simply expanding V ′(r) about r = r0 and taking the coefficient of the quadratic 1/2 term in the expansion. The frequency of small oscillations will then be ωr =(α/m) (where the index r indicates that the oscillation are in the radial direction). The expansion of the potential is 2 1 ∂ V ′(r) 2 3 V ′(r)= V ′(r0)+ (r r0) + O((r r0) ) , (14) 2 ∂r2 − − r=r0 such that ∂2V (r) α = ′ (15) ∂r2 r=r0 3l2 1 k 2 = 4 + βk(k 1)r0− m r0 − 4 k+2 l2 − k+2 3l2 l2 = + βk(k 1) mβk ! m − mβk ! 2 4 l = r− (k + 2) , 0 m and the frequency of small oscillations ωr is α 1/2 l √ ωr = = 2 k +2 . (16) m mr0 (2.c) The ratio of the frequency of small (radial) oscillation, ωr, to the frequency ωθ = θ˙ of the (nearly) circular motion is l ω 2 √k +2 r mr0 √ = l = k +2 . (17) ωθ 2 mr0 The four given cases are: ω k = 1 r =1 (18) − −→ ωθ ω k =2 r =2 −→ ωθ ω k =7 r =3 −→ ωθ 7 ω 1 k = r = −4 −→ ωθ 2 1 which correspond to r making 1,2,3, or respectively 2 oscillation(s) for each complete revolution in θ. Problem 3 (Goldstein 3.11) The reduced system also moves in a circular orbit with some radius r = a (and thereforer ¨ = 0). The corresponding equation of motion is l2 k r¨ =0= . ma3 − a2 We solve this, using l = mr2θ˙: l2 k k = θ˙2 = . ma3 a2 ⇒ ma3 k 2π 2π ma3 θ˙ = ω = = τ = =2π . (19) sma3 τ ⇒ ω s k Here we can note that ω is constant and τ must be the period of both the reduced system and the original circular motion. When the two masses are stopped and then released from rest, they have zero angular mo- mentum l = 0, so they just satisfy a radial motion equation of the form k 1 k = mr˙2 , −a 2 − r which is easily found using conservation of energy. Therefore 2k 1 1 r˙2 = m r − a 2k 1 1 1/2 2k a r 1/2 r˙ = = − . s s − m r − a − m ra Integrating the previous relation between t = 0 and t, we get, 0 dr 2k = t, (20) − a a r s m Z ra− q where t is the time it takes for the two masses to move from r = a to r = 0. Performing this integration: 0 dr 0 r 0 x2 = √a dr =2√a dx a a r a sa r √a √a x2 Z ar− Z − Z − q 0 x 2 a 1 x π = 2√a √a x + sin− = a √a. "− 2 − 2 √a!#√a − 2 In the first line we have changed integration variables with r = x2, and to get to the second line we have used a standard integration table. Thus, from (20) we have π 2k m π a √a = t t = a√a 2 s m ⇒ r2k 2 m π2 τ 2 τ t2 = a3 = t = . 2k 4 32 ⇒ 4√2 To get this final result we have used the period we found in (19). 2 Non-graded Problems Problem 4 (Goldstein 3.19) (Note that the Yukawa potential is a kind of screened Coulomb potential, and can be used to describe some common particle interactions - pion exchange between nucleons, for instance.) The force corresponding to the Yukawa potential (for k,a > 0) is k r/a F (r)= e− . −r2 (4.a) The Lagrangian corresponding to a particle in the Yukawa potential is 1 2 2 2 1 2 2 2 k r/a L = m(r ˙ r θ˙ ) V (r)= m(r ˙ + r θ˙ )+ e− . 2 − 2 r The equation of motion for θ simply gives us conservation of angular momentum: mr2θ˙ = constant := l. The equation of motion for r is 2 k r/a k r/a mr¨ mrθ˙ + e− + a e− =0 − r2 r 2 l k ak r/a mr¨ + + e− =0. − mr3 r2 r ! Using this we can write the energy as: 1 E = m(r ˙2 + r2θ˙2)+ V (r) 2 2 1 2 1 l k r/a 1 2 = mr˙ + e− = mr˙ + V ′(r), 2 2 mr2 − r 2 where V ′(r) is the effective potential (see figure). Asymptotically, this potential has the feature that for both large (r ) and small (r 0) it is dominated by the 1/r2 term. In the middle regions it will depend on→∞ the value of l. → (4.b) The circular orbit condition is verified (for those values of l when V ′(r) has a minimum) if: 2 ∂V ′(r) l k k r/a =0 2 + 2 + e− =0 ∂r ⇒ −mr r ra! 2 l r0/a r0 = r0e− a + . (21) mk a In this case we explain what happens when we examine small deviations from r = r0. Take r(θ)= r0 [1 + δ(θ)] and insert this into the equation for the orbit 2 2 d 1 1 mr mk r/a 2 + = 2 F (r)= 2 e− . dθ r r − l l Using the standard change of variables 1 1 u := = (1 δ), r r0 − 50 l=0.03*K l=K 40 30 20 V'(r) 10 0 -10 0 0.2 0.4 0.6 0.8 1 r/a Figure 1: A graph of the effective Yukawa potential for two different vales of angular momentum. Here we have set k/a = 1 and K = √2mk. we find that 2 d u mk 1/au + u = e− dθ2 l2 2 ⇓ d u mk r0 r a (1+δ) 0 2 + (1 δ) = 2 r0e− a δ dθ − l − a 2 ⇓ d δ mk 2 r0/a mk r0/a + 1 r0e− δ = 1 r0e− . dθ2 − l2a ! − l2 This is the equation for a simple harmonic oscillator (with a constant shift) and frequency 2 mk 2 r0/a r0 1 1 ω =1 r e− =1 = , 2 0 r0 r0 − l − a 1+ a 1+ a where we have used the definition of r0 from (21). Now choose δ to be at maximum when θ = 0, then the next maximum will occur when 2π r r 2 ωθ =2π θ = =2π 1+ 0 + o 0 . ⇒ ω 2a a ! Therefore the apsides advance by πr ∆θ = 0 a each revolution..