PHY 5246: Theoretical Dynamics, Fall Assignment # 5, Solutions. θ = l mr 2 = l

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1 PHY 546: Theoreticl Dynics, Fll 15 Assignent # 5, Solutions 1 Grded Probles Proble 1 (1.) Using the eqution of the orbit or force lw d ( 1 dθ r)+ 1 r = r F(r), (1) l with r(θ) = ke αθ one finds fro which α r + 1 r = r F(r), () l F(r) = (1+α )l 1 r. (3) 3 (1.b) For centrl force otion we hve tht θ = l r = l k e αθ, (4) where l is the gnitude of the conserved ngulr oentu. We cn esily integrte this eqution by seprtion of vribles, i.e. e αθ dθ = l 1 kdt α eαθ +C = l kt, (5) where C constnt of integrtion. Isolting the exponentil ter nd tking the logrith of both l.h.s nd r.h.s. one gets θ(t) θ = 1 [ ] αl α ln k t+c where C = αc is deterined by the initil conditions on θ. Substituting θ(t) into the expression of r(θ) one gets where K = ke αθ. r(t) = K, (6) [ ] 1/ [ ] 1/ αl αl = k t+c t+k C, (7)

2 (1.c) The totl energy of the orbit is where, using Eqs. (6)-(7), we cn clculte the kinetic energy s E = 1 (ṙ +r θ )+V(r), (8) T = 1 (ṙ +r θ ) = (1+α )l while the potentil energy (odulus constnt of integrtion) is nd E = T +V =. Proble (.) V(r) = [ F(r)dr = (1+α )l The proble is esily discussed in ters of the effective potentil l 1 r, (9) ] 1 = (1+α )l r 3 l 1 r, (1) V (r) = 1 r +V(r) = 1 r +βrk. (11) In order for circulr orbit to exist the effective potentil hs to hve iniu for soe finite vlue of r. The iniu condition is V (r) r = l r 3 +βkrk 1 =, (1) which dits rel solution only if β nd k re either both positive or both negtive. In which cse the rdius of the circulr orbit is r = ( l kβ ) 1 k+. (13) (.b) Since bout the equilibriu position r = r the syste behves s liner hronic oscilltor subject to restoring force F(r) = α(r r ), with potentil energy V (r) = V (r )+ 1 α(r r ), we cn find α by siply expnding V (r) bout r = r nd tking the coefficient of the qudrtic ter in the expnsion. The frequency of sll oscilltions will then be ω r = (α/) 1/ (where the index r indictes tht the oscilltion re in the rdil direction). The expnsion of the potentil is V (r) = V (r )+ 1 V (r) r (r r ) +O((r r ) 3 ), (14) r=r

3 such tht α = V (r) r = 3l 1 r 4 ( l = βk (15) r=r +βk(k 1)r k ) 4 ( ) k+ 3l l k+ +βk(k 1) βk = r 4 l (k +), nd the frequency of sll oscilltions ω r is (.c) ( ) α 1/ l ω r = = r k +. (16) The rtio of the frequency of sll (rdil) oscilltion, ω r, to the frequency = θ of the (nerly) circulr otion is l ω r r k + = = k +. (17) The four given cses re: l r k = 1 ω r = 1 (18) k = ω r = k = 7 ω r = 3 k = 7 4 ω r = 1 which correspond to r king 1,,3, or respectively 1 in θ. oscilltion(s) for ech coplete revolution Proble 3 (Goldstein 3.11) The reduced syste lso oves in circulr orbit with soe rdius r = (nd therefore r = ). The corresponding eqution of otion is r = = l 3 k. We solve this, using l = r θ: l 3 = k θ = k 3.

4 k θ = ω = = π 3 τ τ = π ω = π 3 k. (19) Here we cn note tht ω is constnt nd τ ust be the period of both the reduced syste nd the originl circulr otion. When the two sses re stopped nd then relesed fro rest, they hve zero ngulr oentu l =, so they just stisfy rdil otion eqution of the for k = 1 ṙ k r, which is esily found using conservtion of energy. Therefore ṙ = k ( 1 r ) 1 ( k 1 ṙ = r 1 ) 1/ ( k r = r Integrting the previous reltion between t = nd t, we get, dr r r = ) 1/. k t, () where t is the tie it tkes for the two sses to ove fro r = to r =. Perforing this integrtion: dr = r dr r r = x dx r x = [ x x + ( )] x sin 1 = π. In the first line we hve chnged integrtion vribles with r = x, nd to get to the second line we hve used stndrd integrtion tble. Thus, fro () we hve π k = t t = k π t = k 3π 4 = τ 3 t = τ 4. To get this finl result we hve used the period we found in (19). Non-grded Probles Proble 4 (Goldstein 3.19) (Note tht the Yukw potentil is kind of screened Coulob potentil, nd cn be used to describe soe coon prticle interctions - pion exchnge between nucleons, for instnce.) The force corresponding to the Yukw potentil (for k, > ) is F(r) = k r e r/.

5 (4.) The Lgrngin corresponding to prticle in the Yukw potentil is L = 1 (ṙ r θ ) V(r) = 1 (ṙ +r θ )+ k r e r/. The eqution of otion for θ siply gives us conservtion of ngulr oentu: The eqution of otion for r is Using this we cn write the energy s: r θ = constnt := l. r r θ + k r e r/ + k r e r/ = ( r l k r + 3 r + k ) e r/ =. r E = 1 (ṙ +r θ )+V(r) l = 1 ṙ + 1 r k r e r/ = 1 ṙ +V (r), where V (r) is the effective potentil (see figure). Asyptoticlly, this potentil hs the feture tht for both lrge (r ) nd sll (r ) it is dointed by the 1/r ter. In the iddle regions it will depend on the vlue of l. (4.b) The circulr orbit condition is verified (for those vlues of l when V (r) hs iniu) if: V (r) r ( = l k r + r + k ) e r/ = r l ( k = r e r / + r ). (1) In this cse we explin wht hppens when we exine sll devitions fro r = r. Tke nd insert this into the eqution for the orbit Using the stndrd chnge of vribles r(θ) = r [1+δ(θ)] d ( 1 dθ r)+ 1 r = k r F(r) = l l e r/. u := 1 r = 1 r (1 δ),

6 5 l=.3*k l=k 4 3 V (r) Figure 1: A grph of the effective Yukw potentil for two different vles of ngulr oentu. Here we hve set k/ = 1 nd K = k. r/ we find tht d ( δ dθ + 1 k l r e r / d u k +u = dθ d u k +(1 δ) = dθ ) l e 1/u ( l r e r (1+δ) δ = 1 k l r e r /. r δ ) This is the eqution for siple hronic oscilltor (with constnt shift) nd frequency ω = 1 k l r e r / = 1 r 1 1+ r = 1 1+ r, where we hve used the definition of r fro (1). Now choose δ to be t xiu when θ =, then the next xiu will occur when ωθ = π θ = π ( ω = π 1+ r ) ( [r ] ) +o. Therefore the psides dvnce by ech revolution. θ = πr

PHYS 601 HW3 Solution

PHYS 601 HW3 Solution 3.1 Norl force using Lgrnge ultiplier Using the center of the hoop s origin, we will describe the position of the prticle with conventionl polr coordintes. The Lgrngin is therefore L = 1 2 ṙ2 + 1 2 r2