Classical Mechanics Homework set 7, due Nov 8th: Solutions

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1 Classical Mechanics Homewok set 7, due Nov 8th: Solutions 1. Do deivation 8.. It has been asked what effect does a total deivative as a function of q i, t have on the Hamiltonian. Thus, lets us begin with a Lagangian L, and conside L = L + df dt = L + f q i + f t 1) I m using Einsteins convention of summation ove epeated indices i) Now, emembe the definition of the canonical momenta p i = L q i. We thus obtain that p i = p i + f ) The main esult is that the momentum is shifted by a gadient of a function f. O equivalently p i = p i f 3) now, conside the Legende tansfom of L, as given by H = p i q i L = p i + f ) q i L f q i f t = H f t 4) In the function H we should use p i = p i f eveywhee so that H is witten in tems of q i, p i. Notice that the Hamiltonian has been also shifted by a patial deivative of f. Combining the shift in momentum and the Hamiltonian by patial deivatives of f should emind you of how gauge tansfomations wok in electodynamics, whee the vecto and scala potential tansfom with deivatives of a common function. Now notice that p i / p j = δ j i, so that Hamilton s equations fo the q i ae given by q i = H p i = H p i 1 = H p i 5)

2 which is the same as befoe. Now, conside the Hamilton s equation fo p i, which ae given by ṗ i = H 6) = H H + f t 7) Let us use these to compute ṗ i in the new Hamiltonian system. We find that ṗ i = ṗ i d f = ṗ i f q j f dt q j t 8) = H H + f t 9) f q j f q j t = H H f q j ) 10) f q j q j 11) now we can use equation 5) to emembe that H/ = q j to obtain that ṗ i = H ) f + q j f q j = H 1) q j q j This is, we ae able to epoduce the equations of motion that we had peviously.. Poblem 8.3. The Lagangian is given by L = 1 m x ) + e 1 B x) x V ) 13) = 1 m x ) + e 1 x x) B V ) 14) and let us assume that B is oiented along the z axis, so that x x)

3 B = xẏb yẋb. The canonical momenta ae given by The Hamiltonian is given by p x = mẋ 1 eyb 15) p y = mẏ + 1 exb 16) p z = mż 17) H = p x ẋ + p y ṗ y + p z ṗ z L 18) = 1 p x + 1 m eyb) + p y 1 ) exb) + p z + V ) 19) Now, let us define a new coodinate system whee the coodinates ae given by z = z and ) ) ) x cosωt) sinωt) x y = 0) sinωt) cosωt) y equivalently, we get that ) ) ) x cosωt) sinωt) x = y sinωt) cosωt) y 1) so that ẋ ) = ẏ = ) ẋ ) ) ) cosωt) sinωt) sinωt) cosωt) x sinωt) cosωt) ẏ + ω ) cosωt) sinωt) y ) ẋ ) )) cosωt) sinωt) y sinωt) cosωt) ẏ ω x 3) Let us now eplace these values in the Lagangian. Notice that the vectos ẋ, ẏ, etc ae in the 1, so that thei coss-poduct points in the z diection always. This shows that the time dependent tems in the magnetic field cancel, and we get the following tems: L kin = 1 mẋ + ωy ) + ẏ ωx) + ż ) 4) L B = e 1 x x ω x )) B 5) L V = V ) 6) 3

4 A bit of algeba shows that the tems linea in ẋ and ẏ cancel with the choice of ω, and instead we get a magnetic contibution to the potential that is given by 1 mω y ) + x ) ) 7) This is also called Lamo s theoem: that to linea ode in the magnetic field, we get that a unifom otation cancels the magnetic effects, and that at second ode the unifom magnetic field manifests itself as a quadatic potential. 3. Poblem 8.6 The Lagangian is given by L = 1 m q 1 k 1q 1 k a q) = 1 m q 1 k 1 +k )q b) +constant We can dop the constant, so that 8) L = 1 m q 1 k 1 + k )q b) 9) This is time independent, so the Hamiltonian is conseved equal to the enegy). The Hamitonian is tivially given by H = p m + 1 k 1 + k )q b) 30) Now let us do it with q = Q + b sinωt). A possible lagangian is given by L = m Q + ωb cosωt)) 1 k 1 + k )Q + b sinωt) b) 31) This is geneally time dependent, and so will be the case fo the Hamiltonian. Notice that P = Q + ωb cosωt). It is then easy to find the Hamiltonian. Since the Hamiltonian is explicitly time dependent, it will not be conseved. I believe that this poblem would be slightly diffeent it ω was explicitly the fequency of the system of spings. Then one can show that the second lagangian diffes by a total time deivative fom the lagangian of the fist oscillato if we eplace Q by q). Then they ae equivalent and one can find a new conseved Hamiltonian 4

5 4. Poblem 8.33 Let, θ be the pola coodinates of m 1 on the plane, and z the vetical coodinate of m adjusted so that if we fix = 0, then z = 0 is an equilibium position of the ope). The lagangian is then given by L = 1 m 1ṙ + θ ) + 1 m ż m gz 1 kz ) 3) The Hamiltonian is theefoe given by H = p m 1 ) + p z + p θ m m 1 + m gz + 1 kz ) 33) Notice that p θ is cyclic. The conditions fo equilibium ae that we fix p θ = l and that the paticle does not move in z,. This equies that p = p z = 0, and that z H = H = 0. The conditions fo that ae that l + k z) = 0 m ) m g + kz ) = 0 35) equivalently, we get that l /m 1 3 ) = m g = k z), this is, we balance the centifugal foce with foce of gavity on the hanging mass. Now, fo the small oscillations poblem, we expand the above answe to second ode aound the equilibium point. We need the following deivatives: H = 3l m 1 + k = 3m g/ + k = k k 36) H z = k 37) H z = k 38) 5

6 whee we have intoduced a new constant k 1. The small oscillations Hamiltonian is given by H osc = p m 1 + p z m + 1 k 1 + k)δ kδδz + kδz ) 39) = p m 1 + p z m + 1 k 1δ + 1 kδ δz) 40) Notice that the potential is stable, because k 1, k > 0, and indeed, we can see it is a sum of squaes. Notice that z, mix with each othe. Solving fo the chaacteistic fequencies is staightfowad but tedious. The poblem is sufficiently vague on this issue that detemining the stability is all we eally need to do. 5. The geomety of a spheical sta in Geneal elativity is chaacteized by the pope time diffeential given by c dτ = 1 GM ) c dt 1 GM ) 1 d dφ 41) c c whee M is the mass of the sta and whee c is the speed of light. This fom aleady assumes that you ae on the plane of an obit thee is no angle θ in the spheical coodinates to woy about). a) The action of a paticle is given by S = mc dτ 4) Calculate the equations of motion. b) Show that since t and φ ae ignoable coodinates, that thee ae two conseved quantities associated to this motion you can call them enegy and angula momentum). Wite thei explicit fom. c) A cicula obit is an obit whee is a constant of motion. Compute the peiod of a cicula obit the amount of time t that it takes fo the angle φ to go fom 0 to π, given a fixed ). Show that this educes to Keple s thid law in the case whee the obit satisfies >> R s = GM/c indeed, it woks fo all adius due to a coincidence in the choice of coodinates). Also show that thee 6

7 is a adius R fo which it is impossible to have cicula obits if < R The foces in geneal elativity ae moe attactive than in Newtons theoy of gavitation, so that centifugal foce can not always balance them ). d) Find a diffeential equation fo the obit φ) using the conseved quantities. This is the analog of equation 3.39 in the book.) Expand this equation in small oscillations in aound a cicula obit and ague that φ) 0 + ɛ sinβφ) fo some β. e) Exta cedit: Use the obit equation fo almost cicula obits to compute the advance of the peihelion of Mecuy. You job is to compute β fo Mecuy. Compae with data. Hint: This is a classic poblem in Geneal elativity. Thee ae many texts fom whee the wok can be deduced. You might want to look at the book by James B. Hatle called Gavity: An intoduction to Einstein s Geneal Relativity. The complete solution to this poblem can be found in Hatle, Ch. 9. Howeve, on gading you homewok submissions, I found that only a handful of you took the shotest oute to the solution, and many took the long way instead. Because of this, I ll simply outline the key points of the solution below. Fo convenience, set m = c = 1 hencefoth. Intoduce the integation paamete λ and wite the action as whee f = 1 GM S = fdλ 43) ) ) dt 1 GM dλ ) 1 d dλ ) ) dφ. 44) dλ Then vaiation of the action about its extemum gives δs = 1 1 δfdλ. 45) f The key hee is to ecognize that if we now choose the integation paamete to be the pope time, λ = τ, then f = 1 and the poblem 7

8 educes to locating the extemum of the integal I 1 fdτ. 46) This way thee is no need to handle the squae oot that would have made the solution incedibly tedious), and the equations of motion ae simply given by the Eule-Lagange equations fo f: d dτ f ṫ ) = L t, 47) whee ṫ dt/dτ, and similaly fo the othe two coodinates and φ. Because t and φ ae ignoable coodinates, it follows that the two quantities E 1 GM ) ṫ; l φ 48) ae constant of the motion. In paticula, l has the same fom as the angula momentum in Newtonian mechanics. On the othe hand, the equation of motion fo is 1 GM ) 1 1 GM ) GM ṙ = φ GM ṫ. 49) Fo cicula motion at some adius 0, we set = ṙ = 0 above and get 0 φ = GM/ 0)ṫ. Theefoe, the angula velocity is given by ) dφ = φ dt ṫ = GM. 50) 0 3 Thus we ecove Keple s thid law fo the obital peiod T : 3 T = π 0 GM. 51) Note that this esult is independent of the adius 0 of the obit. Some of you did not use the equation of motion fo to compute the peiod, and had to use appoximations such as GM. This is in fact not necessay. 8

9 We now etun to the expession fo the pope time in the fom 1 = 1 GM ) ṫ 1 GM ) 1 ṙ φ, 5) and eaange the tems using the conseved quantities E and l found above. We find E E 1 = ṙ + 1 GM ) l GM = ṙ + l GMl GM 3 ṙ + V eff). 53) V eff ) is the effective one-dimensional potential fo the adial coodinate; it is almost the same as the one fo Newtonian cental-foce motion, except fo the exta tem GMl / 3. Ou machiney fo analyzing adial motion in the Newtonian case still applies. Fo each given angula momentum l, cicula motion is possible at adius given by dv eff /d = 0; thus the adii fo cicula motion ae given by ) 1, = l GM ) GM l It is easy to see that 1 is a maximum of V eff ) while is a minimum; they coespond to unstable and stable cicula obits, espectively. The minimum adius s fo a stable cicula obit is thus given by setting the quantity unde the squae-oot sign above equal to zeo; thus s = 6GM. Howeve, it is still possible to have cicula obits with adius smalle than this; the only issue is that it is unstable with espect to small petubations. The minimum possible adius of a cicula obit is thus given by the lowe bound to 1, which is 3GM. 9