Classical Mechanics Problem 1: Central Potential Solution a) Integrals of motion for a central potential V (r): 2 _ Angular Momentum L = rv¡t = r Á ¢ 1 2 2 1 2 Energy per unit mass E = 2 r_ + vt + V (r) = 2 r_ + Ve®(r) where vt is the tangential velocity and Ve® is de¯ned as L2 V (r) = V (r) + e® 2r2 If the orbit is circular, the distance of the test body from the origin is invariant:r _ = 0, which implies that the body is always at the equilibrium-distance: dV dV L2 v2 e® = 0 ) = = t = rÁ_2 dr dr r3 r then µ ¶ L 1 dV 1=2 Á_ = ! = = Á r2 r dr so for the period we get µ ¶ 2¼ 1 dV ¡1=2 TÁ = = 2¼ !Á r dr b) Write the orbit as in the statement of the problem: dV r(t) = r + ²(t) with e® (r ) = 0 and ²2 ¿ r2: 0 dr 0 0 1 2 The energy per unit mass is now E = 2 ²_ + Ve®(r0 + ²), and since ² is small we may Taylor-expand the potential as 2 dVe® 1 d Ve® 2 3 Ve®(r0 + ²) = Ve®(r0) + (r0)² + 2 (r0)² + O(² ) | dr{z } 2 dr =0 so then 1 1 d2V E ¡ V (r ) = ²_2 + e® (r )²2 + O(²3) = const. e® 0 2 2 dr2 0 In the above equation we readily recognize the equation of the simple harmonic oscil- lator with µ ¶ d2V 1=2 ! = e® r dr2 r=r0 1 and its general solution is p E ¡ Ve®(r0) ²(t) = cos [!r(t ¡ t0)] !r where t0 is an arbitrary constant. Now return to writing !r in terms of V (r) instead of Ve®(r). d2V d2V 3L2 d2V d2V 3 dV !2 = e® = + = + 3!2 = + r dr2 dr2 r4 dr2 Á dr2 r dr µ ¶ · µ ¶¸ d2V 3 dV 1=2 1 d dV 1=2 ! = + = r3 r dr2 r dr r3 dr dr r=r0 r=r0 And the radial period is 2¼ Tr = !r 2 2 c) Stability is determined by the sign of !r . For stability: !r > 0, so µ ¶ 1 d dV r3 > 0 r3 dr dr for the Yukawa-potential GM V (r) = ¡ e¡kr r so the condition is µ ¶ 1 d dV GM £ ¤ £ ¤ r3 = e¡kr 1 + kr ¡ (kr)2 > 0 ) 1 + kr ¡ (kr)2 > 0 r3 dr dr r3 Ãp !Ãp ! £ ¤ 5 ¡ 1 5 + 1 1 + kr ¡ (kr)2 = + kr ¡ kr > 0 2 2 which is satis¯ed only if Ãp ! 5 + 1 kr < 2 Therefore circular orbits are unstable for Ãp ! 5 + 1 kr > 2 ¥ 2 d) The outermost stable circular orbit is at Ãp ! 5 + 1 r = 0 2k its energy per unit mass is µ ¶ 1 1 dV E = V (r ) + (r ! )2 = V (r ) + r 0 2 0 Á 0 2 dr r=r0 Ãp ! 1 GM GM 5 ¡ 1 ¡kr0 ¡kr0 e (kr0 ¡ 1) = e > 0 2 r0 r0 4 If r0 is decreased only slightly, E > 0 still and the orbit is absolutely stable ¥ The e®ective potential for the Yukawa-potential has the form shown in Figure 1. Veff r Figure 1: E®ective potential against distance from the origin 3 Classical Mechanics Problem 2: Planar Double Pendulum Solution q1 l q2 l a) L = T ¡ V The moment of inertia for a uniform rod of length l and mass m is 1 I = ml2 about one of the ends 3 and 1 I = ml2 about the rod's center c 12 The kinetic energy term we can decompose into three parts: T = T1 + T2;rot + T2;trans where T1 is the kinetic energy of the ¯rst rod, T2;trans is the translational energy of the center of mass of the second rod and T2;rot is its rotational energy about its center of mass. Then 1 T = ml2θ_2 1 6 1 1 T = ml2θ_2 2;rot 24 2 and 1 ¡ ¢ T = m x_ 2 +y _2 2;trans 2 c c where xc and yc are the coordinates of the second rod's center of mass, so l x = l sin θ + sin θ c 1 2 2 l y = ¡l cos θ ¡ cos θ c 1 2 2 from which · ¸ 1 x_ 2 +y _2 = l2 θ_2 + θ_2 + θ_ θ_ (sin θ sin θ + cos θ cos θ ) c c 1 4 2 1 2 1 2 1 2 4 The potential energies are simply Vi = mgyc;i, where yc;i are the vertical coordinates of the rods' centers of mass. Since both rods are uniform, yc;i are simply the coordinates of the centers. Thus, µ ¶ l l V = ¡mg cos θ ; V = ¡mg l cos θ + cos θ 1 2 1 2 1 2 2 The full Lagrangian is then L = T1 + T2;rot + T2;trans ¡ V1 ¡ V2 · ¸ · ¸ 2 1 3 1 = ml2 θ_2 + θ_2 + θ_ θ_ cos (θ ¡ θ ) + mgl cos θ + cos θ 3 1 6 2 1 2 1 2 2 1 2 2 b) Expand the Langrangian from part a) for small angles. The only function we have to deal with is 1 cos θ = 1 ¡ θ2 + O(θ4) 2 ^ ^ Since we are going to look for normal modes with θj = θj exp (i!t), where the θj ¿ 1, _ _ we immediately see that in the term θ1θ2 cos (θ1 ¡ θ2), the θ-dependence in the cosine can be dropped, because even the ¯rst θ-dependent term gives a fourth order correction. Then the approximate Lagrangian is · ¸ · ¸ 2 1 3 1 L = ml2 θ_2 + θ_2 + θ_ θ_ ¡ mgl θ2 + θ2 + const. 3 1 6 2 1 2 4 1 4 2 The Euler-Lagrange equations are d @L @L = _ dt @θj @θj so in the speci¯c case: µ ¶ 4 1 g 3 θÄ + θÄ + θ = 0 3 1 2 2 l 2 1 µ ¶ 1 1 g 1 θÄ + θÄ + θ = 0 2 1 3 2 l 2 2 if we now look for normal modes, as mentioned, the above set of equations takes the form " ¡ ¢ #" # 4 2 3 g 1 2 ^ ! ¡ ! θ1 3 2 l ¡ 2 ¢ = 0 1 2 1 2 1 g ^ 2 ! 3 ! ¡ 2 l θ2 Non-trivial solutions exist if the determinant of the matrix on the left is zero. Denoting !2 = ¸g=l, we can write this condition as µ ¶ µ ¶ 4 3 1 1 ¸2 ¸ ¡ ¸ ¡ ¡ = 0; 3 2 3 2 4 5 that is 7 7 3 ¸2 ¡ ¸ + = 0 36 6 4 whose solutions are 6 ¸§ = 3 § p ; 7 so ¯nally ·µ ¶ ¸ 6 g 1=2 !§ = 3 § p 7 l c) To sketch the eigenmodes, ¯nd eigenvectors of the matrix in part b). 2 ² ! = ¸¡g=l (low-frequency mode) µ ¶ 3 8 1 ³ p ´ θ^ = ¡ θ^ = 2 7 ¡ 1 θ^ 2 ¸ 3 1 3 1 p (2 7 ¡ 1)=3 > 0 and real, therefore the two pendula are in phase; 2 ² ! = ¸+g=l (high-frequency mode) µ ¶ 3 8 1 ³ p ´ θ^ = ¡ θ^ = ¡2 7 ¡ 1 θ^ 2 ¸ 3 1 3 1 p (¡2 7 ¡ 1)=3 < 0 and real, therefore the two pendula are perfectly out of phase. a b Figure 2: The low- (a) and high-frequency (b) normal modes of the planar double pendulum. 6 Electromagnetism Problem 1 Solution a) Normal modes are products of harmonic standing waves in the x, y and z directions. For their frequencies, we have · ¸ q ³¼n ´2 ³¼n ´2 ³¼n ´2 1=2 ! = c k2 + k2 + k2 = c x + y + z ; n ; n ; n 2 Z+ x y z a b b x y z Since a > b, the lowest frequency has nx = 1 and either ny = 1; nz = 0 or ny = 0; nz = 1 (note that ny = 0; nz = 0 does not satisfy the boundary condition Ek;at wall = 0). Since we are told to pick the mode with E~ k y^, the boundary conditions require ¼x ¼z E~ (r; t) = E y^sin sin cos !t 0 a b The magnetic induction we can get from Faraday's Law: @B~ ³ ´ @E @E = ¡c r £ E~ =xc ^ y ¡ zc^ y @t @z @x ³ ¼ ¼x ¼z ¼ ¼x ¼z ´ = xcE^ sin cos ¡ zcE^ cos sin cos !t 0 b a b 0 a a b µ ¶ ¼cE x^ ¼x ¼z z^ ¼x ¼z B~ (r; t) = ¡ 0 sin cos ¡ cos sin sin !t ! b a b a a b where the frequency ! is (by the argument above) µ ¶ 1 1 1=2 ! = ¼c + a2 b2 b) At a boundary of media, the discontinuity in the normal component of the electric ¯eld is 4¼ times the surface charge density σ, so Ey(x; 0; z) = 4¼σ E ¼x ¼z σ(x; 0; z) = 0 sin sin cos !t 4¼ a b σ(x; b; z) = ¡σ(x; 0; z) and σ(0; y; z) = σ(a; y; z) = σ(x; y; 0) = σ(x; y; b) ´ 0 Similarly, at the boundary of media the discontinuity of the tangential component of the magnetic ¯eld is given by the surface current ~· 4¼ n^ £ B~ = ~· c 7 wheren ^ is a unit vector normal to the surface, so µ ¶ c2E z^ ¼x ¼z x^ ¼x ¼z ~·(x; 0; z) = 0 sin cos + cos sin sin !t 4! b a b a a b ~·(x; b; z) = ¡~·(x; 0; z) c2E y^ ¼z ~·(0; y; z) = ¡ 0 sin sin !t 4! a b ~·(a; y; z) = ¡~·(0; y; z) c2E y^ ¼x ~·(x; y; 0) = ¡ 0 sin sin !t 4! b a ~·(x; y; b) = ¡~·(x; y; 0) c) Since there is no charge on the b £ b sides, the force there is purely magnetic and is given by Z 1 ³ ´ F~ (t) = ~· £ B~ d2x 2c b£b Zb Zb E2c2¼ ¼z F~ (x = 0; t) = ¡ 0 x^ dy dz sin2 sin2!t 8!2a2 b 0 0 µ ¶ c b 2 = ¡x^ E sin !t 4! a 0 F~ (x = a; t) = ¡F~ (x = 0; t) The forces point outwards from the box on both sides (as is indicated by the sign in the equation above).