QUALIFYING EXAMINATION, Part 1. Solutions. Problem 1: Mathematical Methods. x 2. 1 (1 + 2x 2 /3!) ] x 2 1 2x 2 /3

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1 QUALIFYING EXAMINATION, Part 1 Solutions Problem 1: Mathematical Methods (a) Keeping only the lowest power of x needed, we find 1 x 1 sin x = 1 x 1 (x x 3 /6...) = 1 ) 1 (1 x 1 x /3 = 1 [ 1 (1 + x /3!) ] x = 1 ( ) x = 1 x 3! 3. (b) The sum of finite number of terms is finite. For the rest, the ratio of successive terms test gives the condition (n+1)!) z n+1 lim (n+)! n (n!) z n < 1 (n)! (n + 1) z lim < 1 n (n + )(n + 1) z < 4. (c) We expand Then y = m c m x m. y = m = m x y = m c m x m m(m 1) c m+ x m (m + )(m + 1) (dummy label m m + ) c m x m(m 1) xy = m c m x m m y = m c m x m. 1

2 Collecting the coefficient of x m and setting it to 0, we find (m 1)(m ) = c m (m + 1)(m + ). c m+ Choosing c 0 and c 1, arbitrarily, we find c = 1 and all others 0. Thus the general solution is given by y = c 0 (1 x ) + c 1 x. (d) 0 x sin x x + a dx = 1 ( 1 = I = I = π e a. x sin x dx (even integrand) x + a ) ze iz [ 1 πiiae a ia z + a dz ] (I is imaginary part) (close contour in UHP, pick up residue at z = ia)

3 Problem : Classical Mechanics (a) The Hamiltonian is H = p m ax + λx 4. Hamilton s equations of motion are given by ṗ = H x = ax 4λx3, ẋ = H p = p m. (b) Equilibrium requires ṗ = 0 = x(a 4λx ) so that x = ± a λ and x = 0. Around x = 0 we have U = a so it is unstable since the second derivative is negative. Around x = ± a we have λ a a U(± λ + ɛ) = a(± λ + ɛ) + λ(± a λ + ɛ)4 = a 4λ + aɛ +... Thus, U = 4a > 0 so the equilibrium points are stable and the effective spring constant for the oscillations is k eff = 4a. The oscillation frequency is then ω = keff m (c) = 4a m E>0 0.5 p 0.0 E<0 E< x (d) The principle of least action is a variational principle which states that the system evolves in a way which extremizes the classical action S. 3

4 In the Hamiltonian formulation, the classical action for evolution from t 1 to t is S = t t 1 (pẋ H) and the principle states that the variation is δs = 0 under small changes in the configuration {x(t) x(t) + δx(t), p(t) p(t) + δp(t)}, where the coordinates are held fixed at the endpoints, δx(t 1 ) = δx(t ) = δp(t 1 ) = δp(t ) = 0. (We would also accept a formulation of the principle of least action that leads to Lagrange s equations.) In the situation asked about in the problem, the classical action is S = T 0 T ( ) ( ) a (pẋ H) = U ± = T a = a T 0 λ 4λ 4λ. 4

5 Problem 3: Electromagnetism I (a) The trace of the second rank tensor Q is 3 ( Q ii = 3x i r ) (3r ρ(x ) d 3 x = 3r ) ρ(x ) d 3 x = 0. i=1 i Q ij is real, symmetric, and traceless. Therefore it has 5 independent components. (b) q and p are, respectively, the total charge and dipole moment of the charge distribution. They are given by q = ρ(x ) d 3 x, p = ρ(x )r d 3 x. (c) (d) Thus p = 0. (e) Thus q = e e + e e = 0. p x = ea ea ea ea = 0 p y = ea + ea ea ea = 0 p z = 0. Q xx = (3a a )(e e + e e) = 0 Q yy = (3a a )( e + e + e e) = 0 Q zz = 0 Q xy = 3a ( e) 3a e + 3a ( e) 3a e = 1a Q xz = 0 Q yz = ea 0 Q = 1ea (f) Since q and p are zero, only the quadrupole term survives. We then have Φ(x) = 1 ( 1ea xy ). r 5 The electric field is ( E = 1ea xy ) r 5 = ( 1ea y r 5 60ea x y, 1ea x r 7 60ea xy r 5 r 7, 60ea xyz r 7 ). 5

6 Problem 4: Electromagnetism II (a) Define k = ω/c. The electric and magnetic fields are now given by E = E 0 sin(kz ωt)ˆx, and B = B 0 sin(kz ωt)ŷ. We can relate E 0 and B 0 by Maxwell s equations; in SI units we have B 0 = E 0 /c, while in Gaussian units, we have E 0 = B 0. (b) The Poynting vector is S = 1 µ 0 E B (SIunits) S = c 4π E B (Gaussianunits) Recall that the average of sin over a period is 1/. Substituting and taking timeaverages, we have S = E 0 µ 0 c (SIunits) = ce 0 8π (Gaussianunits) (c) The acceleration of the electron is a = ee/m; plugging this into the Larmor formula (and time averaging) we get P rad = 1 1 e 4 E0 4πɛ 0 3 m ec (SIunits) 3 = 1 e 4 E0 3 m ec (Gaussianunits). 3 (d) The ratio of the previous two parts is σ. This is simplest in Gaussian units, where we immediately find that σ = 8π e 4 3 m ec = 8π 4 3 r e. We get the same result in SI units, using µ 0 ɛ 0 = 1/c. (e) The integral of dσ/dω over all angles must equal σ. This implies 1 πa sin θ d cos θ = 8π 1 3 r e. 6

7 Using sin θ = 1 cos θ, we have which yields A = r e. 1 ( πa (1 cos θ)d cos θ = π ) A = 8π r e (f) For the x polarized wave θ = π/ θ, and so dσ/dω = r e cos α. For the y polarized wave, θ = π/ and we have dσ/dω = r e. 7