QUALIFYING EXAMINATION, Part 1. Solutions. Problem 1: Mathematical Methods. r r r 2 r r2 = 0 r 2. d 3 r. t 0 e t dt = e t

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1 QUALIFYING EXAMINATION, Part 1 Solutions Problem 1: Mathematical Methods (a) For r > we find 2 ( 1 r ) = 1 ( ) 1 r 2 r r2 = 1 ( 1 ) r r r 2 r r2 = r 2 However for r = we get 1 because of the factor in front To calculate the limit r 2 r, we integrate ( ) 2 1 r over a small sphere of radius a surrounding the origin and use Gauss theorem (converting a volume integral to a surface integral) It follows that r a d 3 r 2 ( 1 r ) = = r a r=a ( ) 1 d 3 r r ( ) 1 ds r = r=a ( ) 1 2 = 4πδ 3 ( r) r ( 1 a 2 ) ê r d S = 4π Note: 1 r 4πϵ is the solution to Poisson s equation with a point charge at the origin with charge (b) Γ(1) = t e t dt = e t = 1 (c) Γ(x + 1) = t x e t dt = t x e t + x t x 1 e t dt = + xγ(x), where the surface term vanishes at t = if x It follows that Γ(n + 1) = nγ(n) which, given Γ(1) = 1, means Γ(n + 1) = n! 1

2 (d) n n ln n! = ln [n(n 1)(n 2) 1] = ln m ln xdx = n ln n n + 1, m=1 1 where we can drop the 1 compared to n Exponentiating, we find (e) We have We rescale t = nz to obtain where n! e n ln n n = n n e n n! = Γ(n + 1) = t n e t dt n! = n n+1 z n e nz dz = n n+1 e nz+n ln z dz n n+1 e nf(z) dz, f(z) = z ln z The saddle point z = z is determined by the condition f (z) = z 1/z =, giving z = 1 We expand f(z) to second order around z = 1 to find Γ(n + 1) = n n+1 Using f (z ) = 1 z = 1, we obtain z 2 e nf(z) dz n n+1 e nf(z ) n! = n n e n 2πn (1 + corrections), e n 1 2! f (z )(z z ) 2 dz where we have changed variables to w = z z and extended the limits to w = ± (the integrand is negligible far away from z ) We can then use the familiar result for a Gaussian integral π e αw2 dw = α The initial rescaling was done to get a large factor n in the exponent in front of f(z) (to justify the saddle-point approximation) It is also possible to carry out the saddlepoint approximation directly for f(t) = n ln t t, in which case the saddle-point condition f (t ) = gives t = n 2

3 Problem 2: Classical Mechanics (a) The first Lagrangian is invariant under translations in the x direction x x + a and therefore p x (the generator of translations in the x direction) is conserved It does not depend on time explicitly so L/ t = dh/dt = and the Hamiltonian H = E is conserved The Lagrangian is not invariant under translations in y or under rotations, and therefore p y and L z are not conserved The second Lagrangian is not invariant under translations (in x or y) and therefore p x and p y are not conserved It is rotationally invariant and explicitly time independent, and therefore L z and E are conserved The third Lagrangian is invariant under translations in x but not under translations in y or under rotations Therefore p x is conserved but p y and L z are not conserved Because of the explicit time dependence in L, E is not conserved (b) The potential is time independent, and hence the total energy is conserved: E = E = 1 2 mv2 = const The potential is central and provides no torque, hence the angular momentum around the scattering center is conserved L = L = mv bẑ = const L x = L y = guaranteeing that the motion is in the x-y plane (c) Using the effective potential for the reduced 1D motion in the radial coordinate r, the energy can be written as E = 1 2 mṙ2 + U eff (r) = 1 2 mṙ2 + L2 2mr 2 k r 2 = 1 2 mv2 At the distance r = r min of closest approach ṙ =, and we have (mv b) 2 2mr 2 min k r 2 min = 1 2 mv2 Therefore r min = mv2 b 2 /2 k mv/2 2 = b 2 2k mv 2 Since k >, r min < b Qualitatively this is expected since the potential is attractive 3

4 (d) The horizontal coordinate of the point particle is given by x + X, while its vertical coordinate is y = x tan α The kinetic energy T of the system is then T = 1 2 MẊ m(ẋ + ẋ) mẏ2 = 1 2 (M + m)ẋ2 + mẋẋ m(1 + tan2 α)ẋ 2 The potential energy of the particle is The Lagrangian of the system is therefore: V = mgy = mgx tan α L = T V = 1 2 (M + m)ẋ2 + mẋẋ m(1 + tan2 α)ẋ 2 mgx tan α The equations of motion are: d dt ( ) d L L dt Ẋ X = (M + m)ẍ + mẍ = ( ) L L ẋ x = mẍ + m(1 + tan2 α)ẍ + mg tan α = From the first equation we get Ẍ = m M + mẍ Substituting in the second equation, we find (M + m)g sin α cos α ẍ = M + m sin 2 α We then have Ẍ = mg sin α cos α M + m sin 2 α 4

5 Problem 3: Electromagnetism I (a) Using Coulomb s law V (r, θ) = q 4πϵ ( ) r + r r where (applying the law of cosines) r ± = ( r 2 + a 2 2ra cos θ ) 1/2 (b) Expanding 1/r ± to order a 2 /r 2 (all higher order terms will vanish in the limit), we have 1 1 ( 1 1 a 2 r ± r 2 r ± a 2 r cos θ + 3 a 2 ) 2 r 2 cos2 θ Substituting into the potential, we get V (r, θ) = Q 4πϵ r 3 ( 3 cos 2 θ 1 ) = Q 2πϵ r 3 P 2(cos θ), where we have used the expression for the second Legendre polynomial P 2 (c) Since the conducting sphere is grounded V (r = R) = We also know that V (r ) = The uniqueness of the solution of Laplace s equation with boundary conditions gives us V (r) = outside the sphere (d) Inside the sphere, we use the general solution of Laplace s equation V in = ( A l r l + B ) l P r l+1 l (cos θ) The potential is continuous across r = R Using the result in (c), we have V in (R, θ) = Since this is true for all angles θ, the sum must vanish term by term This gives A l R l = B l R l+1 In the limit r (r R), we match the solution to the result from part This gives us B l 2 =, B 2 = Q 2πϵ Using the first set of relations between A l and B l, we find 5

6 Thus the potential inside is given by V in (r, θ) = A l 2 =, A 2 = B 2 R 5 Q ( ) 1 r5 P 2πϵ r 3 R 5 2 (cos θ) 6

7 Problem 4: Electromagnetism II (a) The voltage per turn has to balance the induced electro-motive force (emf) V = ε = dφ dt = πa2 db dt (b) Using a rectangular Amperian loop outside the solenoid, one can show that the field outside (parallel to the cylinder axis) is independent of the radial distance r Since the field vanishes at r, it must be zero everywhere outside the cylinder (c) From symmetry considerations B is along the cylinder s axis Using a rectangular Amperian loop for the field H with one side inside the solenoid and another side outside, gives H = ni or I = B(t) µn (d) We use a circular loop C around the cylinder with radius r By symmetry the magnitude of A is constant along the loop and is tangential along the same direction as the current through the curled wire Using Stokes theorem with A = B we have C A d l = where Φ is the magnetic flux through the loop We find for r > a S B d S Φ, and for r < a 2πrA(r, t) = B(t)πa 2, A(r, t) = a2 2r B(t), 2πrA(r, t) = πr 2 B(t), A(r, t) = r 2 B(t) (e) The field is uniform inside the torus Since B =, the flux of B is constant around the torus Since the cross section of the torus is the same everywhere, B must be a constant inside the torus Since the gap h is small, B in the gap will be near homogeneous (negligible fringe fields) and will have the same value as inside the torus To find the value of B, we use a circular Ampreian loop of radius Z around the torus We have Hd l = NI Z 7

8 or This gives B µ (2πZ h) + B µ h = NI B = µµ NI µ (2πZ h) + µh µµ NI µ 2πZ + µh The current I in the wire is I = V, giving the final answer R B = µµ NV R(2µ πz + µh) 8