Suggested solutions, FYS 500 Classical Mechanics and Field Theory 2015 fall

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1 UNIVERSITETET I STAVANGER Institutt for matematikk og naturvitenskap Suggested solutions, FYS 500 Classical Mecanics and Field Teory 015 fall Set 1 for 16/17. November 015 Problem 68: Te Lagrangian for a particle in an electromagnetic field is given by Goldstein eq as: L = 1 mv qφ + qa v, were φ is te electric scalar potential qφ is te potential energy, and A te magnetic vector potential. Wit a gauge transformation, A A = A + ψ, φ φ = φ ψ/ t, te transformed Lagrangian is: L = 1 mv qφ + qa v = 1 ψ mv qφ + qa v + q t + ψ v = L + dψ dt, were we ave used tat v = ṙ. But we learned in problem 6 cf. Goldstein derivation 1.8 tat adding a total derivative to te Lagrangian does not cange te equations of motion, and ence te motion of te particle. Problem 69: Te relativistic Lorentz force law, dp µ /dτ = qf µ νu ν in te absence of a magnetic field takes te simple form dp/dt = qe. For a constant electric field we can cose coordinates suc tat E = E 0 k and v 0 = v 0 i. Te equations of motion ten reduce to: dp x dt = 0 = p x = p x 0 dp y dt = 0 = p x = p z 0 dp z dt = qe 0 = p x = qe 0 t + p z 0. From te boundary condition p0 = γ 0 mv 0 i, wit γ 0 = 1/ 1 v 0 /c, we find p x 0 = γ 0 mv 0 p 0 and p y 0 = p z 0 = 0. Tus we ave: γẋ = γ 0 v 0 γż = qe 0 m t, were γ = 1/ 1 ẋ + ż /c, so ẋ +ż = c 1 γ. In order to proceed, we first ave to determine γ = γt from te above equations: γ ẋ + ż = γ c 1 1 γ = c γ 1 = γ0v 0 + qe0 t. m Tis yields γt as: γ = γ0 v 0 c qe0 + t mc + 1 = γ 0 + qe0 mc t, were we ave used tat γ 0 v 0 /c + 1 = γ 0. Te equation for xt tus becomes dx dt = γ 0v 0 γt = γ 0 v 0 γ0 + qe0 t mc = v qe0 γ 0mc t, 1

2 wic can be integrated to: t xt = x 0 + v 0 0 d t 1 + qe0 γ 0mc t = x 0 + γ 0mv 0 c qe0 t arsin qe 0 γ 0 mc x 0 + v 0 t as c. were x 0 = x0, and we ave used lim arsinx x as x 0. Te corresponding z-integration proceeds as follows: dz dt = qe 0t γtm = qe 0 ct γ 0 m c + q E, 0 t wit solution: t dt zt = z 0 + qe 0 c γ 0 m c + q E0 t = z 0 + γ 0mc qe qe0 t 1 γ 0 mc z qe 0 m t as c. Here z 0 = z0, and we ave used lim 1 + x x as x 0. We see tat z z 0 = 1 qe0 x x mv0 0, wic is te equation of a parabola, in te non-relativistic limit. [Remark: As t te z-velocity dz/dt c, as expected, wile dx/dt γ 0 mv 0 c/qe o t 0. Te latter is a consequence of relativistic momentum conservation in te x-direction: p x = γtmv x is conserved, and γt is increasing, so v x is decreasing.] Problem 70: In tis case te equation of motion is dp/dt = qv B, or: ṗ x = dp x dt = qv yb 0 = qb 0 γm p y ṗ y = dp y dt = qv xb 0 = qb 0 γm p x ṗ z = dp z dt = 0. Te equation for te motion in te z-direction ten yields p z = γmv z = p 0 z, a constant. Now, by observing tat 1 d dt p = p ṗ = γmv qv B = 0, by te properties of te vector product, we see tat p = γ m v is a constant of motion, wic means tat v and ence γ are constants. Ten also v z is constant, and te same must be tus true for te magnitude of te transverse velocity, v = v vz. Differentiating te equation of motion, we ten find: d p x dt = qb 0 dp y γm dt = qb0 p x. γm Tis is te equation for armonic oscillations, wit solution: p x t = p cosω c t + δ ω c = qb 0 γm. Here p and δ are constants of integration. Te frequency ω c is te relativistic version of te well known cyclotron frequency. Furtermore, p y = ṗ x /ω c = A cosω c t + δ, so we find p x + p y = p = γ m v, or p = γmv, and we can write ω c = qb 0 /p, just like in te non-relativistic case. Tus te transverse motion of te particles is circular. Te period is T = π/ω, and since for circular motion we must ave v = πr 0 /T, were r 0 is te radius of te orbit, we find r 0 = T v /π = v /ω c = p /qb 0.

3 Problem 71: Exam problem, 013 fall. See separate solution seet. Problem 7: Exam problem, 015 spring. See separate solution seet. Problem 73: a We must ave: a = C b = La = LCb = LCL 1 b C = LCL 1. b Since Tr AB = Tr BA, we ave: C µµ = Tr LCL 1 = Tr L 1 LC = Tr C = C µ µ. c Wit te matrix notation F = F α β we ave from te explicit representation of F see Goldstein 7.71: 0 E x E y E z F = 1 E x 0 cb z cb y c E y cb z 0 cb x E z cb y cb x 0 E ce B x ce B y ce B z F = 1 ce B x Ex c By + Bz E x E y + c B x B y E x E z + c B x B z c ce B y E x E y + c B x B y Ey c Bx + Bz E y E z + c B y B z. ce B z E x E z + c B x B z E y E z + c B y B z Ez c Bx + By According to part b above, te trace of c F is ten invariant a scalar: Problem 74: 1 c Tr F = 1 c F α βf β α = E c B. a From te identity given and te symmetry ε klm = ε lmk we find: a b c = ε ijk ε klm e i a j b l c m = δ il δ jm δ im δ jl e i a j b l c m = ba c ca b. b Te last identity follows immediately from te previous result. Te first identity follows from te property of te vector triple product a b e = a b e = ɛ ijk a i b j e k, wit e = c d. Problem 75: a Since E and B are vectors under rotation, teir tree-space scalar product E B is invariant under rotations. It tus suffices to prove tat it is also invariant under a boost. It ten suffices to consider a boost in te x-direction, L x β, as te result for an arbitary boost can ten be obtained by an additional rotation. Since L 1 β = L β, we ave from Goldstein eq and te fact tat F must transform as BsC in problem 73 above v = cβ: γ γβ E x E y E z γ γβ 0 0 F = 1 γβ γ 0 0 E x 0 cb z cb y γβ γ 0 0 c E y cb z 0 cb x E z cb y cb x γ 1 β E x γe y cβb z γe z + cβb y = 1 γ 1 β E x 0 γ βe y + cb z γ βe z cb y. c γe y cβb z γβe y cb z 0 cb x γe z + cβb y γβe z + cb y cb x 0 3

4 We can ten read off te transformed fields as: E x = γ 1 β E x = E x, E y = γe y vb z, E z = γe z + vb y, B x = B x, B y = γ B y + v c E z, B z = γ B z v c E y. By taking te parallel and perpendicular components of te results given on te problem seet, on finds identical results. Te identity γ β = γ 1 may be useful in proving tis. b Using te identity proven in problem 74b, one finds: ] E B = γ [E γ B γ + 1 γ β γ + 1 β Eβ B 1c v E v B ] = γ [E γ B γ + 1 γ 1 γ + 1 β Eβ B β E B + β Eβ B = E B. It would ave sufficed to prove tis for a boost in a specific dirtection, as te result is obviously rotation invariant. Note tat tis result means tat if E and B are perpendicular in some inertial coordinate system, as tey are for electromagnetic radiation, tey are perpendicular in any frame. c Tis follows trivially from part a. If E vanises in some coordinate system, we ave E c B = c B < 0 in tat frame, and ence in any frame, contrary to te assumption. Problem 76: Exam problem, 014 spring. See separate solution seet. Problem 77: a Since we ave a particle moving in a central force field, angular momentum is conserved, and te motion takes place in a plane, wic we take to be te xy-plane. Introducing polar coordinates r, θ in te usual manner, we can write te Lagrangian as: L = T V = 1 mṙ + 1 mr θ + k r e r/a. Since L is independent of θ, te conjugate momentum, l = p θ, is conserved: Te radial Euler-Lagrange equation ten reads: d L dt ṙ = L r l = p θ = L θ = mr θ. = m r = mr θ k r 1 + r e r/a = dv a dr. Here V r is te effective potential, obtained after first replacing θ by te constant l: V r = V r + l mr = k r e r/a + l mr. Since L is independent of time, te corresponding energy, E, given by te usual expression for te Hamiltonian, is conserved: E = T + V = 1 mṙ + V r = 1 mṙ + l mr k r e r/a. Te discussion of te orbits for fixed E and l ten exactly follows tat of te Kepler problem. A turning point, r t, of an orbit is a point were te radial motion vanises, i.e.: ṙ = 0 r=rt = 0 E = V r t = l mr t 4 k r t e rt/a.

5 We cannot solve tis equation analytically for r t, but a figure sowing V r, like fig. 3.3 in Goldstein, immediately gives te solution. If E 0 tere is only one turning point, and we ave only unbound scattering solutions. Tis is confirmed by noting tat since V r 0 as r, so te particle can only reac infinity if E 0. If V m < 0 is te minimum of V r, we see tat: E = 1 mṙ + V r > V r > V m. Hence, if E < V m < 0 tere is no solution. If V m < E < 0 te figure will sow two turning points, an inner and an outer, and te particle will move in a bound orbit between te two. [Note tat te orbit will not be closed.] Finally, if E = E 0 = V m, te inner and te outer turning point will coincide, and te particle will move in a circular orbit wit r t = ρ, were ρ is te solution of: dv = l dr r=ρ mρ 3 k 1 ρ + ρ e ρ/a = 0. a Again, we cannot find an analytical solution. However, for later use, we note tat we can write tis equation as: k ρ e ρ/a = a l a + ρ mρ. [By inserting tis into te expression for E, one finds tat te energy of te circular orbit can be written: E 0 = V ρ = a ρ l a + ρ mρ. One sees tat E 0 < 0 only if ρ < a. For ρ > a, wic is possible wit te rigt combination of E end l, te circular orbit is a potentially unstable solution, as it is energetically allowed for te particle to escape to infinity]. b If we insert r = ρ + δ in te equation of motion, and exploit tat dv m/dr = 0 for r = ρ, we find by Taylor expanding te equation of motion around r = ρ: m δ = dv dr dv d V r=ρ dr r=ρ dr δ + Oδ mω δ. Tis is an armonic differential equation for δ wit angular frequency ω given by: ω = 1 d V r=ρ m dr = k [ mρ 3 + ρ ρ ] a + e ρ/a + 3l a m ρ 4 = l a + aρ ρ m aρ 4 = a + aρ ρ ω0, a + ρ aa + ρ were ω 0 = l/mρ is te angular velocity for te circular orbit. If we assume tat te particle is at its outer turning point apoapsis at t = 0, te boundary condition is δ0 = δ 0, te amplitude of te oscillations about te circular orbit. Te solution is ten δt = δ 0 cos ωt, so: As for te angular motion, we find: φ = rt ρ + δt = ρ + δ 0 cos ωt. l mρ + δ = ω δ/ρ ω 0 1 δ + O ρ δ. Since δt is known, tis can be immediately integrated, wit te boundary condition φ0 = 0, and we find: φt = ω 0 t δ 0 sin ωt. ωρ Tus after a time τ = π/ω, te particle will be at a distance rτ = r + δ 0 = r0 again, but at an angle: φτ = ω 0 τ = π ω 0 ω. Te difference in angle from precisely a full revolution is: ω0 φ = φτ π = π ω 1, Tis is te advance of te apsides. In te limit ρ/a 0, i.e. in te case were te circular orbit is in a region were te potential deviates only sligtly from a 1/r-potential, we find, wit ξ = ρ/a: φ = π 1 + ξ 1 + ξ ξ 1 Tere is a printing error in tis answer in Goldstein. 5 ρ ξ 0 πξ = π ρ a.

6 Problem 78: We cose coordinates suc tat te poton initially moves along te z-axis wile te electron is at rest. Tus teir initial are p µ γ = [/λ, 0, 0, /λ], and p µ e = [mc, 0, 0, 0] respectively. Te total conserved four-momentum is ten: p µ = p µ γ + p µ e = [mc + /λ, 0, 0, /λ]. Since te total momentum is conserved, we can cose te x-axis suc tat bot particles moves in te xz-plane after te collision. We denote te polar angles after te collision of te poton θ, tat of te electron φ, and let p be te electron s s final tree-momentum. Ten te momenta of te two particles after te collision can be written: p µ γ = [/λ, /λ sin θ, 0, /λ cos θ], p µ e = [ m c + p, p sin θ, 0, p sin θ]. Te conservation of four-momentum i.e. energy-momentum conservation, p µ = p µ γ + p µ e ten yields tree non-trivial equations: sin θ + p sin φ = 0, λ λ cos θ + p cos φ = λ, λ + m c + p = mc + λ. From te first of tese, we ave: sin φ = λ sin θ = cos φ = 1 + p λ sin θ. p Combining te momentum conservation in te spatial directions we also find: p = p sin φ + p cos φ = On te oter and, from energy conservation, we ave: p = From tese two expressions for p, we find: λ sin θ + λ λ cos θ = λ + λ λλ cos θ. λ 1 λ + mc m c = λ 1 1 λ + mc λ 1 λ. 1 λ + 1 λ λλ cos θ = 1 λ λλ + 1 mc + λ λ λ λλ = cos θ = 1 mc λ λ. Using te identity cos θ = 1 sin θ/ and solving for λ λ, we find te wanted result: λ λ = λ C sin θ, were λ C = mc/ is called te Compton wavelengt. Note tat λ > λ, a poton always losses energy wen colliding wit a stationary carged particle, as some energy is transferred to te particle. To find te electron energy, T, after te collision, we rewrite energy conservation as mc + /λ = γmc + /λ. We ten find, since ν = c/λ, te frequency of an electromagnetic wave of wavelengt λ: 1 T = γ 1mc = c λ 1 λ λ λ = c λλ + λ λ = c λ C sin θ λ λ + λ C sin θ = ν λ Cλ sin θ 1 + λ Cλ sin θ. 6