PHYS 561 (GR) Homework 1 Solutions

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1 PHYS 561 (GR) Homework 1 Solutions HW Problem 1: A lightweight pole 20m long lies on the ground next to a barn 15m long. An Olympic athlete picks up the pole, carries it far away, and runs with it toward the end of the barn at speed 0.8c. His friend remains at rest, standing by the door of the barn. Attempt all parts of this question, even if you can t answer some. a. How long does this friend measure the pole to be, as it approaches the barn? b. The barn door is initially open, and immediately after the runner and pole are entirely inside the barn, the friend shuts the door. How long after the door is shut does the front of the pole hit the other end of the barn, as measured by the friend? Compute the interval between the events of shutting the door and hitting the wall. Is it spacelike, timelike, or null? c. In the reference frame of the runner, what is the length of the barn and the pole? d. Does the runner believe that the pole is entirely inside the barn when its front hits the end of the barn? Can you explain why? e. After the collision, the pole and the runner come to rest relative to the barn. From the friend s point of view, the 20m pole is now inside a 15m barn, since the barn door was shut before the pole stopped. How is this possible? Alternatively, from the runner s point of view, the collision should have stopped the pole before the door closed, so the door could not be closed at all. Was or was not the door closed with the pole inside? f. Draw a spacetime diagram from the friend s point of view and use it to illustrate and justify all your conclusions. v c = 0.8, γ = 1 1 v 2 /c 2 = 5 3. a: The friend sees the pole contracted by γ: L = L γ = (20m)(3/5) = 12m. b: When the friend sees the barn door is shut, the front of the pole is d = 15m 12m = 3m from the back of the barn, traveling at 0.8c. Thus the time passed is t = d 0.8c = 3m = 12.5 ns. m/s where 1 ns = 10 9 s. For the requested interval, the spatial separation x = 15m, while the temporal separation is given above. Thus, c 2 ( t) 2 < ( x) 2, indicating a space-like separation. Assuming a metric signature (+,,, ), the spacetime interval is ( s) 2 = c 2 ( t) 2 ( x) 2 = ( ) 3m 2 (15m) 2 ; 0.8 ( s) 2 = m2 = m 2 The proper distance σ = ( s) 2 is given by 3375 σ = m = m. 16 c: The runner sees the pole at rest, so it is at its rest length, 20m. The runner sees the barn moving at 0.8c, and its 15m rest length seen contracted by γ: L = L γ = (15m)(3/5) = 9m. d: The runner sees the pole hit the back of the barn before the pole has fully entered the barn. This is a consequence of the relativity of simultaneity: There is no absolute time ordering for spacelike separated events. Calling the two spacelike separated events A and B, we can always find observers at different relative speeds who see: A before B, B before A, A and B simultaneous (note that the proper distance given above is the spatial distance between A and B in a frame where they occur simultaneously). This is no problem, since there is no causality between spacelike separated events.

2 e: An extended object cannot have an instantaneous reaction over its full length that is triggered by a single event: The fastest the reaction can propagate would be at the speed of light, though in reality the reaction is much slower than that. This undermines the normal intuition of extended rigid objects, but we can still accomodate such a situation. The pole stopping is triggered by its front touching the back of the barn. The back of the pole can t be affected by this event until (at best) enough time has passed for light from this event to reach it. Thus, for a time, the back of the pole continues forward at 0.8c despite the fact that the front of the pole is at rest (!). Thus the pole will compress. This allows the pole to be in the barn with the door closed in both cases. was inside before it reached the back of the barn. f: Two diagrams are drawn to better illustrate the different views. A view at rest with respect to the barn, and a view from an observer travelling at 0.8c to the right (in the pole s rest frame). The horizontal (space) axis is in meters, while the vertical (time) axis is not shown. The grey lines refer to the pole, while the black refers to the barn. The dotted line shows a light signal travelling from the pole-barn collision to the back of the pole. It s clear that in both cases the barn door closes with the pole inside, though they disagree on if the pole HW Problem 2: The 4-velocity u = γ(c, v) corresponds to the 3-velocity v. Express: a. u 0 in terms of v u 0 = γc = c 1 v 2 /c 2 ; v = v. b. u β (β = 1, 2, 3) in terms of v u β = γ v = v 1 v 2 /c 2. c. u 0 in terms of u β u 0 = γc = c 1 + u 2 /c 2 ; u 2 = (u β ) 2 = γ 2 v 2. d. d/dτ in terms of d/dt and v dt = γdτ; d dτ = γ d dt = 1 d 1 v 2 /c 2 dt ; e. v β in terms of u β v β = v = uβ γ = u β 1 + u 2 /c 2.

3 f. v in terms of u 0 v = v = c γ 2 1 γ 2 = c 1 1γ 2 ; v = c 1 c2 (u 0 ) 2. HW Problem 3: Find the matrix for the Lorentz transformation consisting of a boost v x in the x-direction followed by a boost v y in the y-direction. Show that boost performed in the reverse order would give a different transformation. Define β a = v a /c, γ a = (1 βa) 2 1/2 for a = x, y. The matrix for the Lorentz transformation in the x-direction is: γ x γ x β x 0 0 Λ x = γ x β x γ x The matrix for the Lorentz transformation in the y-direction is: γ y 0 γ y β y 0 Λ x = γ y β y 0 γ y 0. The matrix for the Lorentz transformation first in the x-direction then in the y-direction is: γ x γ y γ x β x γ y γ y β y 0 Λ xy = γ x β x γ x 0 0 γ x γ y β y γ x β x γ y β y γ y 0. We can also express this as: Λ xy = Λ y Λ x. The matrix for the Lorentz transformation first in the y-direction then in the x-direction is: Or, in matrix form, Λ yx = Λ x Λ y. γ x γ y γ x β x γ x γ y β y 0 Λ yx = γ x β x γ y γ x γ x β x γ y β y 0 γ y β y 0 γ y 0. We see that Λ yx Λ yx = (Λ xy ) T Λ xy, but we do have that HW Problem 5: Frame S moves with velocity β relative to frame S. A bullet in frame S is fired with velocity v at an angle θ w.r.t. the forward direction of motion. What is the angle θ as measured in S? What if the bullet is a photon? (c = 1 herein) Let s use coordinates such that β is along the +x direction and v is in the x-y plane, i.e., β = βˆx, v = v (ˆx cos θ + ŷ sin θ ). Also, we have, v = v (ˆx cos θ + ŷ sin θ). The law of transformation of velocities can relate components the bullet velocity in frame S (namely, v ) with that in frame S (namely, v): v x = v x + β 1 + v xβ, v y = v y 1 β v xβ, v z = v z 1 β v xβ. In our case, this means v cos θ = v cos θ + β 1 + v β cos θ, v sin θ = v sin θ 1 β v β cos θ. Thus, tan θ = v sin θ 1 β 2 v cos θ + β. If the bullet is a photon, v = v = c = 1, and tan θ = sin θ 1 β 2 cos θ + β.

4 HW Problem 7: (Compton scattering) A photon of wavelength λ hits a stationary electron (mass m e ) and comes off with wavelength λ at an angle θ. Derive the expression: λ λ = (h/m e ) (1 cos θ) 4-vectors are the simplest approach. Let p = (ε, p) be a momentum 4-vector, and use subscript γ, e to indicate photon or electron, and unprimed/primed to indicate pre/post scattering. Conservation of momentum gives: p γ + p e = p γ + p e, and for 4-momentum we always have Thus p 2 = ε 2 p 2 = m 2. (p γ + p e p γ) 2 = (p e) 2 = m 2 e. After distributing the product: p 2 γ + p 2 e + p 2 γ + 2p e p γ 2p e p γ 2p γ p γ = m 2 e We know that p 2 γ = p 2 γ = 0, p 2 e = m 2 e, so we can simplify: p e (p γ p γ) p γ p γ = 0 The 4-momenta in more detail are: Thus or p γ = (h/λ)(1, 1, 0, 0) p γ = (h/λ )(1, cos θ, sin θ, 0) p e = (m e, 0, 0, 0). ( 1 m e h λ 1 ) λ h2 (1 cos θ) = 0 λλ λ λ = h m e (1 cos θ). HW Problem 8: a. When a photon scatters off a charged particle, which is moving with a speed very nearly that of light, the photon is said to have undergone an inverse Compton scattering. Consider an inverse Compton scattering in which a charged particle of rest mass m and total mass-energy (as seen in the lab frame) E m, Collides head-on with a photon of frequency ν (hν m). What is the maximum energy the particle can transfer to the photon? b. If space is filled with black-body radiation of temperature 3 K, how much energy can a proton of energy ev transfer to a 3 K photon? (a.) 4-vectors are again the simplest approach. We re interested only in the energy the particle can transfer to the photon, E = E E = h(ν ν). Following problem 7, we have The 4-momenta are: p m (p γ p γ) p γ p γ = 0 p γ = hν(1, 1, 0, 0) p γ = hν (1, cos θ, sin θ, 0) p m = (E, p, 0, 0) where p = E 2 m 2. So E h(ν ν ) + p h(ν ν cos θ) h 2 νν (1 cos θ) = 0. This gives ν ν = E + p E + p cos θ + hν(1 cos θ). In terms of E = h(ν ν), we have [ ] p(1 cos θ) hν(1 cos θ) E = hν E + p cos θ + hν(1 cos θ) This is zero when θ = 0 (no scattering) and maximum when θ = π (full 180 backscattering). The maximum case gives [ E = 2hν p hν E p + 2hν This formula is exact, but we can use E m hν to simplify this further. First note that p = ( m ) 2 E 2 m 2 = E 1 E Expand p/e in powers of m/e: p E = 1 1 ( m ) 2 ( m ) 4 + O. 2 E E ]

5 Thus E E 2hν [ ] 1 m 2 hν 2E 2 E (m 2. /2E) + 2hν We know that 1 m/e hν/e, so the denominator in brackets is 1. So E E E (m 2 /4hν) + E (b.) For this case, m = 938 MeV 10 9 ev., hν kt = ev, and E = ev. So, E E = 0.106, E = ev Plotted below is E/E as a function of E for the collision of a proton and a 3 K photon. Note that for E m 2 /hν the proton can transfer virtually all of its energy to the photon. HW Problem 9: a. If a rocket has engines that give it a constant acceleration of 1g (relative to its instantaneous inertial frame, of course), and the rocket starts from rest near the Earth, how far from the Earth (as measured in the Earth s frame) will the rocket be in 40 years as measured on the Earth? How far after 40 years as measured in the rocket? b. Compute the proper time for the occupants of a rocket ship to travel the 30, 000 light years from the Earth to the center of the galaxy. Assume they maintain an acceleration of 1g for half the trip and decelerate at 1g for the remaining half. c. What fraction of the initial mass of the rocket can be payload in part (b)? Assume an ideal rocket that converts rest mass into radiation and ejects all the radiation out of the back with 100% efficiency and perfect collimation. We ll solve using 4-vectors, keeping only one spatial dimension. The position 4-vector is The velocity 4-vector is x = (t, r), x 2 = t 2 r 2. u = dx ds = (γ, γv), u2 = 1; where we ve used ds = dt/γ and v = dr/dt = ṙ. The acceleration 4-vector is simple in 1-D: w = du ds = γ( γ, (γv)) = (γ 4 av, γ 4 a), w 2 = γ 6 a 2, w 2 = γ 3 a, where γ = γ 3 av and a = v. We also note that γ 3 a = d dt (γv). In the rocket frame, u = (1, 0); w = (0, g); w 2 = g 2. In the fixed frame, u = (γ, v); d dt (γv) = w 2 = g. where the last equality holds since w 2 is invariant. Thus, γv = gt + const. We set v = 0 at t = 0, so Solving for v, we have v(t) = γv = gt gt 1 + g 2 t 2. We can integrate again to get the position: setting x = 0 at t = 0, this gives x(t) = 1 ( 1 + g g 2 t 2 1).

6 We can also get the proper time from v: τ(t) = t 0 1 [v(t)] 2 dt = 1 g sinh 1 (gt). We ll need the inverted form of this too: t(τ) = 1 g sinh(gτ). (Extra: we can also take a derivative of v to get the non-proper acceleration:) a(t) = g (1 + g 2 t 2 ) 3/2 We ll want distances in light years (lyr) and time in years (yr). In natural units, lyr=yr; that means g = 1.03 ly yr 2 = 1.03yr 1 (a.) The distance travelled after 40yr Earth time is x(t 40yr) = lyr The proper time elapsed on the rocket after 40yr Earth time is τ(t 40yr) = 4.28 yr The Earth time elapsed after 40yr proper time on the rocket is t(τ 40yr) = yr 10 7 (age of universe) and the distance travelled is x(τ 40yr) = lyr 10 7 (size of obs. univ.) (answers here will be fairly sensitive to the significant figures retained in g, etc.) (b.) Let τ 1/2 be the half-trip time: the time to travel d = 15, 000 lyr. The total trip time will be 2τ 1/2. The distance traveled in terms of proper time is x(τ) = 1 [cosh(gτ) 1] g Setting x(τ 1/2 ) = d and solving for τ 1/2 : τ 1/2 = 1 g cosh 1 (gd + 1) = The full trip time is T = 2τ 1/2 : T = 2 g cosh 1 (gd + 1) = lyr. (c.) We only need conservation of energy and momentum. Let M be the fuel mass and m be the payload mass. Let ε be the energy of fuel expended, and note that the momentum of this fuel is ε. Then we can relate the initial energy and momentum with that of any later point on the initial leg of the journey: (E i, p i ) = (M +m, 0) = (E f, p f ) = (γm+ε, γmv ε). From momentum we see that ε = γmv, so M = m(γ + γv 1). Inserting the equations for v(t) into v and γ, and inserting the equation for t(τ) where needed, we get a simple final form: M m = egτ 1. The second leg of the journey follows the same law; this is because the rate of fuel burn needed to generate a 1g proper acceleration doesn t depend on the direction of acceleration: It s the same for acceleration and deceleration. So, we simply need to use τ = T for the full trip: M m = egt 1 = The payload is roughly one billionth of the fuel mass.

7 HW Problem 15: Suppose that E B = 0. Show that there is a Lorentz transformation which makes E = 0 if B 2 E 2 > 0, or one that makes B = 0 if B 2 E 2 < 0. What if B 2 E 2 = 0 in addition to E B = 0? The law of transformation for the fields can be written in vector form (Jackson E&M): ( ) E = γ E + β B γ2 β γ + 1 ( ) β E, ( ) B = γ B β E γ2 β γ + 1 ( β B ). We can assume that E B = 0, and that β is parallel to E B. This leads to a simpler equation describing β: β 1 + β 2 = E B E 2 + B 2. There s a single root satisfying 0 β 2 < 1: 2E β = B E 2 + B 2 + (E 2 B 2 ) E B. 2 If E B 0 and B 2 E 2 0, we have an electromagnetic wrench. If E B = 0 and B 2 E 2 > 0, we have a magnetostatic field: β = 2E B E 2 + B 2 + (E 2 B 2 ) = E B 2 B 2 Plugging back into the transformation law, we have: E = 0, (B ) 2 = B 2 E 2. (Note: This situation is related to a velocity selector, a.k.a. Wien filter) If E B = 0 and B 2 E 2 < 0, we have an electrostatic field: β = 2E B E 2 + B 2 + (E 2 B 2 ) = E B 2 E 2 Plugging back into the transformation law, we have: B = 0, (E ) 2 = E 2 B 2. If E B = 0 and B 2 E 2 = 0, we have situation describing transverse electromagnetic waves in vacuum. We will always have E 2 = B 2 and E B in any frame. The velocity formula gives β = 1, i.e. it fails to find a frame where E B = 0.