Midterm Solutions. 1 1 = 0.999c (0.2)

Transcription

1 Midterm Solutions 1. (0) The detected muon is seen km away from the beam dump. It carries a kinetic energy of 4 GeV. Here we neglect the energy loss and angular scattering of the muon for simplicity. a. We wish to determine the time it took the muon to travel as measured by the physicists in the lab frame. In this frame, the distance traveled by the muon is L lab = km. Since the muon has a kinetic energy of 4 GeV and a mass that is larger than that of the electron by a factor of about 00, m µ = MeV/c, we find: KE = (γ 1)m µ c γ = KE m µ c + 1 = = 38.9 (0.1) In this way, we can find the velocity of the muon v = c 1 1 = 0.999c (0.) γ So, the muon is a highly relativistic particle and hence we can safely approximate its velocity as v c. We then have t lab = L lab c = s = s (0.3) Note that the mean muon lifetime is given by τ µ = s, which is smaller than the elapsed time, so that to the lab observers, the muon will have already decayed by the time they detect it. However, they do see it. To understand why, let us look at the travel-time in the muon s frame. b. Let us now measure the travel-time in the reference frame of the muon. In this frame, the muon itself is at rest and sees the detector moving toward it at v c. Due to length contraction, the distance between the beam dump and the detector is shortened by a factor 1/γ in this frame: t µ = L lab γc = ( ) = s (0.4) Alternatively, due to time dilation, the time elapsed on the muon s clock is shorter than the time on the physicists clock by a factor of γ: t lab = γt µ t µ = t lab γ = s (0.5) 1

2 This is the proper time elapsed on the muon s clock in the muon s frame. Note that this proper time, s, is smaller than the decay lifetime of the muon. So, in this frame, it is clear that the muon reaches the detector before it decays.. (0) The Michelson interferometer provided two discoveries of key importance for the Theory of Relativity. a. First, sketch both experiments. The key apparatus involved in each was the Michelson interferometer. Michelson-Morley experiment (dimensions on the order of several meters): Gravitational Wave LIGO experiment (dimensions on the order of several kilometers):

3 b. The 1887 Michelson-Morley experiment tested the validity of the ether theory, which claimed that light was an electromagnetic wave traveling through a special medium which permeated the entire Universe, the so-called luminiferous ether. This suggestion was a natural one, as every other wave which physicists knew about needed a medium in order to propagate. The Michelson-Morley experiment denounced this hypothesis. In the above setup for the experiment, a source emits a monochromatic light beam which travels toward a beam splitter that sends two beams off at 90 angles. These outgoing beams then bounce off mirrors and get sent back to the beam splitter, where they again partly combine before reaching the screen. Due to the wave nature of light, the final outgoing light results in an interference pattern on the screen. The idea was that if one rotated the entire apparatus around at various angles, then supposing that the ether existed, one would observe an interference shift (a pattern of fringes). This experiment measured the speed of light traveling in different directions along different but equal-length paths, while varying the orientation of the apparatus, in order to measure the speed of the ether relative to the Earth. However, none was detected either during the first attempt or half a year later. The correct conclusion from the 3

4 experiment was that the ether, in fact, did not exist and that light traveled through a vacuum at a constant speed c. The 015/16 experiment used the same sort of setup, a modified Michelson interferometer, to detect gravitational waves. In this experiment, physicists successfully detected the gravitational waves generated during the last instances of the merger of two black holes coalescing into a single, more massive spinning black hole. See the above setup. A linearly polarized gravitational wave coming in orthogonally to the plane of the detector and parallel to the 4-km optical cavities will result in the lengthening of one of the 4-km arms and a shortening of the other during one half-cycle of the wave. In the next half-cycle, these length changes will get switched. What was measured here, then, was the gravitational-wave strain as a difference in the length of the detector s orthogonal arms. c. The significance of the Michelson-Morley experiment for Einstein s theory of special relativity was in its demonstration that light traveled through a vacuum at a constant velocity c. Special relativity correctly explained the results of this experiment. The 015/16 experiment, meanwhile, showed that gravitational waves exist, providing support for Einstein s general theory of relativity (GR), which predicted their existence. The existence of gravitational waves was, in fact, the final unconfirmed prediction of GR. This discovery thus verified the theory s predictions of spacetime distortions on large scales. 3. (0) We are told that the electron is accelerated to a kinetic energy of 40 kev in a TV tube. The mass of the electron is given by m e = MeV/c. a. We first wish to determine the velocity of the electron: Computing the Lorentz factor, KE = (γ 1)m e c γ = KE m e c + 1 = = 1.078, (0.6) 511 we find that v e = c 1 1 γ = 0.374c = m/s m/s (0.7) b. Let us next determine the velocity of the proton of the same kinetic energy. The proton has a mass of m p = 938 MeV/c. For this case, the Lorentz factor is γ = KE m e c + 1 = = (0.8)

5 so that the velocity is v p = c 1 1 γ = c = m/s m/s (0.9) c. The case of the electron still requires a relativistic treatment, because the velocity is about 0.37 of the velocity of light, 37%, which is a sizable fraction of c. If we apply Newtonian mechanics to this case to compute the kinetic energy, we find KE = 1 m ev = 1 (511)(0.37 ) = 35.7 kev 36 kev (0.10) Therefore, for a careful calculation, we need to resort to the fully relativistic treatment here. Meanwhile, the proton s velocity is only about 0.9% of c. The proton is heavy and nonrelativistic at this kinetic energy. This case is fully amenable to a Newtonian nonrelativistic treatment. Computing the kinetic energy using the Newtonian formula, we find KE = 1 m pv = 1 ( )( ) = kev 40 kev (0.11) The results for the velocity from the relativistic and nonrelativistic expressions match up to three decimal places. For a precise calculation valid to all decimal places, we need to use the relativistic expression. However, given that the kinetic energy is only known to one significant digit, a nonrelativistic result is correct for all intents and purposes here and is sufficient for a precise calculation in this situation. 4. (0) Consider an inertial reference frame S that moves with respect to the inertial frame S in the positive x direction with a velocity v = 0.9c. We are told that the coordinate systems of S and S are aligned. An electron moves in the positive x direction in the S frame with velocity v xe = +0.85c. a. We wish to calculate the momentum, energy, and mass of that electron in the frame S. Momentum in S frame: p xe = γm e v xe = m e v xe 1 v xe /c = (0.511)(0.85) MeV/c = 0.8 MeV/c = ev/( m/s) J 1 ev = kg m/s All other components are zero, p ye = p ze = 0, because the electron is moving purely in the x-direction. 5

6 Energy: E e = γm e c = MeV = 0.97 MeV = J (0.1) Mass: The mass of the electron is an invariant quantity. It does not change from frame to frame. So, the mass is simply or in particle physics units, m e = kg (0.13) m e = MeV/c (0.14) b. Let us now use Lorentz transformations to calculate the momentum and energy of the electron in the frame S, which moves with respect to S at a velocity v = 0.9c. The associated Lorentz factor is γ = 1 1 v /c = =.3 (0.15) For the momentum, we then have ( ) ( ) p xe = γ v p xe ve e /c = MeV/c 0.9(0.97 MeV)/c (0.16) = 0.11 MeV/c = kg m/s (0.17) and for the energy ( ) ( ) E e = γ v E e vp xe = MeV 0.9(0.8 MeV) = 0.5 MeV = J (0.18) Alternatively, we can find the velocity of the electron in this frame by applying the velocity-addition formula as follows: u = and use this velocity to determine v xe v 0.85c 0.9c = 1 v xe v/c 1 (0.85)(0.9) = 0.1c (0.19) p xe = γ u m e u = 0.11 MeV/c = kg m/s (0.0) 6

7 and E e = γ u m e c = 0.5 MeV = J (0.1) c. The mass of the electron is invariant. Therefore, it did not change from frame S to frame S. 5. (5) Gravitational waves are transverse waves of spatial strain that propagate at the speed of light. They are (d) traveling distortions in the curvature of spacetime. 7