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Chapter 35: Relativity

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Chapter 35: Relativity Bauer/Westfall: University Physics, 1E Chapter 35: Relativity In-Class Exercises 35.1. c, d, e 35.2. a, c, d, e 35.3. e 35.4. a) True b) False c) True 35.5. a 35.6. b, c, d 35.7. a 35.8. d Multiple Choice 35.1. a 35.2. d 35.3. c 35.4. d 35.5. a 35.6. d 35.7. c 35.8. c Questions 35.9. A direct corollary of Einstein’s special theory of relativity postulates that no entity or interaction in the universe can propagate with a speed greater than the speed of light in vacuum. Therefore, instantaneous effects of events originating at one point in space on another point in space are impossible. The translational motion of a perfectly rigid object would imply that, by moving one end of the object, the other end of the object would also move instantaneously, without any time delay. This contradicts Einstein’s theory. 35.10. The y-axis is the time given in μs and the x-axis is the ‘distance’ x / c given in units of μs also, since the speed of light can be written c 3.00 1082 m/s 3.00 10 m/μs 0.300 km/μs. To hit the target, the world line from t 13 μs of the person (Eddie and/or Martin) must lie inside the past light cone of the target at x 0 and t 0. As seen in the diagram, Eddie’s world line is inside the past light cone of the target from t 13 μs to t 2 km / 0.3 km/μs 20 / 3 μs and so Eddie could hit the target. However, Martin’s world line lies outside of the light cone for all time after t 13 μs and so he could not have hit the target. Eddie and Martin find out the target has been hit at the point where their individual world lines intersect the light cone from the target at the origin at some time after the target is hit at t = 0. As shown in the diagram, Eddie finds out the target has been hit at t 20 / 3 μs and Martin finds out it has been hit at t 5km/0.3km/μs 50/3 μs 1234 Chapter 35: Relativity 35.11. If the lens was situated perfectly, there would be indeed be a halo, since the alignment is typically not exact, we see arcs instead. Likewise, the curvature is a result of the mass, so if the object does not have a uniform mass distribution, different rays would be affected non-uniformly. 35.12. In the relativistic limit, velocities must be added relativistically (using the Lorentz transformation), not classically (using the Galilean transformation), as your friend is suggesting. Let F be the frame of the rocket and F be the frame of the Earth. The torpedo has a speed of uc'2/3 with respect to the rocket (frame F ) and the rocket travels at a speed of vc 2/3 with respect to Earth (frame F ). According to the Lorentz transformation the velocity, u, of the torpedo in the Earth’s frame is uv'12 2/3cc 2/3 uc . 1'/ vu c2 14/9/ cc2213 This is less than the speed of light, so no violation of the theory of relativity occurs. 35.13. Yes, the observer still sees the positive charge attracted to the wire. If the positive charge is moving, with velocity v in the lab frame, parallel to the current, then it is actually moving anti-parallel to electrons, which have velocity u in the lab frame. Since the positive charge sees only a magnetic field, this must mean that the wire is electrically neutral, i.e. there are equal positive charges (ion cores) per unit length as there are negative charges per unit length. When the wire is seen in the reference frame of the positive charge, the positive charge is stationary while the ion cores are moving away from the positive charge with velocity v. The electrons are also moving away from the positive charge with a velocity uv uv'. 1/ vu c2 Both the electrons and ion cores have their separation contracted due to their velocities. Since the electrons are; however, moving faster than the ion cores, their separation is smaller than the separation of 1235 Bauer/Westfall: University Physics, 1E the ion cores, meaning the positive charge now sees a net electric charge in any given length of wire and is therefore, attracted to the wire via an electric force instead of the magnetic force in the lab frame. 35.14. The pilot of the rocket sees the garage length contracted. At the speed of the rocket the value of is: 1/2 1/2 2 v2 0.866c 11 2. cc22 The rocket pilot therefore thinks that the garage has a length that is reduced by the factor of 2; that is, LL/2 / /4. 35.15. Since the rod makes an angle with the x-axis, it has a projected length on both the x and y axes. Since the velocity is in the x-direction, only the projection of the length on the x-axis will be contracted, meaning the y-projection length remains unchanged. Since the angle is given by tan1 yx / , as x decreases, the angle increases as viewed by an observer on the ground. 35.16. The primary reason that this presents no contradiction is that the two observations are made in reference frames that are not equivalent. As such, the measurements cannot be directly compared simply by making comparison of observed dimensions. The Earth’s shape is distorted from the usual spherical shape due to the fact that length contraction that occurs in the direction of the observers motion only – perpendicular to the axis of rotation for the first astronaut and along the axis of rotation for the second astronaut. If the two observers really want to compare what they’ve seen, they must exchange information that includes their own relative speed and direction with respect to the Earth. 35.17. The Lorentz transformation for the positions relating the coordinates in the moving frame (primed coordinates) to our reference frame (unprimed coordinates) takes the form x ', xvt with and z-coordinates unchanged, and given by 1/2 v2 1.2 c Hence, the moving clock at x' 0 has coordinate x vt and the clock at x' l has coordinate x vt l /. The time readings are then related by the Lorentz transformation, vx tt'. c2 For the clock at x' 0 the reading is vvt vtt2 1 tt'1. ttt ttt 0 22 2 2 cc For the clock at x' l the reading is vvtl / vvl2 1 vltlv tt'1. tt t t 1 222222 ccccc These results display two important effects. First, time dilation is apparent, as the advance of the t ' values is slowed compared to the advance of t by factor 1/ . Second, relativity of simultaneity is also manifest, as the readings on the moving clocks – which are synchronized in their own reference frame – differ by lv/ c2 at fixed time t in our reference frame. The clock behind in position is “ahead” in time reading. That is, “the same time” at different positions is a reference-frame-dependent notion. This effect is often overlooked, but most purported relativistic kinematics are resolved unambiguously once it is take into account. 1236 Chapter 35: Relativity 35.18. Velocities are added using the relativistic velocity transformation. Assume that the velocities are along the x-axis. Then the transformation equation is uv x y uc , 1/ uv c2 1 xy where x and y represent the fractions of the speed of light of the two sub-light velocities being added. Now, since x 1, it follows that x2 1. Multiply both sides of this inequality by 1 y2 (which is positive since y 1 ), to obtain x2211.yy 2Expand, and add the negative terms to the opposite sides to get xy221. x 22y Subtract 2xy from both sides, to yield: xxyy22212. xyxy 22Factoring both 22 sides as squares gives the inequality: x yxy1 . Divide both sides by the right-hand side (which 2 xy is positive since xy 1 ) which results in the inequality 2 1. Taking square roots of both sides 1 xy xy preserves the inequality (with absolute values), so 1. It follows that the velocity added 1 xy xy xy relativistically is still less than c, since uccc . 11xy xy 35.19. Classically, conservation of kinetic energy in an elastic collision for identical particles of mass m means that 111 mv2220, mv mv 2221i 1f 2f Where v1i is the velocity before the collision and v2i and v2f are the velocities after the collision. If the particles have the same mass this reduces to vvv222, which can only be true if the velocities are 1i 1f 2f perpendicular (since conservation of momentum requires also that vvv1i 1f 2f ). Let the energy and momentum of the originally moving particle be E and p. Let the two particles have total energies after the collision of E1 and E2 , and momenta after the collision of p1 and p2 , respectively. Energy- momentum conservation implies the relationships: 2 EmcE12 E pp12 p. The term Epc222 is a scalar invariant so it is the same before and after the collision, implying: 2 22222 2 2 Emc pc E12 E c p 12 p c 2 2 24 22 2 2 22 22 2 E222 Emc mc pc E112212 EE E pc pc ppc 12 222 2 24222222 2 E pc222 Emc mc E11 pc E 22 pc EE 1212 ppc Using the term Epcmc22224, this reduces to 24 2 24 2 22mc Emc 222 mc EE12 ppc 12 22 Emc E12 E p 12 p c Hence, the dot product of the momenta p1 and p2 is given by 22 ppc12 EE 12 Emc 22 EEmcE11 Emc.
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