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Time dilation D.3.1Describe the concept of a light clock. D.3.2Define proper time interval. D.3.3Derive the time dilation formula. D.3.4Sketch and annotate a graph showing the variation with relative velocity of the Lorentz factor . D.3.5Solve problems involving time dilation. Option D: Relativity and particle physics D3 Relativistic kinematics

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Time dilation Describe the concept of a light clock. Dobson is standing in a room S 0 having a height L that has a light source on the floor and a mirror on the ceiling. When Dobson turns on the light he can measure the time ∆t 0 it takes the light to travel from the floor to the ceiling and back to the floor again. Since Dobson knows the speed of the light is c 0 and the distance traveled by the flash is L up and L back down, he uses c 0 = 2L/∆t 0 and determines that his stopwatch will measure Option D: Relativity and particle physics D3 Relativistic kinematics (S 0 ) L light clock time in S 0 ∆t 0 = 2L /c 0 This is a “light clock.”

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Time dilation Describe the concept of a light clock. Now suppose Dobson’s room (S 0 ) is traveling to the right at a velocity v while his twin brother Nosbod stands outside the room on the ground (S). Nosbod measures the same flash of light to travel from floor to ceiling and back to the floor to be a time ∆t. (S 0 ) L v Option D: Relativity and particle physics D3 Relativistic kinematics (S) (S 0 ) L v

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Time dilation Derive the time dilation formula. For Nosbod (S) the light beam actually travels a distance of 2D. Thus The base of the triangle is given by v∆t. From the Pythagorean theorem D 2 = L 2 + (v∆t/2) 2 or (2D) 2 = (2L) 2 + (v∆t) 2. Option D: Relativity and particle physics D3 Relativistic kinematics (S) (S 0 ) v L v L D D v∆tv∆t 2 v∆tv∆t v∆tv∆t 2 light clock time in S ∆t = 2D /c D D

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Time dilation Derive the time dilation formula. From (2D) 2 = (2L) 2 + (v∆t) 2 and the two times measured in S 0 and S we can write the following: 2L = c 0 ∆t 0 2D = c∆t (2D) 2 = (2L) 2 + (v∆t) 2 (c∆t) 2 = (c 0 ∆t 0 ) 2 + (v∆t) 2 Option D: Relativity and particle physics D3 Relativistic kinematics light clock time in S ∆t = 2D /c light clock time in S 0 ∆t 0 = 2L /c 0 FYI This last formula gives the relationship between the times measured by the brothers and the speed of light in their respective coordinate systems.

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Time dilation Derive the time dilation formula. According to Newton "Absolute, true, and mathematical time, of itself and from its own nature, flows equably without relation to anything external." Thus for Newton, ∆t = ∆t 0, regardless of speed. From (c∆t) 2 = (c 0 ∆t 0 ) 2 + (v∆t) 2, then, we can cancel the times and we arrive at c 2 = c 0 2 + v 2, which shows that c is greater than c 0. But this is in violation of Einstein’s second postulate, which says that the speed of light is the same in all IRFs. Option D: Relativity and particle physics D3 Relativistic kinematics FYI Thus Newton’s belief in absolute time is incorrect! Time elapses differently in different inertial reference frames!

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Time dilation Derive the time dilation formula. According to Einstein “The speed of light c is the same in all inertial reference frames." Thus for Einstein, c = c 0. Then (c∆t) 2 = (c 0 ∆t 0 ) 2 + (v∆t) 2 becomes (c∆t) 2 = (c∆t 0 ) 2 + (v∆t) 2 (c∆t) 2 - (v∆t) 2 = (c∆t 0 ) 2 c 2 ∆t 2 - v 2 ∆t 2 = c 2 ∆t 0 2 (c 2 - v 2 )∆t 2 = c 2 ∆t 0 2 (1 - v 2 /c 2 )∆t 2 = ∆t 0 2 ∆t 2 = ∆t 0 2 /(1 - v 2 /c 2 ) ∆t = ∆t 0 [ 1/ 1 - v 2 /c 2 ] Option D: Relativity and particle physics D3 Relativistic kinematics time dilation ∆t = ∆t 0 where = 1 1 - v 2 /c 2 The IBO expects you to be able to derive this formula! Sorry :(

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Time dilation Define the proper time interval. From the time dilation formula we see that each IRF has its own time. There is no absolute time! We will explore the differences between the two times in the following examples. But first… We define the proper time ∆t 0 as the time measured in the frame in which the events do not change spatial position. Dobson (S 0 ) is in the proper reference frame for the events (the light beam leaving the floor, and the light beam returning to the floor). Option D: Relativity and particle physics D3 Relativistic kinematics time dilation ∆t = ∆t 0 where = 1 1 - v 2 /c 2 Dobson is in the rest frame of the events. Dobson measures proper time ∆t 0.

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Time dilation Sketch and annotate a graph showing the variation with relative velocity of the Lorentz factor . We call the Lorentz factor. The Lorentz factor shows up in all of the relativistic equations. Option D: Relativity and particle physics D3 Relativistic kinematics time dilation ∆t = ∆t 0 where = 1 1 - v 2 /c 2 EXAMPLE: Sketch and annotate a graph showing the variation with relative velocity of the Lorentz factor . SOLUTION: Note that = 1 when v = 0, and that approaches as v approaches c: Thus v/c = 0 when v = 0 and = 1. Thus v/c = 1 when v = c and = . v/c 0 1 1 Asymptote

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Time dilation Solve problems involving time dilation. Option D: Relativity and particle physics D3 Relativistic kinematics time dilation ∆t = ∆t 0 where = 1 1 - v 2 /c 2 PRACTICE: Suppose S 0 has relative speed of v = 0.75c with respect to S. (a)Find the value of . (b)If Dobson measures the time to cook a 3-minute egg in S 0, how long does Nosbod measure the same event in S? SOLUTION: (a) = [1 - v 2 /c 2 ] -1/2 = [1 - (0.75c) 2 /c 2 ] -1/2 = [1 - 0.75 2 c 2 /c 2 ] -1/2 = [0.4375] -1/2 = 1.5. (b) ∆t = ∆t 0 = 1.5(3 min) = 4.5 min! (S) Note that time elapses more quickly for Nosbod!

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Time dilation Solve problems involving time dilation. Option D: Relativity and particle physics D3 Relativistic kinematics time dilation ∆t = ∆t 0 where = 1 1 - v 2 /c 2 2.2 10 -6 is the proper time because it is measured in the frame of the muon. = 1 /(1 - 0.98 2 ) 1/2 = 1/0.0396 1/2 = 5.025. ∆t = ∆t 0 = 5.025(2.2 10 -6 ) = 1.1 10 -5 s.

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Time dilation Solve problems involving time dilation. Option D: Relativity and particle physics D3 Relativistic kinematics time dilation ∆t = ∆t 0 where = 1 1 - v 2 /c 2 For Earth observer use ∆t = 1.1 10 -5 s. Then d = v ∆t = 0.98(3.00 10 8 )(1.1 10 -5 ) = 3234 m.

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Time dilation Solve problems involving time dilation. Option D: Relativity and particle physics D3 Relativistic kinematics time dilation ∆t = ∆t 0 where = 1 1 - v 2 /c 2 For muon observer use ∆t 0 = 2.2 10 -6 s. Then d = v ∆t = 0.98(3.00 10 8 )(2.2 10 -6 ) = 647 m.

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Time dilation Solve problems involving time dilation. Option D: Relativity and particle physics D3 Relativistic kinematics time dilation ∆t = ∆t 0 where = 1 1 - v 2 /c 2 Time dilation is the relativistic effect that causes time to elapse more slowly in a fast- moving IRF. As the previous example showed, if time dilation did NOT occur, a muon created at 3 km would not live long enough to be detected. But muons ARE detected, providing evidence.

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EXAMPLE: The twin paradox: Suppose Einstein has a twin brother who stays on Earth while Einstein travels at great speed in a spaceship. When he returns to Earth, Einstein finds that his twin has aged more than himself! Explain why this is so. SOLUTION: Since Einstein is in the moving spaceship, his clock ticks more slowly. But his twin’s ticks at its Earth rate. The twin is thus older than Einstein on his return! Time dilation Solve problems involving time dilation. Option D: Relativity and particle physics D3 Relativistic kinematics By the way, this is NOT the paradox. The paradox is on the next slide…

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EXAMPLE: The twin paradox: From Einstein’s perspective Einstein is standing still, but his twin is moving (with Earth) in the opposite direction. Thus Einstein’s twin should be the one to age more slowly. Why doesn’t he? SOLUTION: The “paradox” is resolved by general relativity (which is the relativity of non- inertial reference frames). It turns out that because Einstein’s spaceship is the reference frame that actually accelerates, his is the one that “ages” more slowly. Time dilation Solve problems involving time dilation. Option D: Relativity and particle physics D3 Relativistic kinematics FYI The standard level IB papers do not require more general relativity than is in the twin paradox explanation. General relativity is covered in HL.

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Length contraction D.3.6Define proper length. D.3.7Describe the phenomenon of length contraction. D.3.8Solve problems involving length contraction. Option D: Relativity and particle physics D3 Relativistic kinematics

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Length contraction Define proper length. We define the proper length L 0 as the length of an object as measured in its rest frame. For example, Nosbud is able to measure the proper distance between the cones resting at the ends of the station platform to be L 0 … …but Dobson, on the moving train, is not. Option D: Relativity and particle physics D3 Relativistic kinematics v (S 0 ) (S) L0L0

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Length contraction Describe the phenomenon of length contraction. Both parties agree on the velocity v of the train. Furthermore, the time between the events (train at first cone, then train at second cone) can be timed by both parties. Option D: Relativity and particle physics D3 Relativistic kinematics v (S 0 ) (S) L0L0

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Length contraction Describe the phenomenon of length contraction. As you watch the animation, note that both events occur directly opposite Dobson (S 0 ), but they are spatially separated in Nosbod’s frame (S). Thus Dobson (S 0 ) measures proper time ∆t 0. For Dobson, the platform has length L = v∆t 0. For Nosbod, the platform has length L 0 = v∆t. Option D: Relativity and particle physics D3 Relativistic kinematics v (S 0 ) (S) L0L0

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Length contraction Describe the phenomenon of length contraction. Option D: Relativity and particle physics D3 Relativistic kinematics v (S 0 ) (S) L0L0 PRACTICE: Show that L = L 0 / SOLUTION: Use ∆t = ∆t 0, L = v∆t 0, and L 0 = v∆t. From L 0 = v∆t and ∆t = ∆t 0 we get L 0 = v∆t 0. Since L = v∆t 0 we can then write L 0 = L. Thus Length contraction L = L 0 / Note that this is the same as the Lorentz-FitzGerald contraction.

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Length contraction Solve problems involving length contraction. Option D: Relativity and particle physics D3 Relativistic kinematics v (S 0 ) (S) L0L0 PRACTICE: Nosbod (S) measures the length of the platform to be 15 m and the speed of the train to be 0.75c. What is the length of the platform in Dobson’s IRF (S 0 )? SOLUTION: Use L = L 0 / and = 1.5 (from before). L = (15)/1.5 = 10 m.

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Length contraction Solve problems involving length contraction. Option D: Relativity and particle physics D3 Relativistic kinematics Length contraction is occurring in the muon IRF. Obviously the Earth IRF allows for the muon to make it 3 km (3234 m) and be observed. For the muon, the distance to Earth is just L = L 0 / so that L = 3234/5.025 = 643 m. Length contraction L = L 0 / This is a continuation of a previous problem…

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Length contraction Describe the phenomenon of length contraction. Option D: Relativity and particle physics D3 Relativistic kinematics From L = L 0 / we see that = 2. But = (1 – v 2 /c 2 ) -1/2 so that (1 – v 2 /c 2 ) +1/2 = 0.5. Thus 1 – v 2 /c 2 = 0.5 2 = 0.25. Then v 2 /c 2 = 0.75 so that v = 0.866c.

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