Special Relativity Jan is a railway worker working for South African - PDF document

12/17/19 Test your understanding of simultaneity Te Special Relativity Jan is a railway worker working for South African Railways. He has ingeniously synchronised the clocks on all South Africa’s Presentation to UCT Summer School January 2020 railway stations. Motsi is on a high-speed train travelling from (Part 2 of 3) Cape Town to Johannesburg. As the train passes De Aar at full speed, all the clocks strike noon By Rob Louw According to Motsi when the Cape Town clock strikes noon, what time is it in Johannesburg? (a) noon? (b) before noon? (c) after noon? roblouw47@gmail.com 1 2 1 2 Te Test your understanding of Einstein’s second postulate Ti Time e Dilation and Loren entz gamma ( 𝛿 ) As a very high-speed rocket ship flies past you it fires a flashlight that shines light in all directions An observer aboard the spaceship observes a wave front that spreads away from the spaceship at speed c in all directions What is the shape of the wave front that an earth observer measures a) spherical, b) ellipsoidal with the longest side of the ellipsoid along the direction of the spaceship's movement c) ellipsoidal with the shortest side of the ellipsoid along the direction of the spaceship’s movement d) neither of these? Is the wave front centered on the spaceship? 4 3 4 1

12/17/19 Ti Time e Dilation Th Though ght Ex Exper erimen ent In order to gain a better understanding of what is happening, we clearly need to derive a quantitative relationship that allows us to compare time intervals in different frames of The objective of the experiment is to demonstrate: reference That observers measure any clock to run slow if it This will be done using another thought experiment moves relative to them and as the relative speed approaches the speed of light, the moving clock’s change in time tends to zero 5 6 5 6 Imagine we have a train moving close to the speed of light along a straight stretch of railway track Sarah, sitting in a coach, is riding in frame S’ where she Sarah measures the time interval between two events that occur at Sarah the same point in space (a) on her ‘light clock’ between two events that occur at the same point in space (a) Peter Reference frame S’ 7 8 7 8 2

12/17/19 Mirror Mirror Sarah Sarah d d S’ S’ Light source Light source O’ (Event 1 occurs here) O’ (Event 2 also occurs here) 9 10 9 10 Mirror Mirror Sarah Sarah Sarah measures a round trip Sarah measures a round trip d d time of ∆t 0 for the light beam time of ∆t 0 for the light beam S’ S’ Light source Light source O’ (Events 1 and 2 occur here) O’ (Events 1 and 2 occur here) The light beam travels a total distance of 2d in a time of ∆t 0 and since the speed of light = c, d = c∆t 0 /2 11 12 11 12 3

12/17/19 Peter who is stationary observes the same light pulse following a diagonal path Sarah Sarah Event 1 occurs Event 1 occurs here here Source moves from Source moves from here to here here to here Event 2 occurs here 13 14 13 14 Sarah Sarah Event 1 occurs Event 1 occurs here here Source moves from Source moves from here to here here to here (Distance travelled) Event 2 occurs here Event 2 occurs here Peter measures the round-trip time to be ∆t Peter measures the round-trip time to be ∆t 15 16 15 16 4

12/17/19 Pythagorean theorem Py The Pythagorean theorem states that for a right-angle The round-trip distance for the light beam in reference frame S is 2 ℓ triangle, the square of the hypotenuse ( c ) is equal to the sum Sarah of the squares of the remaining two shorter perpendicular Event 1 occurs sides ( a & b ) here Source moves from Thus c 2 = a 2 + b 2 here to here a c ∴ c = 𝑏 $ + 𝑐 $ (Distance travelled) Event 2 occurs here Peter measures the round-trip time to be ∆t b 17 18 17 18 Sarah Sarah d d u∆t/2 Peter Peter 19 20 19 20 5

12/17/19 We would like to have a relationship between ∆t and ∆t 0 that Using the Pythagorean theorem we can calculate ℓ is independent of d (but is dependent on u and c ) 𝑒 $ + (𝑣 ∆t /2) $ ℓ = Remembering that d = 𝑑 ∆t 0 /2, then by substitution we get The speed of light is the same for both observers so the (𝑑 ∆t 0 /2) $ +(𝑣 ∆t /2) $ ∆t = 2/c round-trip time measured in S is ∆t where Squaring this equation and then solving for ∆t we finally get 𝑒 $ + (𝑣 ∆t/ 2 ) $ ∆t = 2 ℓ /c = 2 /c 1 − 𝑣 $ /𝑑 2 ∆t = ∆t 0 / 21 22 21 22 Since the quantity 1 − 𝑣 $ /𝑑 2 is less than 1, ∆t is always Note that 𝛿 is always ≥ 1 and 1 / 𝛿 is always ≤ 1 ! greater than ∆t 0 If 𝛿 appears in the numerator of any relativistic equation, it Thus Peter measures a longer round-trip time for the light will tend towards infinity as velocity, approaches c pulse than does Sarah Conversely if 𝛿 appears in the denominator of any relativistic equation, it will tend towards zero as velocity, u approaches c The quantity 1/ 1 − 𝑣 $ /𝑑 2 appears so often in relativity that it has its own symbol 𝛿 and is referred to as Lorentz gamma 𝛿 = 1/ 1 − 𝑣 $ /𝑑 2 Lorentz gamma factor 23 24 23 24 6

12/17/19 ∆t = ∆t 0 / 1 − 𝑣 $ /𝑑 2 = 𝛿 ∆t 0 and thus ∆t ≥ ∆t 0 ∆t 0 is called the proper time and is equal to the time interval between two events that occur at the same position The stretching out of time of the time interval is called time Only one inertial frame (S’) measures the proper time and it dilation does so with a single clock that is present at both events The equation Above tells two things: Firstly, if it were possible to travel faster than the speed of An inertial reference frame moving with velocity u relative to light then 1 – u 2 /c 2 would be negative and 1 − 𝑣 $ /𝑑 2 would the proper time frame must use two clocks to measure the time interval: One at the position of the first event and one at be an imaginary number. We don’t have imaginary time! the position of the second event Secondly, a time dilation plot of ∆t/∆t 0 as a function of relative velocity, u will tend to infinity as u approaches c (or By rearranging our earlier equations, the time interval in the in other words as u/c approaches one) frame where two clocks are required is as follows This is illustrated graphically in the following slide 25 26 25 26 Time Ti e dilation on sometimes described by saying that moving clocks run slow. ∆t/∆t 0 = 𝛿 This must be interpreted carefully 8 As u approaches c, 7 ∆t/∆t 0 = 𝜹 = 1/ √(1− u 2 /c 2 ) 𝜹 approaches Time dilation is sometimes described by saying that moving 6 infinity clocks run slow. This must be interpreted carefully 5 4 The whole point of relativity is that all inertial frames are 3 equally valid so there is no absolute sense in which a clock is moving or at rest 2 1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Speed u relative to the speed of light ( u/c ) 27 28 27 28 7

12/17/19 To illustrate this point, this image In this sense the moving clock (the shows two firecracker explosions one that is present at both events) i.e. two events that occur at ‘runs slower’ than the the clocks different positions in the ground that are stationary with respect to frame both events Assistants on the ground need two clocks to measure the time interval More generally, the time interval ∆t between two events is smallest in In the train reference frame the reference frame in which the however a single clock is present at two events occur at the same both events, hence the time position interval measured in the train reference is the proper time ∆t 0 29 30 29 30 In deriving the time dilation equation we made use of a light clock which made our analysis clear and easy Faster than the speed of light? Fa The conclusion is about time itself Any clock, regardless of how it operates (e.g. a grandfather clock, a wind-up wristwatch, digital watch, alarm clock or a super accurate quartz clock) behaves in the same way! 31 31 32 8