Principle of special relativity Their is inconsistency between EM and Newtonian mechanics, as discussed earlier Einstein proposed SR to restore the inconsistency between the two based on two postulates: 1. 11The laws of physics are the same in all inertial reference frames – a generalisation of Newton’s relativity 2. The speed of light in vacuum is the same for all observers independent of the motion of the source – constancy of the speed of light Einstein believes that pure thought is sufficient to understand the world

Postulate 2 simply means that Galilean transformation cannot be applied on light speed. It also explains the Null result of the MM experiment Speed of light is always the same whether one is moving or stationary wrp to the source – its speed doesn’t increase or reduced when the light source is moving

 The notion of absolute frame of reference is discarded (because we could never measure its presence, according to postulate 1) – absolute space is discarded  The Newton notion that time is absolute and flows independently of the state of motion (or the frame of reference chosen) is radically modified – the rate of time flow does depends on the frame of reference.  Simultaneity in one frame is not guaranteed in another frame of reference Einstein’s notion of space-time drastically revolutionarizes that of Newton’s

Time dilation as a consequence of Einstein’s postulate In frames that are moving wrp to the stationary frame, time runs slower We will see how this fact arises in a Gedankan experiment In the Gadankan experiment mentioned below, time and space are measured in terms of light speed (since only c is invariant)

5 O’ (S’ frame in this figure) frame is moving wrp to O frame (S frame) at a relative speed of u Light source is stationary in O’ frame

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Time dilation Due to constancy of light postulate, both observer must agree on c: For observer in O’, c = 2 d /  t’ For observer in O, c = 2 l /  t, where l 2 = d 2 + (u  t/2 ) 2 Eliminating l and d,  t=  t’, where  u 2 /c 2   Lorentz factor, always > or equal 1, so that  t > =  t’,

Proper time Try to discriminate between two kinds of clocks here:  t’, proper time that measures the time interval of the two events at the same point in space (e.g. light emitted and received at the same point in the vehicle) Proper time is the time measured by a clock that is stationary wrp to the events that it measures Note that proper time is always ``shorter’’ compared to non-proper time

The elapsed time  t between the same events in any other frame is dilated by a factor of  compared to the proper time interval  t’ In other words, according to a stationary observer, a moving clock runs slower than an identical stationary clock Chinese proverb: 1 day in the heaven = 10 years in the human plane

10 Example The watch of a student in the class is running at a rate different than that of a student ponteng class to lumba motosikal. The time of the student on the bike ’s is running at a slower rate compared to that of the student in the class

11 To recap The rate of time flowing in one frame is different from the others (frames that are moving with a constant speed relative to a give frame) The relationship between the time intervals of the two frames moving at an non-zero relatively velocity are given by the time dilation formula One must be clear to differentiate which is proper time and which is not

12 Example When you are measuring the time interval between your heartbeats (on your bed in you bedroom) using your watch, you are measuring the proper time interval Say a doctor who is in a car traveling at some constant speed is monitoring your heartbeat by some remote device. The time interval between the heartbeat measured by him, is not a proper time because he is moving wrp to you

Example A spacecraft is moving past the Earth at a constant speed 0.92c. The astronaut measures the time interval between successive ``ticks'' of the spacecraft clock to be 1.0 s. What is the time interval that an Earth observer measures between ``ticks'' of the astronaut's clock?

solution  t’ = 1.0 s is the proper time interval measured by the astronaut Earth observer measures a greater time interval,  t, than does the astronaut, who is at rest relative to the clock The Lorentz factor  u 2 /c 2  -1/2 =   -1/2 = 2.6 Hence,  t =  t’ = 2.6 x 1.0s = 2.6 s

Example: Muon decay lifetime a muon is an unstable elementary particle which has a   = 2 microsecond average lifetime (proper time, measured in the muon rest frame) and decays into lighter particles. Fast muons are created in the interactions of very high-energy particles as they enter the Earth's upper atmosphere. Observation has verified the relativistic effect of time dilation – the muon lifetime as seen by an stationary observer is longer than 2 microsecond

A muon travelling at % the speed of light. has a Lorentz factor  = 22 Hence, to an observer in the rest frame (e.g Earth) the lifetime of the muon is  x   = 22 x 2 microseconds = 44 microseconds Thus the muon would appear to travel for 44 microseconds before it decays The distance it traversed as seen from Earth is D ' = (0.9999c) x 44  s = 13 km If relativistic effect of time dilation not taken into account, it would appear to travel only for a distance c x 2  s = 0.6 km

Twin paradox Two identical twin, E and S, undergo a controlled experiment. S sets out on a journey towards a star many light-years away from Earth in a spaceship that can be accelerated to near the speed of light. After reaching the star, S immediately returns to Earth at the same high speed. Upon arrival on Earth, would S sees his twin brother E (a) becomes much older (b) becomes much younger (c) is still the same age as himself? General theory of relativity effect has to be taken in to account

20 Length contraction Length measured differs from frame to frame – another consequence of relativistic effect Gedankan experiment again!

21 Two observers: O on Earth, O’ traveling to and fro from Earth and alpha centauri with speed u Total distance between Earth - alpha centauri – Earth, according to O (Earth observer), = L 0 O sees O’ return to Earth after  t 0 Observer O’ in a spaceship is heading AC with speed u and returns to Earth after  t’ according to his clock

22 Use some simple logics… In O: 2L 0 = u  t 0 In O’: 2L 0’ = u  t 0’ Due to time dilation effect,  t 0’ is shorter than  t 0, i.e.  t 0 >  t 0’  t 0 is related to  t 0’ via a time dilation effect,  t 0’ =  t 0 / , hence L 0’ / L 0 =  t 0’ /  t 0 = 1 / , or

23 L 0’ = L 0 /  L 0 is defined as the proper length = length of object measured in the frame in which the object (in this case, the distance btw Earth and AC) is at rest L 0’ is the length measured in the O’ frame, which is moving wrp to the object The length of a moving objected is measured to be shorter than the proper length – length contraction

24 If an observer at rest wrp to an object measures its length to be L 0, an observer moving with a relative speed u wrp to the object will find the object to be shorter than its rest length by a foctor 1 / 

25 A stick moves to the right with a speed u. (a) The stick as viewed by a frame attached to it (b) The stick as seen by an observer in a frame at rest relative to the stick. The length measured in the rest frame is shorter than the proper length by a factor 1/ 

26 Length contraction only happens along the direction of motion In 3-D, the length contraction effect is a shortening of length plus a rotation

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28 An observer on Earth sees a spaceship at an altitude of 435 moving downward toward the Earth with a speed of 0.97c. What is the altitude of the spaceship as measured by an observer in the spaceship? Solution One can consider the altitude see by the stationary (Earth) observer as the proper length (say, L'). The observer in the spaceship should sees a contracted length, L, as compared to the proper length. Hence the moving observer in the ship finds the altitude to be L = L' /  = 435 m x [1- (0.97) 2 ] -1/2 = 106 m