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 relativity2. 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 experimentSpeed 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 The notion of absolute frame of reference is discarded (because we could never measure its presence, according to (because we could never measure its presence, according to postulate 1) – absolute space is discardedpostulate 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 Simultaneity in one frame is not guaranteed in another frame of referenceframe of reference

Einstein’s notion of space-time drastically Einstein’s notion of space-time drastically revolutionarizes that of Newton’srevolutionarizes 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 slowerWe will see how this fact arises in a Gedankan experimentIn the Gadankan experiment mentioned below, time and space are measured in terms of light speed (since only c is invariant)

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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 dilationDue 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

u2/c2

Lorentz factor, always > or equal 1, so that t > = t’,

Proper timeTry 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 measuresNote 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

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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

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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

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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 intervalSay 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 astronautEarth observer measures a greater time interval, t, than does the astronaut, who is at rest relative to the clockThe Lorentz factor u2/c2-1/2 = 0.922-1/2 = 2.6Hence, 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 99.99 % 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 microsecondsThus the muon would appear to travel for 44 microseconds before it decaysThe distance it traversed as seen from Earthis D ' = (0.9999c) x 44 s = 13 kmIf relativistic effect of time dilation not taken into account, it would appear to travel only for a distance 0.9999c x 2 s = 0.6 km

Twin paradoxTwo 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

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

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Two observers: O on Earth, O’ traveling to and fro from Earth and alpha centauri with speed uTotal distance between Earth - alpha centauri – Earth, according to O (Earth observer), = L0

O sees O’ return to Earth after t0

Observer O’ in a spaceship is heading AC with speed u and returns to Earth after t’ according to his clock

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Use some simple logics…

In O: 2L0 = ut0

In O’: 2L0’ = ut0’

Due to time dilation effect, t0’ is shorter than t0 , i.e. t0 > t0’

t0 is related to t0’ via a time dilation effect, t0’ = t0 / , hence

L0’ / L0 = t0’ /t0 = 1 / , or

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L0’ = L0 / L0 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

L0’ 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

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If an observer at rest wrp to an object measures its length to be L0 , 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 /

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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/

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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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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 thestationary (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