2. Einstein's postulates in special theory of relativity Imagination is more important than knowledge.

In 1905 the greatest Physicist in 20th century emerged with a great solution to the null result of Michelson Morley experiment. He completely revolutionize our thinking.

Enter Einstein - 1905 In 1905 Albert Einstein proposed that we accept the fact that the speed of light was the same in all reference systems (this was consistent with the M&M result) and was tantamount to doing away with the concept of the ether.

Postulates of the Special Theory of Relativity First Postulate: The laws of physics have the same form in all inertial reference systems. (This is the Relativity Principle) Second Postulate: Light propagates through empty space with a definite speed “c” independent of the speed of the source or of the observer. (Agrees with experiment)

Einstein’s original article on Special Relativity

Special Relativity “Special” : only works for inertial reference frames “Relativity” : there is no unique, absolute reference frame “Special Relativity” : “laws of physics are the same in all (inertial) reference frames”

Classical Relativity 1,000,000 ms-1 1,000,000 ms-1 How fast is Spaceship A approaching Spaceship B? Both Spaceships see the other approaching at 2,000,000 ms-1. This is Classical Relativity. VT = V1 + V2 Mention the Ether

Einstein’s Special Relativity 0 ms-1 300,000,000 ms-1 Moving with respect to the Ether 1,000,000 ms-1 Both spacemen measure the speed of the approaching ray of light. How fast do they measure the speed of light to be?

Special Relativity Stationary man Man travelling at 1,000,000 ms-1 Wrong! The Speed of Light is the same for all observers Mention the Michelson Morley experiment Albert Michelson & Edward Morley Lead Einstein to propose his Theory of Special Relativity

EVENT: Event is defined as an occurrence at a particular point in space at a particular time SIMULTANEITY: A fundamental error in our customary ideas of space and time concerns simultaneity. Two events take place at same instant of time , though not necessarily, at the same point , in space are said to occur simultaneously. COMMON SENSE : TWO EVENTS OCCUER SIMULTANEOUSLY ACCORDING TO ONE PERSON , THEN THOSE EVENTS SHOULD OCCUR SIMULTANEOUSLY WITH RESPECT TO ALL OBSERVERS .

THE RELATIVITY OF SIMULTANEITY Consider two lights which can be switched on at the same time. A B If Dobson is exactly the same distance from bulb A as bulb B, and is not moving, he will see both lights go on at the same time. The event of the lights turning on appears to Dobson to be simultaneous. Now suppose Dobson is moving to the right. At the instant he is at the center the lights are switched on. What will Dobson see? A B The relativity of simultaneity: "Events which are simultaneous in one reference frame may not be simultaneous in another." video

DID YOU SYNCRONISE YOUR CLOCK ??????????

Synchronization of clocks Synchronization of clocks

Synchronization of clocks All these clocks must show the same time. What do we mean by the “same time”? We mean that the clocks are synchronized. The recipe for clock synchronization: 1. Pick the reference clock. 2. Start the reference clock at 0 meters (of time!) 3. Send a signal (radio, optical, microwave, x-ray, γ-ray, death-ray) from the reference clock at that instant, that will spread in all directions 4. When the signal reaches the clock X located x meters away, set the clock X to x meters (of time). 5. Do so for all clocks. Your clocks are now synchronized.

2. Time and Length in Relativity

TIME DILATION (S0) v L Consider two mirrors separated by a distance L and traveling to the right at v as shown: Ranil is inside the apparatus and is moving with it. We will label Ranil's coordinate system (which moves with him) as (S0). When the light goes on, Ranil measures the time t0 it takes the beam to travel from the lower plate to the upper plate, and back to the lower plate: Since the speed of light in Ranil's frame is c0, and the distance covered is 2L, we have © 2006 By Timothy K. Lund 2L t0 c0 = which can be solved for t0: Time in IRF (S0) 2L c0 t0 =

TIME DILATION (S0) v L D D L (S) / 2 vt in a time of t, At the same time Ranil was making his time measurement, his twin brother Anil (in stationary frame S) observed the exact same event from "the ground." He measured the total time to be t. (S0) v L D D L (S) / 2 vt From Anil's perspective the light has traveled farther than from Ranil's perspective. It has traveled a distance of 2D, © 2006 By Timothy K. Lund in a time of t, and at a speed of c. From Anil's perspective we can then write Time in IRF (S) 2D c t = Since the mirrors have traveled a distance of vt in (S), we can calculate D in terms of L, v and t: D2 = L2 + ( vt/2 )2

TIME DILATION (ct)2 = (c0t0)2 + (vt)2 Again, our relationship: (ct)2 = (c0t0)2 + (vt)2 According to Einstein, "It came to me that time was suspect." Invoking Einstein's second postulate: "The speed of light in vacuum has the same value in all inertial reference frames." c = c0 so that (ct)2 = (ct0)2 + (vt)2 Why? (c2 - v2)(t)2 = c2(t0)2 Why? © 2006 By Timothy K. Lund (1 - v2/c2)(t)2 = (t0)2 Why? Time Dilation t = (t0)(1 - v2/c2)-1/2 We can rewrite the time dilation formula like so: Time Dilation t = t0 The Lorentz factor   (1 - v2/c2)-1/2

Since (1-v2/c2)-1/2 is always greater than or equal to 1, the time interval as measured by an observer in S is greater than the time interval as measured by an observer in S0. We define the PROPER TIME as that time measured by an observer who is at rest relative to the events whose time interval is measured. The events in the proper time frame are seen from a single point by the observer. Thus t0 is our proper time. The Lorentz Factor shows up often in relativistic formulations. Einstein and athletics

example: Δt = Δt0[1 – v2/c2]-1/2 Δt = Δt0[1 – 0.64c2/c2]-1/2 -Suppose Ranil is in a spaceship moving at 80% of the speed of light. -Thus, Ranil’s speed is v = .8c. Δt = Δt0[1 – v2/c2]-1/2 Δt = Δt0[1 – 0.64c2/c2]-1/2 Δt = Δt0[0.36]-1/2 Δt = 1.6666Δt0 -Thus, if Ranil observes that his watch his ticking once each second, Δt0 = 1 s. For the stationary observer Anil we have Δt = 1.6666Δt0 Δt = 1.67 s -Thus Ranils twin measures 1.67 s for every 1.00 s measured by Ranil, -Ranil therefore ages more slowly than his twin!

Time Dilation How do we define time? The flow of time each observer experiences is measured by their watch – we call this the proper time Ranil’s watch always displays her proper time Anil’s watch always displays his proper time If they are moving relative to each other they will not agree

Time Dilation A Real Life Example: Lifetime of muons Muon’s rest lifetime = 2.2x10-6 seconds Many muons in the upper atmosphere (or in the laboratory) travel at high speed. If v = 0.999 c. What will be its average lifetime as seen by an observer at rest?

Twin Paradox An amazing consequence of time dilatation

Consider sending an astronaut to a distant star in a spaceship at very high speed. Assume that the astronaut has a twin sibling waiting back on Earth. In the Earth frame of reference, time on the spaceship will be observed to pass more slowly than on the Earth due to time dilation. It may seem as if only a few years have past on the ship while decades pass on the Earth. So the twin of the astronaut waiting on the Earth expects the astronaut to be the much younger of the two upon return. In the spaceship frame of reference, it is the Earth that is moving at a very high speed so time on Earth will be observed to pass more slowly than on the spaceship. Decades will pass on the ship while only a few years pass on Earth. So the astronaut expects that the twin sibling waiting on the Earth to be the much younger of the two upon return. Isn't this a contradiction? What happens when the astronaut comes back to Earth? Which of the twin siblings will be older? How can both observations be correct? This problem is known as the twin paradox.

If v = 0. 99c, need 21 years for a return trip to star Procyon (10 If v = 0.99c, need 21 years for a return trip to star Procyon (10.4 Light years away) according to a twin brother at home. But, astronaut would spend only 3 years in his clock.

Length Dilemma

LENGTH CONTRACTION v Consider a station platform and a train car moving past it at speed v: Ranil is "on the ground" and Einstein is on the train. © 2006 By Timothy K. Lund Both parties decide to calculate the length of the platform using the known speed of the train, and their stopwatches. Both Ranil and Einstein agree on their relative speed v. Both Ranil and Einstein will start their clocks when the red mark on the train is opposite the beginning of the platform. They will stop their watches when the mark is opposite the end of the platform:

LENGTH CONTRACTION………………… : We call L0 the PROPER LENGTH or the REST LENGTH. This is because it is measured in the frame in which it is at rest. v Since   1 we see that L  L0! Thus lengths shorten when measured from any frame other than the rest frame. L0 Einstein measures the proper time t0. Why? Because the events take place at the same spatial position for Einstein. Ranil measures t = t0 from time dilation. © 2006 By Timothy K. Lund Einstein calculates the platform length to be L = vt0 Ranil has the advantage of being able to actually measure the length of the platform to be L0. But he can also calculate the length: L0 = vt Putting these three relations together we get: L0 = vt L0 = vt0 Length Contraction L = L0/ L0 L vt0 vt0 =