Special Relativity Additional reading: Higher Physics for CfE, p.64 – 69. Notes p.38 The idea of relativity goes back a long way … Gallileo was one of the first scientists to consider the idea of relativity. He stated that the laws of physics should be the same in all inertial frames of reference. If you were on a bus moving at constant speed you would experience the laws in the same way as if you were at rest.

Next, Isaac Newton picked up the baton … Newton followed this up by expanding on Gallileo’s ideas. He also introduced the idea of universal time. He believed that it was the same time at all points in the universe as it was on Earth, not an unreasonable assumption you might think!

Then along came a young office clerk whose daydreams blew hundreds of years of theory oot the water! Albert Einstein I wonder, what would happen if I was travelling at the speed of light and looked in a mirror?

Would he see his reflection? If so, then, relative to the ground, the light would be travelling at 6 x 10 8 ms -1. This caused problems because the speed of light should only depend on the medium it was moving through. Einstein finally came up with a solution: He “postulated” that “the speed of light is the same for all observers”. This means he would measure the speed of light as 3 x 10 8 ms -1 while moving and a “stationary” observer would measure the speed of light as 3 x 10 8 ms -1.

But if speed remains constant for light, then distance and time must change. This this was a curved ball right to the heart of scientific understanding in the early 1900’s. To think that time measurements will change, depending on the speed at which a system is moving relative to an observer!!! This is the stuff that fries one’s brain!

Get your head around this … You leave earth and your twin to go on a space mission. (OK That’s the easy bit!) You are in a spaceship travelling at 90% of the speed of light and you go on a journey that lasts 20 years – according to your on flight clock. But … when you get back you will find that 46 years will have elapsed on Earth. Your time had passed more slowly than time on Earth!

This crazy notion has been tested … In October 1971, four caesium-beam atomic clocks were taken aboard commercial airliners. They flew around the world, two eastward, two westward, then the clocks were compared with others that remained at the United States Naval Laboratory. When reunited, the three sets of clocks were found to disagree with one another, and their differences were consistent with the predictions of special and general relativity. … the tests show that our Einstein was correct!

OK, before we get stuck into the maths of it all, let’s warm up our brains. 1.Grandpa on a Train Imagine a little old man walking down the aisle of a train that you are on. He walks past you. How would you describe his speed?

1.Grandpa on a Train Now imagine you are sitting on the bank of the railway as the train passes. You see the old man in the train. How would you describe his speed now?

2.Pouring Coffee on a Train Imagine you are on a train and you are pouring coffee into a cup. What path does the coffee take from jug to cup?

2.Pouring Coffee on a Train Now imagine you are once again on the railway bank. Someone on the train is pouring the coffee. The jug and cup are moving horizontally at 100 mph. What path does the coffee take from jug to cup now?

2.Pouring Coffee on a Train In fact the path of the coffee from jug to cup is now … As the whole system, everything in it, is moving at 100 mph! The distance travelled from jug to cup depends on where the observer is!

These ideas are the basis of Einstein’s Theory of Special Relativity. Imagine you are in a train, carrying out an experiment. You have a special light source that makes a click when you switch it on AND a special mirror that clicks when the light strikes it. mirror click

mirror click The distance from the light source to the mirror is D. D The time measured by the moving observer on the train, between clicks, is t. So, using basic d = v t, we can say that, D = c t where “c” is the speed of light.

Now, let’s say the train is moving with a constant horizontal speed of “v”. An observer sitting on the railway bank (regarding himself as stationary relative to the train) will actually see the following light path: v

So the light path to the mirror, as seen by this “stationary” observer is longer. Let’s call it “h”. We can also determine the horizontal distance travelled by the light source as … d = v t’. d h D So, again using basic d = v t,we can say that : h = c t’ Where t’ is the time measured by the stationary observer.

By substituting h = ct’, D = ct and d = vt’, and applying basic Pythagoras (you can do this but it’s not examinable) we arrive at very important equations. These equations allow us to calculate the “relativistic” time and length of fast moving objects.

Time Dilation Time measured from within a moving system, passes more slowly relative to time measured by a “stationary” observer, outside the moving system, as follows: t’ = t 1 - vcvc 2 t’ = time measured from outside the moving system (s) (usually by an Earth bound “stationary”observer!) t = time measured from within the moving system (s) v = velocity of object observed by “stationary” observer (ms -1 ) c = speed of light (ms -1 )

Worked Example 1 A spacecraft moves with a velocity of 0.3c, relative to Earth, as recorded by a computer on board. The computer also records a time of 2 years for a particular journey. Determine the time for this journey as recorded by an observer on Earth. t’ = ? t = 2 years v = 0.3 c v/c = 0.3 t’ = t 1 - vcvc 2 = 2 / ( ) = 2.1 years

Complete Problems from Tutorial IV Special Relativity Q. 1 – 11 Time DilationQ. 1 – 9 Answers to “Special Relativity” Q a) 1 ms -1, N b) 2 ms -1, S 2.a) 0.8ms -1, E b) 2.8ms -1, E c) 2.2ms -1, W 3.a) 3 ms -1, up b) 2 ms -1, down 4. A = Einstein’s, B = same, C = zero, D = velocity, E = speed of light, F = slow, G = shortened kmh a) NO b) YES 7. 3 x 10 8 ms a) 100 s b) 100 s 9. a) 3 x 10 8 ms -1 b) 3 x 10 8 ms a)3 x 10 7 ms -1 b) 1.5 x 10 8 ms -1 c)1.8 x 10 8 ms -1 d)2.4 x 10 8 ms a) c b) 0.67c c) 0.5c d) 0.33c

Length Contraction In a similar way it can be shown that the length of a moving object is shorter if measured by an “outside” observer, relative to the length measured by an observer within the object’s frame of reference, as follows: l’ = l 1 - vcvc 2 Where l ’ = length measured by “outside” observer (s) l = length measured by observer within the object’s reference frame (m) v = velocity of object observed (ms -1 ) c = speed of light (ms -1 )

NOTE Length contraction only happens in the DIMENSION of MOTION of the moving object!

Worked Example 2 A spacecraft of length 40m is launched and its flight is observed from Earth. During one observation, the spacecraft velocity is 0.6c, relative to Earth. Determine the length of the moving spacecraft as observed from Earth. l ’ = l = 40 m v = 0.6 c v/c = 0.6 = 40 ( ) = 32 m l’ = l 1 - vcvc 2

Verification of Time Dilation So, how do we know this? There is a particle known as a muon that is created in the upper atmosphere. It only exists for a short time, a half-life of 1.56 x s. This means that for every million muons created at a height of 10km only 0.3 should reach the surface of the Earth. In actual fact around 5000 are detected! This is because the ‘muon clock’ runs slowly compared to the observer on Earth and the muon reaches the ground.

Complete Problems from Tutorial IV Special Relativity Q. 1 – 11 Time DilationQ. 1 – 9 Length ContractionQ. 1 – 9 Miscellaneous RelativityQ

Answers to “Special Relativity” Q a) 1 ms -1, N b) 2 ms -1, S 2.a) 0.8ms -1, E b) 2.8ms -1, E c) 2.2ms -1, W 3.a) 3 ms -1, up b) 2 ms -1, down 4. A = Einstein’s, B = same, C = zero, D = velocity, E = speed of light, F = slow, G = shortened kmh a) NO b) YES 7. 3 x 10 8 ms a) 100 s b) 100 s 9. a) 3 x 10 8 ms -1 b) 3 x 10 8 ms a)3 x 10 7 ms -1 b) 1.5 x 10 8 ms -1 c)1.8 x 10 8 ms -1 d)2.4 x 10 8 ms a) c b) 0.67c c) 0.5c d) 0.33c

Answers to “Time Dilation” Q a) 21 h b) 15 y c) 1043 s 2.d) 1.3 x s e) f) 1 x 10 8 ms -1 3.a) b) x y (or 17 billion years) x s x 10 8 ms -1 9.

Answers to “Length Contraction” Q a) b) c) 2.d) e) f) m km x 10 8 ms -1 9.

Answers to “Relativity Miscellaneous” Q a) b) c) 2.a) 0.31 y b) d = 0.95c x 0.31y (in s) c) 3.a) b) 4.a) b) 5.a) b) 6. 7.a) b) 8.a) b) c) 9.a) b) c)