RELATIVITY Einstein published two theories of relativity In 1905 The Special Theory For uniform motion a = 0 In 1916 The General Theory For non-uniform motion a 0.
First we will discuss The Special Theory At the beginning of the century most believed light was a wave. Thus must have something that waves: Sound has air Water has water, etc. 2
Physicists proposed that for waves of light something must wave. They called it the ether for light This ether then must fill the universe. The earth moves through the universe so: How fast are we traveling through the ether? To answer this we will start with a race between two boats (that run at exactly the same speed) in a river. 3
The boat going downstream will have speed c + v The boat going upstream will have speed c v 4
The calculated time for round trip is t L L 2Lc = + = c+ v c v c v // 2 2 The speed for the boat going cross-stream will have speed both ways c v 2 2 Therefore t cross = v 2L c 2 2 Then the ratio of the two times will be 5
t t // cross 1 = > 2 v 1 2 c 1 Therefore we see that the boat that goes across and back wins the race even though both boats travel the same speed relative to the water. The main point is the time to complete the race is different for the two boats. Note: 6
We can measure ratio of times for boats and calculate the speed of river if we know the speed of boats. 7
The Michelson-Morley Experiment Diagram of Apparatus 8
Drawing of actual apparatus We know the speed of light. We want the speed of the earth through the ether. 9
When light arrives at the eye it has traveled two paths to reach the observer. There will be interference either constructive or destructive The resulting image will be a series of lines. 10
Spectrometer lines Since the direction of the ether flow is not known the apparatus must be rotated. 11
First one than the other path will be parallel to the flow of ether. Therefore the interference lines should shift. Michelson and Morley did the experiment very carefully and did not find a shift. The conclusion has to be that: The Ether does not exist or the earth travels along with it. 12
Another experiment shows that we are not moving with it. Stellar Aberration is that experiment While the light travels down the telescope the telescope moves with the earth. 13
The telescope has to be tilted to keep the image in the center. If the ether (the substance that waves to cause the propagation of light) moves with the scope there would be no need to tilt the it. Therefore we conclude the ether does not exist. Classical Relativity The transformation equations before Einstein 14
x = x vt y = y z = z t = t These are the Galilean Equations that allow observers to compare observations in two different frames moving relative to each other with constant velocity. 15
Observer on ground and observer on railroad car moving in x-direction. The observer on the ground observes the birds separated by distance x 2 x 1 The distances are equal.. 16
If an airplane flies over the railroad car traveling in the + x direction at a speed u x measured by the observer on the ground what will be the speed ( u x ) of the airplane measured by the observer on the railroad car? We can use the transformation equation for x x = x vt and the equation for t t = t Differentiate and divide to get 17
dx dx dv = v t dt dt dt u = u v x x 0 if the velocity of the railroad car is constant. If the observer on the ground measures the velocity of the airplane as u x then the person on the railroad car will measure u x 18
What if the person on the ground points a flashlight in the + x direction? What will be the speed of light measured by the observer on the railroad car? 19
We get x u = u v x giving c = c v We must keep this result in mind as we discuss Einstein s Theory. 20
Einstein s postulates for the Special Theory of Relativity: 1. Fundamental laws of physics are identical for any two observers in uniform relative motion. 2. The speed of light is independent of the motion of the light source or observer. 21
These postulates cannot be satisfied using the Galilean Equations, as we will see. However Einstein found that the following equations worked. x = γ ( x vt) y = y z = z vx t = γ ( t ) 2 c where 1 γ = v 1 c 2 2 22
These are the Lorentz Transformation Equations. Now consider the airplane flying over the railroad car in the x-direction. What is the speed of the airplane as measured by the observer on the ground? What is the speed of the airplane as measured by the observer on the railroad car? We need to answer these questions by using the Einstein-Lorenze Equations. 23
x = γ ( x vt) and t vx = γ ( t ) c 2 differentiate dx = ( dx vdt) and γ dt vdx = γ ( dt ) c 2 divide dx dx vdt = dt vdx dt 2 24
divide by dt dx = dt or u x = dx v dt dx v 1 dt c 2 u x 1 v vu c x 2 Thus if an object (an airplane) flies over the railroad car the observer on the ground will 25
measure the speed in the x direction as u x. The observer on the car will find u x. What about the speed of light when a flashlight is pointed in the x-direction? The observer on the ground points a flashlight in the +x direction. What will be the speed of light measured by the observer on the car? u x c v c v c v = = = = vc v 1 1 ( c v)/ c 2 c c c 26
Both observers, even though they are moving relative to each other, measure the same value for the speed of light. This is in agreement with the Second Postulate. LENGTH CONTRACTION Read the section on Length Contraction in the book. We will do it a little differently. 27
The observer on the moving railroad car has a rod moving with him. He measures the length of the rod to be = 2 1 0 x x L Use the Lorentz equations to get x = γ ( x vt ) 2 2 2 x = γ ( x vt ) 1 1 1 Then putting these in the equation 28
L = γ( x vt ) γ( x vt) 0 2 2 1 1 [( ) ( )] L = γ x x vt t 0 2 1 2 1 If the observer on the ground measures the far end and near end of the rod at the same time Then t = t 1 2 L = γ ( x x ) = γ L 0 2 1 or L = L 0 γ and γ > 1 29
So the observer on the ground with the rod moving past in the x direction measures the rod to be shorter than what is measured by the observer at rest relative to the rod and on the car. Length Contraction is a prediction of the Lorentz Equations. TIME DILATION Again we will find time dilation a different way than the book. Then we will analyze the book s method. 30
Observer on railroad car moving in x-direction with firecrackers and another observer on ground. There is a firecracker on the top of the pole at x 1 and one on the top of the pole at x 2. Let s say the firecracker on the top of the pole at explodes and then some time later the one on x 1 31
top of the pole at x 2 explodes. We want to consider the time between the two events. The time interval between the two events as measured by the observer on the ground will be t = t t 2 1 Using the transformation equations we get t vx ( ) c 2 2 = γ t 2 + 2 So t vx ( ) c 1 1 = γ t1 + 2 32
vx2 vx1 t = γ ( t2 + ) γ ( t 2 1 + ) 2 c c or v t = γ t + ( ( x 2 2 x1) c The time interval is different for the two observers. For simplicity consider two firecrackers on the same post at then x = x = x 1 2 33
and v t t ( x x ) = γ + = γ t 2 c γ > 1 2 1 So the time interval for the observer on the ground with the events moving relative to him is longer. MOVING CLOCKS RUN SLOWLY If an observer is moving relative to a clock the interval between ticks will be longer. This is Time Dilation 34
The book demonstration: 35
For the girl on the railroad car, O frame, the light travels up and back to the floor. The distance traveled is 2d The time to travel is t The speed of light is c So t = 2d c 36
The observer on the ground sees the light travel from the floor to the ceiling but during this time the railroad car moves carrying the mirror with it at a speed v. 37
The observer on the ground sees the light travel farther when it goes from floor to ceiling than the observer on the railroad car. For the girl distance traveled = d c t For the boy distance traveled = 2 Both observers must measure the speed of light as c. Therefore, since the distance traveled is longer for the boy than for the girl 38
t > t Consider triangle 39
We see that 2 2 c t v t = + 2 2 d 2 Solve for t or c v t = d 4 4 2 2 2 2 2 t = 2 4d c v 2 2 and 40
2d 2d t = = c v v c 1 c 2 2 2 2 Use t = 2d c to get 2d = c t and put in above equation to get 41
c t t t = = = γ t 2 2 v v c 1 1 2 2 c c 42
Einstein / Lorentz Transformation Equations 1. Satisfy Einstein s Second Postulate 2. Predict Length contraction 3. Predict Time Dilation But are these correct? Must have an experiment to prove!!! 43
Muon lifetime Experiment Muons are particles created high in our atmosphere. They rain down continuously at high velocity at approximately v 8 = 2.994x10 m/s. If at rest they only exist for approximately τ 6 = 210 x seconds. The experiment is to set up a detector at the top of a mountain and stop the muons in the detector and measure how long they exist. 44
Then ask how far down would they have traveled if they had not been stopped in the detector. distance = (speed) x (time they exist) d = x m s x x s 8 6 (2.994 10 / ) (2 10 ) = 600 m. If the mountain is 2000m tall, at bottom should find very few if any muons. 45
Move the apparatus to the bottom of the mountain and measure the number of the same type muons that reach sea level. The experiment showed that as many reached sea level as passed through the atmosphere at the level of the top of the mountain. Why? When moving relative to us the observer they exist not for τ = 2 x 10-6 s but for 46
t τ = γ = = = = 6 6 t 16t 16(2x10 ) s 32x10 s 8 2.994x10 2 1 ( ) 8 310 x Thus distance traveled will be 8 6 d = (2.994x10 m/ s)(32x10 s) = 10, 000m well below sea level. Or viewed from the muon 47
Experiment also confirms length contraction. 48