Relativity. An explanation of Brownian motion in terms of atoms. An explanation of the photoelectric effect ==> Quantum Theory

Relativity

Relativity In 1905 Albert Einstein published five articles in Annalen Der Physik that had a major effect upon our understanding of physics. They included:- An explanation of Brownian motion in terms of atoms An explanation of the photoelectric effect ==> Quantum Theory On the Electrodynamics of Moving Bodies ==> The Special Theory of Relativity

The Luminiferous Ether In 1873 Maxwell formulated his equations that led to the discovery of electromagnetic waves that propagated through space at the speed of light. All waves were believed to be transmitted by a medium. The medium that transmitted light waves was called the Ether (sometimes called, or spelled the Æther). It had to permeate everything, be incredibly resilient (stiff), and yet not impede motion!!

Light was believed to travel at 3 x 10 8 meters per second with respect to the fixed ether. The Earth moved through space (and therefore through the ether) at a speed of about 3 x 10 4 meters per second, and so the speed of light would be slightly different if measured parallel or perpendicular to the velocity of the Earth. Can the motion of the Earth through the ether be detected? It was searched for in an ingenious experiment performed by A.A. Michelson and E.W. Morley.

First an illustrative example:- Alma wind 50 miles 50 miles Lansing Brighton Alma lies 50 miles due north of Lansing. Brighton lies 50 miles due east of Lansing. Ms. A drives to Alma and back at a constant speed of 50 mph. There is a 5 mph wind blowing from West to East. So Mr. B drives to Brighton at a constant speed of 55 mph, and drives back from Brighton at a constant speed of 45 mph.

Q uick uiz Who arrives back first? (a) (b) (c) Ms. A Mr. B They arrive at the same time

Q uick uiz Who arrives back first? (a) (b) (c) Ms. A Mr. B They arrive at the same time Ms. A takes 50/50 + 50/50 = 2 hrs Mr. B takes 50/55 + 50/45 = 0.909 + 1.111 = 2.02 hrs so Ms. A wins by.02 hours (or 1 minute and 13 seconds).

The Michelson-Morley Experiment Mirror A 1/2 Silvered Mirror Light Source * Beam A Beam B Mirror B Telescope An interferometer

The interferometer was set up with one path parallel to the motion of the earth in orbit. They then rotated the apparatus to put the other path parallel to the motion of the earth. The apparatus was sensitive to the size of the effect expected from adding or subtracting the earth's velocity to the velocity of light. But NO effect was observed! Many repetitions and variations of the M&M experiment by many other investigators showed the same null result. It was the most famous null result in history. What was wrong?

The Fitzgerald-Lorentz Contraction One strange interpretation of the bewildering result of the M&M experiment was suggested by the Irish physicist, George F. Fitzgerald (and, simultaneously, by the Dutch physicist, H.A. Lorentz). It was that, as a consequence of moving through the ether, the length of any object is shortened by just the right amount to counteract the change in the light speed. When you do the math, it turns out that the required contraction is that the changed length of any object is L' = 1 v 2 /c 2 x L

You could never measure this contraction because a ruler placed alongside would also be shrunk by the same amount! Obviously the F-L contraction would explain the M&M result. But why? They had no idea! It was a totally adhoc hypothesis and, because of that, the theory wasn't taken seriously. The F-L contraction turns out to be a consequence of Einstein's theory of relativity.....

.... Meanwhile, what was Einstein worrying about? He noticed that Maxwell s equations appeared not to be invariant under a Galilean transformation. (This is just the addition (or subtraction) of velocity that we used in adding the effect of the wind in my example.)

In frame K, two charges at rest. Force is given by Coulomb s law. y K Q 1 Q 2 x z

y K x z In moving frame K, two charges are moving. Since moving charges are currents, Force is Coulomb + Magnetism.

Principle of relativity: The laws of nature are the same in all inertial reference frames Something is wrong! Maxwell s Equations? The Principle of Relativity? Gallilean Transformations?

Q uick uiz So, what is your guess? Which turns out to be incorrect? (A) (B) (C) Maxwell s Equations The Principle of Relativity Galilean Transformations

Einstein decided fi Galilean Transformations are the problem. Einstein s two postulates: 1. The principle of relativity is correct. The laws of physics are the same in all inertial reference frames. 2. The speed of light in vacuum is the same in all inertial reference frames (c = 3 x 10 8 m/s regardless of motion of the source or observer).

The second postulate seems to violate everyday common sense! Rocket Light pulse Observer v=0.5 c v=c Einstein says: observer measures the light as traveling at speed c, not 1.5c.

Gedanken Experiments A light clock: mirror w mirror photocell It ticks every Dt = 2 w/c seconds. One can synchronize ordinary clocks with it.

Time Dilation w O G : Observer on Ground

w v O T : Observer on Truck O T s clock as seen from the ground: c = 3 x 10 8 m/s w ct/2 vt/2 (ct/2) 2 - (vt/2) 2 = w 2

Time for one round trip of light, as seen from the ground: t = (2 w/c) 1 - v 2 /c 2 For v = 0.6c, t = (2 w/c) x 1.25 All of O T s processes slow down compared to O G as seen by O G. Similarly, All of O G s processes slow down compared to O T as seen by O T.

Length Contraction w O G : Observer on Ground

w v O T : Observer on Truck Device on truck makes mark on track each time clock ticks. As seen from ground: Distance between marks = (time between ticks) x v = [(2 w/c) 1 - v 2 /c 2 ] v

As seen from truck: Distance between marks = (time between ticks) x v = (2 w/c) v (To the person on the truck the time between ticks is (2 w/c).) (Distance measured on truck) = 1 - v 2 /c 2 x (distance measured on ground) As seen from a moving frame, rest distances contract. (L-F contraction)

Q uick uiz A clock on Earth has a pendulum with a period of T = 1 second. The clock is put into a rocket with a velocity of v = 0.6c. What now is the period of the clock as measured by an observer on Earth? (A) (B) (C) (D) (E) 0.6 sec 0.75 sec 1.0 sec 1.25 sec 1.67 sec

Simultaneity Events occur at a well defined position and a time (x,y,z,t). But events that are simultaneous (same t) in one inertial frame are not necessarily simultaneous in another frame.

d d A B The light from the two flashes reach O G at the same time. He sees them as simultaneous.

O T passes O G just as the lights flash. v

c c v But light from B reaches O T first. Since both light beams started the same distance from her, and both travel at speed c, she concludes that B must have flashed before A.

Q uick uiz How fast must a rocket ship travel so that its length appears to be half as long to an observer on Earth as it does to an observer on the rocket? (A) 0.5c (B) 0.87c (C) 0.95c (D) c (E) 2.0c

Lorentz Transformations y y v z z x x Flashbulb at origin just as both axes are coincident. Wavefronts in both systems must be spherical: x 2 + y 2 + z 2 = c 2 t 2 and x 2 + y 2 + z 2 = c 2 t 2 Inconsistent with a Galilean transformation Also cannot assume t=t.

Assuming: Principle of relativity linear transformation (x,y,z,t) -> (x,y,z,t ) Lorentz Transformations (section 2.4) x = g ( x - v t ) y = y z = z t = g ( t - v x / c 2 ) With g = 1 / 1 - v 2 /c 2. (Often also define b = v / c. )

Time Dilation (again) Proper time: time T 0 measured between two events at the same position in an inertial frame. O T s friend O T v O G O G s clock: T 0 = t 2 - t 1, (x 2 -x 1 =0) O T s clock: T = t 2 - t 1 t 2 - t 1 = g (t 2 - t 1 - v/c 2 (x 2 -x 1 ) ) T = g T 0 > T 0 Clocks, as seen by observers moving at a relative velocity, run slow.

Length Contraction (again) Proper length: distance L 0 between points that are at rest in an inertial frame. x 1 x 2 v x 1 x 2 O T on truck measures its length to be L 0 = x 2 - x 1. This is its proper length. O G on ground measures its length to be L = x 2 - x 1, using a meter stick at rest (t 2 = t 1 ). Then L 0 = x 2 - x 1 = g (x 2 - x 1 - v (t 2 - t 1 )) = g L O G measures L = L 0 / g < L 0. Truck appears contracted to O G.

An application Muon decays with the formula: N = N 0 e -t/t N 0 = number of muons at time t=0. N = number of muons at time t seconds later. t = 2.19 x 10-6 seconds is mean lifetime of muon. Suppose 1000 muons start at top of mountain d=2000 m high and travel at speed v=0.98c towards the ground. What is the expected number that reach earth? Time to reach earth: t = d/v = 2000m/(0.98 x 3 x 10 8 m/s) = 6.8 x 10-6 s Expect N = 1000 e -6.8/2.19 = 45 muons. But experimentally we see 540 muons! What did we do wrong?

Time dilation: The moving muon s internal clock runs slow. It has only gone through t = 6.8 x 10-6 1-0.98 2 s = 1.35 x 10-6 s So N = 1000 e -1.35/2.19 = 540 muons survive. Alternate explanation: From muon s viewpoint, the mountain is contracted. Get same result.

Addition of velocities Galilean formula (u=u +v) is wrong. Consider object, velocity u as seen in frame of O T who is on a truck moving with velocity v w.r.t the ground. What is velocity u of the object as measured by O G on the ground? u? u v Recall u = Dx/Dt, u = Dx /Dt. Inverse Lorentz transformation formulae: Dx = g ( Dx + v Dt ) Dt = g (Dt + v Dx / c 2 )

Dx g ( Dx + v Dt ) u = = Dt g (Dt + v Dx / c 2 ) u = u + v 1 + v u /c 2 For u and v much less than c: u u + v Velocities in y and z directions are also modified (due to t t, see section 2.6)

Examples: Rocket Light pulse Observer v=0.5 c u =c Observer sees light move at u = 0.5c + c 1+(0.5c)(c)/c 2 = c Light moves at c=3x10 8 m/s in all frames. Rocket Projectile v=0.8 c u =0.5c Observer Observer sees projectile move at 0.5c + 0.8c u = = 0.93c 1+(0.5)(0.8) Massive objects always move at speeds < c.

Q uick uiz Two spaceships travel in exactly the opposite direction as seen from the Earth. The speed of each, as measured by an observer on Earth, is 0.9c. What is the relative speed of the two spaceships as measured by a passenger on one of them? (A) 0 (B) 0.45c (C) 0.9c (D) 0.99c (E) 1.8c

The Twin Paradox Suppose there are two twins, Henry and Albert. Henry takes a rocket ship, going near the speed of light, to a nearby star, and then returns. Albert stays at home on earth. Albert says that Henry s clocks are running slow, so that when Henry returns he will still be young, whereas Albert is an old man. But Henry could just as well say that Albert is the one moving rapidly, so Albert should be younger after Henry returns! Who is right?

The first scenario is the correct one. The situation is not symmetric, because the rocket has to decelerate, turn around and accelerate again to return to earth. Thus, Henry is not in an inertial frame throughout the trip. He does return younger than Albert.

Relativistic Doppler Effect Light source and observer approach each other with relative velocity, v. Light is emitted at frequency n 0. n 0 v Observer sees light at a higher frequency: 1 + b n = n 0 with b = v/c 1 - b If source is receding, the formula still holds but now b is negative. We know that the universe is expanding, because light from distance galaxies is red-shifted, indicating motion away from us. n

Relativistic Momentum Requirement: momentum is conserved in all inertial frames. Assume: p = m v. Elastic scattering in c-o-m frame: A B before A B after p x : mu + m(-u) = 0 Transform to frame of A: A B A B -2u p x : 0 + m( ) m(-2u) 1+u 2 /c 2 It doesn t work!

Relativistic momentum: mv p = g m v = 1-v2 /c 2 Relativistic Kinetic Energy: K = (g - 1) mc 2 1 = ( - 1) mc 2 1-v 2 /c 2 For small velocities, v/c << 1: K = ( 1 + 1/2 (v/c) 2 +... - 1) mc 2 1/2 m v 2 For large velocities v c: K Massive objects always travel at speeds less than c.

KE and velocity: Relativistic vs. Classical Noticeable departures for v/c >.4 or so Starting from v=0, takes infinite KE to get to v=c

Velocity nearly stops changing after KE ~ 4 mc 2 KE = (γ 1) mc 2 γ = 2 (KE = rest) happens for β = (1 1/γ 2 ) =.87 electron has mc 2 =.511 MeV

Relativistic Energy According to Einstein, even a mass at rest has energy: E 0 = m c 2 (rest energy) Thus, the total energy of a moving object is E = K + E 0 = (g-1) mc 2 + mc 2 = g mc 2 It is straightforward to show: E 2 - p 2 c 2 = m 2 c 4 For a massless particle (e.g. a photon): E = p c

In general p c 2 g mv c 2 v = = E g mc 2 For a massless particle this gives v = c Massless particles travel at the speed of light c.