Special Relativity

Classical Relativity 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.

Einstein’s Special Relativity 1,000,000 ms -1 0 ms ,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 – 300,000,000 ms -1 Man travelling at 1,000,000 ms -1 – 301,000,000 ms -1 ? – Wrong! The Speed of Light is the same for all observers

Postulates of special relativity 1)The constancy of the speed of light: The speed of light is the same for all observers, no matter what their relative speeds. 2) The Principle of Relativity: The laws of physics are the same in any inertial (that is, non-accelerated) frame of reference. This means that the laws of physics observed by a hypothetical observer traveling with a relativistic particle must be the same as those observed by an observer who is stationary in the laboratory.

Consequences In relativistic mechanics – There is no such thing as absolute length – There is no such thing as absolute time – Events at different locations that are observed to occur simultaneously in one frame are not observed to be simultaneous in another frame moving uniformly past the first Events which are simultaneous in one frame may not be in another! Each observer is correct in their own frame of reference

Thought Experiment Thought experiment to show events which are simultaneous in one frame may not be in another – A boxcar moves with uniform velocity – Two lightning bolts strike the ends – The lightning bolts leave marks (A’ and B’) on the car and (A and B) on the ground – Two observers are present: O’ in the boxcar and O on the ground – Observer O is midway between the points of lightning strikes on the ground, A and B – Observer O’ is midway between the points of lightning strikes on the boxcar, A’ and B’

Simultaneity – Thought Experiment Results The light reaches observer O at the same time – He concludes the light has traveled at the same speed over equal distances – Observer O concludes the lightning bolts occurred simultaneously By the time the light has reached observer O, observer O’ has moved The light from B’ has already moved by the observer, but the light from A’ has not yet reached him – The two observers must find that light travels at the same speed – Observer O’ concludes the lightning struck the front of the boxcar before it struck the back (they were not simultaneous events)

Moving Clocks Time between ‘ticks’ = distance / speed of light Light in the moving clock covers more distance… – …but the speed of light is constant… – …so the clock ticks slower! Moving clocks run more slowly! V

Moving Clocks – Time is not absolute To see this, imagine another boxcar experiment: Two observers, one in the car, another on the ground

Moving Clocks – Time is not absolute A mirror is fixed to the ceiling of a vehicle The vehicle is moving to the right with speed v An observer, O’, at rest in this system holds a laser a distance d below the mirror The laser emits a pulse of light directed at the mirror (event 1) and the pulse arrives back after being reflected (event 2)

Moving Clocks and Time Dilation Observer O’ carries a clock She uses it to measure the time between the events (t 0 ) She observes the events to occur at the same place so t 0 = distance/speed = (2d)/c Observer O is a stationary observer on the earth He observes the mirror and O’ to move with speed v By the time the light from the laser reaches the mirror, the mirror has moved to the right mirror The light must travel farther with respect to O than with respect to O’ respect Both observers must measure the speed of the light to be c The light travels farther for O so the time interval, t, for O is longer than the time interval for O’, t 0.

Moving Clocks and Time Dilation The person in the train measures c = 2d/ t 0 so t 0 = c/2d The proper time, t 0, is the time interval between events as measured by an observer who sees the events occur at the same position. – You must be able to correctly identify the observer who measures the proper time interval The person on track measures t, which increases as v increases A clock moving past an observer ticks more slowly than an identical clock at rest 0

Sample Problem Imagine that you are an astronaut who is being paid according to the time spent traveling in space as measured by a clock on Earth. You take a long voyage traveling at a speed near that of light. Upon your return to Earth, your paycheck will be: (a) smaller than if you had remained on Earth, (b) larger than if you had remained on Earth, or (c) the same as if you had remained on Earth. (b). Assuming that your on-duty time was kept on Earth, you will be pleasantly surprised with a large paycheck. Less time will have passed for you in your frame of reference than for your employer back on Earth.

The Twin Paradox The Situation A thought experiment involving a set of twins, Speedo and Goslo Speedo travels to Planet X, 20 light years from earth – His ship travels at 0.95c – After reaching planet X, he immediately returns to earth at the same speed When Speedo returns, he has aged 13 years, but Goslo has aged 42 years years,

The Twins’ Perspectives Goslo’s perspective is that he was at rest while Speedo went on the journey Speedo thinks he was at rest and Goslo and the earth raced away from him on a 6.5 year journey and then headed back toward him for another 6.5 years The paradox which twin is the traveler and which is really older?

The Twin Paradox The Resolution Relativity applies to reference frames moving at uniform speeds The trip in this thought experiment is not symmetrical since Speedo must experience a series of accelerations during the journey Speedo Therefore, Goslo can apply the time dilation formula with a proper time of 42 years proper This gives a time for Speedo of 13 years and this agrees with the earlier result the There is no true paradox since Speedo is not in an inertial frame

Length Contraction The measured distance between two points depends on the frame of reference of the observer The proper length, L 0, of an object is the length of the object measured by someone at rest relative to the object The length of an object measured in a reference frame that is moving with respect to the object is always less than the proper length – This effect is known as length contraction L0L0 L0L0 L0L0

Length Contraction Earth observer measures L 0 : The earth observer is at rest with respect to the earth-star line, so he measures the proper length L 0. Astronaut measures  t 0 : Since the astronaut's clock is at the beginning and end of the event—at his departure from earth, and at his arrival at the star-- he measures the proper time  t 0. Each observer measures the same relative speed: v = v L 0 /  t = L /  t 0 L = L 0 (  t 0 /  t)  t =  t 0 (1- v 2 /c 2 ) 1/2 Substituting: L = L 0 (1- v 2 /c 2 ) 1/2

Length Contraction Spaceship Moving at the 10 % the Speed of Light Spaceship Moving at the 86.5 % the Speed of Light

Comparison of typical velocities to the speed of light  = 1 - v 2 /c 2   t =   t 0 L0L0

Sample Problem A box is cubical with sides of proper lengths L 1 = L 2 = L 3 = 2 m, when viewed in its own rest frame. If this block moves parallel to one of its edges with a speed of 0.80c past an observer, (a)what shape does it appear to have to this observer, and (b)what is the length of each side as measured by this observer?

Sample Problem The average lifetime of a p meson in its own frame of reference (i.e., the proper lifetime) is 2.6 × 10 –8 s. If the meson moves with a speed of 0.98c, what is (a)its mean lifetime as measured by an observer on Earth and (b)the average distance it travels before decaying as measured by an observer on Earth? (c)What distance would it travel if time dilation did not occur? a)Recall that the time measured by observer on Earth will be longer then the proper time. Thus for the lifetime ;t = 1.3 x s b)d = vt = (0.98)(3 x 10 8 )(1.3 x s) = 38 m c)d = vt 0 = 7.6 m

Relativistic Momentum To account for conservation of momentum in all inertial frames, the definition must be modified – v is the speed of the particle, m is its mass as measured by an observer at rest with respect to the mass – When v << c, the denominator approaches 1 and so p approaches mv

Relativistic Addition of Velocities Galilean relative velocities cannot be applied to objects moving near the speed of light. Why? Einstein’s modification is – The denominator is a correction based on length contraction and time dilation – V ab = velocity of object A relative to object B – V ad = velocity of object A relative to object D – V db = velocity of object D relative to object B – Note: V ab = - V ba

Sample Problem A spaceship travels at 0.750c relative to Earth. If the spaceship fires a small rocket in the forward direction, how fast (relative to the ship) must it be fired for it to travel at 0.950c relative to Earth? v RS = ? v SE = c, v RE = c and v ES = -v SE = velocity of Earth relative to ship, the relativistic velocity addition equation gives v RS = (v RE + v ES )/(1+ (v RE ·v ES )/c 2 ) = 0.696c

Relativistic Energy The definition of kinetic energy requires modification in relativistic mechanics KE =  mc 2 – mc 2 – The term mc 2 is called the rest energy of the object and is independent of its speed – The term  mc 2 is the total energy, E, of the object and depends on its speed and its rest energy

Relativistic Energy – Consequences A particle has energy by virtue of its mass alone – A stationary particle with zero kinetic energy has an energy proportional to its inertial mass E 0 = mc 2 The mass of a particle may be completely convertible to energy and pure energy may be converted to particles

Pair Production An electron and a positron are produced and the photon disappears – A positron is the antiparticle of the electron, same mass but opposite charge Energy, momentum, and charge must be conserved during the process The minimum energy required can be calculated E = mc 2. It is 2m e = 1.6 x J or 1.04 MeV

Pair Annihilation In pair annihilation, an electron-positron pair produces two photons – The inverse of pair production The minimum energy of the electromagnetic radiation produced can be calculated with E=mc 2 It is impossible to create a single photon – Momentum must be conserved

Energy and Relativistic Momentum It is useful to have an expression relating total energy, E, to the relativistic momentum, p – E 2 = p 2 c 2 + (mc 2 ) 2 When the particle is at rest, p = 0 and E = mc 2 Massless particles (m = 0) have E = pc – This is also used to express masses in energy units mass of an electron = 9.11 x kg = MeV Conversion: 1 u = MeV/c 2 Note: 1 J = 6.24 x eV

Sources 7.htm 7.htm - ppewww.physics.gla.ac.uk/~caitrian/dace- talk/relativity-public.ppt 25.pdf 25.pdf 26.pdf