Chapter 26 Special Theory of Relativity Classical Physics: At the end of the 19 th century, classical physics was well established. It seems that the natural world was very well explained. Newtonian mechanics (motion, fluids, wave, sound) Kinetic theory (gas, heat) Maxwell's theory of EM (electromagnetism, EM wave, light) Modern Physics: New experimental evidences led to the great revolution at the turn of the 20 th century. Theory of relativity (space, time, speed) Quantum theory (atoms, molecules) 26-1 Galilean-Newtonian Relativity Inertial reference frame: is one in which Newton's law of inertia holds. Inertial reference frames can move at constant velocity relative to one another; acceleration reference frames are non-inertial. In classical mechanics, space and time (also mass and force) are considered to be absolute: their measurement doesn't change from one reference frame to another. The relativity principle: the law of physics are the same in all inertial reference frames. No one inertial frame is special in any sense. All inertial reference frames are equivalent Phys1404-Ch26 1
Example: A coin is dropped by a person in a moving car: Car frame: y = 1 2 gt2, x = 0 (a vertical path) Ground frame: y = 1 2 gt2, x = v x0 t (a parabolic path) Nowton's law of motion is valid in both inertial frames. The Great Puzzle: Maxwell predicted the velocity of light c = 3 10 8 m/s, but did not specify in what reference frame the speed is measured. According to classical physics, the velocity of light would be higher in some reference frame and lower in others. Some physicists suggested existence of "ether", a transparent medium fills all space, the velocity of light must be with respect to the ether. Phys1404-Ch26 2
Number of experiments (The Michelson-Morley experiment) found no difference of light speed in any direction. Where is ether? 26-3 Postulates of the Special Theory of Relativity Albert Einstein published his famous "Special Theory of Relativity" in 1905, which is based on two postulates: The relativity principle: the laws of physics are the same in all inertial reference frames. The constancy of the speed of light: the speed of light in vacuum has the same value in all inertial reference frames, independent of the speed of the source or observer. The 2 nd postulate seems violates commonsense. However, our commonsense is based on our experiences in a world where objects typically travel at a speed much less than c. Einstein's theory is consistent with Maxwell's theory. The speed of light in vacuum is the same in any reference frame. In Einstein's theory, the time and space are not absolute. 26-4 Simultaneity Einstein s Thought Experiment: Two lightning bolts strike both ends of a boxcar. Phys1404-Ch26 3
Observer at O 2 : the light from A 2 and B 2 reach O 2 at the some time. So the two bolts of lighting strike simultaneously. Observer at O 1 : the light from B 1 has already reached O 1, whereas the light from A 1 has not yet reached O 1. The lightning struck the front of the boxcar before it struck the back. According to relativity, both observers are right. There is no "best" inertial frame of reference. Two events that are simultaneous in one reference frame are in general not simultaneous in a second from moving relative to the first. That is, simultaneity is not an absolute concept but one that depends upon the state of motion of the observer. Phys1404-Ch26 4
26-5 Time Dilation and the Twin Paradox Time passes differently in one reference frame than in another. Example: A spaceship traveling past Earth at high speed. The observer on the spaceship: The light travels a distance of 2d at speed c, so the time required is Δt 0 = 2d c The observer on Earth: Phys1404-Ch26 5
The light travels a total distance of 2 d 2 + ( vδt 2 )2, Therefore, c = 2 d 2 + v 2 (Δt) 2 /4 Δt 2d Solve for t: Δt = c 1 v 2 /c 2 Δt or Δt = 0 1 v 2 /c = γδt 2 0 (> t 0 ) The time interval between the two events is greater for the observer on Earth than for the observer on the spaceship. Time dilation: Clocks moving relative to an observer are measured by that observer, run more slowly as compared to clocks at rest. Proper time: t 0 is called the proper time. It is the time interval between the two events in a reference frame where the two events occur at the same point in space (in a rest frame). The factor γ: 1 γ = 1 v 2 /c 1 2 For everyday life, v << c, then γ 1 and Δt Δt 0. Phys1404-Ch26 6
Example: The mean lifetime of a muon at rest is 2.2 10-6 s. Assume that it is traveling at v = 0.6c = 1.8 10 8 m/s with respect to a laboratory. (a) What will be the mean lifetime of a muon as measured in the laboratory? To an observer in the lab, the muon lives longer because of time dilation. Δt Δt = 0 1 v 2 /c = 2.2 10 6 s 2.2 10 6 s = = 2.8 10 6 s 2 1 (0.6c) 2 2 /c 0.64 (b) How far does a muon travel in the laboratory, on average, before decaying? With classical physics: d = vt = (1.8 10 8 m / s)(2.2 10 6 s) = 400m With relativity: d = vt = (1.8 10 8 m / s)(2.8 10 6 s) = 500m Space Travel: How long it takes to reach a star 10 light-years away, if a spaceship can travel at a speed of 0.99c? Classical physics: it would take over 10 years. Relativity: Since v = 0.99c, with time dilation, Δt 0 = Δt 1 v 2 /c 2 = (10yr) 1 (0.99c) 2 /c 2 = 1.4yr While 10 years would pass on Earth, only 1.4 years would pass for the astronaut on the trip. Phys1404-Ch26 7
In the spaceship, not only clock, but all physical, chemical, and biological processes slow down relative to an Earth observer. The Famous Twin Paradox Use the above example, according to the Earth-twin, the round trip to the star toke 20 years. But according to the Astronaut-twin, the trip only lasted 2.8 years. When the astronaut-twin returned to the Earth, he should be ~17 years younger than his Earth-twin brother. The paradox is following: From the reference of Astronaut-twin, he was at rest during the trip, but the Earth moved away then came back. So he should be 17 years older than his Earth-brother instead. Which brother is correct? 26-6 Length Contraction Example: A spaceship traveling at very high speed from Earth to Neptune. Earth Observer: Distance = L 0, speed of spaceship = v The time t = L 0 /v Spaceship observer: The spaceship is at rest. Earth and Neptune move with speed v. The time between the departure of Earth and arrival of Neptune is the proper time (since two events occur at the same point in space, i.e., spaceship). Because of time dilation, the time interval is less for spaceship observer than for the Earth observer. Δt 0 = Δt 1 v 2 /c 2 Phys1404-Ch26 8
Since the speed is the same (= v), the distance between the planets as viewed by the spacecraft observer is L = vδt 0 = vδt 1 v 2 /c 2 or L = L 0 1 v 2 /c 2 (< L 0 ) Length contraction: The length of an object is measured to be shorter when it is moving relative to the observer than when it is at rest. Proper length: L 0 is called the proper length. It is the length of the object or distance between two points whose positions are measured at the same time - as measured by observers at rest with respect to it. Length contraction occurs only along the direction of motion. Example: A rectangular painting measured 1m tall and 1.5m wide. It is hung on the sidewall of a spaceship, which is moving past the Earth at a speed of 0.9c. Spaceship observer: the painting is 1m tall and 1.5m wide. Earth Observer: Only the dimension in the direction of motion is shortened. The painting is still 1m tall. But the width is contracted to: L = L 0 1 v 2 /c 2 = 1.5m 1 (0.9c) 2 /c 2 = 0.65m Phys1404-Ch26 9
1m 1m 1.5m 0.65m 26-7 Four-Dimensional Space-Time Example: A person has a meal in spaceship moving at 0.65c. Spaceship observer: Earth observer: 15 minutes, plate diameter = 20 cm. 20 minutes, plate diameter = 15 cm. Time and space can intermix - gain time but lose space. Four-dimensional space-time (x, y, z, t) Space and time cannot be considered as separated entities. Phys1404-Ch26 10
Lorentz Transformation Equation (not required to master): x' = (x vt)/ 1 v 2 / c 2 y' = y z' = z t' = (t vx /c 2 )/ 1 v 2 / c 2 y y' v z x z' x' 26-8 Momentum and Mass Momentum is redefined as p = m 0 v 1 v 2 /c 2 = γm 0v Mass of a object m = m 0 1 v 2 /c 2 = γm 0 m o is called rest mass. In relativity, mass increases with velocity. Phys1404-Ch26 11
At low velocity, v << c, p = mv and m = m o Example: calculate the mass of an electron when it has a speed of 0.98c in an accelerator used for cancer therapy. m = m 0 1 v 2 / c 2 = m 0 1 (0.98) 2 /c 2 = 5.0m 0 26-9 The Ultimate Speed Speed of an object cannot equal or exceed the speed of light. The speed of light in vacuum is a natural upper speed limit in the universe. When v c, m E = mc 2 ; Mass and Energy In Einstein's theory, the mass of an object increases as its speed increase. Certain portion of energy used to accelerate an object turn into mass. Thus, mass is a form of energy. The total energy of a particle: E = mc 2 E = m 0 c 2 + KE KE is kinetic energy KE = (m m 0 )c 2 = (γ 1)m 0 c 2 m 0 c 2 is rest energy, which is the energy of a object at rest. Mass is a form of energy. Phys1404-Ch26 12
Mass can be converted to other form of energy. Other form of energy can also be converted to mass. The inter-conversion of energy and mass has been experimentally confirmed. For example: (1) Neutual pion π 0 with rest mass decays into pure EM radiation (photons). (2) Uranium fuel loss in mass in fission reaction to generate nuclear power. (3) Sun's mass is continually decreasing as it radiates EM energy outward. The energy of atomic particles are often expressed in electron volts: 1 ev =1.60 10-19 J; 1 kev = 1.0 10 3 ev; 1 MeV = 1.0 10 6 ev Note: For particles having zero mass, such as photons, m 0 = 0 and E = pc. Example: A π 0 meson (m 0 = 2.4 10-28 kg) travels at a speed v = 0.8c. (a) What is its kinetic energy? The mass at v = 0.8c m m = 0 2.4 10 = 1 v 2 2 / c 1 (0.8) 2 28 kg = 4.0 10 28 kg Phys1404-Ch26 13
Thus its KE is KE = (γ 1)m 0 c 2 = (m m 0 )c 2 = (4.0 10 28 kg 2.4 10 28 kg)(3.0 10 8 m /s) 2 = 1.4 10 11 J = (1.4 10 11 1eV J) 1.60 10 19 J = 87.5MeV Classical calculation is much less and is incorrect: KE = 1 2 m 0 v2 = 1 2 (2.4 10 28 kg)(2.4 10 8 m / s) 2 = 6.9 10 12 J (b) How much energy would be released if the π 0 is transformed completely into EM radiation? The total energy: E = mc 2 = (4 10 28 kg)(3 10 8 m /s) 2 = 3.6 10 11 J = 3.6 10 11 1eV J( 1.6 10 19 J ) = 230MeV Total energy E can also be expressed in terms of relativistic momentum: E 2 = p 2 c 2 + m 0 2 c 4 Phys1404-Ch26 14
Where p = m 0 v 1 v 2 / c 2 Relativistic Addition of Velocity Classical physics: u = v + u' = 0.6c+ 0.6c = 1.2c (wrong!) No object can travel faster than the speed of light. One cannot simply add relative speeds to obtain the overall speed. Phys1404-Ch26 15
Relativistic u = Thus, u = v + u' 1 + vu' / c 2 0.6c + 0.6c 1 + (0.6c)(0.6c)/ c 2 = 0.88c Let v = 0.6c and u' = c, then 0.6c + c u = 1 + (0.6c)(c)/c 2 = c The speed of light is the same in any inertial reference frame. The impact of Special Relativity The special theory of relativity has been confirmed in many experiments. At a speed much less than c, the relativistic formulas reduce to the old classical one. Relativity does not contradict classical theory Relativity is a more general theory. Classical physics only work at v << c. Phys1404-Ch26 16