The Theory of Special Relativity Ch 26

Two Theories of Relativity Special Relativity (1905) –Inertial Reference frames only –Time dilation –Length Contraction –Momentum and mass (E=mc 2 ) General Relativity –Noninertial reference frames (accelerating frames too) –Explains gravity and the curvature of space time

Classical and Modern Physics Classical Physics – Larger, slow moving Newtonian Mechanics EM and Waves Thermodynamics Modern Physics Relativity – Fast moving objects Quantum Mechanics – very small

ClassicalRelativistic QuantumClassical Speed Size 10% c Atomic/molecular size

Correspondence Principle Below 10% c, classical mechanics holds (relativistic effects are minimal) Above 10%, relativistic mechanics holds (more general theory)

Inertial Reference Frames Reference frames in which the law of inertia holds Constant velocity situations –Standing Still –Moving at constant velocity (earth is mostly inertial, though it does rotate)

Basic laws of physics are the same in all inertial reference frames All inertial reference frames are equally valid

Speed of Light Problem According to Maxwell’s Equations, c did not vary Light has no medium Some postulated “ether” that light moved through No experimental confirmation of ether (Michelson-Morley experiment)

Two Postulates of Special Relativity Einstein (1905) The laws of physics are the same in all inertial reference frames Light travels through empty space at c, independent of speed of source or observer There is no absolute reference frame of time and space

Simultaneity Time always moves forward Time measured between things can vary Lightning strikes point A and B at the same time O will see both at the same time and call them simultaneous

Moving Observers 1 On train O 2 - train O 1 moves to the right On train O 1 – train O 2 moves to the left

Moving Observers 2 Lightning strikes A and B at same time as both trains are opposite one another

Train O 2 will observe the strikes as simultaneous Train O 1 will observe strike B first (not simultaneous Neither reference frame is “correct.” Time is NOT absolute

Time Dilation Consider light beam reflected and observed on a moving spaceship and from the ground

Distance is shorter from the ship Distance is longer from the ground c = D/t Since D is longer from the ground, so t must be too.

On Spaceship: c = 2D/  t o  t o = 2D/c On Earth: c = 2 D 2 + L 2  t v = 2L/  t L = v  t 2

c = 2 D 2 + v 2 (  t) 2 /4  t c 2 = 4D 2 + v 2  t 2  t = 2D c 1 –v 2 /c 2  t =  t o 1 - v 2 /c 2

 t =  t o √ 1 - v 2 /c 2  t o –Proper time –time interval when the 2 events are at the same point in space –In this example, on the spaceship

Is this real? Experimental Proof 1.Jet planes (clocks accurate to nanoseconds) 2.Elementary Particles – muon Lifetime is 2.2  s at rest Much longer lifetime when travelling at high speeds

Time Dilation: Ex 1 What is the lifetime of a muon travelling at 0.60 c (1.8 X 108 m/s) if its rest lifetime is 2.2  s?  t =  t o √ 1 - v 2 /c 2  t = (2.2 X s) = 2.8 X s 1- (0.60c) 2 1/2 c 2

Time Dilation: Ex 2 If our apatosaurus aged 10 years, calculate how many years will have passed for his twin brother if he travels at: a)¼ light speed b)½ light speed c)¾ light speed

Time Dilation: Ex 2 a)10.3 y b)11.5 y c)10.5 y

Time Dilation: Ex 3 How long will a 100 year trip (as observed from earth) seem to the astronaut who is travelling at 0.99 c?  t =  t o 1 - v 2 /c 2  t o =  t 1 - v 2 /c 2  t o = 4.5 y

Time Dilation: Ex 3 If our apatosaurus aged 10 years, and his brother aged 70 years, calculate the apatosaurus’ average speed for his trip. (Express your answer in terms of c). ANS: 0.99 c

Length Contraction Observers from earth would see a spaceship shorten in the length of travel

Only shortens in direction of travel The length of an object is measured to be shorter when it is moving relative to an observer than when it is at rest.

 t o =  t √ 1 - v 2 /c 2 v = L  t o = L/v (L is from spacecraft)  t o  t = L o /v L o =L v v √ 1 - v 2 /c 2 L = L o √ 1 - v 2 /c 2

L o = Proper Length (at rest) L = Length in motion (from stationary observer)

Length Contraction: Ex 1 A painting is 1.00 m tall and 1.50 m wide. What are its dimensions inside a spaceship moving at 0.90 c?

Length Contraction: Ex 2 What are its dimensions to a stationary observer? Still 1.00 m tall L = L o √ 1 - v 2 /c 2 L = (1.50 m)(√ 1 - (0.90 c) 2 /c 2 ) L = 0.65 m

Length Contraction: Ex 3 The apatosaurus had a length of about 25 m. Calculate the dinosaur’s length if it was running at: a)½ lightspeed b)¾ lightspeed c)95% lightspeed

a)21.7 m b)15.5 n c)7.8 m

Four-Dimensional Space-Time Consider a meal on a train (stationary observer) Meal seems to take longer to observer Meal plate is more narrow to observer

Move faster – Time is longer but length is shorter Move slower – Time is shorter but length is longer Time is the fourth dimension

Momentum and the Mass Increase p = m o v 1 - v 2 /c 2 Mass increases with speed m o = proper (rest) mass m = m o 1 - v 2 /c 2

Mass Increase: Ex 1 Calculate the mass of an electron moving at 4.00 X 10 7 m/s in the CRT of a television tube. m = m o 1 - v 2 /c 2 m = 9.11 X kg= 9.19 X kg 1 - (4.00 X 10 7 m/s) 2 /c 2

Mass Increase: Ex 2 Calculate the mass of an electron moving at 0.98 c in an accelerator for cancer therapy. m = m o √ 1 - v 2 /c 2 m = 9.11 X kg= 4.58 X kg (5m o ) √ 1 - (0.98c) 2 /c 2

The Speed Limit Nothing below the speed of light can be accelerated to the speed of light Would require infinite energy –Mass becomes infinite –Length goes to zero –Time becomes infinite

Mass Increase: Ex 2 The apatosaurus had a mass of about 35,000 kg. Calculate the dinosaur’s mass if it was running at: a)½ lightspeed b)¾ lightspeed c)95% lightspeed

a)40,415 kg b)53,915 kg c)112,090 kg

Relativistic Momentum p = m o v 1 - v 2 /c 2

E = mc 2 Particle at Rest E = m o c 2 E = Total Energy m o = Rest mass c = speed of light Moving Particles E 2 = m o 2 c 4 + p 2 c 2

E = m o c 2 + KE rest kinetic KE does not equal ½ mv 2 at energy energy relativistic speeds

E=mc 2 : Ex 1 How much energy would be released if a  0 meson (m o =2.4 X kg) decays at rest. E = mc 2 E = m o c 2 (particle is at rest) E = (2.4 X kg)(3.0 X 10 8 m/s) 2 E = 2.16 X J

E=mc 2 : Ex 2 A  0 meson (m o =2.4 X kg) travels at 0.80 c. a.Calculate the new mass [4 X kg] b.Calculate the relativistic momentum [9.6 X kg m/s] c.Calculate the energy of the particle (E 2 = m o 2 c 4 + p 2 c 2 ) [ANS: 3.6 X J]

What is the kinetic energy of the  0 meson in the former example. E = m o c 2 + KE KE = E – m o c 2 KE = 3.6 X J - (2.4 X kg)(3.0X10 8 m/s) 2 KE = 1.4 X J E=mc 2 + KE: Ex 3

E=mc 2 + KE An electron is moving at 0.999c in the CERN accelerator. a.Calculate the rest energy b.Calculate the relativistic momentum c.Calculate the relativistic energy d.Calculate the Kinetic energy

Electron Volts 1 eV= 1.6 X J 1 MeV = 10 6 eV = 1.60 X J What is the rest mass of an electron in MeV? E = mc 2 E = m o c 2 (particle is at rest) E = (9.11 X kg)(3.0 X 10 8 m/s) 2 E = 8.20 X J

8.20 X J 1 MeV= MeV/c X J

Relativistic Addition of Velocity Classical Addition –Bus moves at 40 mph –You walk to the front at 5 mph –Overall speed = 45 mph

Relativistic Addition –Cannot simply add velocities above ~0.10 c –Length and time are in different reference frames –Formula u = v + u’ 1 + vu’/c 2 u = overall speed with respect to stationary observer v = speed of moving object with respect to st. observer u’ =speed of 2 nd object with respect to moving observer

Relativistic Addition: Ex 1 What is the speed of the second stage of the rocket shown with respect to the earth?

u = v + u’ 1 + vu’/c 2 u = 0.60c c 1 + [(0.60c)(0.60c)/c 2 ] u = 0.88 c (classical addition would give you 1.20c, over the speed of light)

Relativistic Addition: Ex 2 Suppose a car travelling at 0.60c turns on its headlights. What is the speed of the light travelling out from the car? u = v + u’ 1 + vu’/c 2 u = 0.60c + c=1.60c 1 + [(0.60c)(c)/c 2 ]1.60 u = c

Relativistic Addition: Ex 3 Now the car is travelling at c and turns on its headlights. u = v + u’ 1 + vu’/c 2 u = c + c=2c 1 + [(c)(c)/c 2 ]2 u = c