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=mc2)
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
10% c
Speed
Classical
Relativistic
Atomic/molecular size
Size
Quantum
Classical
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 O2 - train O1
moves to the right
On train O1 – train O2
moves to the left
Moving Observers 2
• Lightning strikes A and B at same time as both
trains are opposite one another
• Train O2 will
observe the strikes
as simultaneous
• Train O1 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/Dto
Dto = 2D/c
On Earth:
c = 2 D2 + L2
Dt
v = 2L/Dt
L = vDt
2
c = 2 D2 + v2 (Dt)2/4
Dt
c2 = 4D2 + v2
Dt2
Dt = 2D
c 1 –v2/c2
Dt =
Dto
1 - v2/c2
Dt =
Dto
√ 1 - v2/c2
Dto
– 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 ms 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 ms?
Dt =
Dto
√ 1 - v2/c2
Dt = (2.2 X 10-6 s) = 2.8 X 10-6 s
1- (0.60c)2 1/2
c2
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?
Dt =
Dto
1 - v2/c2
Dto = Dt 1 - v2/c2
Dto = 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.
Dto = Dt √ 1 - v2/c2
v=L
Dto = L/v
Dto
Dt = Lo/v
Lo =
v
L
v √ 1 - v2/c2
L = Lo √ 1 - v2/c2
(L is from spacecraft)
L = Lo √ 1 - v2/c2
Lo = 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 = Lo √ 1 - v2/c2
L = (1.50 m)(√ 1 - (0.90 c)2/c2)
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 = mov
1 - v2/c2
Mass increases with speed
mo = proper (rest) mass
m = mo
1 - v2/c2
Mass Increase: Ex 1
Calculate the mass of an electron moving at 4.00 X
107 m/s in the CRT of a television tube.
m = mo
1 - v2/c2
m = 9.11 X 10-31 kg
1 - (4.00 X 107 m/s)2/c2
= 9.19 X 10-31 kg
Mass Increase: Ex 2
Calculate the mass of an electron moving at 0.98 c
in an accelerator for cancer therapy.
m = mo
√ 1 - v2/c2
m = 9.11 X 10-31 kg
√ 1 - (0.98c)2/c2
= 4.58 X 10-30 kg (5mo)
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 = mov
1 - v2/c2
E=
Particle at Rest
E = moc2
E = Total Energy
mo = Rest mass
c = speed of light
Moving Particles
E2 = mo2c4 + p2c2
2
mc
E = moc2 + KE
rest
kinetic KE does not equal ½ mv2 at
energy energy relativistic speeds
E=mc2: Ex 1
How much energy would be released if a p0 meson
(mo=2.4 X 10-28 kg) decays at rest.
E = mc2
E = moc2
(particle is at rest)
E = (2.4 X 10-28 kg)(3.0 X 108 m/s)2
E = 2.16 X 10-11 J
E=mc2: Ex 2
A p0 meson (mo=2.4 X 10-28 kg) travels at 0.80 c.
a. Calculate the new mass [4 X 10-28 kg]
b. Calculate the relativistic momentum [9.6 X 10-20
kg m/s]
c. Calculate the energy of the particle (E2 = mo2c4 +
p2c2 ) [ANS: 3.6 X 10-11 J]
E=mc2 + KE: Ex 3
What is the kinetic energy of the p0 meson in the
former example.
E = moc2 + KE
KE = E – moc2
KE = 3.6 X 10-11J - (2.4 X 10-28 kg)(3.0X108 m/s)2
KE = 1.4 X 10-11 J
E=mc2 + 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 10-19 J
1 MeV = 106 eV = 1.60 X 10-13 J
What is the rest mass of an electron in MeV?
E = mc2
E = moc2
(particle is at rest)
E = (9.11 X 10-31 kg)(3.0 X 108 m/s)2
E = 8.20 X 10-14 J
8.20 X 10-14 J 1 MeV
1.60 X 10-13 J
= 0.511 MeV/c2
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’/c2
u = overall speed with respect to stationary observer
v = speed of moving object with respect to st. observer
u’ =speed of 2nd 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’/c2
u = 0.60c + 0.60c
1 + [(0.60c)(0.60c)/c2 ]
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’/c2
u = 0.60c + c
1 + [(0.60c)(c)/c2 ]
u=c
=
1.60c
1.60
Relativistic Addition: Ex 3
Now the car is travelling at c and turns on its
headlights.
u = v + u’
1 + vu’/c2
u=c+ c
=
1 + [(c)(c)/c2 ]
u=c
2c
2