Relativity
Classical” or “Newtonian”
physics does a good job of
describing the behavior of
particles at low speeds.
But as speeds approach light
speed we must use The Special
Theory of Relativity proposed
by Einstein in 1905.
Classical physics represents
the “low speed limit” of
relativistic physics.
What do we have to know about relativity
to solve problems in a correct way?
=> Just change Newton’s laws by
introducing a correction factor for the
mass F = d ( mv ) / dt
m0
m=
2
2
1− v / c
The principle of relativity was
first stated by Newton
“The motions of bodies included in a
given space are the same among
themselves, whether that space is at
rest or moves uniformly forward in a
straight line”
Reference Frames
The frame of reference for a person or object is
the coordinate system which moves with the
person or with the object.
Think of the frame of reference as a set of xyz
axes which are attached to the object.
An inertial reference frame is one in which
Newton’s Laws are valid. All inertial reference
frames move with constant velocity relative to
one another. (= Special Theory of Relativity)
*If frames are accelerated (translational or rotational) then General
Relativity applies (A little bit more complex)
The Postulates of Relativity
The Special Theory of Relativity is based on
two postulates:
• The laws of physics are the same in all
inertial reference frames.
• The speed of light in vacuum is ALWAYS
measured to be 3 × 108 m/s, independent of
the motion of the observer or the motion of
the source of light.
Both postulates tested exhaustively, no exceptions found!
The Speed of Light
If we apply Newtonian Mechanics to describe the
propagation of light, then two observers who are in
two different reference frames will measure a
different value for the speed of light if there is relative
motion between the reference frames. (Recall v13 =
v12 + v23)
However, that does
not happen – all
observers measure
the same speed of
light.
We need Relativity
Theory to explain this!
The Ultimate Speed
Experiment by Bertozzi in 1964 accelerated electrons and measured their
speed and kinetic energy independently. Kinetic energy →∞ as speed → c
Fig. 37-2
Ultimate Speed→Speed of Light:
c = 299 792 458 m/s
37-
Testing the Speed of Light Postulate
If speed of light is same for all inertial reference frames, then speed of light
emitted by a source (pion, π0) moving relative to a given frame (for example, a
laboratory) should be the same as the speed light that is emitted by a source
that is at rest in the laboratory).
1964 experiment at CERN (European particle physics lab): Pions moving at
0.9975c with respect to the laboratory decay, emitting two photons (γ).
π0 →γ +γ
The speed of the light waves (γ-rays) emitted by the pions was measured
always to be c in the lab frame (not up to 2c!)→same as if pions were at rest
in the lab frame!
37-
The problems of
“Classical Physics”
•
•
•
•
Transformation of coordinates
Maxwell equations wrong?
Lorentz transformation
Michelson-Morley experiment
The Lorentz Transformation
How are coordinates x, y, z, and t reporting an event in frame S related to the
coordinates x', y', z', and t' reporting the same event in moving frame S'?
Gallilean Transformation Equations
x ' = x − vt
(approximately valid
t'=t
at low speeds)
(37-20)
Origins coincide at t = t' =0
Fig. 37-9
Lorentz Transformation Equations
x ' = γ ( x − vt )
y'= y
(valid at all
z'= z
physically possible speeds)
t ' = γ ( t − vx c 2 )
(37-21)
37-
Fig. 37-9
x ' = x − vt
(approximately valid
t'=t
at low speeds)
(37-20)
dx' dx
=
−v
dt dt
Assume that you travel in your car with a speed of
1 108 m/s and light travels past your car (S, 3 108 m/s) travels past
your car.
The relative speed should be 2 108 m/s !!!!!!!
The Michelson Morley Experiment
"But it can't be so." "It just doesn't make sense."
horizontal path
cthor ,1 = L + Δx = L + vthor ,1
cthor , 2 = L − Δx = L − vthor , 2
L
thor ,1 =
(c − u )
L
2L / c
→ thor ,1 + thor , 2 =
thor , 2 =
(c + u )
1 − v2 / c2
vertical
(ctvertical ) = L + (utvertical ) → tvertical
2
2tvertical =
2
2L
2L / c
=
c2 − u 2
1 − u 2 / c2
L
= 2
c − u2
2tvertical
2L
2L / c
= 2
=
2
c −u
1 − u 2 / c2
thor ,1 + thor , 2
2L / c
=
2
2
1− v / c
Lorentz:
L|| = L0 1 − u / c
2
2
x ' = γ ( x − vt )
y'= y
(valid at all
z'= z
physically possible speeds)
t ' = γ ( t − vx c 2 )
(37-21)
Time is Relative
Time interval measurements depend on the
frame in which they are measured.
Events are any physical occurrence that
happens at a specific location at a specific time.
This a definition.
Two events which appear to be simultaneous in
one reference frame are in general not
simultaneous in a second frame moving with
respect to the first.
Conclusions:
In Relativity Theory, the distance between
two points and the time interval between
two events depend on the frame of
reference in which they are measured!
There is no absolute measure of time.
There is no absolute measure of length.
Time Dilation
Two observers, each in their
own reference frame moving
with a relative velocity will not
agree on how fast time passes.
Each will think the other’s clock
is wrong.
Moving clocks run slow
This effect is known as time
dilation.
2d
Δto =
c
Δt =
2d
c2 − v2
=
2d
v2
c 1− 2
c
The proper time Δt0 is the time interval between
two events as measured by an observer who
sees the events occur at the same place (same
location x,y,z). Let the “proper” frame move with
velocity v with respect another frame.
Time interval in moving frame:
2d
Δt 0 / Δt =
c
Æ Δt = γ Δt0
where
2d / c
1 − v2 / c2
1
γ=
2
2
1− v / c
Example:
The decay of a muon. A muon is a naturally occurring subatomic particle of
nature that is unstable and transforms itself to other parts with a mean time
to decay of 2.20μs (created by cosmic radiation high in Earth’s atmosphere).
The muon(-/+) mass is roughly 310 times that of an electron (positron).
Assume that a muon is created at an altitude of 5.00 km above the surface of the
earth and it travels toward the earth with a speed of 0.995c. The muon the gets a
certain distance into the atmosphere and decays. So there is a time delay between
the two events. (Without time delay a Muon cove 657m before decaying)
Let that time delay be measured form the rest frame of the earth.
2.2 μ s
2.2 μ s
Δt =
=
2
1 − .990025
(.995c)
1−
c2
2.20μ s
Δt =
= 22.02 μ s
−2
9.987492 × 10
The average distance traveled is
(0.995c)x(22.02μs)=6570 m
So the muons make it to the ground.
Speed
Lorentz factor
Reciprocal
β=v/c
γ
1/γ
0.010
1.000
1.000
0.100
1.005
0.995
0.200
1.021
0.980
0.300
1.048
0.954
0.400
1.091
0.917
0.500
1.155
0.866
0.600
1.250
0.800
0.700
1.400
0.714
0.800
1.667
0.600
0.866
2.000
0.500
0.900
2.294
0.436
0.990
7.089
0.141
0.999
22.366
0.045
Movie simlultaneity
Example 1
An astronaut at rest on Earth has a heartbeat rate
of 70 beats/min. When the astronaut is traveling
in a spaceship at 0.90c, what will this rate be as
measured by (a) an observer also in the ship and
(b) an observer at rest on the Earth.
a )70beats / min
b)Δt = γΔt0 = 2.29 min
⇒ 31beats / min
Example 2 An unstable particle called the pion has a mean
lifetime of 25 ns in its own frame. A beam of pions travels
through the laboratory at a speed of 0.60c. (a) What is the
mean lifetime of the pion as measured in the laboratory?
Given: Δt0 = 25 ns and v = 0.60c (β=0.60)
γ=
1
1− β 2
=
1
1 − 0.60 2
= 1.250
Δt = γΔt0 = (1.250)(25 ns ) = 31 ns
(b) How far does a pion travel (as measured by laboratory
observers) during this time?
d = vΔt = (0.6c )(31 ns ) = 5.6 m
Length Contraction
(Lorentz- or Fitzgerald Contraction)
The distance between two points depends on the
frame of reference in which it is measured.
The proper length L0 is the length of the object
measured in the frame of reference in which the
object is at rest.
If the object is moving in a reference frame, its
length L will be measured to be less than its proper
length:
L = L0 / γ = L0 1 − v 2 /c 2
This is known as relativistic length contraction.
Example 3
A friend in a spaceship travels past you at a high speed. He
tells you that his ship is 20 m long and that the identical ship
you are sitting in is 19 m long. According to your observations,
(a) how long is your ship, (b) how long is his ship, and (c)
what is the speed of your friend's ship?
L=
L0
γ
L0 20m
γ= =
= 1.05
L 19m
1
γ=
1− β 2
1
1
β = 1− 2 = 1−
2 = 0.3
γ
⎛ 20 ⎞
⎜ ⎟
⎝ 19 ⎠
v
β = ⇒ v = 0.3c
c
The proper length of one spaceship is three times
that of another. The two spaceships are traveling
in the same direction and, while both are passing
overhead, an Earth observer measures the two
spaceships to have the same length. If the slower
spaceship is moving with a speed of 0.35c,
determine the speed of the faster spaceship.
L1 =
L0 ,1
γ1
L2 =
L1 = L2
γ 2 = 3γ 1 = 3.2
⇒ β = 0.95
L0 , 2
γ2
=
3L0 ,1
γ2
A supertrain of proper length 100 m travels at a
speed of 0.95c as it passes through a tunnel
having proper length 50 m. As seen by a
trackside observer, is the train ever completely
within the tunnel? If so, by how much?
L1 =
L0
γ
for β = 0.95
γ = 3 .2
⇒ L = 100m / 3.2 = 31m
YES! by 19m
The Relativity of Velocities
Δx = γ ( Δx '+ vΔt ')
dx' dx
=
−v
dt dt
Δx
u=
Δt
Δx '
and u ' =
Δt '
u '+ v
u=
1 + u ' v c2
u = u '+ v
v Δx ' ⎞
⎛
Δt = γ ⎜ Δt '+ 2 ⎟
c ⎠
⎝
Δx
Δx '+ vΔt '
=
Δt Δt '+ Δx ' c 2
Δx
Δx ' Δ t ' + v
=
Δt 1 + v ( Δ x ' Δ t ' ) c 2
(relativistic velocity transformation)
(37-29)
(classical velocity transformation) (37-30)
37-
Relativistic Addition of Velocities
In relativity theory we must modify the way we
add velocities: The correct relativistic addition
of velocities is:
v1 + v 2
v =
v 1v 2
1 +
2
c
We are moving on a straight line relative to an object
(for example an asteroid or earth) with a speed v1 of
with respect to that object. We fire an object away from
us at speed relative to us v2. If the two objects move on
the same straight line then, the speed of object 2 as
seen from the asteroid v is given by the equation above
A spaceship, Enterprise, moving at 0.99c
away from the earth launches a spaceship,
Picard, back toward the earth with a speed
relative to the Enterprise of 0.6c. What
speed do we see the Picard moving on the
earth?
v=
v1 + v2
0.99c − 0.6c
.36c
0.36c
=
=
=
= 0.8866995c
v1v2
.99c( −.6c ) 1 − .5940 .4060
1+ 2
1+
c
c2
If instead, if the Picard was launched away
from the Enterprise in the same direction as
its motion, the answer would be
v=
v1 + v2
0.99c + 0.6c
1.59c
1.59c
=
=
=
= .99624c
v1v2
.99c(.6c ) 1 + .5940 1.5960
1+ 2
1+
c
c2
as seen from the earth (assumed to be at rest
in these two examples).
Spaceship R is moving to the right at a speed of
0.70c with respect to the Earth. A second
spaceship, L, moves to the left at the same
speed with respect to the Earth. What is the
speed of L with respect to R?
v1 + v 2
0 .7 c + 0 .7 c
=
= 0 . 94 c
v=
v1 v 2
0 .7 c 0 .7 c
1+ 2
1+
2
c
c
A spaceship is traveling at 0.95c with
respect to the Earth. Inside the spaceship, a
man shines a flashlight in the direction the
spaceship is moving. What is the velocity of
the light from the flashlight with respect to
Earth?
v1 + v2
0.95c + c 0.95 + 1
v=
=
=
c=c
v1v2
0.95c c 1 + 0.95
1 + 2 1+
2
c
c
Doppler Effect for Light
Let f0 represent the proper frequency (frequency in the source's rest frame)
f = f0
1-β
1+β
(source and detector separating)
(37-31)
If source and detector moving towards one another β → - β
Note: Unlike Doppler shift with sound, only relative motion matters since there
is no ether/air to be moving with respect to.
Low Speed Doppler Effect
For β<<1
f = f 0 (1-β + 12 β 2 )
(source and detector separating, β 1) (37-32)
37-
Doppler Effect for Light, cont'd
Astronomical Doppler Effect
f = f 0 (1-β )
(37-33)
Proper wavelength λ0 associated with rest frame frequency f0.
c
λ
=
c
λ0
λ0 = λ (1-β )
(1-β )
λ = λ0 (1-β )
-1
(37-34)
λ − λ0
β=
λ0
(37-35)
Replacing β=v/c and using λ-λ0 = |Δλ| = Doppler shift
v=
Δλ
λ0
c
(radial speed of light source, v c)
(37-36)
37-
Doppler Effect for Light , cont'd
Transverse Doppler Effect
Classical theory predicts no
Doppler shift observed at point D
when source S is at point P.
f = f 0 1-β 2
Fig. 37-12
(transverse Doppler Effect) (37-37)
For low speeds (β<<1)
f = f 0 (1- 12 β 2 )
(low speeds) (37-38)
Transverse Doppler effect another test of time dilation (T=1/f)
T=
T0
1-β
2
=γ T0
(37-39)
Proper period T0=1/f0)
37-
Doppler Effect for Light , cont'd
The NAVSTAR Navigation System
v1
f03
f01
v2
vairplane
v3
f02
Given v1, v2, v3, f01, f02, f03, and measured f1, f2, f3, can determine vairplane,
37-
Relativistic Momentum
In order for momentum to be conserved in
relativistic interactions, we must modify our
classical expression for momentum to read:
p=
m 0v
1 − v 2 /c 2
= γ m 0v
m0 is called the rest mass of the object.
Doing work increase mass
and velocity!!
http://www.aip.org/history/einstein/sound/voice1.wav
Relativistic Energy
The total energy E of an object moving with velocity v is:
2
m
c
0
E = γm0 c 2 =
= mc 2
2 2
1
−
v
/c
Mass is another form of energy!
The energy of an object at rest is E0 = m0c2, which is known as
the rest energy.
The total energy is comprised of the rest energy and the
2
kinetic energy:
Relativistic
kinetic energy
m0 c
2
= m0 c + K
E=
2
2
1 − v /c
m0 c 2
2
−
m
c
K=
0
2
2
1 − v /c
Mass at rest m0 “rest mass”
Mass at a certain velocity m= γ m0
Old answer:
New answer:
E=
m0 c 2
1− β
2
1− β ≈ 1−
2
E=
m0 c 2
1−
β
2
2
E = ½ m v2
E=mc2= γ m0c2
small velocity β = very small
β
2
2
⎛ β
since ⎜⎜1 −
2
⎝
2
1
now
1−
β
2
2
2
⎞
β2 β4
⎟⎟ = 1 - 2
+
2
4
⎠
≈ 1+
β2
2
→ 1 = 1+
β4
4
2
v
2
2
1
β2
2
2
2 β
2
2 c
2
E = m0 c (1 + ) = m0 c + m0 c
= m0 c + m0 c
= m0 c + m0 v 2
2
2
2
2
Energy of a photon?
E=h f = mc2= γ m0c2
∞
0
In nuclear reactions like in the sun,
we convert four hydrogen atoms
into one helium atom. The mass
before and after the reaction is
different – the mass difference is
released as the energy that
drives the solar furnace.
Nuclear Fusion
The fusion of one pound of hydrogen into helium yields as much energy as
the burning of 10,000 tons of coal.
Example 7
Calculate the momentum of a proton
moving with a speed of 0.5c.
Proton rest mass mp: 1.6726231e-27 kg
1.67 ⋅ 10 kg ⋅ 0.5 c
m0v
p=
=
2
2
2
1 − v /c
1 − 0.5
-27
p = 1⋅ 10 kg ⋅ 0.5 c
-27
Example 8
An electron moves with a speed of 0.80c.
Calculate its (a) rest energy, (b) total energy,
and (c) kinetic energy.
Rest mass[electron] = 9.10939*10-31 kg
E rest = m 0 c = 9 . 1 ⋅ 10
2
− 31
kg c
2
Etotal = γ m0c = 1.67 ⋅ 9.1 ⋅10 kg c
2
−31
2
K = γ m0c 2 − m0c 2 = 0.67 ⋅ 9.1 ⋅ 10−31 kg c 2
A New Look at Energy, cont'd
p 2 = 2 Km
Momentum and kinetic energy
( pc )
2
2
= K 2 + 2 Kmc
0
E = ( pc ) + ( mc
0
2
2
2
)
2
(classical)
(37-51)
(37-54)
(37-55)
sin θ = β and cosθ = 1 γ
(37-56)
Fig. 37-15
37-
A proton in a high-energy accelerator is
given a kinetic energy of 50.0 GeV.
Determine (a) the momentum and (b) the
speed of the proton.
1 electron volt = 1.60217646 × 10-19 Joules
Proton rest mass mp: 1.6726231e-27 kg
( pc )
2
= K 2 + 2 0Kmc 2
K = γ m0c − m0c
2
K
γ =
+1
2
m0c
(37-54)
2