Transcript Document

Special Relativity
SPH4U
Review of Scientific “Theories”




Recall discussion from the first day of class
A scientific “theory” is a proposed
explanation/description for observed facts
It is possible for a theory to be a good
approximation or have some usefulness even if
it is not fully correct
One of the best examples is “Newtonian”
Physics vs. Relativity & Quantum Mechanics
Newtonian Physics

Physics principles as
explained by Newton
and others

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Newton’s 3 Laws and
Law of Gravity
Maxwell’s Equations of
Electromagnetism
Equations for motion,
momentum, kinetic
energy, etc. discussed
earlier in this class
Underlying foundations
of space and time as
absolute
Relativity and Quantum Mechanics

New physics as described by Einstein and
others, most of the work done in the early
1900s
Time dilation, length contraction
 Uncertainty principle
 Bohr Theory of the Atom


Different fundamental assumptions about
the Universe
The Special Theory of Relativity

Aimed to answer some burning
questions:
Could Maxwell’s equations for electricity
and magnetism be reconciled with the laws
of mechanics?
 Where was the aether?

The Conflict
Newtonian physics seems to describe the
world as we are used to it
 However, several experiments as well as
some hypothetical arguments signaled
some problems
 Relativity and Quantum Mechanics
improve upon Newtonian physics

Newtonian Physics

Newtonian physics accurately describes
the Universe when…
Speeds are not too large
 Gravity is not too strong
 You are at a macroscopic level, i.e. not
dealing with individual molecules/atoms

Newtonian Physics, cont.

Under the conditions of the previous slide,
there is no reason to use anything other
than Newtonian physics
Equations give the same results to high
accuracy
 Example: Trajectories of satellites and space
probes use Newtonian physics

Relativity


Relativity is a set of physics concepts and laws
deduced by primarily by Albert Einstein
Special Relativity

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Published by Einstein in 1905
“Special” case with no forces/acceleration
General Relativity

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Published by Einstein in 1915
Extension of previous theory to include forces
What ISN’T Relativity?
Relativity does not simply mean
“everything is relative”
 On the contrary, relativity says certain
things are relative, and other things are
absolute
 Relativity also tells us by how much those
certain things are relative and in what way

Experimental and Theoretical Need
for Relativity

Michelson-Morley Experiment


Speed of light is the same regardless of the
Earth’s motion through the aether (“absolute
space”)
Maxwell’s Equations of Electromagnetism

Predict very unusual things, like magnetic
fields with “loose ends”, when speeds are
extremely large
Michelson-Morley
Experiment


For a long time, scientists
believed in an “aether”—
absolute space
In the Michelson-Morley
experiment, the speed of
light was measured “with”
and “across” the “flow of
the aether” as the Earth
moved through it
Michelson-Morley Experiment
Flash
 Contrary to expectations, however, the
speed of light was the same both “with”
and “across”!

Theoretical Foundations of
Relativity

To explain all of these things, Einstein came up
with new laws of physics based on two
assumptions

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
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The laws of physics are the same in all inertial (nonaccelerating) frames
The speed of light is the same as measured by all
observers in all inertial frames
Einstein took these principles “on faith”
The principles and their implications have
passed subsequent experimental testing
Relatively Speaking

What do Einstein’s two assumptions imply?
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All motion is relative
Relativity of simultaneity
Relativistic velocity addition
Time dilation
Length contraction
Relativistic mass increase
E = mc2
Who is moving?
All Motion is Relative, cont.



You and your friend
Jackie like to travel in
bizarre spherical
spaceships
Who is moving?
Who is stationary?
All Motion is Relative, cont.
In spite of our everyday intuition, the only
velocities that can be measured are
relative velocities
 Examples:

Relative to the surface of the Earth
 Relative to the Sun
 Relative to a distant galaxy

Galilean 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 Galilean or Classical Relativity.
1,000,000 ms-1
Einstein’s Special Relativity
0 ms-1
300,000,000 ms-1
1,000,000 ms-1


Both spacemen measure the speed of the approaching ray of light.
How fast do they measure the speed of light to be?
Nothing Can Go Faster Than The
Speed of Light
Addition of Velocities

In normal circumstances,
if you are moving and
throw an object, an
outside observer will see
the object at a different
velocity


Straight-forward velocity
addition
But all observers
measure the speed of
light to be the same
Velocity Additions Do Not Apply to Light

Even if you are
moving away from
your friend at a very
high velocity, you
will both see a light
beam moving at c.
Nice to know formula
v  u
u
vu
1 2
c
v  velocity of object 1 relative to you
u = velcity of object with respect to object 1
Relativistic Velocity Additions



A formula for adding
velocities exists, but it is
not required for the
course.
The formula works such
that you can never get
velocities greater than c
For small velocities, is
approximately the same
as just adding the
velocities
v  u
0.75c  0.75c
1.5
u


 0.96
2
vu
0.75c  0.75c  1  .75 
1 2 1 
c
c2
v  u
u
vu 
1 2
c
0.8c  0.9c

0.8c  0.9c 

1
c2
1.7c

1  .8 .7 
 0.988
 Relativistic
uvelocity of object in rest frame
Velocity Additions

vvelocity of moving frame  uvelocity
of object in moving frame


vvelocity of moving frameuvelocity
of object in moving frame
1
c2
Relativity of Simultaneity
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Two lights an equal distance
from M go off
A passing train carries M’
M’ sees the light from B first
M see the light flashes at the
same time
M’ is moving in the direction
of B
This relativity is determined
by the speed of light and the
relative motion of the
objects/observers
Relativity of Simultaneity
Events which are simultaneous in one
frame may not be in another!
 Each observer is correct in their own
frame of reference

The Lorentz Factor


Calculating length contraction, time
dilation, and other quantities requires
calculating the Lorentz factor

1
2
v
1 2
c

1  2
 = v/c



1
If v = 99% of c, then  = 0.99
 is always < 1    1
The Lorentz Factor, cont.

Some examples:
v = 0.1% of c   = 1.0000005
 v = 1% of c   = 1.00005
 v = 10% of c   = 1.005
 v = 50% of c   = 1.155
 v = 90% of c   = 2.294
 v = 99% of c   = 7.089
 v = 99.9% of c   = 22.37

Distance = Velocity x Time
Time Dilation
d  vt
2D  cts
D
2D
ts 
c
Note:
A clock using
light pulses to
keep time.
Every time the
pulse returns, a
unit of time has
passed
ts is also written as t0
This time is known as Proper Time. Because the clock is
rest the frame of the occurring event. The Proper Time
interval between two events is always the time interval
measured by an observer for whom the two events take
place at the same position.
Time Dilation
L=vt
½ ct
D
½ vt
V
We are now watching the clock
move
2
4v.DWe will examine
2
2
2
horizontally
with
velocity

t



 ctM 
 vtone
M
2
M 
cycle,
more
specifically
one-half
of one

D

v2 
2




1 light
 2 
 2cycle.
 During a cycle cof the
2 photon the
c

 a distance
clock will have moved horizontally
2
2
 ctM   vtM  L, and
2 if we calculate the distance travelled


D
2 D (a upside down

 
 by the light in this one cycle
2D
2
2

t


 

Now since t s 
2
V), the distance Mwould be cvtimes
the time
c
2
1  that
we measured for theccycle,
is ct. So for
2
 tM  2 2
2
c
by the
says a moving clock run

 c  v  D one-half cycle the distance travelledThis
 2 
light is ½ ct.
slow. If ts =1 then you watching


D
 tM 



2
2
 2  c  v 
2
2
tM 
ts
1
2
v
c2
it move would notice it taking
more than 1s (tm >1) on your
clock, so ts runs slow.
Time Dilation Example
You and a friend are having a eating contest. Your friend
is on a train traveling at speed v=0.9 c. By her watch,
she finishes her food in 5 seconds. Determine the time
you measure, if you are standing still at the train
station.
tM 
ts
2
v
1 2
c
5

1
(0.9c)
c2
2
5

1  .81
 11.5 seconds
Since eating is happening on the train, that is the “proper”
time, ts=5.
Time Dilation Example 2
Now it is your turn to eat. According to your watch you
finish your food in 5 seconds. How long does your
friend think it took you to finish the food?
Now eating is happening at the station, so that is the
“proper” time, again ts=5.
tM 
ts
2
v
1 2
c

5
(0.9c)
1
c2
2

5
1  .81
 11.5 seconds
Your friend would consider you to be moving. Remember the proper time is where
the event and clock are together
Both people think they won!
Space Travel
Alpha Centauri is 4.3 light-years from earth. (It
takes light 4.3 years to travel from earth to Alpha
Centauri). How long would people on earth think it
takes for a spaceship traveling v=0.95c to reach
A.C.?
d 4.3 light-years
 4.5 years

t M 
0.95 c
v
How long do people on the ship think it takes?
People on ship have ‘proper’ time since they see earth
leave, and Alpha Centauri arrive. ts
tM 
ts
v2
1 2
c
ts  tM
v2
1 2
c
 4.5 1  .952
ts = 1.4 years
Space Travel
Another approach that solves any special relativity
problem by treating space and time as spacetime.
The only requirement is that both separated units are
recorded in the same units. (i.e.: light seconds, light
minutes, light years, …)
Space Travel
Alpha Centauri is 4.3 light-years from earth. (It takes light 4.3 years
to travel from earth to Alpha Centauri).
How long would people on earth think it takes for a spaceship
traveling v=0.95c to reach A.C.?
d 4.3 light-years
 4.526 years

t M 
0.95 c
v
How long do people on the ship think it takes?
An amazing technique is to place time and space in the same units
then use the following relativistic formula:
 Rocket time interval    Rocket event seperation 
2
2
  Earth time interval    Earth space seperation 
2
 Rocket time interval    0    4.526 years    4.3 years 
2
 Rocket time interval 
2
2
2
 1.998 years 2
Rocket time interval=1.4 year
2
2
Time Dilation Review



Time flows more slowly
in a moving frame as
observed by an outside
observer
But remember motion is
relative
If you and I are moving
past each other


I see your clock moving
more slowly
But you also see mine
moving more slowly…!!!
Length Contraction

Objects moving relative to an outside
observer appear contracted in the direction of
their motion as measured by the observer
Length Contraction, cont.

If you and I move past
each other in some
sweet sports cars



I measure your sports car
as being shorter
You measure my sports
car as being shorter
Only applies to the
direction of motion

We see our sports cars
as still being the same
height
Length Contraction
v=0.1 c
v=0.8 c
v=0.95 c
Length Contraction Example
People on ship and on earth agree on relative velocity v =
0.95 c. But they disagree on the time (4.5 vs 1.4 years).
What about the distance between the planets?
Earth/Alpha:
d0 = v t = .95 (3x108 m/s) (4.5 years)
= 4x1016m (4.3 light years)
Ship:
d = v t = .95 (3x108 m/s) (1.4 years)
= 1.25x1016m (1.3 light years)
Length in moving
frame
Length in object’s
rest frame
v2
LM  Ls 1  2
c
Twin Paradox
Twins decide that one will travel to Alpha Centauri
and back at 0.95c, while the other stays on earth.
Compare their ages when they meet on earth.
Earth twin thinks it takes 2 x 4.5 = 9 years
Traveling twin thinks it takes 2 x 1.4 = 2.8 years
Traveling twin will be younger!
Note: Traveling twin is NOT in inertial frame!
Question
You’re eating a burger at the interstellar café in outer
space - your spaceship is parked outside. A speeder
zooms by in an identical ship at half the speed of
light. From your perspective, their ship looks:
(1)
longer than your ship
(2)
shorter than your ship
(3)
exactly the same as your ship
In the speeder’s reference frame
v2
LM  Ls 1  2
c
Always <1
Ls > LM
In your reference frame
Comparison:
Time Dilation vs. Length Contraction

to = time in same reference frame as event

i.e. if event is clock ticking, then to is in the reference
frame of the clock (even if the clock is in a moving
spaceship).
vv22
ts0  tm 11 22
cc

t > to
Time seems longer
from “outside”
Lo = length in same reference frame as object

length of the object when you don’t think it’s moving.
vv22
Lm  LL0s 11 22
cc
L0 > L
Length seems shorter
from “outside”
Relativistic Mass Increase

Einstein made two other surprising
discoveries…

Mass must increase with speed, as viewed by
an outside observer
 Due

to conservation of momentum
There is “leftover” energy even when the
object is at rest
 Due
to conservation of energy
 E = mc2
Relativistic Mass
mM 
Actually written
m
ms
Rest mass
2
v
1 2
c
m0
2
v
1 2
c
E = mc2
E = mc2 = m0c2
 This E is the total energy of an object
 When the object is at rest…

v=0
=1
 E = m0c2 (“rest mass energy”)


The reason that energy can be released
through fusion/fission
Total Energy
Relativistic kinetic energy is the extra energy an object with mass has as a
result of its motion:
Etotal  Erest  EK
We can solve this for the Kinetic energy of an object:
EK  Etotal  Erest

m0 c 2
v2
1 2
c
 m0 c
2
Relativistic Momentum
Relativistic Momentum
Note: for v<<c p=mv
Note: for v=c
p
p=infinity
mv
v2
1 2
c
Relativistic Energy
Note: for v=0
E = mc2
E
Note: for v<<c E = mc2 + ½ mv2
Note: for v=c E = infinity (if m<> 0)
Objects with mass can’t go faster than c!
mc 2
v2
1 2
c
Question
Calculate the rest energy of an electron (m=9.1x10-31 kg) in joules.
E0  m0 c 2
m

E0   9.11031 kg   3.0 108 
s

 8.2 1014 J
2
Calculate the electron’s Kinetic energy if it is moving at 0.98c.
EK 
m0 c 2
1
2
v
c2
 m0 c 2
EK 
8.2 1014 J
0.98c 

1
c2
 3.3 1013 J
2
 8.2 1014 J
Simultaneous?
A flash of light is emitted from the exact center of a box.
Does the light reach all the sides at the same time?
At Rest
YES
Moving
NO
Simultaneous depends on frame!
Simultaneous?
Many times, questions are concerned with the determination of the spatial
interval and/or the time interval between two events. In this case a useful
technique is to subtract from each other the appropriate Lorentz contraction
describing each event.
tb  t a 
v
x 'b  x 'a 
2 
c
v2
1 2
c
 t 'b  t 'a  
Three Other Effects

3 strange effects of special relativity
Lorentz Transformations
 Relativistic Doppler Effect
 Headlight Effect

Lorentz Transformations
Lorentz Transformations
■ Light from the top of the bar has further to travel.
■ It therefore takes longer to reach the eye.
■ So, the bar appears bent.
■ Weird!
Doppler Effect

The pitch of the siren:
Rises as the ambulance approaches
 Falls once the ambulance has passed.


The same applies to light!
Approaching objects appear blue (Blue-shift)
 Receding objects appear red (Red-shift)

Headlight effect
V



Beam becomes focused.
Same amount of light concentrated in a
smaller area
Torch appears brighter!
Warp

Program used to visualise the three effects
Fun stuff
Eiffel Tower
Stonehenge
Summary
t 
t0
v2
1 2
c
v  u
u
vu 
1 2
c
v2
L  L0 1  2
c
m0
m
v2
1 2
c
m0 v
p
v2
1 2
c
E0  m0 c 2
Etotal 
m0 c 2
1
v
 t 'b  t 'a   2  x 'b  x 'a 
c
tb  t a 
v2
1 2
c
 m0 c  EK
2
2
v
c2
tb  ta 
v
x  xa 
2  b
c
v2
1 2
c
 tb  t a  
Understanding
An observer has a pendulum that has a period of 3.00 seconds. His friend who
happens to own a spaceship (with cool engines), zooms by the stationary
pendulum. If the speedometer of the spaceship says 0.95c, what will the friend
measure are the period of the pendulum?
Since I am with the
pendulum, my
measured time is the
Proper Time.
t
t0
v2
1 2
c

3s
0.95c 

1
c2
2
 9.6s
This makes sense
because a moving clock
would run slower from
my perspective. So the
pendulum would have a
period of 9.6s.
Understanding


Vega is 25 light-years
away
Travel to Vega at 0.999c





The length would appear
contracted to you
About 1 light-year
Make the trip in ~1 lightyear (each way) as
measured by you
Earth would measure 25
years each way
You would spend 2
years (your time)
travelling and arrive 50
years in the future Earth
time.
Understanding

You throw a photon (3x108 m/s). How fast
do I think it goes when I am:

Standing still
3x108 m/s

Running 1.5x108 m/s towards
3x108 m/s

Running 1.5x108 m/s away
3x108 m/s
Strange but True!
Understanding
A 1.0 m long object with a rest mass of 1.0 kg is moving at 0.90c. Find its relative
length and mass
v2
LM  Ls 1  2
c
Use length contraction formula:
LM  1.0m 
0.90c 

1
2
c2
 1.0m  0.4346 
 0.44m
Mass increase formula:
m
m0
2
v
1 2
c

1.0kg
0.90C 

1
c2
2
 2.3kg
Understanding
For a 1.0 kg mass moving at 0.90c. Find the rest energy and kinetic energy of
the object
For rest energy, Use energy formula:
E0  m0c2
2
m


E0  1.0kg   3.0 108 
s

 9.0 1016 J
For Kinetic energy, Use relativistic
energy formula:
E

m0 c 2
v2
1 2
c
9.0 1016 J
0.90c 

1
c2
2
 2.0 1017 J
Now:
E  m0c2  EK
Therefore
EK  E  m0c 2
 2.0 1017 J  9.0 1016 J
 1.11017 J
Understanding
A person’s pulse rate is 65 beats per minute.
a) If the person is on a spaceship moving at 0.10c, what is the pulse rate as
measured by a person on Earth?
b) What would the pulse rate be if the ship were moving at 0.999c?
a) Use time dilation:
a) Use time dilation:
We need
time for a
heart beat
tM 
tM 
ts
v2
1 2
c
ts
2
v
1 2
c
1 1
t 
 0.015
f 65


0.015min
0.10c 

1
c2
2
0.015min
 0.999c 
1
c2
2
 0.015min  65
beats
min
beats
 0.336min  3.0
min
Understanding
A muon at rest has an average lifespan of 2.20 x 10 -6 s
a) What will an observer on Earth measure as its lifespan if it travels at 0.990c?
b) What distance would we observe it travel before disintegrating?
c) What distance would it travel if relativistic effects were not taken into account?
a) Time dilation:
tM 
t0
v2
1 2
c

2.20 106 s
0.990c 

1
2
 1.56 105 s
c2
b) Distance formula
d  vtM   0.990   3.0 108 m  1.56 105 s   4630m
c) Distance formula
d  vts
To Show
consistency

s

8 m
  0.990   3.0 10   2.2 106 s   653m
s

v2
L  L0 1  2   4630m  1  .9902  653m
c
This is the distance
the muon measures
it travels before
disintegrating
Understanding
You measure the length of an object as 100m when it passes you at 0.90c.
What is its length when at rest?
Use length contraction formula:
http://onestick.com/relativity
v2
L  L0 1  2
c
L
L0 
v2
1 2
c
100m 

L0 
2
0.90c 

1
c2
100m 

 0.4346 
 230m
Understanding
As a rocket ship sweeps past the Earth with speed v, it sends out a light pulse
ahead of it. How fast does the light pulse move according to the people sitting
on the Earth?
uobject in rest frame
 in moving frame
vmoving frame  uobject
v  u



 in moving frame
vu
vmoving frameuobject
1 2 1
c
c2
v c
uobject in rest frame 
vc
1 2
c
vc

v
1
c
vc

vc
c
c
Understanding
A train 0.5 km long (as measured by an observer on the train, therefore this is
the proper length) is travelling at 100 km/h. Two lightening bolts strike the ends
of the train simultaneously as determined by an observer on the ground. What
is the time separation as measured by the observer on the train?
Units:
100
km  1000m   1h 
m



27.78
 

h  km   3600s 
s
We are given that tb-ta=0
and what we want to
determine is tb’-ta’
tb  t a 
v
x 'b  x 'a 
2 
c
v2
1 2
c
 t 'b  t 'a  
Understanding
A train 0.5 km long (as measured by an observer on the train, therefore this is
the proper length) is travelling at 100 km/h. Two lightening bolts strike the ends
of the train simultaneously as determined by an observer on the ground. What
is the time separation as measured by the observer on the train?
m
s
500m 
 t 'b  t 'a  
2 

8 m
3.0

10


s


0
27.78
2
m

27
.78


s

1 
2
m


8
3.0

10


s


m
s
500m 
 t 'b  t 'a   
2 

8 m
3.0

10


s


27.78
 t 'b  t 'a   1.54 1013 s
The negative sign reminds us that even a occurred after event b