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5/21/2018
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AP PHYSICS 2
UNIT 7Quantum Physics,
atomic, and nuclear physics
CHAPTER 25Special Relativity
Ether or no ether?
The work of Maxwell and Hertz led to the conclusion that light propagation could be explained by changing electric and magnetic fields that do not require any medium to travel.
Before this work, physicists were searching for ether.
This search produced an unexpected outcome that eventually changed the way we think about space and time.
WHITEBOARD VECTOR ANALYSISAn analogy: A boat race
Consider a process involving two identical
boats in a race on a wide river. Which boat returns to the starting dock first?
The speed of each boat is 10 km/h (relative to the water) and the river flows
at 6 km/h downstream (relative to the shore).
BOAT 1
𝑡𝑡𝑜𝑡𝑎𝑙 = 𝑡𝑢𝑝 + 𝑡𝑑𝑜𝑤𝑛
𝑡𝑡𝑜𝑡𝑎𝑙 =∆𝑑𝑢𝑝𝑣𝑛𝑒𝑡 𝑢𝑝
+∆𝑑𝑑𝑜𝑤𝑛𝑣𝑛𝑒𝑡 𝑑𝑜𝑤𝑛
𝑡𝑡𝑜𝑡𝑎𝑙 =1.6 𝑘𝑚
4𝑘𝑚ℎ
+1.6 𝑘𝑚
16𝑘𝑚ℎ
𝑡𝑡𝑜𝑡𝑎𝑙 = 0.5 ℎ
BOAT 2
𝑡𝑡𝑜𝑡𝑎𝑙 = 𝑡𝑟𝑖𝑔ℎ𝑡 + 𝑡𝑙𝑒𝑓𝑡
𝑡𝑡𝑜𝑡𝑎𝑙 = 2∆𝑑
𝑣𝑛𝑒𝑡 𝑐𝑟𝑜𝑠𝑠𝑖𝑛𝑔
𝑡𝑡𝑜𝑡𝑎𝑙 =3.2 𝑘𝑚
102 − 62𝑘𝑚ℎ
𝑡𝑡𝑜𝑡𝑎𝑙 = 0.4 ℎ
TESTING THE EXISTANCE OF ETHER Testing the existence of etherAlbert Michelson and Edward Morley
experiment - 1887
Imagine that ether fills the solar system andis stationary with respect to the Sun.
Because Earth moves around the Sun at aspeed of about 3.0 x 104 m/s, ether shouldbe moving past Earth at this speed.
Shining light waves parallel andperpendicular to the ether's motion relativeto Earth is similar to sending boats paralleland perpendicular to a flowing river.
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Albert Michelson and Edward Morley experiment - 1887
Albert Michelson and Edward Morley experiment - 1887
Testing the existence of etherAlbert Michelson and Edward Morley
experiment - 1887
Outcome: no matter how the interferometerorientation was changed, the interferencepattern did not change.
Possible conclusions:
There is not ether through which lighttravels.
There is ether, but it is stuck to Earth’ssurface and does not move relative to theinterferometer.
Testing the existence of ether
Physicists were reluctant to accept this result.
Perhaps the ether is at rest relative to Earth.
If the ether is attached to Earth, then as Earth rotates around its axis and
orbits the Sun, the ether should become twisted.
This would cause light coming from stars to be slightly deflected on its way
to Earth
No one observed such an effect.
INVARIANCE
Using Newton's laws yields consistentresults, regardless of the inertial referenceframes (not accelerating) used—a featureknown as invariance.
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If you are on an airplane moving smoothly at constant velocity all the events occur as if the plane was at rest on the ground:
If you drop an object, it will fall on your feet.
If you haul your luggage it will accelerate thesame it does at home.
If you do not have windows and you don’thear the noise, you don’t know you aretraveling at constant speed.
Hey dude? Where did you get
that stationary ceiling fan?
The forward motion of an adjacent
bus/car/train can give you the impression that
your own bus/car/train is moving backward.
If you are in a reference frame that is moving at a
constant velocity, there is no way to find out whether or
not you are moving (because you are not moving, in an
absolute sense! All motion is relative!)
There is no absolute rest frame in the universe.
All motion is relative. One could never say that a reference frame is “at rest” in an absolute sense.
INERTIAL REFERENCE FRAME
An inertial reference frame is one in which an observer sees that the velocity of the system
objects does not change if no other object exerts a force on it or if the sum of all forces
exerted on the system object is zero.
For observers in non inertial reference frames, the velocity of the system object can change
even though the sum of forces exerted on it is zero.
What would you say if you are . . .
Inside the inertial reference frame:
Outside the inertial reference frame:Dude is not moving
Dude moves to the right, at 0.5 m/s
Assume the sidewalk moves
at 0.5 m/s
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What would you say if you . . .
Run to the left at 0.5 m/s:
Run to the right at 0.5 m/sDude is moving away at 1.0 m/s
Dude is getting closer at 1.0 m/s
Assume the sidewalk moves
at 0.5 m/s
EINSTEIN’S THOUGHT EXPERIMENTS
Einstein embarked on an intellectual journey to determine what the speed of light would appear to be
from the reference frame of various observers.
light source
Detector
Assume the speed of light to be “c”
MADE UP EXPERIMENT 0
light source
Detector
v = 0 v = 0
What speed would the detector measure?
𝒗𝒍𝒊𝒈𝒉𝒕 = 𝒄
MADE UP EXPERIMENT 1
light source
Detector
v = c/2 v = 0
What speed would the detector measure? 𝒗𝒍𝒊𝒈𝒉𝒕 =
𝟑𝒄
𝟐
MADE UP EXPERIMENT 2
light source
Detector
v = c/2 v = 0 c
What speed would the detector measure?
𝒗𝒍𝒊𝒈𝒉𝒕 =𝒄
𝟐
MADE UP EXPERIMENT 3
light source
Detector
v = c/2 v = c/2
What speed would the detector measure?
𝒗𝒍𝒊𝒈𝒉𝒕 = 𝟐𝒄
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MADE UP EXPERIMENT 4
light source
Detector
v = c v = 0
What speed would the detector measure?
𝒗𝒍𝒊𝒈𝒉𝒕 = 𝟎Would light
stay in place?
WHAT DID EINSTEIN SAY?
EINSTEIN THOUGHT EXPERIMENT
Maxwell's equations (chapter 24) have to be written differently for different observers, with a
different speed of light in each case.
Einstein's two postulates
Postulate:
A postulate is a statement that is assumed tobe true. It is not derived from anything.
It is the starting point for a logical argument
Einstein's two postulates
1. The laws of physics are the same in allinertial reference frames.
Newton's second law remains the sameregardless of the inertial reference frame inwhich you choose to apply it.
• There is no experiment (mechanics, electricity,magnetism, thermodynamics) that is affected by themotion of an inertial reference frame.
Einstein's two postulates
2. The speed of light in a vacuum is measuredto be the same in all inertial reference frames.
The speed of light in a vacuum measuredby observers in different inertial referenceframes is the same regardless of therelative motion of those observers.
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In contrast to sound waves or a ball, the velocity of the airplane does not add to the velocity of a
flashlight beam to yield the velocity of light relative to the earth. The speed of light is the same for all
observers. (Velocity vectors are not drawn to scale.)
Experimental evidence for the constancy of light speed
The speed of gamma rays is measured inthe lab to be precisely the speed of light,despite being produced by the decayingpion, which was already moving near lightspeed relative to the lab.
This result supports postulate 2.
WHITEBOARD:What would each of the following observers
measure for the speed of light emitted by the bulb?
v = 0.9c
v = 0.9c
v = 0
The speed of light is constant for all observers!
They will ALL measure the speed of light as c.
v = 0.9c
v = 0.9c
v = 0
v = c/2 v = 0 v = c/2 v = 0
v = c v = 0v = c/2 v = c/2
As a result . . . .
What speed would the detector measure?
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The source emits light at speed c.
The detector will observe light traveling at speed c!
Light always travels at speed c, regardless of the observer’s reference frame!
This leads to some very bizarre results about the nature of space and time.
Simultaneity
So far, we have assumed that the timeinterval for an object moving from onepoint to another is independent of thereference frame.
The second postulate of the special theoryof relativity made physicists completelyrethink their ideas about time.
DetectorDetector
Motion of the train
Light Flashes
t0
DetectorDetector
Motion of the train
Light moves toward both detectors from
the point of flash
t1
DetectorDetector
Motion of the train
Light arrives at both detectors at the
same time
t2
DetectorDetector
Motion of the train
Light Flashes
t0
observer on platform
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DetectorDetector
Motion of the train
t1
observer on platform
DetectorDetector
observer on platform
t2Motion of the train
Detector
On the train you see the light flash reach both
detectors at the same time
An observer on platform sees the light reach before it reaches the one on the right
Simultaneity
Events that happen at the same time inone reference frame do not necessarilyoccur at the same time in anotherreference frame.
For the train example the times of eventswont vary more that 10-15 to 10-14 sbetween observers.
IMPLICATIONS OF THE DIFFERENCE IN OBSERVATIONS
This difference gets larger as the speedsinvolved become higher, but the effectbecomes significant only if the speeds area substantial fraction of light speed.
Phenomena that become significant onlyin high-speed circumstances are calledrelativistic effects.
DILATION OF TIME
The Persistence of memory
(Salvador Dalí - 1931).
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(Canela Hedgehog)
(Ginger Bearded Dragon)
Detector
Einstein’s Thought Experiment
A moving train cart is rigged so that a pulse of light starts at its floor and is detected when it reaches the ceiling.
First, let’s consider an observer that is on the train (Ginger the Bearded Dragon), at rest relative to the source of light
d
Detector
Whiteboard: Einstein’s Thought Experiment
Write an expression for d in terms of c and t0.
t0 is the time that it takes the light to reach the ceiling from the reference frame of a person that is at rest (v = 0) relative
to the source.
d
Detector
Whiteboard: Einstein’s Thought Experiment
By the time the light reaches the ceiling, the clock that is at rest relative to the source will have elapsed by an amount t0
d
𝒄 =𝒅
𝒕𝟎𝒅 = 𝒄 ∙ 𝒕𝟎
Now you’re going to need to stretch your conceptionof reality.
We generally think of time as moving forward at aconstant rate, the same for all reference frames.
However, by accepting Einstein’s second postulate, wereach some very surprising conclusions regarding thepassage of time in a reference frame that is in motionrelative to another.
All that is required to achieve the result is Einstein’ssecond postulate and some algebra!
A little bit stranger now!Now, imagine that the train is moving to the right withconstant speed v relative to an observer on the side of thetracks.
The light pulse will still shine on the same part of the ceiling,but... v
According to this observer, (CanelaHedgehog) the light traveled a greater distance to get there!
t
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What will the observer that is outside the train (moving relative to the light source) determine?
(where v is the speed of the train)
d
v
d
v t
d
v
d
v t
Write an expression for d in terms of c, t and v.
t is the time that it takes the light to reach the ceiling from the reference frame of a person that is in motion
relative to the source (Canela).
t
Where t is the amount of time that has elapsed on the observer’s clock that is not
on the train.
d
v t
𝑐 ∙ 𝑡 2 = 𝑑 2 + 𝑣 ∙ 𝑡 2
𝑑 2 = 𝑐 ∙ 𝑡 2 − 𝑣 ∙ 𝑡 2
𝑑 = 𝑐 ∙ 𝑡 2 − 𝑣 ∙ 𝑡 2
𝑑 = 𝑡 ∙ 𝑐2 − 𝑣2
Stationary Observer
Observer Moving Relative to Source
v v
𝒅 = 𝒄 ∙ 𝒕𝟎 𝒅 = 𝒕 ∙ 𝒄𝟐 − 𝒗𝟐
SUMMARY
t0
t
Since the observers will certainly agree on the height of the train (but not the
amount of time that it took light to travel those different distances!)…
d
𝒅 = 𝒄 ∙ 𝒕𝟎 𝒅 = 𝒕 ∙ 𝒄𝟐 − 𝒗𝟐
𝒄 ∙ 𝒕𝟎 = 𝒕 ∙ 𝒄𝟐 − 𝒗𝟐
𝒄 ∙ 𝒕𝟎 = 𝒕 ∙ 𝒄𝟐 − 𝒗𝟐
𝒕𝟎 =𝒕
𝒄∙ 𝒄𝟐 − 𝒗𝟐
Rearrange:
𝒕𝟎 = 𝒕 ∙𝒄𝟐
𝒄𝟐−𝒗𝟐
𝒄𝟐
𝒕𝟎 = 𝒕 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
This is known as the equation for
time dilation.
𝒕 =𝒕𝟎
𝟏 −𝒗𝟐
𝒄𝟐
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We end up with a relationship between the measurements of time that the each observer will
make about the same event, from different reference frames.
Time elapsed according to an observer inside the train
Time elapsed according to an observer outside the train
Speed of the train
Speed of light
𝒕 =𝒕𝟎
𝟏 −𝒗𝟐
𝒄𝟐
Since the speed of light is constant in all reference frames, but different observers will see light travel different distances based on their own
relative motion, time itself must elapse at different rates for different observers.
This means that the person on the tracks (Canela) will measure a larger time interval than the person on the train (Ginger) for the same
event!
Give this some thought!
What does this mean??
𝒕 =𝒕𝟎
𝟏 −𝒗𝟐
𝒄𝟐
Always: v < c
If v > c, then 𝒗𝟐
𝒄𝟐> 1, and
𝟏 −𝒗𝟐
𝒄𝟐will be − 𝐧𝐮𝐦𝐛𝐞𝐫
Can you tell me any special
condition for v?
This scenario is could never happen, because no source or observer could travel at the speed of light!
It is fundamentally forbidden by the laws of physics!
Then what about this case?
light source
Detector
v = c v = 0
According to physics as we currently
know it, no object that has
mass could ever
travel at the speed of light or
faster than the speed of light!
IT IS THE LAW !
𝟐𝟗𝟗, 𝟕𝟗𝟐, 𝟒𝟓𝟖𝒎
𝒔
No matter what the motion of the source or the observer, the observed speed of light will be 3 x 108 m/s.
If this were not true, then it would be possible to determine an “absolute” frame of reference, which is
quite simply not the case in the universe.
The detector will stillmeasure the speed of
light to be 3 x 108 m/s.
v = 0.999c v = 0.999c
Craziness!
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WHITEBOARD
A spaceship moves past Earth at a speed of2.6 x 108 m/s. The ship's captain carries a lightthat flashes each time his heart beats.According to the captain, a flash occurs every1.0 s. Which time interval elapses between
flashes according to an observer on Earth?
WHITEBOARD
Time elapsed according to ship’s captain
Time elapsed according to an observer on Earth
Speed of Earth
Speed of light
𝒕 =𝒕𝟎
𝟏 −𝒗𝟐
𝒄𝟐
The crew in the ship does not know they are moving (constant speed). To them
Earth moves at 2.6 x 108 m/s
WHITEBOARD SOLUTION
𝒕 =𝒕𝟎
𝟏 −𝒗𝟐
𝒄𝟐
𝒕 =𝟏 𝒔
𝟏 −𝟐. 𝟔 ∗ 𝟏𝟎𝟖 𝟐
𝟑 ∗ 𝟏𝟎𝟖 𝟐
𝒕 = 𝟐 𝒔𝒆𝒄𝒐𝒏𝒅𝒔
WHITEBOARD
If an observer on Earth measures the timeinterval between flashes (inside the ship) to be3 seconds, what is the time interval measuredon the ship?
𝒕𝟎 = 𝒕 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝒕𝟎 = 𝟑 ∙ 𝟏 −𝟐. 𝟔 ∗ 𝟏𝟎𝟖 𝟐
𝟑 ∗ 𝟏𝟎𝟖 𝟐
𝒕𝟎 = 𝟏. 𝟓 𝒔𝒆𝒄𝒐𝒏𝒅𝒔
WHITEBOARD
An airplane flies from San Francisco to NewYork (4.8x106 m) at a steady speed of 300m/s. How much time does the trip take,
A) as measured by an observer on theground?
B) as measured by an observer in theplane?
WHITEBOARD SOLUTION
𝒕 = 𝟏𝟔, 𝟎𝟎𝟎 𝒔
Observer on the ground
Observer in the plane
𝒕 =𝒅
𝒗
𝒕 =𝟒. 𝟖𝟎 ∗ 𝟏𝟎𝟔
𝟑𝟎𝟎
𝒕 = 𝟏𝟔, 𝟎𝟎𝟎 𝒔
𝒕𝟎 = 𝒕 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝒕𝟎 = 𝟏𝟔, 𝟎𝟎𝟎 ∙ 𝟏 −𝟑𝟎𝟎 𝟐
𝟑 ∗ 𝟏𝟎𝟖 𝟐
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WHITEBOARD
Ivana boards a spaceship, then zips past Alanon Earth at a relative speed of 0.800 c. At theinstant she passes, both starts timers.
A) At the instant when Alan measures thatIvana has traveled 2.88x108 m past him,
what does Ivana’s timer read?
B) At the instant when Ivana reads 0.720 son her timer, what does Alan’s timer read onhis?
WHITEBOARD SOLUTION
𝒕 =𝒅
𝒗
𝒕 =𝟐. 𝟖𝟖 ∗ 𝟏𝟎𝟖𝒎
𝟎. 𝟖𝟎𝟎 ∙ 𝟑 ∗ 𝟏𝟎𝟖𝒎𝒔
𝒕 = 𝟏. 𝟐 𝒔
A) At the instant when Alan measures thatIvana has traveled 2.88x108 m past him,what does Ivana’s timer read?
𝒕 = 𝟎. 𝟕𝟐 𝒔
𝒕𝟎 = 𝒕 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝒕𝟎 = 𝟏. 𝟐 ∙ 𝟏 −𝟎. 𝟖 𝒄 𝟐
𝒄 𝟐
𝒕𝟎 = 𝟏. 𝟐 ∙ 𝟏 − 𝟎. 𝟖𝟐
Alan’s timer
WHITEBOARD SOLUTION
B) At the instant when Ivana reads 0.720 son her timer, what does Alan’s timer read onhis?
𝒕𝟎 = 𝟎. 𝟒𝟑𝟐 𝒔
𝒕𝟎 = 𝒕 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝒕𝟎 = 𝟎. 𝟕𝟐 ∙ 𝟏 −𝟎. 𝟖 𝒄 𝟐
𝒄 𝟐
𝒕𝟎 = 𝟎. 𝟕𝟐 ∙ 𝟏 − 𝟎. 𝟖𝟐
According to Ivana, she is at rest and
Alan is moving away at 0.8c
Two twins are born simultaneously on Earth inthe year 2500. One of them lives a happy lifeon Earth, and the other is put on a spaceshipand travels the Universe at extremely highspeeds, approaching the speed of light. When
the twin returns to Earth from his journey, hefinds that his brother is 100 years old, while heis only 25 years old!
How fast did the spaceship need to travel inorder for this to happen?
Express your answer in terms of the speed oflight c.
Whiteboard: The Twin Trip Paradox
𝒕𝟎 = 𝒕 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝒕𝟎𝟐 = 𝒕 𝟐 ∙ 𝟏 −
𝒗𝟐
𝒄𝟐
𝒕𝟎𝟐
𝒕 𝟐= 𝟏 −
𝒗𝟐
𝒄𝟐
𝒗𝟐
𝒄𝟐= 𝟏 −
𝒕𝟎𝟐
𝒕 𝟐
𝒗𝟐 = 𝒄𝟐 𝟏 −𝒕𝟎
𝟐
𝒕 𝟐
𝒗 = 𝒄 ∙ 𝟏 −𝒕𝟎 𝟐
𝒕 𝟐
𝒗 = 𝒄 ∙ 𝟏 −𝟐𝟓 𝟐
𝟏𝟎𝟎 𝟐
𝒗 = 𝟐𝟗𝟎, 𝟒𝟕𝟑, 𝟕𝟓𝟏𝒎
𝒔
𝒗 = 𝟎. 𝟗𝟔𝟖 𝒄
Less time elapses in a moving reference frame than a stationary one!
THE TWIN PARADOX
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“When you’re talking to a pretty girl (or a boy), an hour feels like a
second. When you put your hand on a red-hot ember, a
second feels like an hour.
That’s relativity.”
Meet the Muon
Muons are formed when high-energy protons from the solar
wind hit Carbon or Beryllium nuclei in our atmosphere.
After being formed, they fly outward at speeds of up to 99.5% of
the speed of light!
However, muons are very unstable particles. A muon at rest
exists for only about 2.2 x 10-6 seconds before it decays.
Muon Whiteboard: Part 1
A muon is formed by a nuclear reaction in the high atmosphere 5 km above Earth’s surface. The muon
travels straight downward at a speed of 0.995c. From the reference frame of the muon, how far will it travel
in the 2.2 x 10-6 seconds before it decays?
Muon Whiteboard: Part 1
𝒅 = 𝒗 ∙ 𝒕
𝒅 = 𝟎. 𝟗𝟗𝟓 ∙ 𝟑 ∗ 𝟏𝟎𝟖 ∙ 𝟐. 𝟐 ∗ 𝟏𝟎−𝟔
𝒅 = 𝟔𝟓𝟔. 𝟕 𝒎
Muon: Earth is moving towards me at 0.995 c
According to classical physics, the muon will decay long before it reaches
Earth’s surface!
d = 656.7 m
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However, muons from the high
atmosphere can be regularly found
striking the surface of the Earth!
How is this possible?
A muon travels straight downward at a speed of 0.995c. From the reference frame of an observer on Earth, how
far will it travel in the 2.2 x 10-6 seconds before it decays?
Note: The time for the muon to decay applies within the reference frame of the muon itself – not to the
observer!
Muon Whiteboard: Part 2
Time elapsed according to the rest frame
(muon’s ref frame)Time elapsed according to
an outside observer
𝒕 =𝒕𝟎
𝟏 −𝒗𝟐
𝒄𝟐
𝒕 =𝟐. 𝟐 ∗ 𝟏𝟎−𝟔
𝟏 −𝟎. 𝟗𝟗𝟓𝒄 𝟐
𝒄𝟐
𝒕 =𝟐. 𝟐 ∗ 𝟏𝟎−𝟔
𝟏 − 𝟎. 𝟗𝟗𝟓 𝟐
𝒕 = 𝟐. 𝟐𝟎𝟑 ∗ 𝟏𝟎−𝟓 seconds
Time dilation Muons are Evidence of Special Relativity
From the reference frame of the Earth, the muon has plenty of time to reach the
ground!
𝒅 = 𝒗 ∙ 𝒕
𝒅 = 𝟎. 𝟗𝟗𝟓 ∙ 𝟑 ∗ 𝟏𝟎𝟖 ∙ 𝟐. 𝟐𝟎𝟑 ∗ 𝟏𝟎−𝟓
𝒅 = 𝟔𝟓𝟕𝟓. 𝟐 𝒎
But hey, wait a second…
How can it be possible that the muon travels only 656.7 m in its own
reference frame, but travels a whole 6,575.2 m in our reference frame?
It either hits the ground or it doesn’t!
…right?
As it turns out, the 656.7 m that the muon travels in its own
reference frame is equivalent to the 6,575.2 m that it travels
in ours.
Time is not the only
quantity that is relative to
the observer.
Lengths are also relative
In formulating special
relativity, Einstein showed
that space and time are
linked in the most
fundamental way.
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LENGTH CONTRACTION Length contraction
Consider an arrow flying across a lab that movespast a clock at rest with respect to the lab.
The two events occurred at the same place in thelab reference frame, so that is the properreference frame.
Proper length (L0)
The arrow's proper length is the length measured in a reference frame in which the arrow is stationary.
In this case, it is the reference frame defined by the arrow itself.
The proper length is always the longest length measured for a given object.
𝒗 =𝑳𝟎𝒕
Arrow point of view
This is known as the equation for length contraction.
𝒗 =𝑳
𝒕𝟎
“you” point of view
𝑳𝟎𝒕=
𝑳
𝒕𝟎𝑳𝟎𝒕=
𝑳
𝒕 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝑳 = 𝑳𝟎 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
LENGTH CONTRACTION
Length measurement of an observer moving relative to the object being measured
Length measurement of an observer at rest relative to the object being measured
Relative speed of object / observer
Speed of light
Lengths are shorter to observers who are moving relative to the object being
measured.
𝑳 = 𝑳𝟎 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
MUON Whiteboard
What is the distance traveled
(from the muon’s point of
view). If it falls a distance of
5000 m at a sped 0.995c?
𝑳 = 𝟓𝟎𝟎𝟎𝒎 ∙ 𝟏 −𝟎. 𝟗𝟗𝟓 𝒄 𝟐
𝒄𝟐
𝑳 = 𝟒𝟗𝟗. 𝟑𝟕 𝒎
𝑳 = 𝑳𝟎 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
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“It takes the muon 22.0 μs to hit the ground,
which is 5000 m away.”
“It takes the ground 2.2 μs to hit me, starting from just 499.37 m away.”
WHITEBOARD
An arrow flies past a person standing onEarth. When at rest with respect to Earth,the arrow's length was measured to be0.6 m. Determine the arrow's length L asmeasured by the person on Earth when
the arrow moves:
A. At speed 0.90c.
B. At speed 300 m/s.
WHITEBOARD SOLUTION
𝑳 = 𝑳𝟎 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝑳 = 𝟎. 𝟔𝒎 ∙ 𝟏 −𝟎. 𝟗 𝒄 𝟐
𝒄𝟐
𝑳 = 𝟎. 𝟐𝟔𝟏𝟓𝒎
𝑳 = 𝟎. 𝟔𝒎 ∙ 𝟏 −𝟑𝟎𝟎 𝟐
𝒄𝟐
𝑳 = 𝟎. 𝟓𝟗𝟗𝟗𝒎
Whiteboard: Length Contraction
How fast would a meter stick have to move for it to become a half-meter stick from your reference
frame?
Express your answer in terms of the speed of light c
𝑳 = 𝑳𝟎 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝑳 𝟐 = 𝑳𝟎𝟐 ∙ 𝟏 −
𝒗𝟐
𝒄𝟐
𝑳 𝟐
𝑳𝟎 𝟐= 𝟏 −
𝒗𝟐
𝒄𝟐
𝒗𝟐
𝒄𝟐= 𝟏 −
𝑳 𝟐
𝑳𝟎 𝟐
𝒗𝟐 = 𝒄𝟐 𝟏 −𝑳 𝟐
𝑳𝟎 𝟐
𝒗 = 𝒄 ∙ 𝟏 −𝑳 𝟐
𝑳𝟎 𝟐
𝒗 = 𝒄 ∙ 𝟏 −𝟎. 𝟓 𝟐
𝟏. 𝟎 𝟐
𝒗 = 𝟐𝟓𝟗, 𝟖𝟎𝟕, 𝟔𝟐𝟏𝒎
𝒔
𝒗 = 𝟎. 𝟖𝟔𝟔 𝒄
Lengths only contract along the direction of
motion
WHITEBOARDDetermine the length of the rocket (L0 = 50 m). Given the speeds below
L =
L =
L =
L =
50 m
25 m
5 m
0.5 m
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An astronaut is resting on a bed inclined at an angle theta above the floor of a spaceship.
From the reference frame of an observer who is moving to the right with a speed close to c, is the angle that the bed makes with the floor(a) greater than the angle as observed by the astronaut(b) less than the angle as observed by the astronaut(c) equal to the angle as observed by the astronaut
Whiteboard: Laying Down WHITEBOARD: Solution
Lengths contract along the direction of relative motion. This will cause x to contract, while y is
constant.
Therefore, the moving observer will measure a
larger angle than the astronaut!
As x decreases, the angle is closer to be a 90˚ angle
WHITEBOARD
In the year 2500, an astronaut takes a trip
to Vega, a distant star. The trip is a distance
of 25.3 light-years, as measured by an
observer on Earth. The astronaut travels at
a speed of 0.99c
How will the astronaut see this?
From the astronaut’s reference frame, Earth and
Vega are moving at 0.99c, and their ship is at
rest.
The astronaut and the Earth observer will agree on
their relative velocity, but that’s about all they will
agree upon!
Whiteboard: Time & length
Contraction!
a) How much time will the trip take, according to
each of the observers?
b) What is the distance (light years) between
Earth and Vega, according to each of the
observers?
WHITEBOARD SOLUTION
𝒕 = 𝟑. 𝟔𝟏 𝒚𝒆𝒂𝒓𝒔
Time Measured on Earth
Time measured in the ship
𝒕 =𝒅
𝒗
𝒕 =𝒄 ∙ 𝟐𝟓. 𝟑
𝟎. 𝟗𝟗𝟎 ∙ 𝒄
𝒕 = 𝟐𝟓. 𝟓𝟔 𝒚𝒆𝒂𝒓𝒔
𝒕𝟎 = 𝒕 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝒕𝟎 = 𝟐𝟓. 𝟓𝟔 ∙ 𝟏 −𝟎. 𝟗𝟗 𝒄 𝟐
𝒄 𝟐
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WHITEBOARD SOLUTION
Length Measured from Earth
Length measured from the ship
𝒅 = 𝟐𝟓. 𝟑 𝑳𝒊𝒈𝒉𝒕𝒚𝒆𝒂𝒓𝒔
Proper length
𝑳 = 𝑳𝟎 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝑳 = 𝟐𝟓. 𝟑 ∙ 𝟏 − 𝟎. 𝟗𝟗𝟎𝟐
𝑳 = 𝟑. 𝟓𝟕 𝑳𝒊𝒈𝒉𝒕𝒚𝒆𝒂𝒓𝒔
Two Different Stories – Both Correct!
Time:
25.56
years
Time:
3.61
years
The only thing they will agree upon is their relative velocity
WHITEBOARD: LOOKIN’ SLIM
An astronaut is on the moon as a spaceshipflies past a speed of 0.950c relative to theMoon.
If the width of the astronaut on the moonis 0.8 m, what width does the spaceship
crew measure him to be?
A crew member on the spaceshipmeasure its length, obtaining the value600 m. What length does the astronautmeasure on the moon?
WHITEBOARD SOLUTION
Width of astronaut measured from the
spaceship
𝑳 = 𝑳𝟎 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝑳 = 𝟎. 𝟖 ∙ 𝟏 − 𝟎. 𝟗𝟓𝟎𝟐
𝑳 = 𝟎. 𝟐𝟓 𝒎
Length of spaceship Measured from the
Moon
𝑳 = 𝑳𝟎 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝑳 = 𝟔𝟎𝟎 ∙ 𝟏 − 𝟎. 𝟗𝟓𝟎𝟐
𝑳 = 𝟏𝟖𝟕. 𝟑𝟓 𝒎
Both observers will measure the other’s lengths
to be contracted (and both will be correct!)
RELATIVISTIC MOMENTUM
When we use the classical definition ofmomentum to analyze collisions at highspeed, we find that even for an isolatedsystem, the momentum of the system isconstant in some reference frames but not
in others.
To get an improved relativistic expressionfor momentum, we use the proper timeinterval.
p = mv
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RELATIVISTIC MOMENTUM
No speed restriction !𝒑 = 𝒎 ∙ 𝒗
𝒑 = 𝒎 ∙∆𝒙
∆𝒕𝒐
𝒑 = 𝒎 ∙∆𝒙
𝒕 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
𝒑 = 𝒎 ∙∆𝒙
𝒕
𝟏
𝟏 −𝒗𝟐
𝒄𝟐
𝒑 =𝒎 ∙ 𝒗
𝟏 −𝒗𝟐
𝒄𝟐
Relativistic momentum
The relativistic momentum of an object ofmass m in a reference frame where theobject is moving at velocity v is:
Speed restriction !
𝒑 =𝒎 ∙ 𝒗
𝟏 −𝒗𝟐
𝒄𝟐
WHITEBOARD
An electron is moving at a speed of 0.9c.
Compare its momentum as calculated usinga nonrelativistic equation and using arelativistic equation.
p = 2.46x10-22 kgm/s
p = 5.64x10-22 kgm/s
RELATIVISTIC ENERGY
An electron is accelerated through apotential difference of 300,000 V. What isthe final speed of the electron?
𝑼𝑬 = 𝑲𝑬
𝒒 ∙ ∆𝑽 =𝒎 ∙ 𝒗𝟐
𝟐
𝒗 =𝟐 ∙ 𝒒 ∙ ∆𝑽
𝒎𝑬
𝒗 =𝟐 ∙ 𝟏. 𝟔 ∗ 𝟏𝟎−𝟏𝟗 ∙ 𝟑𝟎𝟎, 𝟎𝟎𝟎
𝟗. 𝟏𝟏 ∗ 𝟏𝟎−𝟑𝟏
𝒗 = 𝟑. 𝟐𝟓 ∗ 𝟏𝟎𝟖𝒎
𝒔
SUMMARY OF MATH MODELS
𝒕 =𝒕𝟎
𝟏 −𝒗𝟐
𝒄𝟐
Time dilation
𝑳 = 𝑳𝟎 ∙ 𝟏 −𝒗𝟐
𝒄𝟐
Length contraction
𝒕 = 𝒕𝟎 ∙ 𝜸
𝑳 =𝑳𝟎𝜸
𝜸 =𝟏
𝟏 −𝒗𝟐
𝒄𝟐
𝒑 =𝒎 ∙ 𝒗
𝟏 −𝒗𝟐
𝒄𝟐
Momentum𝒑 = 𝒎 ∙ 𝒗 ∙ 𝜸
RELATIVISTIC ENERGY
A point like object of mass m has so-called rest energy because of its mass
The first postulate of relativity states that the laws of physics are the same in all inertial frames.
Einstein showed that the law of conservation of energy is valid relativistically, if we define energy
to include a relativistic factor:
E0 = mc2 rest energy
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Rest energy of particles
Any object with mass has rest energy:
…. But when objects are not moving atrelativistic speeds, mass and energy areconserved separately.
Rest energy can be converted into other forms ofenergy.
The rest energy of the Sun is being slowlyconverted via nuclear fusion reactions intointernal energy.
Rest Energy E0 = mc2
WHITEBOARD
The mass of an electron is 9.11 x 10−31 kg.The mass of a proton is 1.67 x 10−27 kg.Determine the electron and proton restenergies in joules and in electron volts.
ELECTRON PROTON
E0 = 8.199x10-14 J
E0 = 512437.5 eV
E0 = 1.503x10-10 J
E0 = 9.394x108 eV
WHITEBOARD
On average, each year about 2 x 1010 J ofelectric and chemical potential energy isconverted to cool and warm your home. Ifrest energy could be converted for thispurpose, how much mass equivalent of
rest energy would be needed?
m = 2.22x10-7 kg
1/10 the mass of one of the hairs on your head
E = E0 + KE
RELATIVISTIC ENERGY
Rest Energy
Total Energy
E = E0 + KEKineticEnergy
mc2 mc2 KESpeed restriction !
E = E0 + KE
KINENITC RELATIVISTIC ENERGY
mc2 mc2 KE
KE = mc2 - mc2
KE = mc2 (-1)
RELATIVISTIC ENERGY
A point like object of mass m has so-called rest energy because of its mass
Rest Energy E0 = mc2
Total Energy E = mc2
Kinetic Energy KE = mc2 - mc2
KE = mc2( – 1)
Speed restriction !
(difference between total energy and rest energy)
(relativistic energy)
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Electron volt
An electron volt (1 eV) is the increase in kinetic energy of an electron when it moves
across a 1.0 V potential difference:
1 eV = 1.6x10-19 J
RELATIVISTIC ENERGYPART I
An electron is accelerated through a potentialdifference of 300,000 V. What is the finalspeed of the electron (in terms of c)?
𝑣 = 𝑐 1 −𝑚 ∙ 𝑐2
𝑞 ∙ ∆𝑉 +𝑚 ∙ 𝑐2
2
𝒗 = 𝟐𝟑𝟐, 𝟕𝟗𝟖, 𝟎𝟔𝟗𝒎
𝒔𝒗 = 𝟎. 𝟕𝟕𝟔 𝒄
𝑣 = 3 ∗ 108 1−9.11 ∗ 10−31 ∙ 3 ∗ 108 2
1.6 ∗ 10−19 ∙ 300,000 + 9.11 ∗ 10−31 ∙ 3 ∗ 108 2
2
𝑈𝐸 = 𝐾𝐸
𝑞 ∙ ∆𝑉 = 𝑚 ∙ 𝑐2 ∙ 𝛾 − 1
𝛾 − 1 =𝑞 ∙ ∆𝑉
𝑚 ∙ 𝑐2
𝛾 =𝑞 ∙ ∆𝑉
𝑚 ∙ 𝑐2+ 1
𝛾 =𝑞 ∙ ∆𝑉
𝑚 ∙ 𝑐2+𝑚 ∙ 𝑐2
𝑚 ∙ 𝑐2
1 −𝑣2
𝑐2=
𝑚 ∙ 𝑐2
𝑞 ∙ ∆𝑉 +𝑚 ∙ 𝑐2
1
1 −𝑣2
𝑐2
=𝑞 ∙ ∆𝑉 +𝑚 ∙ 𝑐2
𝑚 ∙ 𝑐2
1 −𝑣2
𝑐2=
𝑚 ∙ 𝑐2
𝑞 ∙ ∆𝑉 +𝑚 ∙ 𝑐2
2
𝑣2
𝑐2= 1 −
𝑚 ∙ 𝑐2
𝑞 ∙ ∆𝑉 +𝑚 ∙ 𝑐2
2
𝑣 = 𝑐 1 −𝑚 ∙ 𝑐2
𝑞 ∙ ∆𝑉 + 𝑚 ∙ 𝑐2
2
𝑣2 = 𝑐2 1 −𝑚 ∙ 𝑐2
𝑞 ∙ ∆𝑉 +𝑚 ∙ 𝑐2
2
RELATIVISTIC ENERGYPART II
The electron (m=9.11*10-31 kg) travels at aspeed v = 0.776 c.
A) What is the rest energy of the electron?
B) What is the total energy of the electron?
C) What is the kinetic energy of the electron?
D) What is the electric potential energy of theelectron?
WHITEBOARDELECTRON’S REST ENERGY
A) What is the rest energy of theelectron?
Rest Energy 𝑬𝟎 = 𝒎 ∙ 𝒄𝟐
𝑬𝟎 = 𝟗. 𝟏𝟏 ∗ 𝟏𝟎−𝟑𝟏 ∙ 𝟑 ∗ 𝟏𝟎𝟖𝟐
𝑬𝟎 = 𝟖. 𝟏𝟗𝟗 ∗ 𝟏𝟎−𝟏𝟒𝑱
WHITEBOARDELECTRON’S TOTAL ENERGY
B) What is the total energy of theelectron?
Total Energy
𝑬 = 𝜸 ∙ 𝒎 ∙ 𝒄𝟐
𝑬 =𝟗. 𝟏𝟏 ∗ 𝟏𝟎−𝟑𝟏 ∙ 𝟑 ∗ 𝟏𝟎𝟖
𝟐
𝟏 − 𝟎. 𝟕𝟕𝟔𝟐
𝑬 = 𝟏. 𝟑 ∗ 𝟏𝟎−𝟏𝟑𝑱
𝑬 =𝒎 ∙ 𝒄𝟐
𝟏 −𝒗𝟐
𝒄𝟐
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WHITEBOARDELECTRON’S KINETIC ENERGY
C) What is the kinetic energy of theelectron?
Kinetic Energy
𝑲𝑬 = 𝜸 ∙ 𝒎 ∙ 𝒄𝟐 ∙ (𝜸 − 𝟏)
𝑲𝑬 = 𝟗. 𝟏𝟏 ∗ 𝟏𝟎−𝟑𝟏 ∙ 𝟑 ∗ 𝟏𝟎𝟖𝟐∙
𝟏
𝟏 − 𝟎. 𝟕𝟕𝟔𝟐− 𝟏
𝑬 = 𝟒. 𝟖 ∗ 𝟏𝟎−𝟏𝟒𝑱
𝑲𝑬 = 𝒎 ∙ 𝒄𝟐 ∙𝟏
𝟏 −𝒗𝟐
𝒄𝟐
− 𝟏
WHITEBOARDELECTRON’S ELECTRIC POT. ENERGY
D) What is the electric potential energyof the electron?
Electric Potential Energy
𝑼𝑬𝒊 = 𝒒 ∙ ∆𝑽𝒊
𝑼𝑬𝒇 = −𝟒. 𝟖 ∗ 𝟏𝟎−𝟏𝟒𝑱
𝑼𝑬𝒊 = 𝟎 𝑱
𝑼𝑬𝒇 = 𝒒 ∙ ∆𝑽𝒇
𝑼𝑬𝒇 = −𝟏. 𝟔 ∗ 𝟏𝟎−𝟏𝟗 ∙ 𝟑𝟎𝟎, 𝟎𝟎𝟎
8.199E-14
0.000E+000.000E+000.000E+00
8.199E-14
4.800E-14
-4.800E-14
-8.200E-14
-6.200E-14
-4.200E-14
-2.200E-14
-2.000E-15
1.800E-14
3.800E-14
5.800E-14
7.800E-14
E0 i KE i Uq i W E0 f KE f Uq f
WORK ENERGY BAR CHART Energy for an electron crossing a large potential difference:
Accelerating a particle