powerpoint presentation michelson and edward morley experiment - 1887 imagine that ether fills the...

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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


light source

Detector

v = c/2 v = 0

What speed would the detector measure? 𝒗𝒍𝒊𝒈𝒉𝒕 =

𝟑𝒄

𝟐


light source

Detector

v = c/2 v = 0 c


𝒗𝒍𝒊𝒈𝒉𝒕 =𝒄

𝟐


light source

Detector

v = c/2 v = c/2


𝒗𝒍𝒊𝒈𝒉𝒕 = 𝟐𝒄

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light source

Detector

v = c v = 0


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


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.


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 . . . .


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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


DetectorDetector


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?


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!


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

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