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Albert Einstein. Theory of Relativity. Physics 100 Chapt 18. watching a light flash go by. v. c. 2k k. The man on earth sees c =  (& agrees with Maxwell). watching a light flash go by. v. c. If the man on the rocket sees c -v , he disagrees with Maxwell.

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theory of relativity

Albert

Einstein

Theory of Relativity

Physics 100

Chapt 18

watching a light flash go by
watching a light flash go by

v

c

2kk

The man on earth sees c =

(& agrees with Maxwell)

watching a light flash go by1
watching a light flash go by

v

c

If the man on the rocket sees c-v,

he disagrees with Maxwell

do maxwell s eqns only work in one reference frame
Do Maxwell’s Eqns only work in one reference frame?

If so, this would be the rest frame

of the luminiferous Aether.

if so the speed of light should change throughout the year
If so, the speed of light should change throughout the year

upstream,

light moves

slower

downstream,

light moves

faster

“Aether wind”

michelson morley
Michelson-Morley

No aether wind detected: 1907 Nobel Prize

einstein s hypotheses
Einstein’s hypotheses:

1. The laws of nature are equally valid

in every inertial reference frame.

Including

Maxwell’s eqns

2. The speed of light in empty space is

same for all inertial observers, regard-

less of their velocity or the velocity of

the source of light.

all observers see light flashes go by them with the same speed
All observers see light flashes go by them with the same speed

v

No matter how fast the guy on the rocket

is moving!!

c

Both guys see the light flash

travel with velocity = c

even when the light flash is traveling in an opposite direction
Even when the light flash is traveling in an opposite direction

v

c

Both guys see the light flash

travel past with velocity = c

gunfight viewed by observer at rest
Gunfight viewed by observer at rest

He sees both shots

fired simultaneously

Bang!

Bang!

viewed by a moving observer1
Viewed by a moving observer

He sees cowboy shoot

1st & cowgirl shoot later

Bang!

Bang!

viewed by a moving observer2
Viewed by a moving observer

He sees cowgirl shoot

1st & cowboy shoot later

Bang!

Bang!

time depends of state of motion of the observer
Time depends of state of motion of the observer!!

Events that occur simultaneously according to one observer can occur at different times for other observers

events
Events

y

(x2,t2)

(x1,t1)

x

x

x1

x2

x

t

same events different observers

Prior to Einstein, everyone agreed

the distance between events depends

upon the observer, but not the time.

Same events, different observers

y’

y’

y

(x2,t2)

(x1,t1)

x

x

(x1’,t1’)

(x2’,t2’)

t’

t’

x1’

x1’

x2’

dist’

x’

x’

x1

x2

x

t

dist

time is the 4 th dimension
Time is the 4th dimension

Einstein discovered that there is no

“absolute” time, it too depends upon

the state of motion of the observer

Einstein

Space-Time

Newton

Space

&

Time

completely

different

concepts

2 different aspects

of the same thing

how are the times seen by 2 different observers related
How are the times seen by 2 different observers related?

We can figure this out with

simple HS-level math

( + a little effort)

catch ball on a rocket ship
Catch ball on a rocket ship

Event 2: girl catches the ball

wt

v= =4m/s

w=4m

t=1s

Event 1: boy throws the ball

seen from earth
Seen from earth

V0=3m/s

V0=3m/s

Location of the 2

events is different

Elapsed time is

the same

The ball appears

to travel faster

d=(3m)2+(4m)2 =5m

w=4m

v0t=3m

dt

v= = 5m/s

t=1s

flash a light on a rocket ship
Flash a light on a rocket ship

Event 2: light flash reaches the girl

wt0

c=

w

t0

Event 1: boy flashes the light

seen from earth1
Seen from earth

V

V

Speed has to

Be the same

Dist is longer

Time must be

longer

d=(vt)2+w2

w

vt

dt

c= =

(vt)2+w2

t

t=?

how is t related to t 0
How is t related to t0?

t= time on Earth clock

t0 = time on moving clock

wt0

c =

(vt)2+w2

t

c =

ct = (vt)2+w2

ct0 = w

(ct)2 = (vt)2+w2

(ct)2 = (vt)2+(ct0)2

 (ct)2-(vt)2= (ct0)2

 (c2-v2)t2= c2t02

1

1 – v2/c2

c2

c2 – v2

 t2 = t02

 t2 = t02

1

1 – v2/c2

 t= t0

 t= g t0

this is called g

properties of g

1

1 – v2/c2

Properties of g =

Suppose v = 0.01c (i.e. 1% of c)

1

1 – (0.01c)2/c2

1

1 – (0.01)2c2/c2

g =

=

1

1 – 0.0001

1

1 – (0.01)2

1

0.9999

=

=

g =

g = 1.00005

properties of g cont d

1

1 – v2/c2

Properties of g = (cont’d)

Suppose v = 0.1c (i.e. 10% of c)

1

1 – (0.1c)2/c2

1

1 – (0.1)2c2/c2

g =

=

1

1 – 0.01

1

1 – (0.1)2

1

0.99

=

=

g =

g = 1.005

other values of g

1

1 – v2/c2

Other values of g =

Suppose v = 0.5c (i.e. 50% of c)

1

1 – (0.5c)2/c2

1

1 – (0.5)2c2/c2

g =

=

1

1 – (0.25)

1

1 – (0.5)2

1

0.75

=

=

g =

g = 1.15

other values of g1

1

1 – v2/c2

Other values of g =

Suppose v = 0.6c (i.e. 60% of c)

1

1 – (0.6c)2/c2

1

1 – (0.6)2c2/c2

g =

=

1

1 – (0.36)

1

1 – (0.6)2

1

0.64

=

=

g =

g = 1.25

other values of g2

1

1 – v2/c2

Other values of g =

Suppose v = 0.8c (i.e. 80% of c)

1

1 – (0.8c)2/c2

1

1 – (0.8)2c2/c2

g =

=

1

1 – (0.64)

1

1 – (0.8)2

1

0.36

=

=

g =

g = 1.67

other values of g3

1

1 – v2/c2

Other values of g =

Suppose v = 0.9c (i.e.90% of c)

1

1 – (0.9c)2/c2

1

1 – (0.9)2c2/c2

g =

=

1

1 – 0.81

1

1 – (0.9)2

1

0.19

=

=

g =

g = 2.29

other values of g4

1

1 – v2/c2

Other values of g =

Suppose v = 0.99c (i.e.99% of c)

1

1 – (0.99c)2/c2

1

1 – (0.99)2c2/c2

g =

=

1

1 – 0.98

1

1 – (0.99)2

1

0.02

=

=

g =

g = 7.07

other values of g5

1

1 – v2/c2

Other values of g =

Suppose v = c

1

1 – (c)2/c2

1

1 – c2/c2

g =

=

1

0

1

1 – 12

1

0

=

=

g =

g = 

Infinity!!!

other values of g6

1

1 – v2/c2

Other values of g =

Suppose v = 1.1c

1

1 – (1.1c)2/c2

1

1 – (1.1)2c2/c2

g =

=

1

1-1.21

1

1 – (1.1)2

1

-0.21

=

=

g =

g = ???

Imaginary number!!!

plot results
Plot results:

1

1 – v2/c2

g =

Never-never land

x

x

x

x

x

v=c

moving clocks run slower
Moving clocks run slower

v

t0

1

1 – v2/c2

t= t0

t

t= g t0

g >1  t > t0

length contraction
Length contraction

v

L0

time=t

L0 = vt

Shorter!

man on

rocket

Time = t0 =t/g

Length = vt0

=vt/g

=L0/g

moving objects appear shorter
Moving objects appear shorter

Length measured when

object is at rest

L= L0/g

g >1  L < L0

V=0.9999c

V=0.1c

V=0.86c

V=0.99c

slide49

change inv

time

mass:

F=m0a

=m0

t0

a

time=t0

m0

Ft0

m0

change inv =

Ft0

change inv

m0 =

mass

increases!!

gFt0

change inv

Ft

change inv

= gm0

m=

=

m = g m0

t=gt0

by a factor g

relativistic mass increase
Relativistic mass increase

m0 = mass of an object when it

is at rest

 “rest mass”

mass of a moving

object increases

g

as vc, m

m = g m0

as an object moves

faster, it gets

harder & harder

to accelerate

by the g factor

v=c

summary
summary
  • Moving clocks run slow
  • Moving objects appear shorter
  • Moving object’s mass increases

By a factor of g

plot results1
Plot results:

1

1 – v2/c2

g =

Never-never land

x

x

x

x

x

v=c

twin paradox

a-centauri

Twin paradox

4.3 light years

Twin brother

& sister

She will travel to

a-centauri (a near-

by star on a special

rocket ship v = 0.9c

He will stay home

& study Phys 100

light year
Light year

distance light travels in 1 year

dist = v x time

=cyr

1cyr = 3x108m/sx3.2x107 s

= 9.6 x 1015 m

We will just use cyr units

& not worry about meters

time on the boy s clock
Time on the boy’s clock

d0=4.3cyr

v=0.9c

v=0.9c

According to the boy

& his clock on Earth:

d0

v

4.3 cyr

0.9c

tout =

=

= 4.8 yrs

d0

v

4.3 cyr

0.9c

tback =

=

= 4.8 yrs

ttotal = tout+tback

= 9.6yrs

what does the boy see on her clock
What does the boy see on her clock?

d=4.3cyr

v=0.9c

v=0.9c

According to the boy

her clock runs slower

tout

g

4.8 yrs

2.3

tout =

=

= 2.1 yrs

tback

g

4.8 yr

2.3

tback =

=

= 2.1 yrs

ttotal = tout+tback

= 4.2yrs

so according to the boy
So, according to the boy:

d=4.3cyr

v=0.9c

v=0.9c

his clock her clock

out: 4.8yrs 2.1yrs

back: 4.8yrs 2.1yrs

  • She ages

less

total: 9.6yrs 4.2yrs

but according to the girl the boy s clock is moving so it must be running slower
But, according to the girl, the boy’s clock is moving &, so, it must be running slower

v=0.9c

According to her, the

boy’s clock on Earth says:

tout

g

2.1 yrs

2.3

tout =

=

= 0.9 yrs

tback

g

2.1 yrs

2.3

tback =

=

= 0.9 yrs

v=0.9c

ttotal = tout+tback

= 1.8yrs

her clock advances 4 2 yrs she sees his clock advance only 1 8 yrs
Her clock advances 4.2 yrs& she sees his clock advanceonly 1.8 yrs,

She should think he has aged less than her!!

A contradiction??

events in the boy s life
Events in the boy’s life:

As seen by her

As seen by him

She leaves

0.9 yrs

4.8 yrs

She arrives

& starts turn

short time

????

Finishes turn

& heads home

0.9 yrs

4.8 yrs

9.6+ yrs

1.8 + ??? yrs

She returns

turning around as seen by her
turning around as seen by her

According to her, these

2 events occur very,very

far apart from each other

He sees her

finish turning

He sees her

start to turn

Time interval between 2 events depends

on the state of motion of the observer

gunfight viewed by observer at rest1
Gunfight viewed by observer at rest

He sees both shots

fired simultaneously

Bang!

Bang!

viewed by a moving observer4
Viewed by a moving observer

He sees cowboy shoot

1st & cowgirl shoot later

Bang!

Bang!

in fact 7 8 years
In fact, ???? = 7.8+ years

as seen by him

as seen by her

She leaves

0.9 yrs

4.8 yrs

She arrives

& starts turn

7.8+ yrs

short time

???

Finishes turn

& heads home

0.9 yrs

4.8 yrs

9.6+ yrs

1.8 + ???yrs

9.6+ yrs

She returns

no paradox both twins agree
No paradox: both twins agree

The twin that

“turned around”

is younger

ladder barn door paradox
Ladder & Barn Door paradox

Stan & Ollie puzzle over how

to get a 2m long ladder thru

a 1m wide barn door

???

1m

2m

ladder

ollie remembers phys 100 the theory of relativity
Ollie remembers Phys 100 & thetheory of relativity

Stan, pick up the ladder & run very fast

1m

tree

2m

ladder

view from ollie s ref frame
View from Ollie’s ref. frame

Push, Stan!

1m

2m/g

V=0.9c

(g=2.3)

Stan

Ollie

view from stan s ref frame
View from Stan’s ref. frame

But it doesn’t fit, Ollie!!

1m/g

V=0.9c

(g=2.3)

2m

Ollie

Stan

slide71
If Stan pushes both ends of theladder simultaneously, Ollie sees the two ends move at different times:

Too late Stan!

Too soon Stan!

1m

clunk

clank

V=0.9c

(g=2.3)

Stan

Ollie

Stan

fermilab proton accelerator
Fermilab proton accelerator

V=0.9999995c

g=1000

2km

stanford electron accelerator
Stanford electron accelerator

v=0.99999999995c

3km

g=100,000

status
status

Einstein’s theory of “special relativity” has been carefully tested in many very precise experiments and found to be valid.

Time is truly the 4th dimension of space & time.

slide75
test

g=29.3