Theory of relativity

75
Theory of Relativity Physics 100 Chapt 18 Albert Einstein

Transcript of Theory of relativity

Page 1: Theory of relativity

Theory of Relativity

Physics 100

Chapt 18

AlbertEinstein

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watching a light flash go by

c

v

The man on earth sees c =(& agrees with Maxwell)

2k

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watching a light flash go by

c

v

If the man on the rocket sees c-v,

he disagrees with Maxwell

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

If so, this would be the rest frame of the luminiferous Aether.

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

upstream,light moves

slower

downstream,light moves

faster “Aether wind”

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Michelson-Morley

No aether wind detected: 1907 Nobel Prize

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Einstein’s hypotheses:

1. The laws of nature are equally valid in every inertial reference frame.

IncludingMaxwell’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.

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

c

v

Both guys see the light flash

travel with velocity = c

No matter how fast the guy on the rocketis moving!!

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Even when the light flash is traveling in an opposite direction

c

v

Both guys see the light flash

travel past with velocity = c

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Gunfight viewed by observer at rest

Bang! Ban

g!

He sees both shotsfired simultaneously

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Viewed by a moving observer

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Viewed by a moving observer

Bang! Ban

g!

He sees cowboy shoot1st & cowgirl shoot later

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Viewed by an observer in theopposite direction

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Viewed by a moving observer

Bang!Ban

g!

He sees cowgirl shoot1st & cowboy shoot later

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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

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Light clock

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Seen from the ground

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Events

x

y

x

t

(x1,t1) x

(x2,t2)

x1 x2

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Same events, different observers

x

y

x

t

(x1,t1) x

(x2,t2)

x1 x2

x’

y’

x1’

(x1’,t1’)

y’

x’x1’ x2’

(x2’,t2’)

t’ t’

Prior to Einstein, everyone agreed the distance between events depends upon the observer, but not the time.

dist’

dist

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Time is the 4th dimension

Einstein discovered that there is no“absolute” time, it too depends uponthe state of motion of the observer

Newton

Space &

Time

Einstein

Space-Timecompletelydifferentconcepts

2 different aspectsof the same thing

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How are the times seen by 2 different observers related?

We can figure this out withsimple HS-level math

( + a little effort)

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Catch ball on a rocket ship

w=4m

t=1s

v= =4m/swt

Event 1: boy throws the ball

Event 2: girl catches the ball

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Seen from earth

w=4m

v0t=3md=(

3m)2 +(4

m)2

=5m

v= = 5m/sdtt=1s

V0=3m/sV0=3m/s

Location of the 2events is different

Elapsed time isthe same

The ball appearsto travel faster

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Flash a light on a rocket ship

w

t0

c= wt0

Event 1: boy flashes the light

Event 2: light flash reaches the girl

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Seen from earth

w

vt

d=(vt

)2 +w2

c= =dtt=?

VV

Speed has toBe the same

Dist is longer

Time must be longer

(vt)2+w2

t

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How is t related to t0?

c =(vt)2+w2

t

t= time on Earth clock

c = wt0

t0 = time on moving clock

ct = (vt)2+w2

(ct)2 = (vt)2+w2

ct0 = w

(ct)2 = (vt)2+(ct0)2 (ct)2-(vt)2= (ct0)2 (c2-v2)t2= c2t02

t2 = t02c2

c2 – v2 t2 = t0

2 1

1 – v2/c2

t = t0 1

1 – v2/c2

this is called

t = t0

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Properties of 11 – v2/c2

11 – (0.01c)2/c2 =

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

11 – (0.01)2c2/c2 =

11 – (0.01)2 =

11 – 0.0001

= 10.9999

=

= 1.00005

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Properties of (cont’d)

11 – v2/c2

11 – (0.1c)2/c2 =

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

11 – (0.1)2c2/c2 =

11 – (0.1)2 =

11 – 0.01

= 10.99

=

= 1.005

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Let’s make a chart

v =1/(1-v2/c2)

0.01 c 1.00005

0.1 c 1.005

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Other values of

11 – v2/c2

11 – (0.5c)2/c2 =

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

11 – (0.5)2c2/c2 =

11 – (0.5)2 =

11 – (0.25)

= 10.75

=

= 1.15

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Enter into chart

v =1/(1-v2/c2)

0.01 c 1.00005

0.1 c 1.005

0.5c 1.15

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Other values of

11 – v2/c2

11 – (0.6c)2/c2 =

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

11 – (0.6)2c2/c2 =

11 – (0.6)2 =

11 – (0.36)

= 10.64

=

= 1.25

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Back to the chart

v =1/(1-v2/c2)

0.01 c 1.00005

0.1 c 1.005

0.5c 1.15

0.6c 1.25

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Other values of

11 – v2/c2

11 – (0.8c)2/c2 =

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

11 – (0.8)2c2/c2 =

11 – (0.8)2 =

11 – (0.64)

= 10.36

=

= 1.67

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Enter into the chart

v =1/(1-v2/c2)

0.01 c 1.00005

0.1 c 1.005

0.5c 1.15

0.6c 1.25

0.8c 1.67

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Other values of

11 – v2/c2

11 – (0.9c)2/c2 =

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

11 – (0.9)2c2/c2 =

11 – (0.9)2 =

11 – 0.81

= 10.19

=

= 2.29

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update chart

v =1/(1-v2/c2)

0.01 c 1.00005

0.1 c 1.005

0.5c 1.15

0.6c 1.25

0.8c 1.67

0.9c 2.29

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Other values of

11 – v2/c2

11 – (0.99c)2/c2 =

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

11 – (0.99)2c2/c2 =

11 – (0.99)2 =

11 – 0.98

= 10.02

=

= 7.07

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Enter into chart

v =1/(1-v2/c2)

0.01 c 1.00005

0.1 c 1.005

0.5c 1.15

0.6c 1.25

0.8c 1.67

0.9c 2.29

0.99c 7.07

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Other values of

11 – v2/c2

11 – (c)2/c2 =

Suppose v = c

11 – c2/c2 =

11 – 12 =

10

= 10

=

= Infinity!!!

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update chartv =1/(1-v2/c2)

0.01 c 1.00005

0.1 c 1.005

0.5c 1.15

0.6c 1.25

0.8c 1.67

0.9c 2.29

0.99c 7.07

1.00c

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Other values of

11 – v2/c2

11 – (1.1c)2/c2 =

Suppose v = 1.1c

11 – (1.1)2c2/c2 =

11 – (1.1)2 =

11-1.21

= 1

-0.21 =

= Imaginary number!!!

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Complete the chart

v =1/(1-v2/c2)

0.01 c 1.00005

0.1 c 1.005

0.5c 1.15

0.6c 1.25

0.8c 1.67

0.9c 2.29

0.99c 7.07

1.00c Larger than c Imaginary number

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Plot results:

11 – v2/c2

v=cx x xx

x

Never-

never

lan

d

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Moving clocks run slower

t

t0

t = t0

t = t0 1

1 – v2/c2

v

>1 t > t0

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Length contraction

man onrocket

Time = t0 =t/Length = vt0

L0

v

time=tL0 = vt

Shorter!

=vt/ =L0/

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Moving objects appear shorter

L = L0/

>1 L < L0

Length measured when object is at rest

V=0.1cV=0.86cV=0.99cV=0.9999c

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Length contraction

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mass:

m0

aF=m0a

change in vtime

Ft0

m0

change in v =

time=t0

t=t0

Ftchange in v

m =

= m0

mass increases!!

t0

= m0

Ft0

change in vm0 =

Ft0

change in v

=m = m0

by a factor

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Relativistic mass increase

m0 = mass of an object when it is at rest “rest mass”

m = m0

mass of a movingobject increases

by the factor

as vc, m

as an object movesfaster, it gets

harder & harderto accelerate

v=c

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summary

•Moving clocks run slow

•Moving objects appear shorter

•Moving object’s mass increasesBy

a fa

ctor o

f

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Plot results:

11 – v2/c2

v=cx x xx

x

Never-

never

lan

d

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Twin paradox

Twin brother& sister

She will travel to

-centauri (a near-by star on a specialrocket ship v = 0.9cHe will stay home

& study Phys 100

-centauri

4.3 lig

ht years

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Light year

distance light travels in 1 year

dist = v x time

1cyr = 3x108m/s x 3.2x107 s

= 9.6 x 1015 m

We will just use cyr units& not worry about meters

= c yr

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Time on the boy’s clock

v=0.9cd 0

=4.3 cyr

tout =d0

v4.3 cyr

0.9c= = 4.8 yrs

According to the boy& his clock on Earth:

tback =d0

v4.3 cyr

0.9c= = 4.8 yrs

ttotal = tout+tback

v=0.9c

= 9.6yrs

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What does the boy see on her clock?

v=0.9cd=4.3 cyr

tout =tout

4.8 yrs

2.3= = 2.1 yrs

According to the boyher clock runs slower

tback =tback

4.8 yr

2.3= = 2.1 yrs

ttotal = tout+tback

v=0.9c

= 4.2yrs

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So, according to the boy:

v=0.9cd=4.3 cyr

v=0.9c

She ages

less

his clock her clock

out: 4.8yrs 2.1yrsback: 4.8yrs 2.1yrs

total: 9.6yrs 4.2yrs

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But, according to the girl, the boy’s clock is

moving &, so, it must be running slower v=0.9c

tout =tout

2.1 yrs

2.3= = 0.9 yrs

According to her, theboy’s clock on Earth says:

tback =tback

2.1 yrs

2.3= = 0.9 yrs

ttotal = tout+tbackv=0.9c

= 1.8yrs

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Her clock advances 4.2 yrs

& she sees his clock advance

only 1.8 yrs,

She should think he has aged less than her!!

A contradiction??

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As seen by him

Events in the boy’s life:As seen by her

She leaves

She arrives& starts turn

Finishes turn & heads home

She returns

4.8 yrs

4.8 yrs

short time

9.6+ yrs

0.9 yrs

????

0.9 yrs

1.8 + ??? yrs

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turning around as seen by her

He sees herstart to turn

He sees herfinish turning

According to her, these2 events occur very,veryfar apart from each other

Time interval between 2 events dependson the state of motion of the observer

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Gunfight viewed by observer at rest

Bang! Ban

g!

He sees both shotsfired simultaneously

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Viewed by a moving observer

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Viewed by a moving observer

Bang! Ban

g!

He sees cowboy shoot1st & cowgirl shoot later

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as seen by him

In fact, ???? = 7.8+ years

as seen by her

She leaves

She arrives& starts turn

Finishes turn& heads home

She returns

4.8 yrs

4.8 yrs

short time

9.6+ yrs

0.9 yrs

???

0.9 yrs

1.8 + ???yrs

7.8+ yrs

9.6+ yrs

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No paradox: both twins agree

The twin that“turned around”

is younger

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Ladder & Barn Door paradox

1m

2m

???

ladder

Stan & Ollie puzzle over howto get a 2m long ladder thrua 1m wide barn door

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Ollie remembers Phys 100 & thetheory of relativity

1m

2m

ladder

Stan, pick up the

ladder & run very

fast

tree

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View from Ollie’s ref. frame

1m

2m/

Push, Stan!

V=0.9c(=2.3)Ollie Stan

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View from Stan’s ref. frame

2m

1m/

V=0.9c(=2.3)

Ollie Stan

But it doesn’t

fit, Ollie!!

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If Stan pushes both ends of theladder simultaneously, Ollie sees the

two ends move at different times:

1mToo soon Stan!

V=0.9c(=2.3)Ollie Stan

clunk

Stan

clank

Too late

Stan!

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Fermilab proton accelerator

2km

V=0.9999995c

=1000

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Stanford electron accelerator

3km=100,000

v=0.99999999995 c

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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.

Page 75: Theory of relativity

test

=29.3