theory of relativity

Theory of Relativity

Physics 100

Chapt 18

AlbertEinstein

watching a light flash go by

c

v

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

2k

watching a light flash go by

c

v

If the man on the rocket sees c-v,

he disagrees with Maxwell

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

upstream,light moves

slower

downstream,light moves

faster “Aether wind”

Michelson-Morley

No aether wind detected: 1907 Nobel Prize

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.

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!!

Even when the light flash is traveling in an opposite direction

c

v

Both guys see the light flash

travel past with velocity = c

Gunfight viewed by observer at rest

Bang! Ban

g!

He sees both shotsfired simultaneously

Viewed by a moving observer

Viewed by a moving observer

Bang! Ban

g!

He sees cowboy shoot1st & cowgirl shoot later

Viewed by an observer in theopposite direction

Viewed by a moving observer

Bang!Ban

g!

He sees cowgirl shoot1st & cowboy shoot later

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

Light clock

Seen from the ground

Events

x

y

x

t

(x1,t1) x

(x2,t2)

x1 x2

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

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

How are the times seen by 2 different observers related?

We can figure this out withsimple HS-level math

( + a little effort)

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

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

Flash a light on a rocket ship

w

t0

c= wt0

Event 1: boy flashes the light

Event 2: light flash reaches the girl

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

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

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

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

Let’s make a chart

v =1/(1-v2/c2)

0.01 c 1.00005

0.1 c 1.005

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

Enter into chart

v =1/(1-v2/c2)

0.01 c 1.00005

0.1 c 1.005

0.5c 1.15

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

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

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

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

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

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

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

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

Other values of

11 – v2/c2

11 – (c)2/c2 =

Suppose v = c

11 – c2/c2 =

11 – 12 =

10

= 10

=

= Infinity!!!

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

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!!!

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

Plot results:

11 – v2/c2

v=cx x xx

x

Never-

never

lan

d

Moving clocks run slower

t

t0

t = t0

t = t0 1

1 – v2/c2

v

>1 t > t0

Length contraction

man onrocket

Time = t0 =t/Length = vt0

L0

v

time=tL0 = vt

Shorter!

=vt/ =L0/

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

Length contraction

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

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

summary

•Moving clocks run slow

•Moving objects appear shorter

•Moving object’s mass increasesBy

a fa

ctor o

f

Plot results:

11 – v2/c2

v=cx x xx

x

Never-

never

lan

d

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

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

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

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

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

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

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??

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

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

Gunfight viewed by observer at rest

Bang! Ban

g!

He sees both shotsfired simultaneously

Viewed by a moving observer

Viewed by a moving observer

Bang! Ban

g!

He sees cowboy shoot1st & cowgirl shoot later

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

No paradox: both twins agree

The twin that“turned around”

is younger

Ladder & Barn Door paradox

1m

2m

???

ladder

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

Ollie remembers Phys 100 & thetheory of relativity

1m

2m

ladder

Stan, pick up the

ladder & run very

fast

tree

View from Ollie’s ref. frame

1m

2m/

Push, Stan!

V=0.9c(=2.3)Ollie Stan

View from Stan’s ref. frame

2m

1m/

V=0.9c(=2.3)

Ollie Stan

But it doesn’t

fit, Ollie!!

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!

Fermilab proton accelerator

2km

V=0.9999995c

=1000

Stanford electron accelerator

3km=100,000

v=0.99999999995 c

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.

test

=29.3