Theory of relativity
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Transcript of 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