Transcript pptx - Florida State University

RELATIVITY 2
Harrison B. Prosper
Florida State University
YSP
Topics
Part 1
 Recap
 Mapping Spacetime
 When Is Now?
Part 2
 Distances in Spacetime
 Paradoxes
 Summary
Recap
Einstein’s theory is based on two postulates:
 Principle of relativity:
The laws of physics are the same in all inertial (that is, nonaccelerating) frames of reference.
 Constancy of the speed of light:
The speed of light in vacuum is independent of the motion
of the light source.
Spacetime Diagrams
Events can be represented as points in a
spacetime diagram
Time
A
B
Space
Events with the
same time values,
such as events A and
B, are said to be
simultaneous
Event: A place at a given time
Spacetime: The set of all events
Earth’s Time Axis
3000 AD
C now
D
B
(t,x,y,z)
2500 AD
A
x
O
y
2000 AD
MAPPING SPACETIME
6
Mapping Spacetime – I
C
tC
Starship’s worldline
v
= speed
= BD / OB
= x / tB
Earth’s
worldline
B
tB
c = BD / AB
tA
A
D
tD
tB = γ tD
tD = κ tA
O
Mapping Spacetime - II
tC C
tB B
tA A
O
D
tD
κ-factor
Relates elapsed times
from a common event O
to two events A and D
that can be connected
by a light ray.
γ-factor
Relates elapsed times
from a common event O
to two events B and D
that are judged
simultaneous in one
of the frames.
κ and γ Factors
1 

1 
Relativistic Doppler Factor
Dilation Factor

v

c
1
1 
2
κ and γ Factors
1 

1 
Relativistic Doppler Factor
Problem 1: derive the formula for κ.
Problem 2: light of 500 nm wavelength is emitted by a
starship, but received on Earth at a wavelength of 600 nm.
What is the relative speed between the Earth and the
starship?
When is Now ?
Δt = tB - tE
B
E
O
tB
tE
Events B and D
are simultaneous for Earth
so
tD = t B / γ
tD
D
But events D and E
are simultaneous
for the starship
so
tE = t D / γ
When Is Now? - II
“Nows” do not coincide!
Δt = tB - tE
tB
B
D
tE
E
O
tD
Writing distance between
B and D as x = BD
the temporal discrepancy
is given by
 v  x 
 x
t         
 c  c
 c
Problem 3: derive Δt
Problem 4: estimate Δt
between the Milky Way and Andromeda, assuming
a relative speed between the galaxies of 120 km/s
Worlds in Collision
T. J. Cox and Abraham Loeb
DISTANCES IN SPACETIME
14
The Metric
P
A
dl
dy
O
dx
B
The distance between points O and P is given
by:
OB2 + BP2 = OP2 = OA2 + AP2
OP2 is said to be invariant.
The formula
dl2 = dx2 + dy2
for computing dl2 is called a metric
In 3-D, this becomes
dl2 = dx2 + dy2 + dz2
The Metric
z
The metric in spherical polar coordinates
(r, θ, φ)
Consider the spatial plane θ = 90o
AC
CB
AB
r
θ
φ
Δφ
x
A
= r dφ
= dr
= dl
dl 2  dr 2  r 2 d 2
y
C
B
The Interval
Q
Suppose that O and Q are events.
How far apart are they in spacetime?
First guess
ds2 = (cdt)2 + dl2
ds
dl
Unfortunately, this does not work!
O
In 1908, Hermann Minkowski showed
that the correct expression is
ds2 = (cdt)2 – dl2
ds2 is called the interval
Hermann Minkowski
1864 -1909
cdt
P
The Interval
In general, the interval ds2 between any two events is either
timelike
ds2 = (cdt)2 – dl2
cdt > dl
or
spacelike
ds2 = dl2 – (cdt)2
dl > cdt
or
null
ds2 = (cdt)2 – dl2 = 0
dl = cdt
1. Which is the longest side and
which is the shortest side?
C
ct
F
3
B
A
5
6
D
x
from Gravity by James B. Hartle
2. Which path is longer,
D to F or D to E to F?
3
E
units: light-seconds
PARADOXES
The Pole and the Barn Paradox
A 20 m pole is carried so fast that it contracts to 10 m in
the frame of reference of a 10 m long barn with an
open front door.
Consequently, the pole can fit within the barn for an
instant, whereupon the back door is swung open.
But in the pole’s frame of reference, the barn is only 5
m long, so the pole cannot possibly fit in the barn!
Resolve the paradox (hint: draw a spacetime diagram)
21
Betty and Ann
2060
Ann’s Now in 2060
 x
t    
 c
Δt = (0.8) (8y)
= 6.4 years
2053.6
2050
Ann’s Now in 2050
8 light years (ly)
β
1/γ
Example from About Time by Paul Davies
= 0.8
= 0.6
Twin Paradox
Ann’s Now in 2070
2070
E-mail sent by Betty
in 2056, Betty’s time,
received by Ann in
2068, Ann’s time.
2068
2060
Ann’s Now in 2060
E-mail sent by Ann
in 2052, Ann’s time,
received by Betty in
2056, Betty’s time.
2053.6
2052
2050
8 light years
Ann’s Now in 2050
Temporal Paradox
2060
Ann’s Now in 2060
Super-luminal signal sent
to star A in 2050, arriving
in 2056 according to Betty.
But for Ann, signal sent in
2050 arrives in 2047!
2053.6
8 light years
8 light years
2050
Ann’s Now in 2050
2048
A
2047.2
Ann and Betty’s Now in 2047.2
Super-luminal signal sent from A
arrives in 2048, preventing signal
sent in 2050!
How To Make A Time Machine!
t
t2
Earth’s worldline
The Two Spaceship Problem
During the acceleration, do the captains
measure a fixed, or changing, distance
between the spaceships? Explain!
Captain
Vivian
Captain
Luke
t1
x
26
Summary
 There is no absolute “now”. Each of us has our own
“now” determined by how we move about
 Super-luminal travel, within a simply-connected
spacetime, would lead to temporal paradoxes
 The time between events depends on the path taken
through spacetime, with an inertial (non-accelerating)
path yielding the longest time
APPENDIX
Proper Distance
By definition:
C
D
The proper distance is the
spatial separation between
two simultaneous events.
x’
E
C & D are simultaneous in the Red
frame of reference.
x
A
B
E & D are simultaneous in the Yellow
frame of reference.
The Lorentz Transformation
t
Define
t'
What is the
interval
between
event O
and
event P?
P
Δt
Δx
= tBC + tCP
= xOA + xAB
Δx'
Δt'
= xOQ
= tQP
tCP = tQP /γ
x'
C
Q
O
A
tBC = β Δx / c
B
xOA = xOQ / γ
xAB = v Δt
x
The Lorentz Transformation
t
We obtain the Lorentz transformation
(Δx, Δt) → (Δx', Δt')
t'
Δx'
cΔt'
P
tCP = tQP / γ
Q
x'
C
tBC = v Δx/ c2
O A B
xOA = xOQ / γ
x
xAB = v Δt
= γ (Δx – β cΔt)
= γ (cΔt – β Δx)
The interval from O to P is
OP2 = (cΔt)2 – (Δx)2
= (cΔt')2 – (Δx’)2
Problem 5
Compute the spacetime distance (ds) between the
following events:
1. event 1: solar flare on Sun (in Earth’s) now.
event 2: a rainstorm here, 7 (Earth) minutes later.
(Give answer in light-minutes.)
2. event 1: the fall of Alexandria in 640 AD
event 2: Tycho’s supernova seen in 1572 AD
(the star was then 7,500 ly from Earth).
(Give answer in light-years.)