Download Life in the fast lane: the kinematics of Star Trek

Life in the fast lane:
the kinematics of Star Trek
Last week we learned that length and time are not absolute, but instead,
depend on the frame of reference of the observer.
We derived the relations:
T
T'
1 v
Lp
L
2
1 v
c2
time dilation
2
c2
length contraction
Using only geometrical arguments and applying Pythagorean's theorem.
cT/2
vT /2
2
d
in blue
 frame
2d  cT / 2
in red frame
 d 2  cT /2
2
To an observer at rest in one frame, the
light in the other frame travels further
because the other frame is moving with
respect to the observer. Since c is
constant in all frames, to each observer
time appears to be moving more slowly
and length appears to be contracted in a
reference frame moving with respect to the
observer.
Everything is slowed/contracted by a
factor of:

1
1 v
2
c2
in a frame moving with respect to the observer.
Time always runs slower when measured by an observer moving
with respect to the clock.
The length of an object is always shorter when viewed by an
observer who is moving with respect to the object.
Hendrik Lorentz: a guy who was almost as smart as Einstein
Maxwell’s highly successful equations…
Gauss’ law
no magnetic
monopoles
In the 19th century, these equations
were thought to hold only in the
luminiferous ether!
Faraday’s law
Ampere’s law
Contain a constant
velocity!
Lorentz wanted to make these equations
covariant (the same in all reference
frames). He came up with the Lorentz
transformation:
1

2
1 v 2
c
Recognize this?
• Postulate 1 – The laws of nature are
the same in all inertial frames of
reference
• Postulate 2 – The speed of light in a
vacuum is the same in all inertial
frames of reference.
• Let’s start to think about the
consequences of these postulates.
• We will perform “thought experiments”
(Gedanken experiment)…
S
v
S’
x
Galilean transformation from S to S’:
x’
x  x  vt
Guess that the relativistic version has a similar form but differs by some
dimensionless factor G:
x  G ( x  vt)
(we know that this must reduce to the Galilean transformation as v/c ->0)
The transformation from S’ to S must have the same form:
x  x  vt 
x  G ( x  vt )
From first postulate of relativity-laws of physics must have the same form in S and S’
substitute
x  G ( x  vt)
solve for t’, you get:
x  G ( x  vt )
into



t' Gt  1 2 1
G

x 

v 
Now we need an expression for the velocity dx’/dt’ in the moving frame:
Take derivatives of:
x  G ( x  vt)
dx'  G (dx  vdt)

u 



x 
t' Gt  1 2 1 
G
v 

1
dx 

dt '  G dt  ( 2  1) 
G
v

dx
u v

dt  1  (1 G 2  1)(u / V )
where u=dx/dt
From second postulate of relativity- the speed of light must be the same for an
observer in S and S’
u=dx/dt=c u’=dx’/dt’=c
Plug this into: u  
Solve for G:
dx
u v

dt  1  (1 G 2  1)(u / V )
G  
1
1 v
2
c2
and get:
c
cv
1  (1 G 2  1)(c / V )
The Lorentz transformations!
S
v
S’
x
The
transformation:
x’
To transform from S’ back to S:
Coordinate systems
Let’s look at this another way: thinking Newtonianly…
x
y
y
x
Professor Hoffman rolls a cat toy
across the floor towards her cat.
By Pythagorean’s theorem, the toy
rolls:
d  x2  y2
The same cat, the same cat toy,
different (arbitrary) choice of
coordinate systems.
Easier visualization, easier calculation—
Professor Hoffman is happy (although her
cat probably doesn’t care as long as he
gets his toy.)
This is an example of rotating your coordinate axes in space-distance is preserved.
Time, the fourth dimension?
“Spacetime”
In the previous slide, the two space dimensions were shown to be interchangeable. A
similar relationship can be used to express the relationship between space and time in
relativity.
ct
Light propagating in one
dimension in a spacetime
coordinate system as viewed from
a frame S. The distance traveled
is equal to the speed of light times
the time elapsed.
x=ct
x
Space-time Diagrams
The red and blue spaceships viewed in spacetime coordinates.
The red and blue
spaceships are at
x=0, t=0 when one
emits a pulse of
light.
At that instant one of
the spaceships starts
to move away from
the other with
velocity v.
The lightcone- the distance
light has traveled since x=0, t=0
as a function of time.
Who perceives him/herself to be at the center of the
lightcone? A passenger on the red spaceship or a
passenger on the blue spaceship?
The red spaceship’s reality.
The blue spaceship’s reality.
They are both at the center of the light cone!
This can be achieved by rotating their coordinate axes as
Professor Hoffman did when playing with her cat, except
one thing…notice that one set of axes is not orthogonal!
You need to add a Lorentz boost.
ct
The speed of light is absolute, therefore the axes must be drawn to preserve
the speed of light!
x  ct in frame S
the enterprise is
Newtonianly:
in frame S’
traveling in a
reference frame
with v<c w/r to S
ct’
ct
x’
x
Vt
x  x Vt
t  t
ct’
x  ct 
a light pulse is
sent out
x=ct
x’
x
x
ct’ and x’ axes must be symmetrically
placed with respect to the light pulse!
The velocity of a particle in spacetime coordinates is:
x
x
c
particle velocity 
c

t
ct slope
ct
ct’
x=ct
x’
x
x  x 
V
ct
c
ct   ct 
V
x
c
The axes aren’t perpendicular but
are scaled by some factor. They
must be symmetric w/r to the
light pulse. What is this factor?
Spacetime is invariant!
ct
ct’
E2
To find the space and time coordinates of
an event in a specific frame, draw lines
from the event parallel to the axis of that
frame.
x=ct
True, observers will differ on the length
of objects or the time events occurred,
however, they will always agree on how
far apart two events are in spacetime.
E1
x’
s  ct  x   c t2  t1  x 2  x1
2
2
2
s  ct   x   c t2 t1  x 2  x1
2

2
2
2
2
2
x
s
2


 s
2
!
2
Could an event at O cause A?
where light that is
here now may go in
the future
x=-ct
ct
A
B
x=ct
O
ct  x
Could an event at O cause B?
C
here,now
Yes, because a “messenger” at O
would not have to travel at a speed
greater than the speed of light to get
there.
x
A light signal sent from O could
reach B.
ct  x
Could an event at O cause C?
where light that is
here now may have
been in the past
No, the spacetime distance between
O and C is greater than could be
covered by light. It would require
time travel.
ct  x
Warp speed
revisited
Can we achieve warp speed?
It’s beginning to become clear
that we cannot.
The Lorentz factor approaches 1
asymptotically, but never
exceeds it:
 1
We will gain more insight into why
this happens when we talk about
relativistic mass next time.
One more thought about the kinematics of Star Trek: a worked
example
The enterprise is speeding along at some “realistic” velocity (less than
the speed of light). Let’s pretend that these stars emit light of a
single wavelength. Will a passenger on the enterprise perceive the
star to be the same color as a passenger on a Klingon ship that is
hovering near the star? (Hint, think about the waves in the still
versus the moving stream from the last lecture.)
Measuring the Universe
Quarknet
J. Goodman Р July 2007