Special Relativity
Jim Wheeler
Physics: Advanced Mechanics
Postulates of Special Relativity
1. Spacetime is homogeneous and isotropic
2. All inertial frames are equivalent
3. The paths and speed of light are universal
Postulate 1: Homogeneity and Isotropy
We assume that, in empty space,
any two points are equivalent and
any two directions are equivalent.
In particular, this means that there
exists a class of inertial frames
which move with constant relative
velocities.
Further, the spatial origins of
these frames may be represented
by straight lines in a spacetime
diagram.
Postulate 2: Inertial Frames
The equivalence of inertial frames
implies that there is no absolute
motion or absolute rest.
Only relative velocities can be of
physical relevance.
One consequence is that there is
no such thing as “vertical” or
“horizontal” in a spacetime
diagram.
Postulate 3: The speed of light
In agreement with Maxwell’s
theory of electromagnetism, and
in conflict with Newton’s laws of
mechanics, we assume the speed
of light in vacuum relative to any
observer in an inertial frame takes
the same value,
c = 2.998 x 108 m/s
= 1 light second / second
This is particularly striking, given
the equivalence of inertial frames.
A picture of spacetime
Time flows sort of upward
Space goes side to side
How we think about it
What we all agree on
Light
(moving right)
Postulate 3:
Constancy of the speed of light
What we all agree on
more light
(moving left)
Postulate 3:
Constancy of the speed of light
We can choose our units so that light moves at 45 degrees.
(e.g., light seconds & seconds)
Lotsa light
Postulate 3: Every observer sees light move the same
distance in a given amount of time
An observer
moving along in
spacetime
Suppose some
observers move in
straight lines
Observers move
slower
than light.
Postulate 1:
Spacetime is
homogeneous and
isotropic
World line of
an observer
Observers
move along in
spacetime
Postulate 2:
Equivalence of inertial
frames
An observer
moving along in
spacetime
How might an
observer label
points in
spacetime?
An observer
marking time
Some observers
have watches
They can mark progress along
their world line. A watch gives a
perfectly good way to label
points along an observer’s
world line.
Postulate 1:
Homogeneity
and isotropy.
We assume
“good” clocks
give uniform
spacing.
Some observers
have watches
An observer
marking time
Observer A
How might an
observer label
other points in
spacetime?
Observer Al
How might an
observer label
other points in
spacetime?
They can send
out light signals.
Observer Ali
There needs to
be dust or
something, so
some light comes
back.
How might an
observer label
other points in
spacetime?
They can send
out light signals.
Observer Alic
+2 s
Suppose they
send out a signal
two seconds
before noon…
Noon = 0 s
Labeling points in
spacetime
dust
…and the signal
reflects and
returns at two
seconds after
noon.
-2 s
Observer Alice
+2 s
(0, x)
0s
The time of a
remote event.
-2 s
The observer
assumes the
reflection
occurred at noon.
Observe Alice
+2 s
(0, 2)
0s
The distance of a
remote event
The light took 2
seconds to go
out, and two
seconds to come
back.
The dust must be
2 light seconds
away at t = 0.
-2 s
(Postulate 3:
Constancy of the
speed of light)
O serve Alice
+2 s
(0 s, 2 ls)
0s
Spacetime
coordinates
-2 s
serve Alice
+2 s
(0, 2)
0s
Spacetime
coordinates
-2 s
The observer can
send out other
signals, at various
times, in various
directions.
Spacetime coordinates
serv Alice
+2 s
(0, 2)
0s
-2 s
Sometimes there
will be dust in just
the right places.
Spacetime coordinates
ser Alice
+2 s
(0, 0)
(0, -2)
(0, 1)
(0, 2)
(0, 3)
(0, -1)
-2 s
We assign
coordinates to each
point where a
reflection occurs.
Spacetime coordinates
se Alice
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
+2 s
(0, 0)
(0, -2)
(0, -1)
-2 s
(0, 1)
(0, 2)
(0, 3)
Spacetime coordinates
s Alice
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
+2 s
(0, 0)
(0, -2)
(0, -1)
-2 s
(0, 1)
(0, 2)
(0, 3)
Spacetime coordinates
Alice
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
+2 s
(0, 0)
(0, -2)
(0, -1)
-2 s
(0, 1)
(0, 2)
(0, 3)
Spacetime coordinates
Alice
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
+2 s
(0, 0)
(0, -2)
(0, -1)
-2 s
(0, 1)
(0, 2)
(0, 3)
Spacetime coordinates
Alice
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
+2 s
(0, 0)
(0, -2)
(0, -1)
-2 s
(0, 1)
(0, 2)
(0, 3)
Spacetime coordinates
Alice
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
(2, 0)
(0, 0)
(0, -2)
(0, -1)
(-2, 0)
(0, 1)
(0, 2)
(0, 3)
Spacetime coordinates
Alice
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
(2, 0)
(1, 0)
(0, 0)
(0, -2)
(0, -1)
(-1, 0)
(-2, 0)
(0, 1)
(0, 2)
(0, 3)
Spacetime coordinates
Alice
(3, 0)
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
(2, 0)
(1, 0)
(0, 0)
(0, -2)
(0, -1)
(-1, 0)
(-2, 0)
(0, 1)
(0, 2)
(0, 3)
Spacetime coordinates
Alice
(4, 0)
(3, 0)
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
(2, 0)
(1, 0)
(0, 0)
(0, -2)
(0, -1)
(-1, 0)
(-2, 0)
(0, 1)
(0, 2)
(0, 3)
Spacetime coordinates
(5, 0)
Alice
(4, 0)
(3, 0)
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
(2, 0)
(1, 0)
(0, 0)
(0, -2)
(0, -1)
(-1, 0)
(-2, 0)
(0, 1)
(0, 2)
(0, 3)
Spacetime coordinates
(5, 0)
Alice
(4, 0)
(3, 0)
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
(0, -3)
(2, 0)
(1, 0)
(0, 0)
(0, -2)
(0, -1)
(-1, 0)
(-2, 0)
(0, 1)
(0, 2)
(0, 3)
(0, 4)
Spacetime coordinates
(5, 0)
Alice
(4, 0)
(3, 0)
In this way
we find all
points that
are labeled
by t = 0, but
different
values of x.
This is what
we mean by
the x-axis.
(0, -4)
(2, 0)
(1, 0)
(0, 0)
(0, -3)
(0, -2)
(0, -1)
(-1, 0)
(-2, 0)
(0, 1)
(0, 2)
(0, 3)
(0, 4)
Spacetime coordinates
(5, 0)
Alice
(4, 0)
(3, 0)
We assign
coordinates
to other
points in the
same way.
(1, -4)
(2, 0)
(1, 1)
(1, -3)
(1, -2)
(1, -1)
(0, -4)
(0, -3)
(0, -1)
(-1, 0)
(-2, 0)
(1, 4)
(1, 0)
(0, 0)
(0, -2)
(1, 2)
(1, 3)
(0, 1)
(0, 2)
(0, 3)
(0, 4)
Spacetime coordinates
(5, 0)
Alice
(4, 0)
(3, 0)
(2, 0)
(1, 0)
(0, 1)
(0, -4)
(0, -3)
(0, -2)
(0, -1)
(0, 2)
(0, 3)
(0, 4)
(0, 0)
(-1, 0)
(-2, 0)
We now have
coordinate labels for
each point in
spacetime.
Spacetime coordinates
(5, 0)
Alice
(4, 0)
(3, 0)
(2, 0)
(1, 0)
(0, 1)
(0, -4)
(0, -3)
(0, -2)
(0, -1)
(0, 2)
(0, 3)
(0, 4)
(0, 0)
(-1, 0)
(-2, 0)
Notice that the paths
of light move one light
second per second
Alice’s coordinate axes
t
Alice
x
Alice and Bill’s
coordinate axes
t
t’
Alice
Bill
Since the coordinate
system is constructed
using only the postulates,
similar coordinates can be
constructed for any inertial
frame
x
x’
Spacetime terminology
Events
Alice
Generic points in
spacetime are called
events.
An event is
characterized by both
time and place
The light cone
Alice
The light cone
Alice
The t and x axes
make equal angles
with light paths.
j
j
The light cone
t
Alice
If we add another
spatial dimension
the light cone really
looks like a cone.
y
x
The light cone
t
Alice
Think of the light
cone as the
surface of an
expanding sphere
of light.
y
x
The light cone
t
Alice
Think of the light
cone as the surface
of an expanding
sphere of light.
y
x
The light cone
t
Alice
Think of the light
cone as the surface
of an expanding
sphere of light.
y
x
The light cone
t
Alice
Think of the light
cone as the surface
of an expanding
sphere of light.
y
x
The light cone
t
Alice
Think of the light
cone as the surface
of an expanding
sphere of light.
y
x
The light cone
t
Alice
Think of the light
cone as the surface
of an expanding
sphere of light.
y
x
The light cone
t
Alice
Think of the light
cone as the surface
of an expanding
sphere of light.
y
x
The light cone
t
Alice
Think of the light
cone as the surface
of an expanding
sphere of light.
y
x
The light cone
t
Alice
Think of the light
cone as the surface
of an expanding
sphere of light.
y
x
Forward light cone
t
Alice
The forward light
cone includes all
places that can
receive light
from Alice’s
origin.
y
x
The past light cone
t
Alice
The past light cone includes
all places that can send light
to Alice’s origin.
y
x
Timelike separated events
Whenever events are
timelike separated, it is
possible for an observer to
be present at both events.
Spacelike separated events
t
Whenever events are
spacelike separated, they
are simultaneous for some
observer.
x
Lightlike separated events
Whenever events are
lightlike separated, light can
travel from one to the other.
Return to
Alice’s coordinate axes
Alice
We want to show that
spacetime is a vector
space
Consider an event
Alice
Coordinates for the
event
Alice
(3, 2)
Can we associate a
vector with this point?
Alice
(3, 2)
Vectors must add linearly.
u
Consider a pair of
these “vectors”
Alice
(3, 2)
u
(1,-3)
v
Their sum is the
“vector” at the event
(4,-1)
Alice
(4,-1)
(3, 2)
u+v
u
(1,-3)
v
The sum is the sum of
the components, as
required
Alice
This works
because we have
assumed that
Alice’s
coordinates are
uniform.
Postulate 1: Spacetime is
homogeneous and isotropic.
If clocks don’t tick
uniformly, we need to
reassess what happens.
t
Alice
Events along the x-axis
no longer line up.
x
Like inertial frames in
Newtonian dynamics, we
are restricted to special
“inertial frames” or “frames
of reference”
Both Newton’s 2nd Law and
Special Relativity allow
generalizations to arbitrary
coordinates.
As we have seen, the
definition of vectors becomes
more subtle.
Alice
For now, we assume
that there exists a set
of observers with
uniform clocks.
Alice
(3,2)
(1,-3)
For these observers,
the components of
spacetime events add
as vectors.
Multiple observers
Consider two observers . . .
Alice . . .
t
Alice
x
. . . and Bill
t
t’
Bill
Alice
x
x’
What do
Alice and
Bill agree
on?
t
t’
Bill
Alice
x
x’
What do
Alice and
Bill agree
on?
t
t’
Bill
Alice
The light cone
x
x’
What do
Alice and
Bill agree
on?
t
t’
Bill
Alice
The light cone
Timelike,
null, and
spacelike
separations of
events.
x
x’
Do Alice
and Bill
agree on
anything
else?
t
t’
Bill
Alice
The light cone
Timelike,
null, and
spacelike
separations of
events.
x
x’
t
Spacetime
is a vector
space!
t’
Bill
Alice
x
x’
t
t’
Bill
Alice
Does a spacetime
vector have an
invariant length?
s
x
x’
In Euclidean
space, the
Pythagorean
theorem gives an
invariant length.
y
L2 = Dx2 + Dy2
Dy
Dx
x
y
The invariant
length can be
expressed as a
quadratic
expression in
terms of the
coordinates.
(x2, y2)
L2 = (x2 - x1)2 + (y2 - y1)2
Dy = y2 - y1
(x1, y1)
Dx = x2 - x1
x
t
t’
Bill
Alice
Here’s a plan:
1. Relate the (x,t)
components of
vectors to the
(x’,t’) components.
s
2. Try to find an
invariant quadratic
form.
x
x’
t
How are the
coordinates
(x, t) and
(x’, t’)
related?
t’
Bill
Alice
x
x’
How are the
coordinates
(x, t) and
(x’, t’)
related?
The coordinates (t, x) for Alice and the
coordinates (t’, x’) for Bill are the
components of the same spacetime
vector.
They must therefore be related by a
linear transformation.
x’ = ax + bt
t’ = dx + et
The
coordinates
transform
linearly
The coefficients a,b,c and d may be functions of velocity.
x’ = a(v)x + b(v)t
t’ = d(v)x + e(v)t
The inverse transformation must simply replace v by -v.
x = a(-v)x’ + b(-v)t’
t = d(-v)x’ + e(-v)t’
The Galilean transformation
In Newtonian physics, the result is the
Galilean transformation
x’ = x + vt
t’ = t
and it’s inverse,
x = x’ - vt’
t = t’
For special relativity, the linear map is called the
Lorentz transformation
The Lorentz transformation
First, consider the special form of the inverse transform.
x = a(-v)x’ + b(-v)t’
t = d(-v)x’ + e(-v)t’
The inverse
transform
In general, the inverse of a linear transformation is given by:
x = l (e(v)x’ - b(v)t’)
t = l (-d(v)x’ + a(v)t’)
where l is the inverse determinant, l = 1/(ae -bd).
By adjusting the units in the two systems we can impose l = 1.
Therefore,
x = a(-v)x’ + b(-v)t’ = e(v)x’ - b(v)t’
t = d(-v)x’ + e(-v)t’ = -d(v)x’ + a(v)t’
Equating like terms:
a(-v) = e(v)
b(-v) = - b(v)
d(-v) = - d(v)
The Lorentz
transform
We may now write the Lorentz transformation in the form:
x’ = a(v)x + b(v)t
t’ = -d(v)x + a(-v)t
where b(v) and d(v) are antisymmetric.
The Lorentz transformation
We need to consider clocks in
detail to derive the full Lorentz
transformation
Let’s study
a simple
clock.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
Imagine a
pulse of light
bouncing
between a
pair of
mirrors.
A simple
clock.
One full
cycle of the
clock takes
time
Dt = 2L/c
L meters
v=c
Now let’s
watch the
same clock
as it moves
past us with
velocity v
v
L meters
v
This time, the light
appears to us to travel
further.
Find the cycle time, Dt’, of the
moving clock
L
vDt’
Find the cycle time, Dt’, of the
moving clock
L
L2 + (vDt’/2)2
vDt’
= cDt’/2
Find the cycle time, Dt’, of the
moving clock
L2 + (vDt’/2)2
= cDt’/2
L2 + v 2 (Dt’)2 /4 = c 2 (Dt’) 2 /4
(c 2 - v2)(Dt’) 2 = 4L2
Dt’ = 2L/(c 2 - v2)1/2
Find the cycle time, Dt’, of the
moving clock
Since Dt = 2L/c,
Dt’ =
Dt
1 - v2 /c2
Now substitute into the linear
Lorentz transformation
Let
g =
1
1 - v2 /c2
Then
Dt’ = t’2 - t’1
= d(v)(x2 - x1) + a(-v)(t2 - t1)
= a(-v) Dt
Since Dt’ = g Dt we have a(v) = a(-v) = g
Constancy of the speed of light
Now imagine an expanding sphere of light.
In Alice’s frame, the x-coordinate of the sphere is
given by x = ct.
In Bill’s frame, the x’-coordinate of the sphere is
given by x’ = ct’.
Substitute these conditions into
the Lorentz transformation
equations.
An expanding
sphere of
light
Set x = ct and x’ = ct’:
ct’ = gct + bt
= (gc + b)t
t’ = dct + gt
= (dc + g)t
Then combining,
c(dc + g) = gc + b
d= b/c2
Velocity
The Lorentz transformation must be of the form:
x’ = gx + bt
t’ = bx/c2 + gt
An object at rest in Alice’s frame has dx/dt = 0.
In Bill’s frame, the same object moves with velocity dx’/dt’ = v.
Therefore, differentiating:
dx’ = a dx + b dt
dt’ = b dx/c2 + g dt
So:
a dx + b dt
v = dx’/dt’ =
b dx/c2 + g dt
=
b
g
The Lorentz transformation
The Lorentz transformation must be of the form:
x’ = g(x + vt)
t’ = g (t + vx/c2)
We assume that the relative motion of the frames
is in the x direction. This also leads to
y’ = y
z’ = z
With a bit of algebra, it is not too hard to find the form of the
transformation for an arbitrary relative velocity.
The invariant interval
We now seek a quadratic form that is independent
of the observer’s frame of reference.
x’ = g(x + vt)
y’ = y
z’ = z
t’ = g (t + vx/c2)
We already know the form of the invariant for
Euclidean space:
L2 = x2 + y2 + z2
This must still be invariant when t = 0.
The invariant interval
The invariant interval must therefore be of the
form:
s2 = f t2 + x2 + y2 + z2
+ gt (x + y + z)
Invariance requires
s2 = s’2 = f t’2 + x’2 + y’2 + z’2
+ gt’ (x’ + y’ + z’)
The invariant interval
Substitute the Lorentz transformation:
s’2 = f t’2 + x’2 + y’2 + z’2
+ gt’ (x’ + y’ + z’)
= f g2 (t + vx/c2)2
+ (g2(x + vt)2 + y2 + z2)
+ g g2 (t + vx/c2)(x + vt + y + z)
= g2 [ f (t2 + 2vxt/c2 + v2x2/c4 )
+ x2 + 2xvt + v2t2] + y2 + z2
+ g2 g (xt + vx2/c2 + vt2 + v2tx/c2 + ty
+ vxy/c2 + tz + vxz/c2]
The invariant interval
Now compare:
s2 = f t2 + x2 + y2 + z2
+ g t (x + y + z)
s’2 = g2 [ f (t2 + 2vxt/c2 + v2x2/c4 )
+ x2 + 2xvt + v2t2] + y2 + z2
+ g2 g (xt + vx2/c2 + vt2 + v2tx/c2 + ty
+ vxy/c2 + tz + vxz/c2]
Since s has no xy, xz or yz terms, we require g = 0.
The invariant interval
With g = 0,
s’2 = g2 [ f (t2 + 2vxt/c2 + v2x2/c4 )
+ x2 + 2xvt + v2t2] + y2 + z2
s’2 = g2 [ (f + v2)t2 + (fv2/c4 + 1)x2
+ 2(f/c2 + 1)vxt]
+ y2 + z2
Then
0 = s’2 - s2
= [ (g2f + g2v2 -f )t2
+ (g2fv2/c4 + g2 - 1)x2
+ 2(f/c2 + 1)vxt
The invariant interval
This gives the three conditions
0 = (g2f + g2v2 -f )
0 = (g2fv2/c4 + g2 - 1)
0 = f/c2 + 1
All three equations are solved by the
single condition
f = - c2
The invariant interval finally takes the form
s2 = - c2t2 + x2 + y2 + z2
The invariant interval
s2 = - c2t2 + x2 + y2 + z2
s is called the proper length
The invariant interval
It is also convenient to define the proper time, t, by
c2t2 = c2t2 - x2 - y2 - z2
Both s and t are Lorentz invariant. We use whichever is convenient.
Event separation and the
invariant interval
s2 = - c2t2 + x2 + y2 + z2
For events lying in the x-t plane, we can set
y=z=0
Then, with c = 1, we write
s2 = - t2 + x2
t2 = t2 - x2
Timelike separated
events have s < 0
Whenever events are
timelike separated, it is
possible for an observer to
be present at both events.
s2 = - t2 + x2 < 0
t2 = t2 - x2 > 0
Spacelike separated
events have s > 0
t
Whenever events are
spacelike separated, they
are simultaneous for some
observer.
x
s2 = - t2 + x2 > 0
t2 = t2 - x2 < 0
Lightlike separated
events have s = 0
Whenever events are
lightlike separated, light can
travel from one to the other.
s2 = - t2 + x2 = 0
t2 = t2 - x2 = 0
Some paths are
longer than others
Some paths are
longer than others
t
Alice
B
This is due to the
familiar “triangle
inequality” property of a
vector space
Let’s check…
C
A
x
Some paths are
longer than others
t
Alice
(6,0)
B
Assign coordinates to
all relevant events.
(3,2)
C
Notice that it doesn’t
matter which frame of
reference we choose!
A
x
(0,0)
Some paths are
longer than others
t
Alice
(6,0)
B
Now compute the invariant
length of each vector.
tA2 = (3-0)2 - (2-0)2 = 5
(3,2)
C
tB2 = (6-3)2 - (0-2)2 = 5
A
tC2 = (6-0)2 - (0-0)2 = 36
x
(0,0)
Some paths are
longer than others
t
Alice
(6,0)
B
The triangle inequality
holds with the
inequality reversed:
tA = tB =
5
= 2.24
(3,2)
C
tC = 6
A
tC > tA + tB
x
(0,0)
Some paths are
longer than others
t
Alice
(6,0)
B
tC > tA + tB
The change from < to >
is because of the minus
sign in the invariant
interval.
(3,2)
C
A
x
(0,0)
Some paths are
longer than others
t
Alice
(6,0)
B
tC > tA + tB
(3,2)
Now consider
the physical
interpretation
C
A
x
(0,0)
Some paths are
longer than others
t
(6,0)
tC > tA + tB
C is Alice’s world
tPath
C is just the elapsed
Alice
Alice sees Path B as
During his return, Sid’s
the world line of the
clock advances another
Bfriend returning at 2/3
2.24 years
the speed of light.
(3,2)
line. In
reference
time
onher
Alice’s
watch.C
frame,ages
she by
is at
rest.
Alice
6 years.
Alice
sees Path
A asyears
the
Sid
covers
two light
A
world
line of
friend
Sid
in
3 years.
Atarest
in his
moving
to the
2/3
own
frame,
Sidright
agesat2.24
the speed of light.
years.
Alice ages 6 years
Sid ages 4.48 years
x
(0,0)
The result is
independent of
frame
t
Alice
(6,0)
(3,2)
Alice ages 6 years
Sid ages 4.48 years
x
(0,0)
Look in
Carlos’
frame
t’
Carlos
t
Alice
(6,0)
B
Let Carlos
move to
the left at
.25c
relative to
Alice
(4,-1)
(3,2)
A
Carlos
moves
His
x’ axis
is
to x = -1 by
symmetrically
time t = 4
placed
x
(0,0)
x’
Look in
Carlos’
frame
t’
Carlos
t
(6,0)
Alice
(6g,- 1.5g)
B
(3,2)
Locate the key
events in
Carlos’ frame
(3.5g,2.75g)
A
g = (16/15)1/2
x
(0,0)
(0,0)
x’
Compute the
proper times
t’
Carlos
t
(6,0)
tA2 = (3.5g)2 - (2.75g)2
= (196 - 121)/15
=5
Alice
(6g, 1.5g)
B
tB2 = (2.5g)2 - (1.25)2
= (100 - 25)/15
=5
g = (16/15)1/2
(3,2)
(3.5g,2.75g)
tC2 = (6g)2 - (1.5g)2
= 36g2 - 2.25g2
= 33.75 x 16/15
= 36
A
x
(0,0)
(0,0)
x’
Compute the
proper times
t’
Carlos
t
(6,0)
Alice
(6g, 1.5g)
tA2 = 5
B
g = (16/15)1/2
tB 2 = 5
tC2 = 36
(3,2)
(3.5g,2.75g)
Now consider
the
physical
interpretation.
A
x
(0,0)
(0,0)
x’
Compute the
proper times
t’
Carlos
t
(6,0)
Alice
(6g, 1.5g)
B
In Carlos’ frame
Alice ages 6 years
while Sid ages 4.48
years.
g = (16/15)1/2
(3,2)
(3.5g,2.75g)
A
x
(0,0)
(0,0)
x’
t’
The result is
independent of
frame
t
(6,0)
Alice
(6g, 1.5g)
B
Carlos
In Carlos’ frame:
Alice ages 6 years
Sid ages 4.48 years.
(3,2)
(3.5g,2.75g)
In Alice’s frame:
Alice ages 6 years
Sid ages 4.48 years.
A
x
(0,0)
(0,0)
x’
Physical properties
may always be
characterized by
invariant quantities.
2.24 yr
6 yr
2.24 yr
Compute invariant spacetime
quantities in whichever
reference frame is most
convenient.
Lotsa light!