Special Relativity

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Special Relativity

• Jim Wheeler

Physics Advanced Mechanics
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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

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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.
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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.
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Postulate 3 The speed of light
In agreement with Maxwells theory of
electromagnetism, and in conflict with Newtons
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.
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A picture of spacetime
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How we think about it
Time flows sort of upward
Space goes side to side
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What we all agree on
Light (moving right)
Postulate 3 Constancy of the speed of light
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What we all agree on
more light (moving left)
Postulate 3 Constancy of the speed of light
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Postulate 3 Every observer sees light move the
same distance in a given amount of time
We can choose our units so that light moves at 45
degrees. (e.g., light seconds seconds)
Lotsa light
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Suppose some observers move in straight
linesPostulate 1 Spacetime is homogeneous and
isotropic
An observer moving along in spacetime
World line of an observer
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Postulate 2 Equivalence of inertial frames
Observers move along in spacetime
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An observer moving along in spacetime
How might an observer label points in spacetime?
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Some observers have watches
An observer marking time
They can mark progress along their world line. A
watch gives a perfectly good way to label points
along an observers world line.
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Some observers have watches
An observer marking time
Postulate 1 Homogeneity and isotropy. We assume
good clocks give uniform spacing.
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How might an observer label other points in
spacetime?
Observer A
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How might an observer label other points in
spacetime?
Observer Al
They can send out light signals.
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How might an observer label other points in
spacetime?
Observer Ali
There needs to be dust or something, so some
light comes back.
They can send out light signals.
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Labeling points in spacetime
Observer Alic
2 s
Suppose they send out a signal two seconds before
noon and the signal reflects and returns at
two seconds after noon.
Noon 0 s
dust
-2 s
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The time of a remote event.
Observer Alice
2 s
The observer assumes the reflection occurred at
noon.
(0, x)
0 s
-2 s
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The distance of a remote event
Observe Alice
2 s
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. (Postulate 3 Constancy
of the speed of light)
(0, 2)
0 s
-2 s
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Spacetime coordinates
O serve Alice
2 s
(0 s, 2 ls)
0 s
-2 s
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Spacetime coordinates
serve Alice
2 s
The observer can send out other signals, at
various times, in various directions.
(0, 2)
0 s
-2 s
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Spacetime coordinates
serv Alice
2 s
(0, 2)
0 s
Sometimes there will be dust in just the right
places.
-2 s
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Spacetime coordinates
ser Alice
2 s
(0, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
We assign coordinates to each point where a
reflection occurs.
-2 s
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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, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
-2 s
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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, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
-2 s
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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, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
-2 s
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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, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
-2 s
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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, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
-2 s
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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, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
(-2, 0)
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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, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
(-1, 0)
(-2, 0)
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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, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
(-1, 0)
(-2, 0)
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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, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
(-1, 0)
(-2, 0)
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Spacetime coordinates
Alice
(5, 0)
(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, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
(-1, 0)
(-2, 0)
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Spacetime coordinates
Alice
(5, 0)
(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, 4)
(0, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
(0, -3)
(-1, 0)
(-2, 0)
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Spacetime coordinates
Alice
(5, 0)
(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, 4)
(0, 3)
(0, 2)
(0, 1)
(0, 0)
(0, -1)
(0, -2)
(0, -3)
(0, -4)
(-1, 0)
(-2, 0)
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Spacetime coordinates
Alice
(5, 0)
(4, 0)
(3, 0)
We assign coordinates to other points in the same
way.
(2, 0)
(1, 0)
(0, 0)
(-1, 0)
(-2, 0)
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Spacetime coordinates
Alice
(5, 0)
(4, 0)
(3, 0)
(2, 0)
(1, 0)
(0, 0)
(-1, 0)
We now have coordinate labels for each point in
spacetime.
(-2, 0)
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Spacetime coordinates
Alice
(5, 0)
(4, 0)
(3, 0)
(2, 0)
(1, 0)
(0, 0)
(-1, 0)
Notice that the paths of light move one light
second per second
(-2, 0)
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Alices coordinate axes
t
Alice
Alice
x
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Alice and Bills coordinate axes
t
Alice
Alice
Bill
t
Since the coordinate system is constructed using
only the postulates, similar coordinates can be
constructed for any inertial frame
x
x
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Spacetime terminology
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Events
Alice
Generic points in spacetime are called events.
An event is characterized by both time and place
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The light cone
Alice
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The light cone
Alice
The t and x axes make equal angles with light
paths.
j
j
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The light cone
t
Alice
If we add another spatial dimension the light
cone really looks like a cone.
y
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The light cone
t
Alice
Think of the light cone as the surface of an
expanding sphere of light.
y
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The light cone
t
Alice
Think of the light cone as the surface of an
expanding sphere of light.
y
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The light cone
t
Alice
Think of the light cone as the surface of an
expanding sphere of light.
y
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The light cone
t
Alice
Think of the light cone as the surface of an
expanding sphere of light.
y
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The light cone
t
Alice
Think of the light cone as the surface of an
expanding sphere of light.
y
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The light cone
t
Alice
Think of the light cone as the surface of an
expanding sphere of light.
y
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The light cone
t
Alice
Think of the light cone as the surface of an
expanding sphere of light.
y
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The light cone
t
Alice
Think of the light cone as the surface of an
expanding sphere of light.
y
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The light cone
t
Alice
Think of the light cone as the surface of an
expanding sphere of light.
y
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Forward light cone
t
Alice
The forward light cone includes all places that
can receive light from Alices origin.
y
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The past light cone
t
Alice
The past light cone includes all places that can
send light to Alices origin.
y
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Timelike separated events
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Spacelike separated events
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Lightlike separated events
Whenever events are lightlike separated, light
can travel from one to the other.
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Return toAlices coordinate axes
Alice
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We want to show that spacetime is a vector space
Alice
Consider an event
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Coordinates for the event
Alice
(3, 2)
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Can we associate a vector with this point?
Alice
(3, 2)
Vectors must add linearly.
u
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Consider a pair of these vectors
Alice
(3, 2)
u
(1,-3)
v
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Their sum is the vector at the event(4,-1)
Alice
(4,-1)
(3, 2)
uv
u
(1,-3)
v
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The sum is the sum of the components, as required
Alice
This works because we have assumed that Alices
coordinates are uniform.
Postulate 1 Spacetime is homogeneous and
isotropic.
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If clocks dont tick uniformly, we need to
reassess what happens.Events along the x-axis
no longer line up.
Alice
t
x
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Like inertial frames in Newtonian dynamics, we
are restricted to special inertial frames or
frames of reference
Alice
Both Newtons 2nd Law and Special Relativity
allow generalizations to arbitrary
coordinates. As we have seen, the definition of
vectors becomes more subtle.
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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.
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Multiple observers
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Consider two observers . . .
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Alice . . .
t
Alice
x
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. . . and Bill
t
Alice
Bill
t
x
x
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What do Alice and Bill agree on?
t
Alice
Bill
t
x
x
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What do Alice and Bill agree on?
t
Alice
Bill
t
The light cone
x
x
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What do Alice and Bill agree on?
t
Alice
Bill
t
The light cone Timelike, null,
and spacelike separations of events.
x
x
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Do Alice and Bill agree onanything else?
t
Alice
Bill
t
The light cone Timelike, null,
and spacelike separations of events.
x
x
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Spacetime is a vector space!
t
Alice
Bill
t
x
x
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Does a spacetime vector have an invariant length?
t
Alice
Bill
t
s
x
x
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In Euclidean space, the Pythagorean theorem gives
an invariant length.
y
L2 Dx2 Dy2
Dy
Dx
x
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The invariant length can be expressed as a
quadratic expression in terms of the coordinates.
y
(x2, y2)
L2 (x2 - x1)2 (y2 - y1)2
Dy y2 - y1
(x1, y1)
Dx x2 - x1
x
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Heres a plan1. Relate the (x,t) components of
vectors to the (x,t) components.2. Try to
find an invariant quadratic form.
t
Alice
Bill
t
s
x
x
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How are the coordinates (x, t) and (x, t)
related?
t
Alice
Bill
t
x
x
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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
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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
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The Galilean transformation
In Newtonian physics, the result is the Galilean
transformation
x x vt t t
and its inverse,
x x - vt t t
For special relativity, the linear map is called
the Lorentz transformation
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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
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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)
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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.
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The Lorentz transformation
We need to consider clocks in detail to derive
the full Lorentz transformation
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Lets study a simple clock.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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Imagine a pulse of light bouncing between a pair
of mirrors.
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A simple clock.
One full cycle of the clock takes time Dt 2L/c
L meters
v c
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Now lets watch the same clock as it moves past
us with velocity v
v
v
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This time, the light appears to us to travel
further.
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Find the cycle time, Dt, of the moving clock
L
vDt
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Find the cycle time, Dt, of the moving clock
L
L2 (vDt/2)2 cDt/2
vDt
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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
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Find the cycle time, Dt, of the moving clock
Since Dt 2L/c,
Dt
Dt
1 - v2 /c2
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Now substitute into the linear Lorentz
transformation
1
g
Let
1 - v2 /c2
Then
Dt t2 - t1 d(v)(x2 - x1)
a(-v)(t2 - t1) a(-v) Dt
Since Dt g Dt we have a(v) a(-v) g
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Constancy of the speed of light
Now imagine an expanding sphere of light. In
Alices frame, the x-coordinate of the sphere is
given by x ct. In Bills frame, the
x-coordinate of the sphere is given by x ct.

Substitute these conditions into the Lorentz
transformation equations.
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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
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Velocity
The Lorentz transformation must be of the form
x gx bt t bx/c2 gt
An object at rest in Alices frame has dx/dt
0. In Bills frame, the same object moves with
velocity dx/dt v. Therefore,
differentiating
dx a dx b dt dt b dx/c2 g dt
So
b

g
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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.
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The invariant interval
We now seek a quadratic form that is independent
of the observers 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.
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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 s2 f t2 x2 y2 z2
gt (x y z)
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The invariant interval
Substitute the Lorentz transformation
s2 f t2 x2 y2 z2
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
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The invariant interval
Now compare
s2 f t2 x2 y2 z2 g
t (x y z)
s2 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.
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The invariant interval
With g 0,
s2 g2 f (t2 2vxt/c2 v2x2/c4 )
x2 2xvt v2t2 y2 z2 s2 g2
(f v2)t2 (fv2/c4 1)x2
2(f/c2 1)vxt y2 z2
Then
0 s2 - s2 (g2f g2v2 -f )t2
(g2fv2/c4 g2 - 1)x2
2(f/c2 1)vxt
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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
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The invariant interval
s2 - c2t2 x2 y2 z2
s is called the proper length
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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.
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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
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Timelike separated events have s lt 0
s2 - t2 x2 lt 0 t2 t2 - x2 gt 0
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Spacelike separated events have s gt 0
s2 - t2 x2 gt 0 t2 t2 - x2 lt 0
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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
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Some paths are longer than others
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Some paths are longer than others
t
Alice
B
This is due to the familiar triangle inequality
property of a vector spaceLets check
C
A
x
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Some paths are longer than others
t
Alice
(6,0)
B
Assign coordinates to all relevant
events.Notice that it doesnt matter which
frame of reference we choose!
(3,2)
C
A
x
(0,0)
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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 5tB2 (6-3)2
- (0-2)2 5tC2 (6-0)2 - (0-0)2 36
(3,2)
C
A
x
(0,0)
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Some paths are longer than others
t
Alice
(6,0)
B
The triangle inequality holds with the inequality
reversed
(3,2)
C
2.24
5
tA tB
tC 6
A
tC gt tA tB
x
(0,0)
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Some paths are longer than others
t
Alice
(6,0)
B
tC gt tA tB
The change from lt to gt is because of the minus
sign in the invariant interval.
(3,2)
C
A
x
(0,0)
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Some paths are longer than others
t
Alice
(6,0)
B
tC gt tA tB
(3,2)
Now consider the physical interpretation
C
A
x
(0,0)
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Some paths are longer than others
Alice
t
(6,0)
Alice sees Path B as the world line of the friend
returning at 2/3 the speed of light.
During his return, Sids clock advances another
2.24 years
B
tC gt tA tB
Path C is Alices world line. In her reference
frame, she is at rest.
tC is just the elapsed time on Alices watch.
Alice ages by 6 years.
(3,2)
C
Alice sees Path A as the world line of a friend
Sid moving to the right at 2/3 the speed of light.
Sid covers two light years in 3 years. At rest in
his own frame, Sid ages 2.24 years.
A
Alice ages 6 years Sid ages 4.48 years
x
(0,0)
183
The result is independent of frame
Alice
t
(6,0)
(3,2)
Alice ages 6 years Sid ages 4.48 years
x
(0,0)
184
Look in Carlos frame
Alice
t
t
Carlos
(6,0)
B
Let Carlos move to the left at .25c relative to
Alice
(4,-1)
(3,2)
Carlos moves to x -1 by time t 4
His x axis is symmetrically placed
A
x
(0,0)
x
185
Look in Carlos frame
Alice
t
t
Carlos
(6g,- 1.5g)
(6,0)
B
Locate the key events in Carlos frameg
(16/15)1/2
(3,2)
(3.5g,2.75g)
A
x
(0,0)
(0,0)
x
186
Alice
t
t
Compute the proper times
Carlos
(6g, 1.5g)
(6,0)
tA2 (3.5g)2 - (2.75g)2 (196 - 121)/15
5tB2 (2.5g)2 - (1.25)2 (100 -
25)/15 5tC2 (6g)2 - (1.5g)2
36g2 - 2.25g2 33.75 x 16/15 36
B
g (16/15)1/2
(3,2)
(3.5g,2.75g)
A
x
(0,0)
(0,0)
x
187
Alice
t
t
Compute the proper times
Carlos
(6g, 1.5g)
(6,0)
tA2 5tB2 5tC2 36
B
g (16/15)1/2
(3,2)
(3.5g,2.75g)
Now consider the physical interpretation.
A
x
(0,0)
(0,0)
x
188
Alice
t
t
Compute the proper times
Carlos
(6g, 1.5g)
(6,0)
In Carlos frame Alice ages 6 yearswhile Sid
ages 4.48 years.
B
g (16/15)1/2
(3,2)
(3.5g,2.75g)
A
x
(0,0)
(0,0)
x
189
Alice
t
t
The result is independent of frame
(6g, 1.5g)
(6,0)
B
Carlos
In Carlos frame Alice ages 6 yearsSid ages
4.48 years.In Alices frameAlice ages 6
yearsSid ages 4.48 years.
(3,2)
(3.5g,2.75g)
A
x
(0,0)
(0,0)
x
190
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.
191
Lotsa light!