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PHY3221 Detweiler
Lecture #4
Relativity
Lorentz transformation
A coordinate rotation:
t0
x0
y0
z0
=
=
=
=
t
x cos θ + y sin θ
−x sin θ + y cos θ
z0
Notice that distance is conserved under a coordinate rotation rotation:
x02 + y 02 =
=
=
=
(x cos θ + y sin θ)2 + (−x sin θ + y cos θ)2
x2 cos2 θ + 2xy cos θ sin θ + y 2 sin2 θ + x2 sin2 θ − 2xy cos θ sin θ + y cos2 θ
x2 (cos2 θ + sin2 θ) + y 2 (sin2 θ + cos θ)
x2 + y 2 .
This might seem obvious because the mathematics just confirms your everyday experiences.
The Galilean transformation:
t0
x0
y0
z0
=
=
=
=
t
x − vt
y
z
Inverse transformation:
t
x
y
z
=
=
=
=
t0
x0 + vt0
y0
z0
The Lorentz transformation:
t − vx/c2
t0 = p
1 − v 2 /c2
x − vt
x0 = p
1 − v 2 /c2
y0 = y
z0 = z
2
Inverse transformation:
t0 + vx0 /c2
t = p
1 − v 2 /c2
x0 + vt0
x = p
1 − v 2 /c2
y = y0
z = z0
Notice that in the limit that v/c → 0, but v remains finite, the Lorentz transformations
approach the Galilean transformation. So, only when v is comparable to c are the effects of
special relativity revealed.
Derive time dilation from the Lorentz transformations:
Two events, #1 at (t1 , x1 ) and #2 at (t2 , x2 ), with occur at the same place (x1 = x2 ) in the
t, x coordinate system. Thus the proper time between the events is
∆T0 = t2 − t1 .
The time between the events in the primed coordinate system is
∆T 0 = t02 − t01
t1 − vx1 /c2
t2 − vx2 /c2
−p
= p
1 − v 2 /c2
1 − v 2 /c2
t2 − vx2 /c2 − t1 + vx1 /c2
p
=
1 − v 2 /c2
t2 − t1
= p
because x1 = x2
1 − v 2 /c2
∆T0
= p
,
1 − v 2 /c2
which is the correct expression for time dilation.
Derive length contraction from the Lorentz transformations:
A stick is at rest in the unprimed t, x coordinate system. Two events, #1 at (t1 , x1 ) and #2
at (t2 , x2 ), occur at different times t1 6= t2 but at either end of the stick. And because the
stick is at rest in the unprimed coordinate system, the proper length of the stick is
L0 = x2 − x1
It has been arranged that these same two events occur at the same time in the primed
coordinate system, t01 = t02 . So, in the primed coordinates the length of the stick is measured
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to be L0 = x02 − x01 . Now, the proper length of the stick is
L 0 = x2 − x1
x0 − vt02
x0 − vt01
= p2
−p1
1 − v 2 /c2
1 − v 2 /c2
x0 − vt0 − x01 + vt01
= 2 p 2
1 − v 2 /c2
x0 − x01
= p 2
because t01 = t02
2
2
1 − v /c
L0
= p
and, finally
1 − v 2 /c2
p
L0 = L0 1 − v 2 /c2
which is the correct expression for length contraction.
Consider the Galilean addition of velocities:
With the Lorentz transformations in hand, we can now see how velocities are viewed from
different coordinate systems. First consider the Galilean addition of velocities. Consider a
bird flying along, and note two nearby events along the bird’s path—perhaps the events are
two flaps of the bird’s wings. These events are noted to occur at (t1 , x1 , y1 ) and (t2 , x2 , y2 ) in
the unprimed coordinate system, and at (t01 , x01 , y10 ) and (t02 , x02 , y20 ) in the primed coordinate
system. The components of the speed of the bird in the unprimed system are
ux =
x2 − x1
∆x
=
t2 − t1
∆t
uy =
y2 − y1
∆y
=
t2 − t1
∆t
u0y =
∆y 0
y20 − y10
=
t02 − t01
∆t0
and in the primed coordinates
u0x =
∆x0
x02 − x01
=
t02 − t01
∆t0
Now use the Lorentz transformations to relate these different speeds:
x2 − x1
∆x
=
t2 − t1
∆t
p
(∆x0 + v∆t0 )/ 1 − v 2 /c2
p
=
(∆t0 + v∆x0 /c2 )/ 1 − v 2 /c2
∆x0 /∆t0 + v
=
and, finally
1 + v∆x0 /∆t0 c2
u0x + v
,
ux =
1 + vu0x /c2
ux =
4
and
uy =
=
∆y
y2 − y1
=
t2 − t1
∆t
∆y 0
p
(∆t0 + v∆x0 /c2 )/ 1 − v 2 /c2
∆y 0 /∆t0
p
=
(1 + v∆x0 /∆t0 c2 )/ 1 − v 2 /c2
p
u0y 1 − v 2 /c2
uy =
.
1 + vu0x /c2
If v and u0x and u0y are all less than c then ux , uy , and also
this out with v = u0x = u0y = 9c/10.
and, finally
p 2
ux + u2y are all less than c. Try