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1.6
Fourdimensional Space
The Lorentz transformation treats the
x
i
with
i
=1
,
2
,
3 as equivalent variables. Let’s
introduce time as simply a forth coordinate:
x
0
=
ct
(85)
This fourdimensional “space” is called Minkowski space. The Lorentz transformation now
becomes
x
±
0
=
γ
(
x
0

βx
1
)
(86)
x
±
1
=
γ
(
x
1

0
)
(87)
x
±
2
=
x
2
(88)
x
±
3
=
x
3
(89)
and the inverse transformation
x
0
=
γ
(
x
±
0
+
±
1
)
(90)
x
1
=
γ
(
x
±
1
+
±
0
)
(91)
x
2
=
x
±
2
(92)
x
3
=
x
±
3
(93)
(94)
which we can write as a matrix
x
±
0
x
±
1
x
±
2
x
±
3
=
γ

βγ
00

γ
0
0
1 0
0
0
0 1
x
0
x
1
x
2
x
3
(95)
Which we can write, letting Greek letters run from 0 to 3 as customary, as
x
±
μ
=
3
±
ν
=0
a
μν
x
ν
,μ
=0
,
1
,
2
,
3
(96)
where
a
is the transformation matrix elements.
example:
Suppose that there are three inertial reference frames:
S
,
S
±
and
S
±
, which
are in collinear motion along their respective
x
1
axes. Let the velocity of
S
±
w.r.t.
S
be
v
1
,
and the velocity of
S
±
w.r.t.
S
±
v
2
. What is the speed
u
of
S
±
relative to
S
?
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This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.
 Spring '10
 RogerMelko

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