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1
Physics 200a Relativity crib sheet Shankar (2006)
Here are some very basic things you should know to do this problem set.
X
= (
ct,x
) = (
x
0
,x
1
)
Under a LT
x
0
1
=
x
1

βx
0
√
1

β
2
(1)
x
0
0
=
x
0

βx
1
√
1

β
2
,
(2)
x
0
2
=
x
2
(3)
x
0
3
=
x
3
(4)
where
β
=
u
c
. Let us forget
x
2
,x
3
.
Note that we use
x
0
=
ct
and not just
t
since
x
0
has the same units as
x
and the LT looks nice and symmetric. You can check that
X
·
X
≡
X
2
=
x
2
0

x
2
1
=
x
0
2
0

x
0
2
1
=
s
2
is the same for all observers, an invariant.
A particle of mass
m
and velocity
v
has energy
E
and momentum
p
given
by
E
=
mc
2
q
1

v
2
/c
2
p
=
mv
q
1

v
2
/c
2
(5)
The energymomentum vector is
P
= (
P
0
,P
1
) = (
E
c
,p
)
It is a pity we need to bring in
E
c
and not
E
but this is to make sure both
components have the same units (recall
x
0
=
ct
) and LT has the same form
as for components of
X
:
P
0
1
=
P
1

βP
0
√
1

β
2
(6)
P
0
0
=
P
0

βP
1
√
1

β
2
,
(7)
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It follows
P
2
=
P
2
0

P
2
1
=
P
0
2
0

P
0
2
1
is an invariant. What is this invariant value? You can show by explicit
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 Fall '08
 RAMAMURTISHANKAR
 Physics

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