Environmental Engineering Reference
In-Depth Information
(B 0 ,B x ,B y ,B z ) .
The question is 'can we combine these two four-vectors to produce a pure number
in such a way that the result does not depend upon the choice of inertial frame?'.
The answer is in the affirmative for we can define the scalar product as follows:
Let us take two four-vectors, A
=
(A 0 ,A x ,A y ,A z ) and B
=
A
·
B
=
A 0 B 0
A x B x
A y B y
A z B z .
(11.31)
This is very similar to the way we multiply vectors in three dimensional space
except for the fact that one of the terms has opposite sign to all of the others (it is
the term multiplying the components of the two vectors in the time direction). 6 We
can easily check that this definition does indeed yield a result which is the same
in S and S since
A ·
B =
A 0 B 0
A x B x
A y B y
A z B z
=
(A 0 cosh θ
A x sinh θ)(B 0 cosh θ
B x sinh θ)
(A x cosh θ
A 0 sinh θ)(B x cosh θ
B 0 sinh θ)
A y B y
A z B z
=
A 0 B 0
A x B x
A y B y
A z B z
=
A
·
B
(11.32)
and we have made use of cosh 2 θ
sinh 2 θ
1. Thus we have a recipe for com-
bining two four-vectors into a four-scalar. If we take the scalar product of the
position four-vector of an event with itself we obtain
=
c 2 t 2
x 2
y 2
z 2 .
X
·
X
=
(11.33)
In space-time language, this is the squared distance of the event from the origin.
Similarly,
the squared length of the momentum four-vector in space-time is
given by
E 2 /c 2
p x
p y
p z .
P
·
P
=
(11.34)
Since all inertial observers must agree upon the value of this quantity, we can
evaluate it in the inertial frame where the momentum of the particle is zero in
which case E
mc 2 and thus we know that P
m 2 c 2 ,i.e.
=
·
P
=
m 2 c 2
E 2 /c 2
2
=
p
E 2
c 2
2
m 2 c 4 .
p
=
(11.35)
This is none other than the result we presented first in Eq. (7.36). Viewed this
way, the mass of a particle is simply the length of the particle's momentum
four-vector (divided by c ).
6 The overall sign is a matter of convention, as in fact it is when we define the scalar product in three
dimensions.
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