Environmental Engineering Reference
In-Depth Information
Our first guess might be to assume the familiar expressions, i.e.
p
=
m
v
(7.1)
for the momentum and
1
2
mv
2
K
=
(7.2)
for the kinetic energy. However these will not do. To see why, we need to notice that
if the total momentum or energy is conserved in one inertial frame then it cannot
also be conserved in all other inertial frames, and this is a violation of the prin-
ciple of relativity (Einstein's 1
st
postulate). To prove that the quantities defined in
Eqs. (7.1) and (7.2) cannot be conserved in all inertial frames one only has to look at
the velocity transformation formulae we derived earlier, i.e. Eqs. (6.33) and (6.34).
As an example consider an inertial frame
S
in which two particles, of mass
m
1
and
m
2
, scatter elastically off each other. For simplicity we'll consider that the
particles always move in the
x
-direction. Before they scatter, the particles have
speeds
u
1
and
u
2
and afterwards
v
1
and
v
2
. The same collision is recorded by an
observer in
S
which, once again, is moving with speed
U
along the positive
x
-axis.
In this case the particles have speeds
u
1
and
u
2
before the collision and
v
1
and
v
2
after it. In classical theory, the total momentum before and after the collision as
recorded in
S
is
m
1
u
1
+
m
2
u
2
=
m
1
v
1
+
m
2
v
2
.
(7.3)
u
1
−
Using the classical law of addition of velocities, i.e.
u
1
=
U
etc., this can be
rewritten as
m
1
u
1
+
m
2
u
2
−
m
1
v
1
+
m
2
v
2
−
+
=
+
(m
1
m
2
)U
(m
1
m
2
)U
i.e.
m
1
u
1
+
m
2
u
2
=
m
1
v
1
+
m
2
v
2
.
(7.4)
Therefore we see that so long as momentum is conserved in
S
so it is also conserved
in
S
. The same can be said for kinetic energy since in
S
we have
1
2
m
2
u
2
=
1
2
m
1
v
1
+
1
2
m
2
v
2
,
1
2
m
1
u
1
+
(7.5)
which can be rewritten as
1
2
m
1
u
1
1
2
m
2
u
2
1
2
(m
1
+
(m
1
u
1
+
m
2
u
2
)U
m
2
)U
2
+
−
+
1
2
m
1
v
1
1
2
m
2
v
2
1
2
(m
1
(m
1
v
1
+
m
2
v
2
)U
m
2
)U
2
=
+
−
+
+
