Environmental Engineering Reference
In-Depth Information
algebra then you might skip to the end result, Eq. (7.19), otherwise we write
1
γ(v
A
)
=
1
v
A
/c
2
−
=
1
v
Ay
/c
2
−
1
/
2
v
Ax
/c
2
−
−
1
−
1
/
2
v
Ay
/c
2
U)
2
/c
2
(v
Ax
−
1
γ(U)
2
=
−
Uv
Ax
/c
2
)
2
−
(
1
−
(
1
−
Uv
Ax
/c
2
)
2
Uv
Ax
/c
2
1
−
=
1
U
2
v
A
/c
4
v
A
/c
2
+
−
−
U
2
/c
2
Uv
Ax
/c
2
1
−
=
1
v
A
/c
2
U
2
(
1
v
A
/c
2
)/c
2
−
−
−
Uv
Ax
/c
2
1
−
=
(
1
v
A
/c
2
)(
1
−
−
U
2
/c
2
)
Uv
Ax
/c
2
).
=
γ(v
A
)γ (U)(
1
−
(7.19)
Eq. (7.18) then becomes
p
Ay
γ(v
A
)m
A
v
Ay
=
=
m
A
γ(v
A
)v
Ay
=
p
Ay
.
(7.20)
This is an interesting result. It tells us that the
y
-component of momentum does
not change as we move from
S
to
S
and because of this property it is evident that
if the
y
component of momentum is conserved in
S
then it must also be conserved
in
S
. Now we turn our attention to the
x
component of the momentum. Again we
focus (rather arbitrarily) on particle A:
p
Ax
=
γ(v
A
)m
A
v
Ax
v
Ax
−
U
γ(v
A
)m
A
·
=
Uv
Ax
/c
2
.
(7.21)
1
−
Now we can once again use Eq. (7.19) to write
p
Ax
=
m
A
γ(U)γ(v
A
)(v
Ax
−
U)
=
γ (U)(p
Ax
−
γ(v
A
)m
A
U).
(7.22)
Substituting this (and the corresponding expressions for the other particles) into
the
x
component part of Eq. (7.17) gives
γ (U)(p
Ax
+
p
Bx
−
p
Cx
−
p
Dx
)
−
γ(U)U
[
γ(v
A
)m
A
+
γ(v
B
)m
B
−
γ(v
C
)m
C
−
γ(v
D
)m
D
]
=
0
.
(7.23)
