Environmental Engineering Reference
In-Depth Information
components in
S
. In this case we must use the inverse transformation matrix, which
you can easily check is equal to
cosh
θ
sinh
θ
00
sinh
θ
cosh
θ
00
0
−
1
=
.
(12.9)
0
1 0
0
0
0 1
Thus we can write
=
c
u
x
u
y
u
z
c
u
x
u
y
u
z
cosh
θ
sinh
θ
00
sinh
θ
cosh
θ
00
0
γ(u
)
γ(u)
,
(12.10)
0
1 0
0
0
0 1
V/c
and
V
is the relative speed between
S
and
S
. These are the
equations which explain how to transform velocities from
S
to
S
but to make the
link with Eqs. (6.33) and (6.34) explicit we would like to write down formulae for
u
x
,
u
y
and
u
z
in terms of
V
and the velocity in
S
. The factor of
γ(u)
on the left
hand side prevents us from writing the result immediately however the first of the
four equations encoded in Eq. (12.10) tells us that
where tanh
θ
=
γ(u
)(
cosh
θ
sinh
θu
x
/c)
γ(u)
=
+
(12.11)
and we can use this in the remaining three equations. Doing so gives
1
u
x
cosh
θ),
u
x
=
sinh
θu
x
/c)
(c
sinh
θ
+
(12.12)
(
cosh
θ
+
1
sinh
θu
x
/c)
u
y
,
u
y
=
(12.13)
(
cosh
θ
+
and a similar equation for the
z
component. Using tanh
θ
=
V/c
and cosh
θ
=
γ(V)
gives the final answer:
1
u
x
)
and
u
x
=
Vu
x
/c
2
)
(V
+
(12.14)
(
1
+
1
γ(V)
1
Vu
x
/c)
u
y
,
u
y
=
(12.15)
(
1
+
which are identical to Eqs. (6.33) and (6.34).
12.2 THE WAVE FOUR-VECTOR
Let us now consider a travelling wave whose equation in
S
is written
x
=
k
·
x
−
f(
,t)
A
cos
(
ωt).
(12.16)
It might correspond to a disturbance in some medium or it might describe the
propagation of a plane polarised light wave. Before asking how such a wave looks
