Environmental Engineering Reference
In-Depth Information
as before. Recall that we want to express the co-ordinates in
S
in terms of those
measured in
S
. Again in order for the 1st postulate to remain valid the transforma-
tions must be of the form
x
=
ax
+
bt,
(6.19a)
t
=
dx
+
et.
(6.19b)
Notice that we have not assumed that there exists a unique time variable, i.e. we
allow for
t
=
t
. Our goal is to solve for the coefficients
a, b, d
and
e
. As with the
derivation of the Galilean transforms we require that the origin
O
(i.e. the point
x
=
0) move along the
x
-axis according to
x
=
vt
. Substituting this information
into Eq. (6.19a) yields
−
b/a
=
v.
(6.20)
Similarly we require that the origin
O
move along the line
x
=−
vt
.From
0 satisfies
x
=
bt
and
t
=
et
such that
x
=−
vt
implies
Eqs. (6.19) the point
x
=
that
−
b/e
=
v.
(6.21)
Eqs. (6.20) and (6.21) imply that
e
=
a
and
b
=−
av
. Substituting these into
Eqs. (6.19) gives
x
=
ax
−
avt,
t
=
dx
+
at.
(6.22)
We have two unknowns,
a
and
d
, remaining and have two postulates to implement.
Let us first implement the 2nd postulate. We shall do this by considering a pulse
of light emitted at the origins
O
and
O
when they are coincident, i.e. when
0. We know that this pulse must travel outwards along the
x
and
x
axes
such that it satisfies
x
t
=
t
=
ct
, i.e. it travels out at the same speed
c
in
both frames. These two equations must be simultaneous solutions to Eqs. (6.22)
and so we require that
ct
and
x
=
=
ct
=
act
−
avt,
t
=
dct
+
at.
(6.23)
From which it follows directly that
av
c
2
.
d
=−
(6.24)
It only remains to determine the value of
a
. Let us summarise progress so far. We
have reduced Eqs. (6.19a) and (6.19b) to
x
=
a(x
−
vt),
(6.25a)
a
t
c
2
.
vx
t
=
−
(6.25b)
