98
10 Measuring-Rods and Clocks in Motion
system . What is the length of the moving rod in ? The length of a moving rod is
a new concept that has to be defined. This is done as follows:
Take the coordinates x 1 (t) and x 2 (t) of both end points of the rod in motion at one and the
same moment t in . Then x 2 (t) − x 1 (t) is the coefficient of measure for the length of the
moving rod in .
We see, in order to be able to measure the length of a moving rod, we would have to
distribute (identical) clocks throughout our reference system . These clocks would
have to be started in such a manner that they run synchronically. Thus, the idea of a
length of a moving rod is connected with the idea of the synchronisation of clocks.
Our preferred frame o is the static crystal lattice. In this lattice, we have derived
the sine-Gordon equation, with which we can directly calculate the length of a moving measuring-rod. This leads us to Eq. (112). Thus, we have principally solved the
problem of measuring lengths in this preferred frame o . If however the reference
system
has a velocity v with respect to the preferred frame o , what is the length
that we measure for the rod in
if the rod moves relative to
? According to our
definition of a moving length, we now need synchronised clocks in
. Finding a
solution to this problem will cost us a lot of hard work. However, the answer will be
very simple and will give us a very deep insight into the physics of the considered
phenomena of crystals. We will concern ourselves with this in the Chaps. 12–15.
We now calculate that q
I
(x, t) is in fact just as much a solution of the sine-Gordon
equation (88) as is q
I
o (x). In order to do this, we put
u :=
x − vt
γ
.
(114)
According to (111) then it is
q
I
(x, t) = q
I
o (u) ,
(115)
and we get
∂
2
∂x 2 q
I
(x, t) =
1
γ 2
∂
2
∂u 2 q
I
o (u) ,
1
c 2
o
∂
2
∂t 2 q
I
(x, t) =
1
γ 2
v
2
c 2
o
∂
2
∂u 2 q
I
o (u) ,
thus, because q
I
o (u) fulfils the time-independent sine-Gordon equation with u in the
place of x,
∂ 2
∂x 2 q
I (x, t)−
1
c 2
o
∂ 2
∂t 2 q
I (x, t) =
1
γ 2
1−
v 2
c 2
o
∂ 2
∂u 2 q
I
o (u) =
∂ 2
∂u 2 q
I
o (u) =
D
σ
sin
2π
a
q
I
o (u)
and thus
∂
2
∂x 2 q
I
(x, t) −
1
c 2
o
∂
2
∂t 2 q
I
(x, t) =
D
σ
sin
2π
a
q
I
(x, t)
,
10 Measuring-Rods and Clocks in Motion
system . What is the length of the moving rod in ? The length of a moving rod is
a new concept that has to be defined. This is done as follows:
Take the coordinates x 1 (t) and x 2 (t) of both end points of the rod in motion at one and the
same moment t in . Then x 2 (t) − x 1 (t) is the coefficient of measure for the length of the
moving rod in .
We see, in order to be able to measure the length of a moving rod, we would have to
distribute (identical) clocks throughout our reference system . These clocks would
have to be started in such a manner that they run synchronically. Thus, the idea of a
length of a moving rod is connected with the idea of the synchronisation of clocks.
Our preferred frame o is the static crystal lattice. In this lattice, we have derived
the sine-Gordon equation, with which we can directly calculate the length of a moving measuring-rod. This leads us to Eq. (112). Thus, we have principally solved the
problem of measuring lengths in this preferred frame o . If however the reference
system
has a velocity v with respect to the preferred frame o , what is the length
that we measure for the rod in
if the rod moves relative to
? According to our
definition of a moving length, we now need synchronised clocks in
. Finding a
solution to this problem will cost us a lot of hard work. However, the answer will be
very simple and will give us a very deep insight into the physics of the considered
phenomena of crystals. We will concern ourselves with this in the Chaps. 12–15.
We now calculate that q
I
(x, t) is in fact just as much a solution of the sine-Gordon
equation (88) as is q
I
o (x). In order to do this, we put
u :=
x − vt
γ
.
(114)
According to (111) then it is
q
I
(x, t) = q
I
o (u) ,
(115)
and we get
∂
2
∂x 2 q
I
(x, t) =
1
γ 2
∂
2
∂u 2 q
I
o (u) ,
1
c 2
o
∂
2
∂t 2 q
I
(x, t) =
1
γ 2
v
2
c 2
o
∂
2
∂u 2 q
I
o (u) ,
thus, because q
I
o (u) fulfils the time-independent sine-Gordon equation with u in the
place of x,
∂ 2
∂x 2 q
I (x, t)−
1
c 2
o
∂ 2
∂t 2 q
I (x, t) =
1
γ 2
1−
v 2
c 2
o
∂ 2
∂u 2 q
I
o (u) =
∂ 2
∂u 2 q
I
o (u) =
D
σ
sin
2π
a
q
I
o (u)
and thus
∂
2
∂x 2 q
I
(x, t) −
1
c 2
o
∂
2
∂t 2 q
I
(x, t) =
D
σ
sin
2π
a
q
I
(x, t)
,