246
22 Particles and Fields
We therefore receive the energy–momentum tensor T belonging to the solution q =
q(x, t) of the sine-Gordon equation according to
T =
⎛
⎝
σ
2
∂q
∂x
∂q
∂x +
1
c 2
o
∂q
∂t
∂q
∂t
+ A cos
2π
a q
− A
−
σ
c 2
o
∂q
∂x
∂q
∂t
σ
∂q
∂x
∂q
∂t
−
σ
2
∂q
∂x
∂q
∂x +
1
c 2
o
∂q
∂t
∂q
∂t
+ A cos
2π
a q
− A
⎞
⎠ .
(290)
We check that the energy–momentum tensor, which is built from the field equation
belonging to the Lagrangian density L, in other words the sine-Gordon equation (88)
fulfils the energy–momentum conservation (278),
div T = 0 .
(291)
Indeed,
∂
∂x
σ
2
∂q
∂x
∂q
∂x
+
1
c 2
o
∂q
∂t
∂q
∂t
+ A cos
2π
a
q
− A
+
∂
∂t
−
σ
c 2
o
∂q
∂x
∂q
∂t
=
σ
∂q
∂x
∂ 2 q
∂x 2 +
1
c 2
o
∂q
∂t
∂ 2 q
∂x∂t
−
2π A
a
∂q
∂x
sin
2π
a
q
−
σ
c 2
o
∂q
∂x
∂ 2 q
∂t 2 +
∂q
∂t
∂ 2 q
∂x∂t
=
∂q
∂x
σ
∂
2 q
∂x 2 −
1
c 2
o
∂
2 q
∂t 2
−
2π A
a
sin
2π
a
q
= 0 ,
Here, we have used the sine-Gordon equation (88) and taking (285) into consideration.
The following is just as valid,
∂
∂x
σ
∂q
∂x
∂q
∂t
+
∂
∂t
−
σ
2
∂q
∂x
∂q
∂x
+
1
c 2
o
∂q
∂t
∂q
∂t
+ A cos
2π
a
q
− A
=
σ
∂q
∂x
∂ 2 q
∂x∂t
+
∂q
∂t
∂ 2 q
∂x 2
− σ
∂q
∂x
∂ 2 q
∂x∂t
+
1
c 2
o
∂q
∂t
∂ 2 q
∂t 2
−
2π A
a
∂q
∂t
sin
2π
a
q
=
∂q
∂t
σ
∂
2 q
∂x 2 −
1
c 2
o
∂
2 q
∂t 2
−
2π A
a
sin
2π
a
q
= 0 .
We now only need to pick out the individual solutions of (88), insert them into (290)
and check if a tensor of the form (279) is generated.