22 Particles and Fields
245
α = σ ,
A =
a D
2π
.
⎫
⎬
⎭
(285)
If we also introduce our measuring-rod L o according to (103), thus
A
σ
=
a
2
4 L 2
o
,
(286)
we discover for the Lagrangian density of our sine-Gordon field in a crystal
L = −
σ
2
∂q
∂x
∂q
∂x
−
1
c 2
o
∂q
∂t
∂q
∂t
+
σ a
2
a L 2
o
cos
2π
a
q
− 1
.
(287)
The field equation for q = q(x, t) can be determined from L using the following
rule,
∂ L
∂q
−
∂
∂x
∂ L
∂
∂q
∂x
−
∂
∂t
∂ L
∂
∂q
∂x
= 0 .
(288)
We check that (288) together with L does in fact give us our sine-Gordon equation,
doing this using the parameters α and A out of reasons of simplicity,
∂ L
∂q
=
2π A
a
sin
2π
a
q
.
∂ L
∂
∂q
∂x
=− α
∂q
∂x
=⇒ −
∂
∂x
∂ L
∂
∂q
∂x
= α
∂
2 q
∂x 2 ,
∂ L
∂
∂q
∂x
=
α
c 2
o
∂q
∂x
=⇒ −
∂
∂t
∂ L
∂
∂q
∂t
= −
α
c 2
o
∂
2 q
∂t 2 ,
hence finally
−
2π A
a
sin
2π
a
q
+ α
∂
2 q
∂x 2 −
α
c 2
o
∂
2 q
∂t 2 = 0 .
(88
)
This is actually our sine-Gordon equation (88) if we insert the parameters from
Eq. (285) in place of α and A.
The individual components of the energy–momentum tensor (276) can be determined directly from the Lagrangian density according to the following rule:
−t = L + σ
∂q
∂x
∂q
∂x
, p = −
σ
c 2
o
∂q
∂x
∂q
∂t
,
−s = σ
∂q
∂x
∂q
∂t
,
e = L −
σ
c 2
o
∂q
∂t
∂q
∂t
.
⎫
⎪ ⎪ ⎬
⎪ ⎪ ⎭
(289)