23 A Particle Solution—The Inertia of Energy
249
ρ(x − vt) =
1
c 2
o
1
γ 2
4 σa
2
L 2
o γ 2
exp
2
π(x − vt)
L o γ
1 + exp
2
π(x − vt)
L o γ
2 ,
(292)
we receive for the first component −t of the energy–momentum tensor
− t = v
2
ρ(x − vt) .
(293)
We also find
−
σ
c 2
o
∂q
I
∂x
∂q
I
∂t
= −
1
c 2
o
(−v)
4 σa
2
L 2
o γ 2
exp
π(x − vt)
L o γ
1 + exp
2
π(x − vt)
L o γ
2
and thus
p = v p
(293a)
and according to (290) also
− s = −v c
2
o ρ .
(293b)
Finally, the following is derived
−
σ
2
∂q I
∂x
∂q I
∂x
+
1
c 2
o
∂q I
∂t
∂q I
∂t
+
σa 2
4L 2
o
cos
2π
a
q I
− 1
= −
σ
2
4 a 2
L 2
o γ 2
1 +
v 2
c 2
o
exp
2
π(x − vt)
L o γ
1 + exp
2
π(x − vt)
L o γ
2 −
σa 2
4L 2
o
8
exp
2
π(x − vt)
L o γ
1 + exp
2
π(x − vt)
L o γ
2
=
exp
2
π(x − vt)
L o γ
1 + exp
2
π(x − vt)
L o γ
2
2σa 2
L 2
o γ 2
1 −
v 2
c 2
o
− 2
−
2σa 2
L 2
o
= −
4 σa 2
L 2
o γ 2
exp
2
π(x − vt)
L o γ
1 + exp
2
π(x − vt)
L o γ
2