248
23 A Particle Solution—The Inertia of Energy
= −8
tan
2
arctan exp
π(x − vt)
L o γ
1 + tan 2
arctan exp
π(x − vt)
L o γ
2 ,
cos
2π
a
q
I
− 1 = −8
exp
2
π(x − vt)
L o γ
1 + exp
2
π(x − vt)
L o γ
2 ,
and from
∂q
I
∂x
=
2a
L o γ
exp
π(x − vt)
L o γ
1 + exp
2
π(x − vt)
L o γ
as well as
∂q
I
∂t
= −v
∂q
I
∂x
we find, taking (285) and (286) into consideration,
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
v 2
c 2
o
−
2σa 2
L 2
o
=
v 2
c 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 .
If we introduce another function ρ = ρ(x − v t) according to