From the post "Not A Wave But Work Done!",
¨x(2−i˙xc)∂ψ∂tc=ic(1+i˙xc)˙x2∂2ψ∂x2+2¨x∂V∂tc --- (*)
derived under the assumption that ⃛x=0, a time invariant field.
Multiply (*) by i,
i¨x(2−i˙xc)∂ψ∂tc=−c(1+i˙xc)˙x2∂2ψ∂x2+2i¨x∂V∂tc
Multiply by (1−i˙xc),
i¨x(2−i˙xc)(1−i˙xc)∂ψ∂tc=−c(1−˙x2c2)˙x2∂2ψ∂x2+2i(1−i˙xc)¨x∂V∂tc
c(1−˙x2c2)˙x2∂2ψ∂x2=−i¨x(2−˙x2c2−3i˙xc)∂ψ∂tc+2(˙xc+i)¨x∂V∂tc
c(1−˙x2c2)˙x2∂2ψ∂x2=−¨x{3˙xc+i(2−˙x2c2)}∂ψ∂tc+2(˙xc+i)¨x∂V∂tc
Equating Real terms,
c(1−˙x2c2)˙x2∂2ψ∂x2=−3˙xc¨x∂ψ∂tc+2˙xc¨x∂V∂tc
c(1−˙x2c2)˙x∂2ψ∂x2=¨xc{−3∂ψ∂tc+2∂V∂tc}
Let's define
1γ2=(1−˙x2c2)
We have,
c2γ2˙x∂2ψ∂x2=¨x{−3∂ψ∂tc+2∂V∂tc} --- (**)
Equating Imaginary terms,
−¨x(2−˙x2c2)∂ψ∂tc+2¨x∂V∂tc=0
¨x≠0
(2−˙x2c2)∂ψ∂tc=2∂V∂tc
∂ψ∂tc+(1−˙x2c2)∂ψ∂tc=2∂V∂tc
(1+1γ2)∂ψ∂tc=2∂V∂tc --- (***)
Substitute the above into (**),
c2γ2˙x∂2ψ∂x2=−¨x{3∂ψ∂tc−(1+1γ2)∂ψ∂tc}
c2γ2˙x∂2ψ∂x2=−¨x{2−1γ2}∂ψ∂tc --- (1)
¨x∂ψ∂tc=−c2{2γ2−1}˙x∂2ψ∂x2
Equivalently from (1),
(1+˙x2c2)¨x∂ψ∂tc=−c2(1−˙x2c2)˙x∂2ψ∂x2
This suggests that when ˙x=c, 1γ2=0
2¨x∂ψ∂tc=0
Then either,
¨x=0 and ∂ψ∂tc=∂V∂tc
or,
∂ψ∂tc=0,
that the total energy of the system is a constant in time tc and is at an extrema. In this case ψ has a stable point when ˙x=c. From (***),
∂ψ∂tc=2∂V∂tc=0
when ˙x=c, and
∂2ψ∂t2c=2∂2V∂t2c
ψ is minimum when V is at its local maximum. This shows that the system, as far as ψ, the total energy is concerned, can be stable at light speed, ˙x=c. At that point V is at its local maximum.
We now consider the case when ˙x<<c,
¨x∂ψ∂tc=−c2˙x∂2ψ∂x2
m¨x∂ψ∂tc=−c2m˙x∂2ψ∂x2
With F=m¨x and p=m˙x,
F∂ψ∂tc=−c2p∂2ψ∂x2
Since, −∂ψ∂x=F=m¨x
∂ψ∂x∂ψ∂tc=c2p∂2ψ∂x2
with tc=1√2t.e−iπ/4,
∂ψ∂x∂ψ∂t=c2p√2.∂2ψ∂x2.e−iπ/4
in time t.
There should be more implications from this equation to prove its validity or disprove it.