Consider the wave equation,
∂2ψ∂t2=c2∂2ψ∂x2=∂x∂t∂x∂t∂2ψ∂x∂x
when one of the space dimension has been replaced by a time dimension. The particle exists along the charge time tc axis or the gravitational time tg axis, from which its velocity is based. And as a wave, travels down the other orthogonal time axis.
∂2ψ∂t2c=∂x∂tc∂tg∂tc∂2ψ∂tg∂x
∂2ψ∂t2g=∂x∂tg∂tc∂tg∂2ψ∂tc∂x
Since,
tc=1√2t.e−iπ/4 and tg=1√2t.e+iπ/4
tc=tge−iπ/2 --- (a)
tc=−itg and ∂tc∂tg=−i
tg=itc and ∂tg∂tc=i
and,
∂x∂tc=∂x∂tg=ic
this definition of velocity has a explicit direction notation i to be consistent with the explicit orthogonality between the time axes in (a) above
The equations for 'time' waves are then given by,
∂2ψ∂t2c=ic.i∂2ψ∂x∂tg=ici∂2ψ∂x∂(itc)=ic∂2ψ∂x∂tc
∂2ψ∂t2g=ic.(−i)∂2ψ∂x∂tc=ic.(−i)∂2ψ∂x∂(−itg)=ic∂2ψ∂x∂tg
from the post "How Time Flies". tc and tg are equivalent. A mass particle is associated with tg and an electron is associated with tc.
Consider the Lagrangian,
L=T−V
ddt(∂L∂˙x)=∂L∂x
If all dimension are equivalent, then the Lagrangian and Euler-Lagrange Equation is still applicable, when t is either tc or tg.
ψ=T+V=12mp˙x2+V
L=12mp˙x2−(ψ−12mp˙x2)
where ψ is the total energy between the two space dimensions and ˙x the velocity of the particle.
L=mp˙x2−ψ
We consider existence along tc,
ddtc(∂L∂˙x)=ddtc(∂∂˙x{mp˙x2−ψ})
ddtc(∂L∂˙x)=ddtc(2mp˙x−∂ψ∂˙x)=2mp¨x−∂∂˙x{dψdtc}
and,
∂L∂x=∂∂x{mp˙x2−ψ}=−∂ψ∂x
So,
∂∂˙x{dψdtc}=2mp¨x+∂ψ∂x --- (*)
Differentiate with respect to x,
∂∂x{∂∂˙x{dψdtc}}=∂2ψ∂x∂x
∂∂˙x{∂2ψ∂tc∂x}=∂∂˙x{−ic∂2ψ∂t2c}=∂2ψ∂x2
∂∂˙x{∂2ψ∂t2c}=ic∂2ψ∂x2 --- (b)
but,
∂ψ∂x=∂ψ∂˙xd˙xdx0+∂ψ∂tcdtcdx
So (*) becomes,
∂∂˙x{dψdtc}=2mp¨x+∂ψ∂tcdtcdx
Consider,
∂∂tc∂∂˙x{dψdtc}=∂∂˙x{∂2ψ∂t2c}
=∂∂tc{2mp¨x+∂ψ∂tcdtcdx}=2mp⃛x+∂2ψ∂t2cdtcdx+∂ψ∂tc∂∂tc{1dxdtc}
∂∂˙x{∂2ψ∂t2c}=2mp⃛x+∂2ψ∂t2cdtcdx−∂ψ∂tc{1(dxdtc)2d2xdt2c}
So, since ic is a constant, substitute in (b)
ic∂2ψ∂x2=2mp⃛x+∂2ψ∂t2cdtcdx−∂ψ∂tc{1(dxdtc)2d2xdt2c}
ic∂2ψ∂x2=2mp⃛x+∂2ψ∂t2c1˙x−∂ψ∂tc{1˙x2¨x}
Rearranging the terms,
¨x∂ψ∂tc=−ic˙x2∂2ψ∂x2+˙x∂2ψ∂t2c+2mp˙x2⃛x
If the field forces are constants in time given location x, ⃛x=0
¨x∂ψ∂tc=−ic˙x2∂2ψ∂x2+˙x∂2ψ∂t2c --- (1)
Consider,
∂2ψ∂t2c=∂∂tc{∂ψ∂tc}=∂∂x{∂ψ∂tc}dxdtc+∂∂˙x{∂ψ∂tc}d˙xdtc
From previously,
∂∂˙x{dψdtc}=2mp¨x+∂ψ∂tcdtcdx
Therefore,
∂2ψ∂t2c=∂∂x{∂ψ∂tc}dxdtc+{2mp¨x+∂ψ∂tcdtcdx}d˙xdtc
∂2ψ∂t2c=∂∂tc{∂ψ∂x}˙x+2mp¨x2+∂ψ∂tc¨x˙x
˙x∂2ψ∂t2c=∂∂tc{∂ψ∂x}˙x2+2mp˙x¨x2+¨x∂ψ∂tc
˙x∂2ψ∂t2c=∂∂tc{∂ψ∂x}˙x2+¨x∂∂tc{mp˙x2}+¨x∂ψ∂tc
Since,
ψ=12mp˙x2+V
mp˙x2=2(ψ−V)
We have,
˙x∂2ψ∂t2c=∂∂tc{∂ψ∂x}˙x2+2¨x∂∂tc{ψ−V}+¨x∂ψ∂tc
˙x∂2ψ∂t2c=−ic∂2ψ∂t2c˙x2+2¨x∂∂tc{32ψ−V}
˙x∂2ψ∂t2c(1+i˙xc)=2¨x∂∂tc{32ψ−V}
˙x∂2ψ∂t2c=2¨x(1+i˙xc)∂∂tc{32ψ−V}
And (1) becomes,
¨x∂ψ∂tc=−ic˙x2∂2ψ∂x2+2¨x(1+i˙xc)∂∂tc{32ψ−V}
¨x∂ψ∂tc{1−3(1+i˙xc)}=−ic˙x2∂2ψ∂x2−2¨x(1+i˙xc)∂V∂tc
¨x∂ψ∂tc{−2+i˙xc(1+i˙xc)}=−ic˙x2∂2ψ∂x2−2¨x(1+i˙xc)∂V∂tc
¨x(2−i˙xc)∂ψ∂tc=ic(1+i˙xc)˙x2∂2ψ∂x2+2¨x∂V∂tc --- (**)
This is not a wave in space and time t.
When ¨x=0,
ic(1+i˙xc)˙x2∂2ψ∂x2=0
which implies,
∂2ψ∂x2=0
which means,
∂ψ∂x=constant=A --- (2)
ψ(x)=Ax+C
ψ(x) is linear in space! If at x=0, ψ(0)=C=0
ψ(x)=Ax
A is then a force and we have simply,
Energy=ψ(x)=A.x=Force.distance=Work
We started with a particle that might be responsible for gravity or electrostatic force and ended with a simple work done equation. This means the energy of this particle is just work done against a force.
Here the force is given the definition as rate of change of energy, ψ with distance x, from (2).
When ˙x=c, which is the case for charges, expression (**) becomes,
¨x(2−i)∂ψ∂tc=ic3(1+i)∂2ψ∂x2+2¨x∂V∂tc
¨x(2−i)∂ψ∂tc=c3(−1+i)∂2ψ∂x2+2¨x∂V∂tc
Comparing real and imaginary parts,
2¨x∂ψ∂tc=−c3∂2ψ∂x2+2¨x∂V∂tc --- (2)
−i¨x∂ψ∂tc=ic3∂2ψ∂x2
¨x∂ψ∂tc=−c3∂2ψ∂x2 substitute into (2)
∂ψ∂tc=2∂V∂tc and ∂ψ∂t=2∂V∂t
Since, ψ=T+V
∂ψ∂tc=2∂T∂tc and ∂ψ∂t=2∂V∂t
∂ψ∂tc is divided evenly between ∂T∂tc and ∂V∂tc
From tc=1√2t.e−iπ/4 and ¨x∂ψ∂tc=−c3∂2ψ∂x2
¨x∂ψ∂t=−c3√2∂2ψ∂x2e−iπ/4
This expression is unit dimension correct. However, in space and time t, ψ is not a wave along x even when ˙x=c.
