∂ψ1∂tc=√2∂ψ1∂teiπ/4
∂ψ2∂(−tc)=−√2∂ψ2∂teiπ/4
given, tc=1√2t.e−iπ/4
When they are exchanging energy along the two orthogonal space dimension,
∂ψ1∂tc=∂ψ2∂(−tc)
∂ψ1∂t=−∂ψ2∂t
Here ψ is a wave, oscillating between two space dimensions and traveling down a third dimension that is the time dimension, tg. The wave exist in a second time dimension, tc by which velocities and other rate of change in time are measured.
With these relationships we formulate,
¨x∂ψ1∂t=¨x.(−∂ψ2∂t)=−¨x∂ψ2∂t
Since,
¨x∂ψ1∂t=−c3√2∂2ψ1∂x2e−iπ/4
and
−¨x.∂ψ2∂t=c3√2∂2ψ2∂x2e−iπ/4
We have,
−c3√2∂2ψ1∂x2e−iπ/4=c3√2∂2ψ2∂x2e−iπ/4
and so,
−∂2ψ1∂x2=∂2ψ2∂x2
The gradient of the forces which are indicative of their directions,
∂F1∂x=∂2ψ1∂x2=−√2c3¨x∂ψ1∂teiπ/4
∂F2∂x=∂2ψ2∂x2=√2c3¨x∂ψ1∂teiπ/4
These forces satisfy the change in energy required of both particles as ψ oscillates in tc time. ψ however is not oscillating in t. In the diagram above, we consider a point on ψ2 along the line joining the two centers of the particles. The resultant force F is towards ψ1 drawing the particles closer and we see that the forces are attractive; that particles in opposing directions on the time axis tc attract each other. It is assumed that the force decreases monotonously with x.
In a similar way,
when both particle are traveling in the same direction,
∂F1∂x=∂2ψ1∂x2=−√2c3¨x∂ψ1∂teiπ/4
∂F2∂x=∂2ψ2∂x2=−√2c3¨x∂ψ1∂teiπ/4
The resultant force F is away from ψ1 pushing the particles closer and we see that the forces are repuslive; that particles in the same directions on the time axis tc repel each other. It is assumed that the force decreases monotonously with x; that ∂ψ∂t does not flip sign.
It is better to solve for the force,
Fψ=−∂ψ∂x
directly. Just from the above analysis, attractive force is greater than the repulsive force.