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Wednesday, November 19, 2014

Passing Opposite In Love

Consider two similar particles one along the positive time line, tc and the other on the negative time line, tc,

ψ1tc=2ψ1teiπ/4

ψ2(tc)=2ψ2teiπ/4

given, tc=12t.eiπ/4

When they are exchanging energy along the two orthogonal space dimension,

ψ1tc=ψ2(tc)

ψ1t=ψ2t

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,

¨xψ1t=¨x.(ψ2t)=¨xψ2t

Since,

¨xψ1t=c322ψ1x2eiπ/4

and

¨x.ψ2t=c322ψ2x2eiπ/4

We have,

c322ψ1x2eiπ/4=c322ψ2x2eiπ/4

and so,

2ψ1x2=2ψ2x2

The gradient of the forces which are indicative of their directions,

F1x=2ψ1x2=2c3¨xψ1teiπ/4

F2x=2ψ2x2=2c3¨xψ1teiπ/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,

F1x=2ψ1x2=2c3¨xψ1teiπ/4

F2x=2ψ2x2=2c3¨xψ1teiπ/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 ψt does not flip sign.

It is better to solve for the force,

Fψ=ψx

directly.  Just from the above analysis, attractive force is greater than the repulsive force.