If the velocity of the particle is \(c\) then,
\(v=c.sin(\theta)\)
And the distortion to the particle and/or its path, due to the force acting in the direction opposite to \(v\) is zero when \(\theta=0\) because \(v=0\) according to the above. The force that distorts the particle is along in the plane of the particle's circular motion. As is the case when \(\theta=90^o\).
The mid point between these two equivalent extremes is at \(\theta=45^o\).
Is this the same squeeze as before? Maybe...
