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Tuesday, May 16, 2017

Squawky Maths, Brain Dead

Yes, at x=xf along the line that passes through the focus of the parabola,


ψx|xf=0

2ψx2|xf=0

on the parabola profile.  But at the focus, the effect of ψ, and the rate of change of ψ,

ψt0

We are looking at the solutions of a partial differential equation not an algebraic equation.  Strictly speaking,

2ψx2|xf=0

cannot be substituted into

ψxψt=c2p2.2ψx2.eiπ/4

before solving the equation.  But intuitively,

ψxψt=0

seems valid along x=xf and ψ varies due to a change in t, time only.