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Sunday, November 23, 2014

Something Old, Something New, A Shoulder To Rely On

From the post "We Have A Problem, Coulomb's Law",

F=2mc2xeiπ/2

So,

ψ=2mc2xeiπ/2dx

ψ=2mc2ln(x)eiπ/2


Similarly ψ is delimited at x=xa, where ψ=0.  However, without A (cf. post "Not Exponential, But Hyperbolic And Positive Gravity!"), Coulomb's Law applies for all x>0.

F=mc26πx2eiπ/2

Nice to know there's something old to fall back on.