Size Of Basic Particle From Its Spectrum

From the post "Discreetly, Discrete $\lambda$ And Discrete Frequency, $f$", we have an expression for discrete $\lambda$,

$x=e^{ -\cfrac { Ax+C }{ n^{ 2 } }  }$

From the post "Standing Waves, Particles, Time Invariant Fields", ψ is a circular standing wave about a center,

$2\pi x=n\lambda$

where $n=1,2,3...$ when $x=0$, $n=0$.

Substitute for $x$,

$2\pi e^{ -\cfrac { An\lambda  }{ 2\pi n^{ 2 } } -\cfrac { C }{ n^{ 2 } }  }=n\lambda$

Since,  $f\lambda =c$

$2\pi e^{ -\cfrac { Anc }{ 2\pi fn^{ 2 } } -\cfrac { C }{ n^{ 2 } }  }=n\cfrac { c }{ f }$

$f=\cfrac { nc }{ 2\pi  } e^{ \cfrac { Anc }{ 2\pi fn^{ 2 } } +\cfrac { C }{ n^{ 2 } }  }$

$f=\cfrac { nc }{ 2\pi  } e^{ \cfrac { A }{ 2\pi \, n } \cfrac { c }{ f }  }.e^{ \cfrac { C }{ n^{ 2 } }  }$

From the post "Folding Up, Peekapoo, What's Under $a_\psi$?", where $a_\psi$ is the radius of the particle mass,

$C=-n_{ o }^{ 2 }ln(a_{ \psi  })-Aa_{ \psi  }$

Substitute this into $C$,

$f=\cfrac { nc }{ 2\pi  } e^{ \cfrac { A }{ 2\pi \, n } \cfrac { c }{ f }  }.e^{ \cfrac { -n_{ o }^{ 2 }ln(a_{ \psi  })-Aa_{ \psi  } }{ n^{ 2 } }  }$

$f=\cfrac { nc }{ 2\pi  } e^{ \cfrac { A }{ 2\pi \, n } \cfrac { c }{ f }  }.a_{ \psi  }^{ -\cfrac { n_{ o }^{ 2 } }{ n^{ 2 } }  }e^{ -\cfrac { Aa_{ \psi  } }{ n^{ 2 } }  }$

$ln(\cfrac { 2\pi f }{ nc } )=\cfrac { A }{ 2\pi \, n } \cfrac { c }{ f } +ln(a_{ \psi  }^{ -\cfrac { n_{ o }^{ 2 } }{ n^{ 2 } }  })-\cfrac { Aa_{ \psi  } }{ n^{ 2 } }$

$n^{ 2 }ln(\cfrac { 2\pi f }{ nc } )=n^{ 2 }\left\{ \cfrac { A }{ 2\pi \, n } \cfrac { c }{ f } -\cfrac { n_{ o }^{ 2 } }{ n^{ 2 } } ln(a_{ \psi  }) \right\} -Aa_{ \psi  }$

$n^{ 2 }ln(\cfrac { 2\pi f }{ nc } )=A\cfrac { nc }{ 2\pi f } -n_{ o }^{ 2 }ln(a_{ \psi  })-Aa_{ \psi  }$

If we let $x=\cfrac { 2\pi f }{ nc }$

$n^{ 2 }ln(x)=A\cfrac { 1 }{ x } -n_{ o }^{ 2 }ln(a_{ \psi  })-Aa_{ \psi  }$

where each $n$ has a corresponding $f$ from the spectrum of the particles.

So, a plot of $n^{ 2 }ln(x)$ verses  $\cfrac { 1 }{ x }$ will give $A$ as the gradient and $-n_{ o }^{ 2 }ln(a_{ \psi  })-Aa_{ \psi  }$ as the y-intercept.  from which we obtain the size of the particle $a_\psi$.

If we assume that $f$ is the minimum at $a_\psi$ then $n_o=1$.

Various such plots can be obtained by varying $A$ by changing temperature $T$.  Since $f$ can be folded upwards as shown in the posts "Discreetly, Discrete $\lambda$ And Discrete Frequency, $f$" and "Another Take On Discrete", $T$ should be high such that $A$ is low and does not fold the values of $f$ corresponding to lower values of $n$ upwards.  The assignment of $f$ to $n$ is then simply 1 to 1, starting from the lowest value of $f$, $f_{min}$ to $n_o=1$.

We are assuming that $T$ does not change $a_\psi$.  Which would be very interesting if it does.

$T$ does effect $a_\psi$.  Increasing $T$ reduces the resistance to $\psi$, more of a particle is manifested as $\psi$ and its mass decreases, thus $a_\psi$ decreases.