More Squeeze Rhythmically

From he post "$\psi_{max}$ Is New And Important" dated 13 Jul 2015,

$F=\psi_{n}-\psi(x)=m\cfrac{d^2x}{dt^2}$

when we substitute for $\psi(x)$,

$\psi(x)=-i{ 2{ mc^{ 2 } } }\,ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }(x-x_z)))+c$ --- (*)

from the post "Not Quite The Same Newtonian Field" dated 23 Nov 2015.  Assuming that at $x=0$,  $\psi(0)=0$,

$c=i{ 2{ mc^{ 2 } } }\,ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }(x_z)))$

and so,

$\psi(x_z)=\psi_{max}=i{ 2{ mc^{ 2 } } }\,ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }(x_z)))$

Therefore,

$m\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\psi _{ n }+i2{ mc^{ 2 } }ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z)))-\psi_{max}$

$m\cfrac { d^{ 2 }x }{ dt^{ 2 } } =i2{ mc^{ 2 } }ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z)))+\psi_{c}$

where

$\psi_c=\psi_n-\psi_{max}$

In cases where,

$\psi_d=\psi_{max}-\psi_n$

we have a particle of opposite sign embedded in another particle.

Also,

$\psi_n=\psi _{ photon }+\psi_{o}$

$\psi_n$ is the elevated energy level of the particle from its ground state $\psi_{o}$ after receiving the photon $\psi _{ photon }$.

In the case when,

$\psi _{ photon }=0$,  $\psi_c=\psi_o-\psi_{max}$

and $\psi_c$ is a constant.  Since,

$F=m\cfrac { d^{ 2 }x }{ dt^{ 2 } }$

$F=i2{ mc^{ 2 } }ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z)))+\psi_{c}$

We now consider the effect of exerting an external force on the system, the resulting change in $x$, the relative displacement of the embedded particle and the containing particle can be obtained from the expression,

$\cfrac{dF}{dx}=iG \sqrt { 2{ mc^{ 2 } } }.tanh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z))$

which is just the force density expression we formulated in the post "Not Exponential, But Hyperbolic And Positive Gravity!" dated 22 Nov 2014.  More explicitly,

$F_{\rho}=e^{ i3\pi /4 }D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o).e^{ i\pi /4 } \right)$

where we insist that,

$G=D.e^{ i\pi /4 }$

is real.

The term $i$ originates from the expression for $\psi$, which is a wave in the $ix$ direction.  From the post "Opps Lucky Me" dated 25 May 2015, the Newtonian force $F_{\small{N}}$

$F_{\small{N}}=-\psi$ 

and the post "From The Very Big To The Very Small" dated 16 Jul 2015.

$F_{x}=\int_0^x{F_{\rho}}dx$

so,

$F_{\small{N}}=\int{F_{\rho}}dx$

where $x=0$ when $F_{\small{N}}=0$.  Combining all these, the Newtonian force $F_{\small{N}}$ when applied resulting in a deformation $\Delta x$ is given by,

$F_{\small{N}}=\cfrac{dF}{dx}|_{x=x_o}.\Delta x=i2{ mc^{ 2 } }ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z)))|_{x=x_o}.\Delta x$  --- (*)

where $F_{\small{N}}$ can be the result of an applied electric field, $E$ or magnetic field $B$.

When $\Delta x$ changes $\theta$ around $\theta=45^o$, an EMW together with an photon are emitted.  The EMW can be amplified and be used as the driving force (positive feedback) and thus generates sustained oscillations.

And we have a problem with $i$ on the RHS of the expression for $F_{\small{N}}$, otherwise we have an indication for electrostriction/magnetostriction and piezoelectric.

The problem is $i$ does make sense,

where the deformation, $\Delta x$ is in the direction perpendicular to $F_{\small{N}}$.  This is the result of $\psi$ being perpendicular to the direction of $x$, from the expression (*).  Since the total volume of the material ($\psi$), is approximately a constant, there are corresponding deformations in the two perpendicular directions with respect to $\Delta x$ too.

With the issue of $i$ aside,

we differentiate between  piezoelectricity and its linear dependence on the applied force $F_{\small{N}}$, and electrostriction and its quadratic dependence on the applied force, base on the same expression (*).  Piezoelectricity requires that $\small{\Delta x}$ changes $\small{\theta}$ around $\small{\theta=45^o}$.