General Field Equation (factor 2 correction)

In the case for gravity, time slows down in denser space, but the total energy of the body is conserved, so a decrease in time speed leads to an increase in space speed.  The body accelerates, without the action of a immediate force.  No touch, touch.  Similarly, the present of a charge slows down time... an assumption of course.  In both repulsion and attraction time slows down, and the charge body experiences acceleration in the space dimension.  The question remains however, as to which direction is this acceleration.  ( the anser is in the post "i'm Imagining It...")  The initial condition for the specific case (same sign charges or different sign charges) will provide for the direction of the acceleration.  Combining both scenarios, we have the following postulation:

If the field effects such as gravity and charge, where, spreaded out in a region of space, a body experience acceleration without a contact force, is due only to time variation expressed through the conservation of energy equation across the time and space dimensions,

$qv^2_{t} + \frac{1}{2}qv^2_c=\frac{1}{2}qc^2=constant$

$2v^2_{t} + v^2_c=c^2$

The factor of two before time velocity,  $v_t$,  is due to the fact that the kinetic energy along the time dimension at speed is not

$E=\frac{1}{2}mv^2_t$  but

$E=mv^2_t$

Differentiating with respect to $x$,

$4v_{ t }\cfrac { dv_{ t } }{ dx } +2v_{ c }\cfrac { dv_{ c } }{ dx } =0$

$2 v_{ t }\cfrac { dv_{ t } }{ dx } +v_{ c }\cfrac { dv_{ c } }{ dt } \cfrac { dt }{ dx } =0$

$2 v_{ t }\cfrac { dv_{ t } }{ dx } +a_{ c }=0$  ----(1)

$2v_{ t }\cfrac { dv_{ t } }{ dt } \cfrac { dt }{ dx } +a_{ c }=0,$

$2\cfrac { v_{ t } }{ v_{ c } } \cfrac { dv_{ t } }{ dt } +a_{ c }=0$

Differentiating (1) again with respect to $x$,

$2(\cfrac { dv_{ t } }{ dx } )^{ 2 }+2v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } } +\cfrac { da_{ c } }{ dx } =0$

$-\cfrac { da_{ c } }{ dx } =2(\cfrac { dv_{ t } }{ dx } )^{ 2 }+2v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } }$

From the post "Calculus We Have A Problem",

$v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } } =\cfrac { v_t }{ v^2_c }\cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } +\cfrac { v_t }{ a_c }\cfrac { dv_{ t } }{ dt }$

So,

$-\cfrac { da_{ c } }{ dx } =2(\cfrac { dv_{ t } }{ dt } \cfrac { dt }{ dx } )^{ 2 }+2\cfrac { v_t }{ v^2_c }\cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } +2\cfrac { v_t }{ a_c }\cfrac { dv_{ t } }{ dt }$

$-\cfrac { da_{ c } }{ dx } =2(\cfrac { 1 }{ v_{ c } } )^{ 2 }(\cfrac { dv_{ t } }{ dt } )^{ 2 }+2\cfrac { v_t }{ v^2_c }\cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } +2\cfrac { v_t }{ a_c }\cfrac { dv_{ t } }{ dt }$ ----(*)

Differentiating the conservation of energy equation with respect to time $t$ twice,

$4v_{ t }\cfrac { dv_{ t } }{ dt } +2v_{ c }\cfrac { dv_{ c } }{ dt } =0,$

$2v_{ t }\cfrac { dv_{ t } }{ dt } +v_{ c }.a_{ c }=0$

$2(\cfrac { dv_{ t } }{ dt } )^{ 2 }+2v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } +a^{ 2 }_{ c }+v_{ c }\cfrac { da_{ c } }{ dt } =0$

$2\cfrac { v_{ t } }{ v^2_{ c } } \cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } +\cfrac { 2 }{ v^2_{ c } } (\cfrac { dv_{ t } }{ dt } )^{ 2 }=-\cfrac{a^2_{ c }}{v^2_c}-\cfrac { 1}{ v_{ c } } \cfrac { da_{ c } }{ dt }$

Substitute into (*)

$-\cfrac { da_{ c } }{ dx } =2\cfrac { v_t }{ a_c }\cfrac { dv_{ t } }{ dt }-\cfrac{a_{ c }}{v^2_c}-\cfrac { 1}{ v_{ c } } \cfrac { da_{ c } }{ dt }$

$-\cfrac { da_{ c } }{ dx } =-\cfrac { da_{ c } }{dt}\cfrac{ dt }{dx}=-\cfrac{1}{v_c}\cfrac { da_{ c } }{dt}$

$2\cfrac { v_t }{ a_c }\cfrac { dv_{ t } }{ dt }=\cfrac{a^2_{ c }}{v^2_c}$

since, ${v}^2_{c}={c}^{2}-{v}^{2}_{t}$

$({c}^{2}-{v}^{2}_{t})2{ v_t }\cfrac { dv_{ t } }{ dt }=a^3_{ c }$

$({c}^{2}-{v}^{2}_{t})\cfrac { d(v^2_{ t }) }{ dt }=a^3_{ c }$

$({c}^{2}-{v}^{2}_{t})\cfrac { d(v^2_{ t }) }{ dx }\cfrac{dx}{dt}=a^3_{ c }$

$({c}^{2}-{v}^{2}_{t})v_c\cfrac { d(v^2_{ t }) }{ dx }=a^3_{ c }$

$({c}^{2}-{v}^{2}_{t})\sqrt{{c}^{2}-{v}^{2}_{t}}\cfrac { d(v^2_{ t }) }{ dx }=a^3_{ c }$

but, from (1)

$2v_t\cfrac{d{v}_{t}}{dx}=-{a}_{c}$

$\cfrac{d({v}^2_{t})}{dx}=-{a}_{c}$

$({c}^{2}-{v}^{2}_{t})^{\cfrac{3}{2}}(-a_c)=a^3_{ c }$

$a^2_c=-({c}^{2}-{v}^{2}_{t})^{\cfrac{3}{2}}$

$a^2_c=-c^3(1-{\gamma}^{2})^{\cfrac{3}{2}}$

$a_c=ic^\cfrac{3}{2}(1-{\gamma}^{2})^{\cfrac{3}{4}}$

which is true because  $a_c$  is perpendicular to  $v_t$.

$|a_{c max}|=c^\cfrac{3}{2}$    when   $\gamma=0$

Maximum  $a_c$ not  $c^2$  but  $c^\cfrac{3}{2}$.