Other Time Dimensions Counts

Consider the expression,

$x.mv$

Differentiating wrt $t$,

$\cfrac { d\left( x.mv\right)  }{ dt } =\cfrac { dx }{ dt } mv+x\cfrac { d\left(mv\right) }{ dt } =mv^{ 2 }+x.F$

where,

$F=\cfrac { d\left(mv\right) }{ dt }$,    $v=\cfrac { dx }{ dt }$

When $x.mv$ is periodic over period $T$,

$x.mv=\int _{ 0 }^{ T }{ mv^{ 2 }+x.F } dt=\left( mv^{ 2 }+x.F \right) T$

where $mv^{ 2 }+x.F$ is assumed constant over $t$ and

$f=\cfrac{1}{T}$

Let,

$h=x.mv$

then,

$hf=E=mv^{ 2 }+x.F$

But,

 $E\ne mv^{ 2 }+x.F$.

$E=KE+PE$

What happened?  From the post "Not A Wave, But Work Done!" when $v=c$,

$\cfrac { \partial \, \psi  }{ \partial \, t } =2\cfrac { \partial V\,  }{ \partial \, t }=2\cfrac { \partial T\,  }{ \partial \, t }$

We have

$V=T+C$

where $C$ is the constant of integration after integrating over $t$.

If $C=0$, ie there is no bias in oscillation as $\psi$ interchange between $V$ and $T$.  Then,

$V=T$

If a particle is in circulation at velocity $v=c$,

$T=\cfrac{1}{2}mc^2$

and so,

$U=T=\cfrac{1}{2}mc^2$  and so,

$\psi=U+T=mc^2$

In which case,

$hf=mc^{ 2 }+x.F=\psi+x.F$

$E=\psi+x.F$

That is to say, $E$ is the energy of the particle plus work done on it.  As long as $F$ acts within the period $T$, the integration over $t=T$ is valid.  A collision or a photon passing at light speed are within this assumption.

The assumption that $C=0$, also applies to $E$, as such $C=0$ is just assuming a reference potential of zero in a system.

Why is $x.F$ called the work function before it was worked out as work done by a force?