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?
