\(p=\int { \cfrac { \partial \, t }{ \partial x^{ ' } } } \partial \, E\)
\( \Delta p=\int _{ E }^{ E+\Delta E }{ \cfrac { \partial \, t }{ \partial x^{ ' } } } \partial \, E\)
As \( \Delta E\rightarrow0\),
\( \cfrac { \partial \, p }{ \partial \, E } =\cfrac { \partial \, t }{ \partial x^{ ' } } \)
and
\( \cfrac { \partial \, E }{ \partial \, p } =\cfrac { \partial \, x^{ ' } }{ \partial \, t } =v\)
If the particle is in circular motion in \(B\) orbits, the centripetal force is provided by,
\( \pi E=\pi\cfrac { F }{ q } =\cfrac {\pi mv^{ 2 } }{ q x} \)
then
\( \cfrac { \partial \, E }{ \partial \, v } =\cfrac { 2mv }{ q x } \)
Let,
\( p=\cfrac { 2mv }{ q x} \)
such that,
\( \cfrac { \partial \, E }{ \partial \, p } =\cfrac { q x}{ 2m } \cfrac { \partial \, E }{ \partial \, v } =v\)
Wait a minute, does not \(E\) depends on \(x\)? In this case the integral is taken over changing values of \(E\). \(x\) may or may not change over the integral interval. It is not a variable that the integral is dependent on.
\( \cfrac { \partial \, p }{ \partial \, E } =\cfrac{ 2m } { q x}\cfrac{ \partial \, v } { \partial \, E } =\cfrac { \partial \, t }{ \partial x^{ ' } }\)
and so,
\(\int { \cfrac { \partial \, t }{ \partial x^{ ' } } } \partial \, E=\int{\cfrac{ 2m } { q x}}{ \partial \, v }=\cfrac{ 2m } { q x}v\)
From the previous post "Maybe Not",
\( \cfrac { \partial B^{ 2 } }{ \partial \, t } =-i2\int { \cfrac { \partial \, t }{ \partial x^{ ' } } } \partial \, E\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } } \right\} \)
\( \cfrac { \partial B^{ 2 } }{ \partial \, t } =-i2\cfrac{ 2m } { q x}v\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } } \right\} \)
\( \cfrac { 1 }{ 2\mu _{ o } } \cfrac { \partial B^{ 2 } }{ \partial \, t } =-i\cfrac { 1 }{ 2\mu _{ o } }2\cfrac{ 2m } { q x}v\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } } \right\} \)
\( \cfrac { 1 }{ 2\mu _{ o } } \cfrac { \partial B^{ 2 } }{ \partial \, t } =-i\cfrac{ 2m } { \mu _{ o }q x}v\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } } \right\} \)
Let \(\psi=\cfrac{1}{2\mu_o}B^2=-\cfrac{1}{2}\varepsilon_oE^2\),
\(\cfrac{\partial\,\psi}{\partial\,t}=-\cfrac{ 2m } { \mu _{ o }q x}(iv)\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } } \right\} \)
where \(iv\) is along the \(x\) direction. And \(x\) is the orbit of the particle.
This equation shows the rate of change in total energy required as a particle subjected to a B field and is in circular motion along the B field line attains a velocity \(v\), perpendicular to its orbit at the present orbital radius of \(x\).
