91影视

Write the expression for the Lorentz force law in terms of potentials.

$\mathit{F}\mathbf{=}\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{q}\left(-鈭�V-\frac{鈭�A}{鈭�t}+v脳\left(鈭�脳A\right)\right)$ 鈥︹�� (1)

Here, F is the force on the charged particle, V is the potential, A is the vector potential, q is the charge of the particle, and vis the velocity of the particle.

Take the dot product of v on both sides of equation (1) and solve.

$$\begin{gathered} v路F=qv路\left(-鈭�V-\frac{鈭�A}{鈭�t}+v脳\left(鈭�脳A\right)\right) \\ v路F=q\left(-v路鈭�V-v\frac{鈭�A}{鈭�t}+v路v脳\left(鈭�脳A\right)\right) \\ v路F=q\left(-v路鈭�V-v路\frac{鈭�A}{鈭�t}+0\right) \\ v.F=q\left(-v路鈭�V-v路\frac{鈭�A}{鈭�t}\right) \end{gathered}$$

Add $q\frac{dV}{dt}$on both the sides of the above expression.

$$\begin{gathered} v路F+q\frac{qV}{dt}=q\left(-v路鈭�V-v路\frac{鈭�A}{鈭�t}\right)+q\frac{dV}{dt} \\ v路m\frac{dv}{dt}+q\frac{dV}{dt}=q\left(-v路鈭�V-v路\frac{鈭�A}{鈭�t}\right)+q\frac{dV}{dt} \\ \frac{1}{2}\frac{d\left(m{v}^2\right)}{dt}+q\frac{dV}{dt}=q\left(-v路鈭�V-v路\frac{鈭�A}{鈭�t}\right)+q\frac{dV}{dt} \\ \frac{d}{dt}\left(\frac{1}{2}m{v}^2+qV\right)=q\left(-v路鈭�V-v路\frac{鈭�A}{鈭�t}\right)+q\frac{dV}{dt} \end{gathered}$$

Since, the kinetic energy is given as $T=\frac{1}{2}m{v}^2$, rewrite the above equation.

$\frac{d}{dt}\left(T+qV\right)=q\left(-v路鈭�V-v路\frac{鈭�A}{鈭�t}\right)+q\frac{dv}{dt}.....\left(2\right)$

Write the total derivative $\frac{dV}{dt}$ of V in terms of convective derivative.

$\frac{dV}{dt}=q\left[\left(v路鈭�\right)V+\frac{鈭�V}{鈭�t}\right]$

Substitute $\frac{dV}{dt}=q\left[\left(v路鈭�\right)V+\frac{鈭�V}{鈭�t}\right]$in equation (2).

role="math" localid="1653740234755" $$\begin{gathered} \frac{d}{dt}\left(T+qV\right)=q\left(-v路鈭�V-v路\frac{鈭�A}{鈭�t}\right)+q\left[\left(v路鈭�\right)V+\frac{鈭�V}{鈭�t}\right] \\ \frac{d}{dt}\left(T+qV\right)=q\left[\left(-v路鈭�V\right)+\left(v路鈭�\right)V+\frac{鈭�V}{鈭�t}-v路\frac{鈭�A}{鈭�t}\right] \\ \frac{d}{dt}\left(T+qV\right)=\frac{鈭�}{鈭�t}q\left(V-v路A\right) \end{gathered}$$

Rewrite the equation as,

$\frac{d}{dt}\left(T+qV\right)=\frac{鈭�}{鈭�q}\left[q\left(V-v路A\right)\right]$

Therefore, it is proved that$\frac{d}{dt}\left(T+qV\right)=\frac{鈭�}{鈭�q}\left[q\left(V-v路A\right)\right]$.