Question

Derive Eq. 10.23. [Hint: Start by dotting v into Eq. 10.17.]

Short answer

It is proved that$\frac{d}{dt}\left(T+qV\right)=\frac{\partial }{\partial q}\left[q\left(V-v·A\right)\right]$.

Step-by-step solution

Expression for the Lorentz force law in terms of potentials:

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

$\mathit{F}\mathbf{=}\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{q}\left(-\nabla V-\frac{\partial A}{\partial t}+v\times \left(\nabla \times A\right)\right)$ …… (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.

Determine the expression for the rate of change of the particle’s energy (Eq. 10.23):

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

$$\begin{gathered} v·F=qv·\left(-\nabla V-\frac{\partial A}{\partial t}+v\times \left(\nabla \times A\right)\right) \\ v·F=q\left(-v·\nabla V-v\frac{\partial A}{\partial t}+v·v\times \left(\nabla \times A\right)\right) \\ v·F=q\left(-v·\nabla V-v·\frac{\partial A}{\partial t}+0\right) \\ v.F=q\left(-v·\nabla V-v·\frac{\partial A}{\partial t}\right) \end{gathered}$$

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

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

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

$\frac{d}{dt}\left(T+qV\right)=q\left(-v·\nabla V-v·\frac{\partial A}{\partial t}\right)+q\frac{dv}{dt}.....\left(2\right)$

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

$\frac{dV}{dt}=q\left[\left(v·\nabla \right)V+\frac{\partial V}{\partial t}\right]$

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

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

Rewrite the equation as,

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

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