Chapter 13, Problem 13.18P

(a)

To determine

The Hamiltonian for a charge in an electromagnetic field.

Answer to Problem 13.18P

The Hamiltonian for a charge in an electromagnetic field is $H=\frac{{\left(P-qA\right)}^2}{2m}+qV$

Explanation of Solution

Write the Hamiltonian equation for charge particle in an electromagnetic field,

  $H=p\cdot \dot{r}-L$        (I)

Here, $p$ is the momentum, $L$ is the langrangian of the system and $r$ is the position of the charge.

Write the generalized momentum equation for charge particle in an electromagnetic field,

  $p=\frac{dL}{d\dot{r}}$        (II)

Write the lagrangian equation for a charge $q$ in an electromagnetic field is,

  $L=\frac{1}{2}m{\dot{r}}^2-q\left(V-\dot{r}\cdot A\right)$        (III)

Here, $m$ is the mass of the charge of particle, $\dot{r}$ is the velocity of the charge particle, $V$ is the scalar potential and $A$ is the vector potential.

Conclusion:

Substitute expression (III) in (II), and write in terms of $\dot{r}$

  $\begin{array}{c}P=\frac{\partial }{\partial \dot{r}}\left(\frac{1}{2}m{\dot{r}}^2-q\left(V-\dot{r}\cdot A\right)\right)\\ =\left(\frac{1}{2}m2\dot{r}+q\left(A\right)\right)\\ P=m\dot{r}+qA\\ \dot{r}=\frac{P-qA}{m}\end{array}$        (IV)

Substitute the expression (IV), expression for $P$ and expression (III) in equation (I)

  $\begin{array}{c}H=p\cdot \dot{r}-L\\ =\left(m\dot{r}+qA\right)\cdot \dot{r}-\left(\frac{1}{2}m{\dot{r}}^2-q\left(V-\dot{r}\cdot A\right)\right)\\ =m{\dot{r}}^2-\frac{1}{2}m{\dot{r}}^2+qV\\ =\frac{1}{2}m{\dot{r}}^2+qV\\ =\frac{{\left(P-qA\right)}^2}{2m}+qV\end{array}$        (V)

The Hamiltonian for a charge in an electromagnetic field is $H=\frac{{\left(P-qA\right)}^2}{2m}+qV$

(b)

To determine

The Hamilton’s equations are equivalent to the Lorentz force equation.

Answer to Problem 13.18P

Yes the Hamilton’s equation is equivalent to the Lorentz force equation $m\ddot{r}=q\left(E+v\times B\right)$

Explanation of Solution

Write the Hamiltonian equation in terms of generalized coordinates along the  $x$  axis is,

  $\dot{x}=\frac{\partial H}{\partial p_x}$        (I)

Here, $p_x$ is the component of the generalized coordinate along the axis.

Substitute expression (V) from sub part (a) for $H$ 

  $\begin{array}{c}\dot{x}=\frac{\partial }{\partial p_x}\left(\frac{{\left(P-qA\right)}^2}{2m}+qV\right)\\ =\frac{\left(P-qA\right)}{m}\left(2\right)+0\\ =\frac{\left(P-qA\right)}{m}\end{array}$        (II)

Write the second Hamiltonian equation,

  $\dot{P}=-\frac{\partial H}{\partial x}$        (III)

Substitute expression (V) from sub part (a) for $H$ 

  $\begin{array}{c}\dot{P}=-\frac{\partial }{\partial x}\left(\frac{{\left(P-qA\right)}^2}{2m}+qV\right)\\ =-\left(\frac{2\left(P-qA\right)}{2m}\left(-q\frac{\partial A}{\partial x}\right)\right)-q\frac{\partial V}{\partial x}\\ =\left(q{v}_x\frac{\partial A}{\partial x}\right)-q\frac{\partial V}{\partial x}\end{array}$        (IV)

Conclusion:

Derive Lorentz force equation from Hamiltonian equation,

  $m\ddot{x}=m\frac{d\dot{x}}{dt}$        (V)

Substitute the expression (II) and (IV) in (V) solve

  $\begin{array}{c}m\ddot{x}=m\left(\frac{d}{dt}\left(\frac{\left(P-qA\right)}{m}\right)\right)\\ =\left(q{v}_x\cdot \frac{\partial A}{\partial x}\right)-q\frac{\partial V}{\partial x}-q\frac{dA\left(x,t\right)}{dt}\\ =\left(q{v}_x\cdot \frac{\partial A}{\partial x}\right)-q\frac{\partial V}{\partial x}-q\left(\frac{dA\left(x,t\right)}{dt}+\left(v\cdot \nabla \right)A\left(x,t\right)\right)\end{array}$        (VI)

Similarly $m\ddot{y}$ and $m\ddot{z}$ can be written as,

  $m\ddot{y}=\left(q{v}_y\cdot \frac{\partial A}{\partial y}\right)-q\frac{\partial V}{\partial y}-q\left(\frac{dA\left(y,t\right)}{dt}+\left(v\cdot \nabla \right)A\left(y,t\right)\right)$        (VII)

And

  $m\ddot{z}=\left(q{v}_z\cdot \frac{\partial A}{\partial z}\right)-q\frac{\partial V}{\partial z}-q\left(\frac{dA\left(z,t\right)}{dt}+\left(v\cdot \nabla \right)A\left(z,t\right)\right)$        (VIII)

Add the expression $m\ddot{x}$, $m\ddot{y}$ and $m\ddot{z}$

  $\begin{array}{c}m\ddot{x}+m\ddot{y}+m\ddot{z}=\left\{\begin{array}{l}\left(q{v}_x\cdot \frac{\partial A}{\partial x}\right)-q\frac{\partial V}{\partial x}-q\left(\frac{dA\left(x,t\right)}{dt}+\left(v\cdot \nabla \right)A\left(x,t\right)\right)\\ +\left(q{v}_y\cdot \frac{\partial A}{\partial y}\right)-q\frac{\partial V}{\partial y}-q\left(\frac{dA\left(y,t\right)}{dt}+\left(v\cdot \nabla \right)A\left(y,t\right)\right)\\ +\left(q{v}_z\cdot \frac{\partial A}{\partial z}\right)-q\frac{\partial V}{\partial z}-q\left(\frac{dA\left(z,t\right)}{dt}+\left(v\cdot \nabla \right)A\left(z,t\right)\right)\end{array}\right\}\\ =\left\{\begin{array}{l}\left(q{v}_x\cdot \frac{\partial A}{\partial x}\right)+\left(q{v}_y\cdot \frac{\partial A}{\partial y}\right)+\left(q{v}_z\cdot \frac{\partial A}{\partial z}\right)\\ -q\left(\frac{dA\left(x,t\right)}{dt}+\frac{dA\left(y,t\right)}{dt}+\frac{dA\left(z,t\right)}{dt}\right)\\ -q\left(\frac{\partial V}{\partial x}+\frac{\partial V}{\partial y}+\frac{\partial V}{\partial z}\right)-q\left(v\cdot \nabla \right)A\end{array}\right\}\\ =\left\{\begin{array}{c}\left(q{v}_x\cdot \frac{\partial A}{\partial x}\right)+\left(q{v}_y\cdot \frac{\partial A}{\partial y}\right)+\left(q{v}_z\cdot \frac{\partial A}{\partial z}\right)-q\nabla \cdot A\\ -q\left(\frac{\partial V}{\partial x}+\frac{\partial V}{\partial y}+\frac{\partial V}{\partial z}\right)-q\left(v\cdot \nabla \right)A\end{array}\right\}\end{array}$        (IX)

Here, the resultant area vector around a point in a space at certain time is zero, therefore the differential will be is equal to zero.

    $\nabla \cdot A=0$

Re-write the expression (IX),

  $\begin{array}{c}m\ddot{r}=\left\{\begin{array}{c}\left(q{v}_x\cdot \frac{\partial A}{\partial x}\right)+\left(q{v}_y\cdot \frac{\partial A}{\partial y}\right)+\left(q{v}_z\cdot \frac{\partial A}{\partial z}\right)-q\left(v\cdot \nabla \right)A\\ -q\left(\frac{\partial V}{\partial x}+\frac{\partial V}{\partial y}+\frac{\partial V}{\partial z}\right)\end{array}\right\}\\ =q\nabla \left(v\cdot A\right)-q\left(v\cdot \nabla \right)A+qE\end{array}$        (X)

Write the expression for vector triple product and re-write the expression (X) in terms of that

  $v\times \left(\nabla \times A\right)=\nabla \left(v\cdot A\right)-\left(v\cdot \nabla \right)A$

Thus,

    $m\ddot{r}=q\left(v\times \left(\nabla \times A\right)\right)+qE$

Here, the curl $A$ is equivalent to the magnetic field intensity $B$

Thus, re-write the above expression, and here $\nabla \times A=B$

  $m\ddot{r}=q\left(v\times \left(B\right)\right)+qE$

Therefore the Hamilton’s equation is equivalent to the Lorentz force equation $m\ddot{r}=q\left(E+v\times B\right)$