Exercise 4

We start with the Hamiltonian of a charged particle in a magnetic field: \[ H = \sum_j \frac{1}{2m} (p_j - \frac{e}{c} A_j)^2\]

and compute Hamilton’s equations of motion. \[ \dot{x}_i = \frac{\partial H}{\partial p_i} = \frac{\partial}{\partial p_i} \sum_j \frac{1}{2m} (p_j - \frac{e}{c} A_j)^2 \]

We need the chain rule, since \(p_j\) arises in a quadratic term. Also notice \(\frac{\partial p_j}{\partial p_i} = \delta_{ji}\), hence the sum over \(j\) boils down to the one term with \(i\): \[ \dot{x}_i = \frac{1}{m}(p_i - \frac{e}{c} A_i) \]

\[ \dot{p}_i = - \frac{\partial H}{\partial x_i} = - \frac{\partial}{\partial x_i} \sum_j \frac{1}{2m} (p_j - \frac{e}{c} A_j)^2 \] We apply the chain rule twice, since \(A_j\) arises in a quadratic term and is a function of the \(x_i\): \(A_j = A_j \, (x_1, x_2, x_3)\).

\[ \dot{p}_i = - \sum_j \frac{1}{m}(p_j - \frac{e}{c}A_j) (-\frac{e}{c}) \frac{\partial A_j}{\partial x_i} = \sum_j \dot{x}_j \frac{e}{c} \frac{\partial A_j}{\partial x_i} \]

Now let’s derive an expression for the Lorentz force. Newton’s law states that: \[ F_i = m \ddot{x}_i = m \frac{\mathrm{d}}{\mathrm{d}t} \dot{x}_i \]

Now, for \(\dot{x}_i\) we plug in the expression from Hamilton’s equation of motion: \[ F_i = \frac{\mathrm{d}}{\mathrm{d}t} (p_i - \frac{e}{c} A_i) = \dot{p_i} - \frac{e}{c} \sum_j \frac{\partial A_i}{\partial x_j} \dot{x}_j\]

Same for \(\dot{p}_i\): \[ F_i = \sum_j \dot{x}_j \frac{e}{c} \frac{\partial A_j}{\partial x_i} - \frac{e}{c} \sum_j \frac{\partial A_i}{\partial x_j} \dot{x}_j = \frac{e}{c} \sum_j \dot{x}_j \, \left( \frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j} \right) \]

Spock would say: fascinating!

Let’s compute the \(x_1\) component of the Lorentz force for \(i=1\): \[ F_1 = \frac{e}{c} \left[ \dot{x}_1 \left( \frac{\partial A_1}{\partial x_1} - \frac{\partial A_1}{\partial x_1} \right) + \dot{x}_2 \left( \frac{\partial A_2}{\partial x_1} - \frac{\partial A_1}{\partial x_2} \right) + \dot{x}_3 \left( \frac{\partial A_3}{\partial x_1} - \frac{\partial A_1}{\partial x_3} \right) \right] \]

Using the definition of the magnetic field \(\vec{B} = \nabla \times \vec{A}\) we get: \[F_1 = \frac{e}{c} ( \dot{x}_2 \, B_3 - \dot{x}_3 \, B_2 ) \]

The other components of the Lorentz force are computed using analogous considerations, hence: \[ \vec{F} = \frac{e}{c} ( \vec{v} \times \vec{B}) \]