### The simplest case

Recall, we said that the simplest case would be to set $$\begin{align*}\vec{B} &= -B\hat{z}\\\vec{v} &= \dot{x}\hat{x} + \dot{y}\hat{y}\\\vec{E} &= E_x\hat{x} + E_y\hat{y}\end{align*}$$ so that all motion will happen in the $x,y$-plane and the velocity in the $z$-direction is zero, $$\dot{z} = 0$$

Thus, neither the electric or magnetic potential need have any dependence on $z$, $$\begin{align*} V &= V(x,y) \\ \vec{A} &= \vec{A}(x,y) \end{align*}$$

Hence, in this case, the Lagrangian simplifies to $$L = \tfrac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 \right) - qV(x,y) + q \left( A_x(x,y) \dot{x} + A_y(x,y) \dot{y}\right)$$

### Gauges

For simplicity, we set the electric field to zero for now.

Each gauge (or vector potential) we can find for a given magnetic force, $\vec{B}$, will lead to a conserved quantity (this is not proved).

Given that $$\vec{B} = - B \hat{z}$$ we want to find $\vec{A}$ such that $$-B = \frac{\partial A_y}{\partial x} -\, \frac{\partial A_x}{\partial y}$$

We pluck two gauges out of the air - the two most primitive gauges, it seems.

#### Gauge 1

$$A_x = By, \quad A_y = 0$$

First we check that this is a valid gauge, $$\frac{\partial A_y}{\partial x} -\, \frac{\partial A_x}{\partial y} = 0 - B = -B$$ as required. The Lagrangian is now (neglecting electric fields) $$L = \tfrac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 \right) + q By\dot{x}$$

The canonical momenta are given by $$\begin{align*}p_x &= m \dot{x} + qBy\\p_y &= m \dot{y}\end{align*}$$

Notice that $p_x$ doesn't change under the translation $$x \to x + \epsilon$$ hence is a conserved quantity. The same is true for $p_y$, but the equation for $p_x$ involves $x,y$. However, translation in the $y$-direction is not symmetric.

#### Gauge 2

$$A_x = 0, \quad A_y = -Bx$$

Again check validity, $$\frac{\partial A_y}{\partial x} -\, \frac{\partial A_x}{\partial y} = -B - 0 = -B$$ as required. The Lagrangian is $$L = \tfrac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 \right) - q Bx\dot{y}$$ and the (conserved) canonical momentum is $$p_y = m \dot{y} - qBx$$

So, the information gained from considering the two separate gauges are the two conserved quantites, $$\begin{align*}m \dot{x} + qBy &= p_x\\m \dot{y} - qBx &= p_y\end{align*}$$

We could rewrite this as $$\begin{align*}\dot{x} &= -\frac{qB}{m}(y - y_0)\\\dot{y} &= +\frac{qB}{m}(x - x_0)\end{align*}$$ where the conserved momenta have been replaced by the constants, $$\begin{align*}p_x &= +\frac{qB}{m}y_0\\p_y &= -\frac{qB}{m}x_0\end{align*}$$

Noting that $$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}t}\left(x - x_0 \right) &= \dot{x}\\\frac{\mathrm{d}}{\mathrm{d}t}\left(y - y_0 \right) &= \dot{y}\end{align*}$$ we could substitute $$\begin{align*}X &= x - x_0\\Y &= y - y_0\end{align*}$$ in the equations to give the particularly simple form, $$\begin{align*}\dot{X} &= -\frac{qB}{m}Y\\\dot{Y} &= +\frac{qB}{m}X\end{align*}$$

To see that this is circular motion, simply plug the transformation $$\begin{align*}X &= R \cos(\omega t)\\Y &= R \sin(\omega t)\end{align*}$$ into the pair of equations to get $$\omega = \frac{qB}{m}$$ which is called the **cyclotron frequency** associated with the magnetic field ($R$ is left undetermined).

Finally put back the original $x,y$ coordinates to give $$\begin{align*}x(t) &= x_0 + R \cos(\omega t)\\y(t) &= y_0 + R \sin(\omega t)\end{align*}$$ hence we see that the conserved momenta, $$\begin{align*}p_x &= +\omega y_0\\p_y &= -\omega x_0\end{align*}$$ determine the coordinates of the center of the circle around which the charge moves in a magnetic field. The conservation law tells us that this location doesn't change over time.

### Including the electric field

For simplicity, choose a constant electric force in the $x$-direction, $\vec{E} = E\hat{x}$, so that the field potential is $$V = -Ex$$ and the Lagrangian is $$L = \tfrac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 \right) + qEx + q \left( A_x \dot{x} + A_y \dot{y}\right)$$

We find the Euler-Lagrange equations of motion using both of the gauges. For **gauge 1**, where $A_x = By, \quad A_y = 0$, the Lagrangian is $$L = \tfrac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 \right) + qEx + qBy \dot{x}$$ so the equation is then $$\begin{align*}0 &= \frac{\mathrm{d} }{\mathrm{d}t}\left( \frac{\partial L}{\partial \dot{x}} \right) - \frac{\partial L}{\partial x}\\&= \frac{\mathrm{d} }{\mathrm{d}t}\left( m \dot{x} + qBy \right) - qE\\&= m \ddot{x} + qB\dot{y} - qE\end{align*}$$

For **gauge 2**, where $A_x = 0, \quad A_y = -Bx$, the Lagrangian is $$L = \tfrac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 \right) + qEx - qBx \dot{y}$$ so the Euler-Lagrange equation is then $$\begin{align*}0 &= \frac{\mathrm{d} }{\mathrm{d}t}\left( \frac{\partial L}{\partial \dot{y}} \right) - \frac{\partial L}{\partial y}\\&= \frac{\mathrm{d} }{\mathrm{d}t}\left( m \dot{y} - qBx \right) - 0\\&= m \ddot{y} - qB\dot{x}\end{align*}$$

In summary, $$\begin{align*}m \ddot{x} + qB\dot{y} &= qE\\m \ddot{y} - qB\dot{x} &= 0\end{align*}$$

We can solve these equations for different conditions.

For example, we can take the case where there is no acceleration, which leads to $$\begin{align*}\dot{x} &= 0\\\dot{y} &= \frac{E}{B}\end{align*}$$

So, an electric field in the $x$-direction and a magnetic field in the (negative) $z$-direction means that a charge moving in the $y$-direction (perpendicular to both fields), and with speed, $E/B$, is a solution.

This is called the **Hall effect**.