# Magnetic fields

We will study both electric and magnetic fields in the next lecture, but we first need to consider velocity dependent forces, like magnetism.

The force acting on a particle, with charge $q$, and velocity $\vec{v}$ in a magnetic field $\vec{B}$, is given by $$m \vec{a} = \vec{F} = q \left( \vec{v} \times \vec{B} \right)$$

### Velocity dependent forces

Before, we have thought of forces as derivatives of scalar potential functions of position, $$- \nabla U(x, y, z) = \vec{F}(x, y, z)$$ but we will have to adjust this to accommodate forces that depend on velocities as well.

Not all velocity dependent forces are valid - friction, for example - because they don't satisfy certain conditions. A valid velocity dependent force must ...

• ... be derivable from the principle of least action
• ... have a Lagrangian formulation
• ... satisfy conservation of energy
• ... have a Hamiltonian formulation

### Vector potential

Up until now, we have been able to separate the Lagrangian into the kinetic energy associated with the system, and the scalar potential of a position dependent force, $$L = T - U$$ but we will have to add a new term, involving the vector potential of $\vec{B}$, which is defined as any vector, $\vec{A}$, that satifies $$\vec{B} = \nabla \times \vec{A}$$

Somewhat analagously to the definition of the scalar potential, we imagine that magnetism is the vector potential, $\vec{A}$, and the field, $\vec{B}$ is derived from that. Thus, \begin{align*}B_x &= [\nabla \times \vec{A}]_x = \partial_y A_z - \partial_z A_y = \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}\\B_y &= [\nabla \times \vec{A}]_y = \partial_z A_x - \partial_x A_z = \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}\\B_z &= [\nabla \times \vec{A}]_z = \partial_x A_y - \partial_y A_x = \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\end{align*}

The components of the force can then be written in terms of the potential, \begin{align*}F_x &= [ \vec{v} \times \vec{B}]_x = \dot{y} B_z - \dot{z} B_y = \dot{y} \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) - \dot{z} \left( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right)\\F_y &= [ \vec{v} \times \vec{B}]_y = \dot{z} B_x - \dot{x} B_z = \dot{z} \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right) - \dot{x} \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right)\\F_z &= [ \vec{v} \times \vec{B}]_z = \dot{x} B_y - \dot{y} B_x = \dot{x} \left( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right) - \dot{y} \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right)\end{align*}

### Action due to the magnetic potential

We just make a guess and show that it leads to the correct equations. The guess is to add $$q \int \vec{A} \cdot \mathrm{d}\vec{x}$$ to the action we had previously defined. This is an integral along the trajectory.

If there are no other forces acting on the charge then the total action is \begin{align*}\int L \mathrm{d}t &= \int \tfrac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) \mathrm{d}t + q \int \vec{A} \cdot \mathrm{d}\vec{x}\\&= \int \tfrac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) \mathrm{d}t + q \int \vec{A} \cdot \vec{v} \mathrm{d}t\\&= \int \tfrac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) \mathrm{d}t + q \int \left( A_x \dot{x} + A_y \dot{y} +A_z \dot{z} \right) \mathrm{d}t\\&= \int \tfrac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) + q \left( A_x \dot{x} + A_y \dot{y} +A_z \dot{z} \right)\mathrm{d}t\end{align*} and if this is the correct action then the Lagrangian is given by $$L = \tfrac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) + q \left( A_x \dot{x} + A_y \dot{y} +A_z \dot{z} \right)$$

We want to show that this equivalent to Newton's form, $$m \vec{a} = \vec{F} = q \left( \vec{v} \times \vec{B} \right)$$

The conjugate momentum (in the $z$-direction) is given by $$p_z = \frac{\partial L}{\partial \dot{z}} = m \dot{z} + qA_z$$ which isn't just the mechanical momentum this time. From this, we get $$\dot{p_z} = m \ddot{z} + q \left( \frac{\partial A_z}{\partial x}\dot{x} + \frac{\partial A_z}{\partial y}\dot{y} + \frac{\partial A_z}{\partial z}\dot{z}\right)$$

We also need $$\frac{\partial L}{\partial z} = q \left( \frac{\partial A_x}{\partial z}\dot{x} + \frac{\partial A_y}{\partial z}\dot{y} + \frac{\partial A_z}{\partial z}\dot{z} \right)$$

The Euler-Lagrange equations are then \begin{align*}0 &= \dot{p_z} -\, \frac{\partial L}{\partial z}\\&= m \ddot{z} + q \left( \frac{\partial A_z}{\partial x}\dot{x} + \frac{\partial A_z}{\partial y}\dot{y} + \frac{\partial A_z}{\partial z}\dot{z} \right) - q \left( \frac{\partial A_x}{\partial z}\dot{x} + \frac{\partial A_y}{\partial z}\dot{y} + \frac{\partial A_z}{\partial z}\dot{z} \right)\\&= m \ddot{z} - q \left\{ \dot{x} \left( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}\right) - \dot{y} \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \right) \right\}\\&= m \ddot{z} - q \left [ \vec{v} \times \left( \nabla \times \vec{A} \right) \right ]_z\\&= m \ddot{z} - F_z\end{align*}

Similar equations result for the $x$ and $y$ components. Thus, the extra action we defined was correct, and we recovered Newton's form from the principle of least action. We shall consider energy in the next section.