General Mechanics/Velocity Dependent Forces

We've seen that a Hamiltonian of the form


 * $$H=\frac{\mathbf{p}\cdot\mathbf{p}}{2m}+V(\mathbf{r})$$

describes motion under a conservative force, but this is not the most general possible Hamiltonian.

Many waves are described by another simple form, in the geometrical optics limit,


 * $$H=f(|\mathbf{p}|)$$

What happens if we consider other forms for the Hamiltonian?

Suppose we have


 * $$H=\frac{\mathbf{p}\cdot\mathbf{p}}{2m} +

\mathbf{p} \cdot \mathbf{A}(\mathbf{r})$$

The dot product term is the simplest scalar we can add to the kinetic energy that's dependent on momentum. We'll see that it acts like a potential energy.

Using Hamilton's equations we can immediately write down equations of motion.


 * $$\begin{matrix}

\dot{x}_i & = & \frac{\partial H}{\partial p_i} & = & \frac{p_i}{m} + A_i \\ \dot{p}_i & = & -\frac{\partial H}{\partial x_i} & = & -p_j \frac{\partial A_j}{\partial x_i} \end{matrix}$$

Note that we are using the summation convention here.

Now that we've got the equations of motion, we need to work out what they mean.

The first thing we notice is that the momentum is no longer mv. There is an additional contribution from the potential field A. We can think of this as a form of potential momentum, analogous to potential energy.

The force on the particle is


 * $$F_i = m \ddot{x}_i = \dot{p}_i + m\frac{d}{dt}A_i(\mathbf{r})$$

Using the chain rule, and substituting for dp/dt with the second equation of motion, we get


 * $$F_i = -p_j \frac{\partial A_j}{\partial x_i} +

m\dot{x}_j \frac{\partial A_i}{\partial x_j} $$

Using the first equation, this becomes


 * $$F_i = m \dot{x}_j

\left( \frac{\partial A_i}{\partial x_j} -\frac{\partial A_j}{\partial x_i} \right) + m A_j \frac{\partial A_j}{\partial x_i} $$

The term in brackets is recognizable as being the type of thing we see in cross products. With a little manipulation, using the Kronecker delta and alternating symbol, we can write


 * $$\begin{matrix}

F_i & = & \dot{x}_j \left( \delta_{il}\delta_{jk}-\delta_{ik}\delta_{jl} \right) \frac{\partial A_l}{\partial x_k} & + & m A_j \frac{\partial A_j}{\partial x_i} \\ & = & \dot{x}_j \epsilon_{nij} \epsilon_{nlk} \frac{\partial A_l}{\partial x_k} & + & m A_j \frac{\partial A_j}{\partial x_i} \\ & = & -\epsilon_{ijn} \dot{x}_j \left( \nabla \times \mathbf{A} \right)_n & + & m A_j \frac{\partial A_j}{\partial x_i} \\ & = & -\epsilon_{ijn} \dot{x}_j \left( \nabla \times \mathbf{A} \right)_n & + & \frac{1}{2} \frac{\partial}{\partial x_i} A_j A_j \end{matrix}$$

so, writing the last line as a vector equation,


 * $$ \mathbf{F} = -m \mathbf{v} \times (\nabla \times \mathbf{A}) +

\frac{1}{2} m \nabla A^2 $$

The force from this potential has a component perpendicular to its curl and to the velocity, and another component which is the gradient of a scalar.

If we add a carefully chosen potential energy to the Hamiltonian we can cancel out this second term.


 * $$H=\frac{p^2}{2m} + \mathbf{p} \cdot \mathbf{A}(\mathbf{r}) +

\frac{1}{2} m A^2 \qquad \mathbf{F}= -m \mathbf{v} \times \left (\nabla \times \mathbf{A} \right) $$

This Hamiltonian can be simplified to


 * $$H = \frac{1}{2m} \left( \mathbf{p}+ m\mathbf{A} \right)^2 $$

where the term in bracket is mv, written as a function of momentum.

This simple modification to the Hamiltonian gives us a force perpendicular to velocity, like magnetism. Because the force and velocity are perpendicular, the work done by the force is always zero.

We can use potential fields to describe velocity dependent forces, provided that they do no work.

The coefficient of A in these expression is arbitrary. Changing it amount merely to measuring A in different units so we can equally well write


 * $$H = \frac{1}{2m} \left( \mathbf{p}+ \alpha\mathbf{A} \right)^2 \qquad

\mathbf{F}= -\alpha \mathbf{v} \times \left (\nabla \times \mathbf{A} \right) $$

Having &alpha;=m means A is measured in units of velocity, which may sometimes be convenient, but we can use any other constant value of &alpha; that suits our purpose. When we come to study relativity, we'll use &alpha;=-1.

Notice that the force depends only on the curl of A, not on A itself. This means we can add any function with zero curl to A without changing anything, just as we can add a constant to the potential energy.

Furthermore, it is a standard result of vector calculus that any vector field is the sum of two components, one with zero curl, the other with zero divergence. Since the zero curl component does not affect the total force, we can require it to be zero; i.e. we can require A to have zero divergence.


 * $$ \nabla \cdot \mathbf{A}= 0 $$

This is called a gauge condition. For the moment, it can be considered to define especially natural values for the integration constants.