Four Tests for a Conservative Force

There are four well-known, equivalent tests to determine if a force is conservative: the curl is zero, a potential function whose gradient is the force exists, all closed path integrals are zero, and the path integral between any two points is the same no matter what the path chosen. In this notebook, quaternion operators perform these tests on quaternion-valued forces.

1. The Curl Is Zero

To make the discussion concrete, define a force quaternion F.

F =  \(0, - k x, - k y, 0\)

The curl is the commutator of the differential operator and the force. If this is zero, the force is conservative.

The commutator of the operator\(d by dt,Del\) acting on a function F  = 0

Let the differential operator quaternion act on the force, and test if the vector components equal zero.

\(d by dt, Del \) acting on F = \(2 k, 0,  0,

2. There Exists a Potential Function for the Force

Operate on force quaternion using integration. Take the negative of the gradient of the first component. If the field quaternion is the same, the force is conservative.

F  = integral F times \(dt, dx, dy,

= integral\(k x dx +k y dy, - k x dt + k y dz, - k y dt - k x dz,

=\(k x squared  over 2 + k y squared  over 2, -k t x + k y z, - k t y - k x
z, 0\)

\(d by dt,Del\) acting on \(k x squared  over 2 + k y squared  over 2, 0\) =
\(0, - k x, - k y, 0\)

This is the same force as we started with, so the scalar inside the integral is the scalar potential of this vector field. The vector terms inside the integral arise as constants of integration. They are zero if t=z=0. What role these vector terms in the potential quaternion may play, if any, is unknown to me.

3. The Line Integral of Any Closed Loop Is Zero

Use any parameterization in the line integral, making sure it comes back to go.

path =  \(0, r cosine \(t\), r sine \(t\),

integral from 0 to 2 pi F dt = 0

4. The Line Integral Along Different Paths Is the Same

Choose any two parameterizations from A to B, and test that they are the same. These paths are from (0, r, 0, 0) to (0, -r, 2 r, 0).

path 1 = \(0, r cosine \(t\), 2 r sine \(t over 2\),

integral from 0 to 2 pi along path 1 F dt = - 2 k r squared

path 2 = \(0, - t r + r, t r, 0\)

integral from 0 to 2 pi along path 1 F dt = - 2 k r squared

The same!


The four standard tests for a conservative force can be done with operator quaternions. One new avenue opened up is for doing path integrals. It would be interesting to attempt four dimensional path integrals to see where that might lead!