Four tests of for a conservative force

doug  <sweetser@world.com>

The tests
Definitions of quaternion operators
1. The curl is zero
2. There exists a potential function
3. The line integral of any closed loop is zero
4. The line lintegral along different paths is the same
Implications

The tests

There are four well-known, equivalent tests to determine if a force is conservative:  the curl must be 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.

Definitions of quaternion operators

Consider how a quaternion composed of differential operators acts the another quaternion composed of functions.

\(dq\  = \ q[d\/dt, \ d\/dx, \ d\/dy, \ d\/dz]; 

    fq\  = \ q[f\_0, \ f\_1, \ f\_2, \ f\_3]; \)

dq  =  ((d/dt),  (d/dx),  (d/dy),  (d/dz))fq  =  ((Subscript[f, 0]),  (Subscript[f, 1]),  (Subscript[f, 2]),  (Subscript[f, 3]))

Multiply the quaternion operator and the function together, and then evaluate each component.

dq . fq . {1, 0, 0, 0}\({\(d\ f\_0\)\/dt - \(d\ f\_1\)\/dx - \(d\ f\_2\)\/dy - \(d\ f\_3\)\/dz, 
      \(d\ f\_0\)\/dx + \(d\ f\_1\)\/dt - \(d\ f\_2\)\/dz + \(d\ f\_3\)\/dy, 
      
	\(d\ f\_0\)\/dy + \(d\ f\_1\)\/dz + \(d\ f\_2\)\/dt - 
        \(d\ f\_3\)\/dx, 
      \(d\ f\_0\)\/dz - \(d\ f\_1\)\/dy + \(d\ f\_2\)\/dx + \(d\ f\_3\)\/dt}
      \)

\(\((\(d\ f\_0\)\/dt - \(d\ f\_1\)\/dx - \(d\ f\_2\)\/dy - 
        \(d\ f\_3\)\/dz, 
      \(d\ f\_0\)\/dx + \(d\ f\_1\)\/dt - \(d\ f\_2\)\/dz + \(d\ f\_3\)\/dy, 
      
	\(d\ f\_0\)\/dy + \(d\ f\_1\)\/dz + \(d\ f\_2\)\/dt - 
        \(d\ f\_3\)\/dx, 
      \(d\ f\_0\)\/dz - \(d\ f\_1\)\/dy + \(d\ f\_2\)\/dx + \(d\ f\_3\)\/dt)
      \)\)

If fo = 0, then the first term is the negative of the divergence.  If f is not a function of t, then the next three terms compose the curl.  In general, 16 operations are required to make up the final product.

1.  The curl is Zero

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

F =  q[0,  -k x,  -k y,  0];

F =  q[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.

[((d/dt), (Overscript[∇, ⇀])), 
        (Overscript[F, ⇀])] =0

[((d/dt), (Overscript[∇, ⇀])), 
        (Overscript[F, ⇀])] =0

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

oponq[D,  F] . {1, 0, 0, 0}{2 k, 0, 0, 0}

((d/dt), ∇) F  = (2 k, 0, 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.

(fint  =  oponq[Integrate,  F]).{1, 0, 0, 0}\({\(k\ x\^2\)\/2 + \(k\ y\^2\)\/2, \(-k\)\ t\ x + k\ y\ z, 
        \(-k\)\ t\ y - k\ x\ z, 0}
 - 
      opinq[D, fint[\([1, 1]\)]\ q[1, 1, 1, 1]]\ .\ {1, 0, 0, 0}\){0, -k x, -k y, 0}

\(F\ \  = \ 
      \(∫\ F \((dt, dx, dy, dz)\)
 = \ 
        \(∫\((k\ x\ dx\  + k\ y\ dy, \ 
              \(-k\)\ x\ dt\  + \ k\ y\ dz, 
			\ 
              \(-k\)\ y\ dt\  - \ k\ x\ dz, 0)\)
 = 
          \((\(k\ x\^2\)\/2 + \(k\ y\^2\)\/2, \(-k\)\ t\ x + k\ y\ z, 
            \(-k\)\ t\ y - k\ x\ z, 0)\)\)\)
\)((d/dt), (Overscript[∇, ⇀])) 
        (((k (x^2))/2) + ((k (y^2))/2), (Overscript[0, ⇀]))  =  
      (0, -k x, -k y, 0)

This is the same force as we started with, so the first term of fint is the scalar potential of this vector field.  The vector terms in fint arrise 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\  = \ \ q[0, \ r\ Cos[t], \ r\ Sin[t], \ 0]; 

    lineintegral[F, path, \ 0, \ 2\ π]\ .\ {1, 0, 0, 0}\){0, 0, 0, 0}

path  =   q[0,  r Cos[t],  r Sin[t],  0](∫_0^( 2 π))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).

\(path1\  = \ q[0, \ r\ Cos[t], \ 2\ r\ Sin[t\/2], \ 0]; 

    \(lineintegral[f\_test, \ path1, \ 0, \ π]\)[\([1, 1]\)]
 - 
      2\ k\ r\^2\)\(path2\  = \ q[0, \ \(-\ t\)\ r\  + \ r, \ t\ r, \ 0]; 

    \(lineintegral[f\_test, \ path2, \ 0, \ 2]\)[\([1, 1]\)]
 - 
      2\ k\ r\^2\)

\(path1\  = \ q[0, \ r\ Cos[t], \ 2\ r\ Sin[t\/2], \ 0]; 

    \(∲\_0\%π\)\+\(path\ 1\)F\ dt\  = 
      \(-2\)\ k\ r\^2
\)\(path2\  = \ q[0, \ \(-\ t\)\ r\  + \ r, \ t\ r, \ 0]; 

    \(∲\_0\%2\)\+\(path\ 2\)F\ dt\  = 
      \(-2\)\ k\ r\^2\)

The same!

Implications

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!