Last modified on 17 July 2009, at 20:42

Physics Using Geometric Algebra/Relativistic Classical Mechanics/The classical spinor

The proper velocity can be expressed in terms of the associated Lorentz transformation spinor called eigenspinor.

The proper time derivative of the eigenspinor can be expressed in terms of the spacetime rotation biparavector  \Omega as


\frac{d \Lambda}{d \tau} = \frac{1}{2} \Omega \Lambda

The spacetime rotation biparavector contains the information of the interaction that the particle is subject to.

The dynamics of a classical particle can be expressed in terms of a differential equation involving the eigenspinor.

  \frac{d p}{d \tau}  = mc \frac{d }{d\tau} 
( \Lambda   \Lambda^\dagger ) =
mc ( \frac{d \Lambda }{d\tau} \Lambda^\dagger 
+ \Lambda \frac{d \Lambda^\dagger}{d \tau}   ),

using the definition of the spacetime rotation biparavector


mc \frac{1}{2}( \Omega \Lambda \Lambda^\dagger + 
\Lambda \Lambda^\dagger \Omega ^\dagger ) =
mc \langle \Lambda \Omega \rangle_{\Re{ }},

that finally gives the dynamic equation


\frac{d p}{d \tau } = mc \langle \Omega \Lambda \rangle_{\Re{ }},

where the role of the force is now played by the spacetime rotation biparavector.

Once the spacetime rotation is identified with the electromagnetic force, the result is the spinorial form of the Lorentz force.


\frac{d \Lambda}{d \tau } = \frac{e}{2 m c}  F  \Lambda

Constant Electric FieldEdit

The eigenspinor form of the Lorentz force can be solved analytically in many interesting cases. One of the simplest is the case of the constant electric field. If the electric field is directed along the z direction, the equation to solve is


\frac{d \Lambda}{d \tau } = \frac{e}{2 m c}  E \mathbf{e}_3 \Lambda

and the solution is


\Lambda = e^{ \frac{e}{2 m c}  E \mathbf{e}_3 \tau }\Lambda_0.

The proper velocity is calculated as


u = e^{ \frac{e}{2 m c}  E \mathbf{e}_3  \tau} u_0e^{ \frac{e}{2 m c}  E \mathbf{e}_3 \tau}.

If the particle starts at rest, the proper velocity is simply


u = e^{ \frac{e}{m c}  E \mathbf{e}_3 \tau}.

An additional integration gives the spacetime position


x = x_0 +  \frac{m c^2}{e E} \mathbf{e}_3 ( e^{ \frac{e}{m c}  E \mathbf{e}_3 \tau}-1)