# Modern Physics/Potential Momentum

## Potential Momentum

In classical physics we know that kinematics can often be described by a potential energy alone. Now we've seen that in relativity the energy is just the temporal component of the momentum 4-vector, so we should expect the same of the potential energy. To see how this works, we'll reason by analogy from the classical case.

For a free, non-relativistic particle of mass m, the total energy E equals the kinetic energy K and is related to the momentum Π of the particle by

${\displaystyle E=K={\frac {|{\boldsymbol {\Pi }}|^{2}}{2m}}\qquad {\mbox{(free, non-relativistic)}}.}$

In the non-relativistic case, the momentum is Π= mv, where v is the particle velocity.

If the particle is not free, but is subject to forces associated with a potential energy U(x,y,z), then the equation must be modified to account for the contribution of U to the total energy:

${\displaystyle E-U=K={\frac {|{\boldsymbol {\Pi }}|^{2}}{2m}}\qquad {\mbox{(non-free, non-relativistic)}}.}$

The force on the particle is related to the potential energy by

${\displaystyle \mathbf {F} =-\left({\frac {\partial U}{\partial x}},{\frac {\partial U}{\partial y}},{\frac {\partial U}{\partial z}}\right).}$

For a free, relativistic particle, we have

${\displaystyle E=(|{\boldsymbol {\Pi }}|^{2}c^{2}+m^{2}c^{4})^{1/2}\qquad {\mbox{(free, relativistic)}}.}$

The obvious way to add forces to the relativistic case is by rewriting this equation with a potential energy:

${\displaystyle E-U=(|{\boldsymbol {\Pi }}|^{2}c^{2}+m^{2}c^{4})^{1/2}\qquad {\mbox{(incomplete!)}}.}$

However ${\displaystyle {\boldsymbol {\Pi }}=(\Pi ,E/c)}$  is a four-vector, so an equation with something subtracted from just one of the components of this four-vector is not relativistically invariant. In other words, this equation doesn't obey the principle of relativity, and therefore cannot be correct!

How can we fix this problem? One way is to define a new four-vector with U/c being its timelike part and some new vector Q being its spacelike part:

${\displaystyle {\underline {Q}}\equiv (\mathbf {Q} ,U/c)\qquad {\mbox{(potential four-momentum)}}.}$

We then subtract Q from the momentum Π. When we do this, equation (13.5) becomes

${\displaystyle E-U=(|{\boldsymbol {\Pi }}-\mathbf {Q} |^{2}c^{2}+m^{2}c^{4})^{1/2}\qquad {\mbox{(non-free, relativistic)}}.}$

The quantity Q is called the potential momentum and Q is the potential four-momentum.

If |Π-Q| is much smaller than mc, this becomes approximately

${\displaystyle E=mc^{2}+{\frac {1}{2m}}\left({\boldsymbol {\Pi }}-\mathbf {Q} \right)^{2}}$

This expression for the energy has the same form as the Hamiltonian we looked at for classical velocity dependent forces, so we know it predicts a force perpendicular to the velocity, when the condition is met. It turns out to be perpendicular even when the condition is met.

In classical physics the potential momentum is an optional extra. In relativity it is a necessary part of any potential field.

Some additional terminology is useful. We define

${\displaystyle \mathbf {p} \equiv {\boldsymbol {\Pi }}-\mathbf {Q} \qquad {\mbox{(kinetic momentum}})}$

as the kinetic momentum since in the classical case it reduces to mv. In order to avoid confusion, we rename Π the total momentum. Thus, the total momentum equals the kinetic plus the potential momentum, in analogy with energy.