# Kinetic momentum

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In physics, in particular electromagnetism, the kinetic momentum is the momentum

${\mathbf {p}}=m{\mathbf {v}}$ of a charged particle of mass m and velocity v moving in an electromagnetic field.

## Non-relativistic dynamics

### Lagrangian formulation

The kinetic momentum is different to the canonical momentum (synonymous with the generalized momentum) P found from the Lagrangian L of a free charged particle,

${\mathbf {P}}={\frac {\partial L}{\partial {\mathbf {\dot {r}}}}},\quad L={\frac {m}{2}}\mathbf {\dot {r}} \cdot \mathbf {\dot {r}} +e\mathbf {A} \cdot \mathbf {\dot {r}} -e\phi \,\!$ where = v is the velocity (see time derivative) and e is the electric charge of the particle, because P includes a contribution from the electric potential φ and vector potential A:

${\mathbf {P}}=m{\mathbf {v}}+e{\mathbf {A}}$ so the kinetic momentum is also

$m{\mathbf {v}}={\mathbf {P}}-e{\mathbf {A}}$ The kinetic momentum mv is not always the result this derivative. See also: Momentum in electromagnetism (section)

### Hamiltonian formulation

The classical Hamiltonian H for a particle in any field equals the total energy of the system - the kinetic energy T = p2/2m plus the potential energy V:

$H=T+V={\frac {{\mathbf {p}}^{2}}{2m}}+V,$ where p2 = p·p (see dot product).

For a particle in an electromagnetic field, the potential energy is V = , and since the kinetic energy T always corresponds to the kinetic momentum p, replacing the kinetic momentum by the above equation (p = PeA) leads to the Hamiltonian:

$H={\frac {({\mathbf {P}}-e{\mathbf {A}})^{2}}{2m}}+e\phi .$ ## Canonical commutation relations

The kinetic momentum (p above) satisfies the commutation relation:

$\left[p_{j},p_{k}\right]={\frac {i\hbar e}{c}}\epsilon _{jk\ell }B_{\ell }$ where: j, k, l are indices labelling vector components, Bl is a component of the magnetic field, and εkjl is the permutation tensor, here in 3-dimensions.

## Relativistic dynamics

In relativity, the Lagrangian for the particle interacting with the field is

$L=-m{\sqrt {1-\left({\frac {\dot {\mathbf {r}}}{c}}\right)^{2}}}+e{\mathbf {A}}({\mathbf {r}})\cdot {\dot {\mathbf {r}}}-e\phi ({\mathbf {r}})\,\!$ The action is the relativistic arclength of the path of the particle in space time, minus the potential energy contribution, plus an extra contribution which quantum mechanically is an extra phase a charged particle gets when it is moving along a vector potential.

The momentum conjugate to r, that is the canonical momentum P, is defined from the variation of the lagrangian:

${\mathbf {P}}={\frac {\partial L}{\partial {\dot {\mathbf {r}}}}}={\frac {m{\dot {\mathbf {r}}}}{\sqrt {1-\left({\frac {\dot {\mathbf {r}}}{c}}\right)^{2}}}}+e{\mathbf {A}}\,\!$ The kinetic momentum is the relativistic momentum of a particle moving with velocity v = , still (PeA), so we have:

${\mathbf {P}}-e{\mathbf {A}}={\frac {m{\dot {\mathbf {r}}}}{\sqrt {1-\left({\frac {\dot {\mathbf {r}}}{c}}\right)^{2}}}}\,$ The Hamiltonian equals total energy (kinetic plus potential), and is the usual relativistic expression for the energy. So in terms of the kinetic momentum:

$H={\mathbf {P}}\cdot {\dot {\mathbf {r}}}-L={m \over {\sqrt {1-\left({\frac {\dot {\mathbf {r}}}{c}}\right)^{2}}}}+e\phi ={\sqrt {({\mathbf {P}}-e{\mathbf {A}})^{2}+(mc^{2})^{2}}}+e\phi \,$ The equations of motion derived by extremizing the action (see matrix calculus for the notation):

${\frac {\mathrm {d} {\mathbf {P}}}{\mathrm {d} t}}={\frac {\partial L}{\partial {\mathbf {r}}}}=e{\partial {\mathbf {A}} \over \partial {\mathbf {r}}}\cdot {\dot {\mathbf {r}}}-e{\partial \phi \over \partial {\mathbf {r}}}\,\!$ ${\mathbf {P}}-e{\mathbf {A}}={\frac {m{\dot {\mathbf {r}}}}{\sqrt {1-\left({\frac {\dot {\mathbf {r}}}{c}}\right)^{2}}}}\,$ are the same as Hamilton's equations of motion:

${\frac {\mathrm {d} {\mathbf {r}}}{\mathrm {d} t}}={\frac {\partial }{\partial {\mathbf {p}}}}\left({\sqrt {({\mathbf {P}}-e{\mathbf {A}})^{2}+(mc^{2})^{2}}}+e\phi \right)\,\!$ ${\frac {\mathrm {d} {\mathbf {p}}}{\mathrm {d} t}}=-{\partial \over \partial {\mathbf {r}}}\left({\sqrt {({\mathbf {P}}-e{\mathbf {A}})^{2}+(mc^{2})^{2}}}+e\phi \right)\,\!$ both are equivalent to the noncanonical form:

${\frac {\mathrm {d} }{\mathrm {d} t}}\left({m{\dot {\mathbf {r}}} \over {\sqrt {1-\left({\frac {\dot {\mathbf {r}}}{c}}\right)^{2}}}}\right)=e\left({\mathbf {E}}+{\mathbf {v}}\times {\mathbf {B}}\right).\,\!$ This formula is the Lorentz force, representing the rate at which the EM field adds relativistic momentum to the particle.