# What is the difference between generalized momentum and ordinary momentum?

I'm studying about motion equation of charge in electromagnetic field.

Lagrangian of charge in E.M field is

$L=-mc^2\sqrt{1-v^2/c^2}+\frac{e}{c}\mathbf{A}\cdot \boldsymbol{v}-e\phi$ .

Thus generalized momentum is

$\boldsymbol{P}=\frac{\partial L}{\partial \boldsymbol{v}}=\frac{m \boldsymbol{v}}{\sqrt{1-v^2/c^2}}+\frac{e}{c}\boldsymbol{A}=\boldsymbol{p}+\frac{e}{c}\boldsymbol{A}$ ,

and it's obvious that generalized momentum has different value with ordinary momentum.

So here's my question. What is the difference between generalized momentum and ordinary momentum?

• "What is the difference...": they're different quantities, that coincide sometimes, but not always. Do you have something specific in mind when you ask for differences? – Samarth Dec 24 '17 at 14:09

## 1 Answer

They're (in principle) different quantities, that coincide for velocity-independent potentials in a Cartesian coordinate system. The Lagrangian and the generalized momenta in such a system are

\begin{align} L(x, y, z, v_x, v_y, v_z; t) &= -mc^2\sqrt{1-\frac{v^2}{c^2}} - V(x, y, z) \\ \frac{\partial L}{\partial v_i} &= \frac{mv_i}{\sqrt{1 - \frac{v^2}{c^2}}} \end{align}

The generalized momentum corresponds to the ordinary 3-momentum in this case.

The invariance of the Lagrangian under translation of a coordinate leads to conservation of the generalized momentum corresponding to that coordinate, through Lagrange's equation of motion. There is no conservation law for the ordinary momentum of a system.

The momentum conservation of the Newtonian formulation is a special case of this general conservation law, as is Newtonian angular momentum conservation.