What is the conserved canonical momentum for a relativistically moving charge in a static Coulomb electric field?

The canonical momentum is a fundamental conserved quantity from Noether's theorem for translational invariance of the Lagrangian. Yet I'm finding it very difficult to see its derivation, or even a statement of what it is for something as fundamental as a relativistically moving charge in a static Coulomb electric field.

Can anyone state what it is, or even give a derivation if it's not too much trouble?

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@Qmechanic why does the Lorentz force need to vanish? The Lagrangian only needs to be independent of coordinates. Adding $\nabla\Lambda$ to $A = 0$ compensates for $\Phi$'s dependence on the coordinates. – John McVirgo Jul 5 '11 at 16:12
The magnetic term depends on the velocity of the particle, while the electric term depends only on the position of the particle; I don't see how changing the vector potential can possibly help. – Willie Wong Jul 5 '11 at 18:05
In other words, what Qmechanic is saying is that the Hamiltonian equation of motion for the canonical momentum will reduce to the Lorentz force law + the analogue of Newton's second law for relativistic particles. – Willie Wong Jul 5 '11 at 18:11
@Willie I'm only interested in making the Lagrangian independent of position for the sake of using it to find the conserved canonical momentum for my particular example. It won't affect the physics of the problem, as expected. – John McVirgo Jul 5 '11 at 18:29
Why should the canonical momentum be conserved at all? You seem to be assuming that your Lagrangian can be made to be independent of position, but if you postulate a non-trivial electric potential, your Lagrangian will depend on position. – Willie Wong Jul 5 '11 at 19:01

Put speed of light $c=1$ and use sign convention $(-,+,+,+)$. To have that the canonical momentum

$$\vec{p}~=~\frac{\partial L}{\partial d_t \vec{x}}~=~ \gamma m_{0}d_t \vec{x} +q\vec{A}$$

is conserved $d_t \vec{p}=0$, and a $\vec{x}$-translational symmetry of the Lagrangian

$$L~=~T-U, \qquad T~=~-\frac{m_0}{\gamma}, \qquad U~=~ q (\phi -\vec{A}\cdot d_t \vec{x}) ,\qquad \gamma = \frac{1}{\sqrt{1- d_t \vec{x}\cdot d_t \vec{x}}},$$

we must have that the Lagrangian $L$ is independent of the position $\vec{x}$, and we must therefore have that the 4-gauge potential

$$A^{\mu}=(\phi,\vec{A})$$

is independent of position $\vec{x}$. Hence $A_{\mu}=A_{\mu}(t)$ is a function of time only. And hence

$$\vec{E}~=~ -\vec{\nabla} \phi -\partial_t \vec{A} ~=~-\partial_t \vec{A}.$$

But $\vec{E}$ is assumed to be static, i.e., independent of $t$. Hence $\vec{A}$ is just an affine function of time, and $\vec{E}$ is independent of both space and time. It follows that the conservation law for the canonical momentum $d_t \vec{p}=0$ is just the relativistic Newton's 2nd law

$$d_t(\gamma m_{0}d_t \vec{x})~=~q \vec{E}$$

for a charged particle in a constant $\vec{E}$-field.

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How do you know it's possible to create a Lagrangian independent of coordinates for any $\Phi(\vec x,t)$ ? – John McVirgo Aug 3 '11 at 13:43

The Lagrangian for a relativistic particle moving in an electromagnetic field is given by $$L = - \frac {mc^2} \gamma + q~( \vec u(t)\cdot \vec A(\vec x,t) -\Phi(\vec x,t))$$

Conservation of the canonical momentum requires the Lagrangian to be independent of coordinates, whereas in general it isn't. We therefore need to show it's possible to create this independence with some suitable gauge transformation of the potentials $\Phi$ and $A$. For the case of a static electric field with $\vec A=0~$, the EM gauge transformations $$\vec A= \nabla\lambda(\vec x,t),\quad\phi(\vec x)\to\phi(\vec x) - \frac d {dt} \lambda(\vec x,t)$$ gives the Lagrangian $$L = -\frac{m_0c^2}\gamma + q~( \vec u\cdot\nabla\lambda(\vec x,t) - \phi(\vec x) + \frac d {dt}\vec\lambda(\vec x, t))$$

It isn't generally possible to make this expression independent of coordinates for every $\Phi(\vec x)$ including that for a static Coulomb electric field. However, lets take the simplest case of an electric field directed along the x axis giving a potential and defining a gauge as$$\Phi(\vec x) = -Ex\qquad \lambda(\vec x, t) = -Ext$$to give a coordinate independent Lagrangian$$L = -\frac{m_0c^2}\gamma - q\vec u\cdot Et$$ and the conserved canonical momentum $$\frac{\partial L}{\partial\vec u} = (\gamma mu_x - qEt, 0, 0)$$ Since this is constant over time, we're finally left with just the Lorentz force on a relativistic charge in a static electric field along the x direction $$\frac d {dt}\gamma m\vec u = q(E,0,0)$$

This can be generalised to any time and coordinate dependent electric field from the principle of locality, causality and memorylessness:

• The dynamics of a charge depends upon the field at its coordinates only

• The dynamics of a charge depends upon the field at time t, and not the past or the future.

Therefore, for any electric field $\vec E(\vec x,t)$, the Lorentz force and associated conserved expression is$$\frac d {dt}\gamma m\vec u = q\vec E(\vec x,t),\qquad\gamma m\vec u - q\int\vec E(\vec x,t)dt$$ and can finally conclude using the gauge $\lambda(\vec x, t) = -\int\Phi(\vec x, t)dt$, canonical momentum is conserved for a relativistic charge in a electric field $\vec E(\vec x, t)~$ and given by $$\frac{\partial L}{\partial\vec u} = \gamma m\vec u - q\int\vec E(\vec x,t)dt$$

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