# Lorentz force law from action

1. Jun 19, 2014

### erogard

Hi, I am trying to derive the Lorentz force law in the following form:
$$q \frac{dw^\mu}{d\tau} = q w^\mu \partial_\nu A_\sigma \epsilon_\mu^{\nu \sigma}$$
by varying the following Lagrangian for a classical particle:
$$S = \int d^3 x \left( -m \int d\tau \delta(x-w(\tau) ) + q \int d\tau \frac{dw^\mu}{d\tau} A_\mu \delta(x - w(\tau) ) \right)$$
where w tracks the position of the particle as a function of proper time. Note that there may be a couple terms/indices missing from the above expressions (still trying to figure that out).

I've read on a different thread that Feynman has that derivation in "QM and Path Integrals", which I have handy right now, however I couldn't find the derivation I am looking for.

Any push in the right direction would be more than appreciated.

2. Jun 20, 2014

### vanhees71

The covariant formalism is a bit tricky. The problem is that you cannot use the proper time directly in the Lagrangian, because it implicitly contains the constraint that $\mathrm{d} \tau=\mathrm{d} t \sqrt{1-v^2}$ (I'm setting the speed of light to 1 here).

You can, however use an arbitrary "world parameter" and write down a parameter-independent action functional. For the free particle it's
$$S_0[x]=-m \int \mathrm{d} \lambda \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}}.$$
I'm using the "west-coast convention" for the pseudometric, i.e., $\eta_{\mu \nu}=\text{diag}(1,-1,-1,-1)$.
The interaction with an external electromagnetic field is determined by
$$S_i[x]=-q \int \mathrm{d} \lambda \dot{x}^{\mu} A_{\mu}(x).$$
Note that the total Lagrangian,
$$L=-m \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}} - q \dot{x}^{\mu} A_{\mu}(x)$$
is a homogeneous function of degree 1 wrt. $\dot{x}$. This implies the parameter independence. Now you can derive the equations of motion from the Hamilton principle of least action as usual, using the Euler-Lagrange equations. The canonical momenta are given by
$$p_{\mu}=\frac{\partial L}{\partial \dot{x}^{\mu}}=-m \frac{\dot{x}_{\mu}}{\sqrt{\dot{x}_{\nu} \dot{x}^{\nu}}}-q A_{\mu}$$
and thus the equation of motion
$$\dot{p}_{\mu}=\frac{\partial L}{\partial x^{\mu}} \; \Rightarrow \; -m \frac{\mathrm{d}}{\mathrm{d} \lambda} \left (\frac{\dot{x}_{\mu}}{\sqrt{\dot{x}_{\nu} \dot{x}^{\nu}}} \right ) - q \dot{x}^{\nu} \partial_{\nu} A_{\mu}(x)=-q \dot{x}^{\nu} \partial_{\mu} A_{\nu}.$$
Now you can bring this into a more familiar form by rearranging the terms to
$$m \frac{\mathrm{d}}{\mathrm{d} \lambda} \left (\frac{\dot{x}_{\mu}}{\sqrt{\dot{x}_{\nu} \dot{x}^{\nu}}} \right ) = q \dot{x}^{\nu} (\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu})=q F_{\mu \nu} \dot{x}^{\nu}.$$
Now you can choose for $\lambda$ whatever parameter you like. It is crucial to note that the equation of motion, if derived from a Lagrangian that is homogeneous in $\dot{x}^{\mu}$ of degree one, automatically fulfills the constraint equation
$$\dot{x}^{\mu} \frac{\mathrm{d}}{\mathrm{d} \lambda} \left (\frac{\dot{x}_{\mu}}{\sqrt{\dot{x}_{\nu} \dot{x}^{\nu}}} \right )=0,$$
which implies that the 4 equations of motion are not independent from each other and thus you can choose the parameter $\lambda$ as you like after the variation is done, i.e., on the level of the equations of motion.

If you choose the proper time, $\lambda=\tau$, then you get $\dot{x}_{\mu} \dot{x}^{\mu}=1$ and thus
$$m \frac{\mathrm{d} u_{\mu}}{\mathrm{d} \tau}=q F_{\mu \nu} u^{\nu} \quad \text{with} \quad u^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
You can as well chose a coordinate time, $t$ of an inertial frame. Then the equation of motion appears in a form that is not longer manifestly covariant, but you get, because of $$\dot{x}^{\mu} \dot{x}_{\nu}=1-\vec{v}^2, \quad \vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t},$$
for the spatial part of the equations of motion (writing the Faraday tensor in terms of the three-dimensional notation with $\vec{E}$ and $\vec{B}$)
$$m \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{\vec{v}}{\sqrt{1-\vec{v}^2}} \right ) = q (\vec{E}+\vec{v} \times \vec{B}),$$
and the time component is just the energy-work relation, following from this:
$$m \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{1}{\sqrt{1-\vec{v}^2}} \right ) = \vec{v} \cdot \vec{E}.$$