# How can the Lorentz force law be derived using the action principle?

• erogard
In summary, the Lorentz force law can be derived in the covariant formalism by using an arbitrary world parameter in the action functional, which leads to a homogeneous Lagrangian of degree 1 in the velocities. This results in equations of motion that automatically fulfill a constraint equation, and choosing the proper time as the world parameter leads to the familiar form of the Lorentz force law.
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.

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:
$$m \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{1}{\sqrt{1-\vec{v}^2}} \right ) = \vec{v} \cdot \vec{E}.$$

1 person

## 1. What is the Lorentz force law?

The Lorentz force law is a fundamental law of electromagnetism that describes the force exerted on a charged particle moving through an electric and magnetic field.

## 2. How does the Lorentz force law relate to the principle of action and reaction?

The Lorentz force law is a manifestation of the principle of action and reaction, as it states that for every action, there is an equal and opposite reaction. In other words, the force exerted on a charged particle by the electric and magnetic fields is equal and opposite to the force exerted on the fields by the charged particle.

## 3. What is the significance of the Lorentz force law in electromagnetic theory?

The Lorentz force law plays a crucial role in understanding and predicting the behavior of charged particles in electric and magnetic fields. It is also an essential component in the development of other laws and principles in electromagnetic theory, such as Maxwell's equations.

## 4. How is the Lorentz force law derived from the principle of least action?

The Lorentz force law can be derived from the principle of least action, which states that the actual path taken by a particle will minimize the action, or the integral of the Lagrangian, along that path. By applying this principle to a charged particle moving through an electric and magnetic field, we can arrive at the Lorentz force law.

## 5. Can the Lorentz force law be applied to all types of charged particles?

Yes, the Lorentz force law can be applied to any charged particle, regardless of its mass or speed. It is a universal law that governs the behavior of all charged particles in electric and magnetic fields.

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