# Lagrangian density for Lorentz force of continuous charge distribution in external field?

It's frequently an exercise to derive the Lorentz force law for a particle with charge $q$ in an external electromagnetic field given by the following Lagrangian:

$$L = -mc^2\sqrt{1-\frac{\dot{r}^2}{c^2}} - q \phi + q \mathbf{\dot{r}} \cdot \mathbf{A}$$

Which leads to the relativistic Lorentz force law:

$$\mathbf{F} = \frac{d}{dt} \Bigg(\frac{m \mathbf{\dot{r}}}{\sqrt{1-\frac{\dot{r}^2}{c^2}}} \Bigg) = q(\mathbf{E} + \mathbf{\dot{r}} \times \mathbf{B})$$

For continuous distributions, we have:

$$\mathbf{f} = \rho \mathbf{E} + \mathbf{J} \times \mathbf{B}$$

I'm trying to find the corresponding Lagrangian density which results in this force. I know if the charge distribution is treated as the source you can use the standard Lagrangian Density for electromagnetism but doing so will not give you the Lorentz Force Equation. However in my specific case I'm ignoring the self-field of the charge distribution, the fields are purely external, I don't need the electromagnetism Lagrangian density for my problem. Naively one might replace all instances of $m$ with a mass density term, $\rho_m$, and all instances of $q$ with a charge density term, $\rho_q$, where $\mathbf{J} = \rho_q \mathbf{\dot{r}}$.

$$\mathcal{L} = -\rho_mc^2\sqrt{1-\frac{v(\mathbf{r},t)^2}{c^2}} - \rho_q \phi + \rho_q \mathbf{v}(\mathbf{r},t) \cdot \mathbf{A} = -\rho_mc^2\sqrt{1-\frac{v(\mathbf{r},t)^2}{c^2}} - \rho_q \phi + \mathbf{J} \cdot \mathbf{A}$$

However the densities are also a function of the coordinates and moreover the mass densities and charge densities are related to each other in some unknown way. If we assume all of our particles are electrons then we can scale the densities by the electron mass and charge. If we take the variation of this Lagrangian density with respect to the charge density then we get the following:

$$\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{\rho}} + \frac{d}{dx}\frac{\partial \mathcal{L}}{\partial \rho_x} + \frac{d}{dy}\frac{\partial \mathcal{L}}{\partial \rho_y} + \frac{d}{dz}\frac{\partial \mathcal{L}}{\partial \rho_z}= \frac{\partial \mathcal{L}}{\partial \rho}$$

The LHS is clearly zero since we have no dependence on derivatives of the density in our Lagrangian density. The RHS just gives us:

$$0 = -\frac{mc^2}{e}\sqrt{1-\frac{\dot{r}^2}{c^2}} - \phi + \mathbf{v}(\mathbf{r},t) \cdot \mathbf{A}$$

So clearly this Lagrangian density is not correct or we should not be varying with respect to the density. Similarly one can take the variation with respect to the velocity field but this also does not result in the correct equation. I feel like I'm having a fundamental misunderstanding here but I can't find any reference that works through this. What is the correct Lagrangian Density? What is the correct quantity to vary the action?

1. OP is essentially asking about the Lagrangian field-theoretic formulation of a relativistic fluid in an external electromagnetic background $A_{\mu}$.
3. Here is the simplest Lagrangian Lagrangian relativistic formulation (with Lorentz symmetry but without manifest Lorentz symmetry). This essentially boils down to replacing discrete sums in point mechanics with continuous integrals in field theory. We put the speed of light $c=1$ to one for simplicity. The 3-position field ${\bf r}:\mathbb{R}^3\times \mathbb{R}\to\mathbb{R}^3$ depends on a continuous labelling variable ${\bf a}\in\mathbb{R}^3$ and time $t\in\mathbb{R}$. The action becomes $$S[{\bf r}] ~=~ \left. \int \! dt~ d^3a~ {\cal L}({\bf r}({\bf a},t),{\bf v}({\bf a},t),{\bf a} ,t) \right|_{{\bf v}=\dot{\bf r}} \quad,\tag{1}$$ where the Lagrangian density is $${\cal L}({\bf r}({\bf a},t),{\bf v}({\bf a},t),{\bf a} ,t) ~=~ -\frac{\mu({\bf a})}{\gamma({\bf v}({\bf a},t))} -\rho({\bf a})~\phi({\bf r}({\bf a},t),t)+\rho({\bf a})~{\bf v}({\bf a},t)\cdot{\bf A}({\bf r}({\bf a},t),t) ,\tag{2}$$ and where $$\gamma({\bf v})~:=~\frac{1}{\sqrt{1-{\bf v}^2}}\tag{3}$$ is the gamma factor. The rest mass $m(\Omega)$ and the charge $Q(\Omega)$ in a region $\Omega\subseteq \mathbb{R}^3$ of labelling 3-space is given by $$m(\Omega) ~=~ \int_{\Omega} \! d^3a~\mu({\bf a})\quad\text{and}\quad Q(\Omega) ~=~ \int_{\Omega} \! d^3a~\rho({\bf a}),\tag{4}$$ respectively.
• I see, I had considered the Lagrangian fluid view but did not cast the position field $\mathbf{r}$ as a function of the configuration vector $\mathbf{a}$. As such my Lagrangian was nonsense having inconsistent references to each vector. This makes complete sense as the position vector is now itself a field and is clearly the parameter the action should be varied with respect to. Thank you for the concise and clear answer. I would be curious to see the Eulerian form if you have time but you have satisfactorily answered my question already and I will not be offended if you don't. – Daniel Kerr Jun 7 '16 at 21:10
• If I'm not mistaken, if you allow the densities to vary in time as well, then the resulting force equation has an explicit dependence on the vector potential as $-\dot{\rho}\mathbf{A}$, correct? Is there a reason you've fixed the densities in time, is it not correct to allow them to be time dependent as well? – Daniel Kerr Jun 8 '16 at 15:43
• The mass and charge densities, $\mu({\bf a})$ and $\rho({\bf a})$, do not depend on $t$ in the Lagrangian picture. – Qmechanic Jun 8 '16 at 16:28
• Ah right, would have to express them in terms of $\mathbf{r}$ to get their dependence. My bad. – Daniel Kerr Jun 8 '16 at 16:29