A Point Charge in the Electromagnetic Field

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In this article we want to derive a Lagrange function for a charged particle in an electromagnetic field. Starting from the Lorenz force, the empirical law for the force acting on a charged particle in an electric and magnetic field, we rewrite the Lorentz force in terms of the vector potential $\mathbf{A}$ to identify the Lagrangian of the system. The Lorentz force is given by the equation. \begin{align} \mathbf{F}=e \left(-\nabla V + \frac{\mathbf{v}}{c} \times \mathbf{B} \right) \end{align} The first term results from the electric field \begin{align} \mathbf{F}_E=e (-\nabla V)= e \cdot \mathbf{E} \end{align} the second from the magnetic field \begin{align} \mathbf{F}_M=e \left(\frac{\mathbf{v}}{c} \times \mathbf{B} \right) \end{align} From Maxwell's Equations we know that \begin{align} \tag{1} \nabla\cdot\mathbf{B}&= 0\\ \tag{2} \nabla\times\mathbf{E}&=-\frac{1}{c}\frac{\partial\mathbf{B}}{\partial t} \end{align} Equation (1) states, that exist no magnetic monopoles. It is convenient to introduce a vector potential \begin{align} \tag{3} \mathbf{B}=\nabla\times\mathbf{A} \end{align} since the divergence of a curl is always zero, equation (1) is automatically satisfied. \begin{align} \tag{4} \nabla \cdot \mathbf{B}=\nabla \cdot (\nabla\times\mathbf{A})=0 \end{align} If we plug in (3) into the Maxwell equation (2) we get

\begin{align} \nabla\times\mathbf{E}+\frac{1}{c}\frac{\partial (\nabla\times\mathbf{A})}{\partial t}=0 \\ \nabla \times \left( \mathbf{E}+\frac{1}{c}\frac{\partial \mathbf{A}}{\partial t} \right) =0\\ \end{align} Because it is always true that the curl of a gradient is zero, the term in the brackets can be written as the gradient of a potential $\varphi$ \begin{align} \mathbf{E}+\frac{1}{c}\frac{\partial \mathbf{A}}{\partial t}=- \nabla \varphi \\ \end{align} and we can write the electric field in terms of the vector potential \begin{align} \mathbf{E}=-\frac{1}{c}\frac{\partial \mathbf{A}}{\partial t}- \frac{1}{c}\nabla \varphi \\ \end{align} and the Lorentz force in terms of the vector potential \begin{align} \mathbf{F}=e \left[ - \frac{1}{c}\nabla \varphi - \frac{1}{c}\left(\frac{\partial \mathbf{A}}{\partial t} - \mathbf{v} \times (\nabla\times\mathbf{A}) \right) \right] \end{align} If we apply the "bac-cab" rule from vector calculus $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})=\mathbf{b} \cdot (\mathbf{a} \cdot \mathbf{c})- \mathbf{c} \cdot (\mathbf{a} \cdot \mathbf{b} ) $ we find \begin{align} \mathbf{v} \times (\nabla\times\mathbf{A})= \nabla (\mathbf{v} \cdot \mathbf{A}) - (\mathbf{v} \cdot \nabla ) \mathbf{A}) \end{align} This seems to make the force look worse than before, but if we write down the total time derivative of $\mathbf{A}(\mathbf{x},t)$ \begin{align} \frac{d\mathbf{A}}{dt}=\frac{\partial \mathbf{A}}{\partial t} + \sum_i \frac{\partial \mathbf{A}}{\partial x_i} \frac{\partial x_i}{\partial t}=\frac{\partial \mathbf{A}}{\partial t} + (\mathbf{v} \cdot \nabla ) \mathbf{A} \end{align} we see that if we plug into the equation for $\mathbf{F}$, two term cancel \begin{align} \mathbf{F}=\frac{-e}{c} \left[ \nabla c\varphi + \left(\frac{d \mathbf{A}}{d t}+(\mathbf{v} \cdot \nabla ) \mathbf{A} - \nabla (\mathbf{v} \cdot \mathbf{A}) - (\mathbf{v} \cdot \nabla ) \mathbf{A} \right) \right] \end{align} so we finally have \begin{align} \mathbf{F}=\frac{-e}{c} \left[ \frac{d \mathbf{A}}{d t}+ \nabla \left( c\varphi -(\mathbf{v} \cdot \mathbf{A}) \right) \right] \tag{5} \end{align} If just an electric field is present, the Lagrangian for a charged particle can be easily written down with $L=T-V$ \begin{align} L(x,\dot{x})=\frac{1}{2}m \dot{x}^2 - e \cdot \varphi \end{align} But if we have an additional potential arising from the magnetic field, how does the magnetic field contribute to the potential? Equation (5) has an additional potential term in the gradient brackets. This suggests the candidate $\mathbf{v} \cdot \mathbf{A}$ as magnetic contribution. We try the Lagrangian \begin{align} L(x,\dot{x})=\frac{1}{2}m \dot{x}^2 - e \cdot (\varphi+\frac{1}{c}\mathbf{v} \cdot \mathbf{A}) \end{align} and see if it really leads to the right force if we plug into the Euler-Lagrange Equation \begin{align} \frac{d}{dt}\nabla_{\dot{\mathbf{x}}} \, L(\mathbf{x},\dot{\mathbf{x}})=m \ddot{\mathbf{x}}=\mathbf{F}=\nabla_{\mathbf{x}} \, L(\mathbf{x},\dot{\mathbf{x}})=\frac{e}{c} \cdot \nabla (\varphi+\mathbf{v} \cdot \mathbf{A}) \end{align} One term $\frac{1}{c} \frac{d \mathbf{A}}{d t}$ is missing, but we can show that this term results from the non-uniqueness of the gauge and does not affect the equations of motion. If we were to choose different potentials \begin{align} \mathbf{A}'(\mathbf{x},t)=\mathbf{A}(\mathbf{x},t)+ \nabla \Lambda(\mathbf{x},t) \end{align} \begin{align} \varphi ' =\varphi - \frac{1}{c} \dot{\Lambda}(\mathbf{x},t) \end{align} we would get from the new Lagrangian \begin{align} \mathbf{F}&=e \cdot \nabla (\varphi'+\frac{1}{c}\mathbf{v} \cdot \mathbf{A}')\\ &=e \cdot \nabla \left( \varphi - \dot{\Lambda}(\mathbf{x},t)+\frac{1}{c}\mathbf{v} \cdot \left( \mathbf{A}(\mathbf{x},t)+ \nabla \Lambda(\mathbf{x},t) \right) \right) \end{align} with $- \frac{1}{c} \dot{\Lambda}(\mathbf{x},t)+ \nabla \Lambda(\mathbf{x},t)$ being the total time derivative of $\Lambda$. We now have for Lorentz force \begin{align} \mathbf{F}=e \left[ - \frac{1}{c} \frac{d \Lambda }{d t}- \nabla \left( \varphi -\frac{1}{c}(\mathbf{v} \cdot \mathbf{A}) \right) \right] \end{align} We could also argue with the general statement of the non-uniqueness of a Lagrangian, only defined up to multiplication of a constant and addition of total time derivatives.

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