# Lorentz force equation from relativistic Lagrangian

The relativistic Lagrangian is given by $$L = - m_0 c^2 \sqrt{1 - \frac{u^2}{c^2}} + \frac{q}{c} (\vec u \cdot \vec A) - q \Phi$$ I need to derive, $\displaystyle \frac{d\vec p}{dt} = q \left( \vec E + \frac 1 c (\vec u \times \vec B)\right)$. Everything (in my note) is fine until there is one expression which I don't understand. It writes, $$\frac 1 c \nabla (\vec u \cdot \vec A ) = \nabla \Phi$$ to make $\displaystyle \vec E = - \nabla \Phi - \frac 1 c \frac{\partial \vec A}{\partial t}$. Why would dot product of $\vec u$ and $\vec A$ be $c$ times the scalar potential?

Here is the derivation as requested:

The equation of motion is given by $$\frac{d}{dt} \left( \frac{\partial L}{\partial u_i} \right ) - \frac{\partial L}{\partial x_i} = 0$$ Since the relativistic Lagrangian is transitionally invariant, there is no dependence on coordinates so the last term is zero. taking derivative w.r.t $u_i$ $$\frac{d}{dt} \left( \frac{m_0 u_i}{\sqrt{1 - \frac{u^2}{c^2}}} + \frac{q}{c} A_i \right ) = 0$$ Adding $i$'s, we get $$\frac{d}{dt} \left( \gamma m_0 \vec u + \frac{q}{c} \vec A \right ) = 0$$ $$\implies \frac{d \vec p}{dt} + \frac{q}{c} \frac{d\vec A }{dt} = 0$$ \begin{align*} \implies \frac{d \vec p}{dt} &= - \frac{q}{c} \left( \frac{\partial \vec A }{\partial t} + (\vec u \cdot \nabla)\vec A\right )\\ &= - \frac{q}{c} \left( \frac{\partial \vec A }{\partial t} + \nabla (\color{Red}{\vec u \cdot \vec A})- \vec u \times \left( \nabla \times \vec A \right )\right )\\ &= - \frac{q}{c} \left( \frac{\partial \vec A }{\partial t}+ \nabla \color{Red}{(c\Phi)} - \vec u \times \vec B \right )\\ &= q \left( -\nabla \Phi - \frac 1 c \frac{\partial \vec A }{\partial t} + \frac 1 c \vec (u \times \vec B) \right)\\ &= q\left ( \vec E + \frac 1 c \vec (u \times \vec B) \right ) \end{align*}

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Could you explain the derivation a bit more fully? It might help if I knew the preceding argument in the notes that led up to the equation it gives. –  gj255 Jan 13 at 13:57
@gj255 the preceding arguments has been added. –  Mula Ko Saag Jan 13 at 14:33

This is not correct: "Since the relativistic Lagrangian is transitionally invariant, there is no dependence on coordinates so the last term is zero." In fact, $$\frac{\partial L}{\partial {\bf r}} = q\nabla\left[ \frac{1}{c} \left({\bf u} \cdot {\bf A}\right) - \phi \right] = q\left\{\frac{1}{c} \left[\left({\bf u} \cdot \nabla \right){\bf A} + {\bf u} \times \left(\nabla \times {\bf A}\right) \right]- \nabla\phi\right\}.$$ Setting this equal to $$\frac{d}{dt}\left(\frac{\partial L}{\partial {\bf u}} \right)= \frac{d}{dt} \left[ \frac{\partial }{\partial {\bf u}}\left( -m c^2 \sqrt{1 - \frac{u^2}{c^2}} + \frac{q}{c} {\bf u} \cdot {\bf A}\right) \right]= \frac{d}{dt}\left(\gamma m {\bf u}\right) +\frac{q}{c} \frac{d{\bf A}}{dt},$$ and noting that $$\frac{d{\bf A}}{dt} = \frac{\partial {\bf A}}{\partial t} + \left({\bf u} \cdot \nabla\right) {\bf A},$$ we get $$\frac{d}{dt}\left(\gamma m {\bf u}\right) = q\left[- \nabla\phi - \frac{1}{c}\frac{\partial{\bf A}}{\partial t} + \frac{{\bf u}}{c} \times \left(\nabla \times {\bf A}\right) \right].$$
how do you evaluate that $$\left({\bf u} \bullet \nabla\right) {\bf A}$$ –  Mula Ko Saag Jan 13 at 15:44
@MulaKoSaag: $\left(\mathbf{u}\cdot\nabla\right)\mathbf{A} = \left(u_x\partial_x+u_y\partial_y+u_z\partial_z\right)\mathbf{A}$ where each derivative term acts on each vector component of $\mathbf{A}$ (normal scalar-times-a-vector thing). –  Kyle Kanos Jan 13 at 17:16
@MulaKoSaag: Your question "how do you evaluate that" has nothing to do any scalar potential. Eric Angle seems to answer your second question here (identical to your original posted answer at the top of the page) by saying, "Your assumption that $\partial L/\partial{x_i}=0$ is wrong" (please re-read his very first sentence). –  Kyle Kanos Jan 13 at 18:44