# Lorentz force in terms of potential

$$\boldsymbol{F}=q\left[- \boldsymbol{\nabla}(\phi-\boldsymbol{v} \cdot \boldsymbol{A})-\frac{d\boldsymbol{A}}{dt}\right]$$

which can take the convenient Euler–Lagrange form $$\boldsymbol{F}=q \left[- \boldsymbol{\nabla}_x(\phi-\boldsymbol{\dot{x}}\cdot \boldsymbol{A})+\frac{d}{dt}\boldsymbol{\nabla}_\dot{x}(\phi-\boldsymbol{\dot{x}}\cdot\boldsymbol{A})\right]$$

How did the author get from the former to the latter and what is the rationale? It seems like the author just find a convenient way to get to the end result by simply replacing the terms.

• All are in the link you give. What exactly you don't understand or you don't know ? Commented Jan 24, 2017 at 15:55
• More on the velocity-dependent potential for the Lorentz force: physics.stackexchange.com/q/77325/2451 and links therein. Commented Jan 24, 2017 at 18:00

Since we know that $$\nabla_{\dot{x}}(\phi) = 0$$
$$\nabla_{\mathbf{\dot{x}}} \left(\mathbf{\dot{x} \cdot A}\right) = \mathbf{A}$$
$$\frac{d}{dt}\nabla_{\mathbf{\dot{x}}}\left(\phi - \mathbf{\dot{x} \cdot A}\right) = \frac{d}{dt}\mathbf{A}$$
• I get this. Just that the author replacing $\nabla(\phi-v.A)$ & A with $\nabla_x(\phi-\dot{x}.A)$ & $\nabla_{\dot{x}}(\phi-\dot{x}.A)$ respectively feels like he is doing it to make the equation more convenient. Commented Jan 25, 2017 at 15:00