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I'm studying Lagrangian Mechanics from Goldstein's Classical Mechanics. My question concerns Section 1.5 which talks about velocity-dependent potentials. I am actually unsure about how Equation 1-64' was derived: $$F_x = g \left[ - \frac{\partial}{\partial x} \left( \phi - v \cdot A \right) - \frac{d}{dt} \left( \frac{\partial A \cdot v}{\partial v_x} \right) \right]$$ Using equation 1-64: $$F = g \left[ - \nabla \phi - \frac{\partial A}{\partial t} + \left( v \times \left( \nabla \times A \right) \right) \right]$$

If the x-component of 1-64 is to be the same as 1-64', some algebra tells us that the following must be true:

$$\frac{dA_x}{dt}=\frac{d}{dt} \left( \frac{\partial A \cdot v}{\partial v_x} \right)$$

Why is this so? Does this have something to do with the definition of $A$, because I haven't done my Electrodynamics course yet. I'd be grateful for any help.

geofisue
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Ambica Govind
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1 Answers1

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This is because the magnetic gauge potential ${\bf A}({\bf r},t)$ does not depend on the velocity ${\bf v}$. Note that the position ${\bf r}$ and the velocity ${\bf v}$ are here independent variables, cf. e.g. this Phys.SE post.

Qmechanic
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