Generalizing the Lorentz force law in 4-vector notation In Goldstein's Classical Mechanics chapter on special relativity page 298 the Lorentz force law:
$$\textbf{f}=q\left\{-\nabla\phi + \frac{1}{c}\frac{\partial \textbf{A}}{\partial t}+\textbf{v}\times(\nabla\times\textbf{A})\right\} \tag{1}$$
Is generalised in 4 vector notation as
$$\frac{dP_{\mu}}{d\tau}=q\left[\frac{\partial(u^{\nu}A_{\nu})}{\partial x^{\mu}}-\frac{dA_{\mu}}{d\tau}\right] \tag{2}$$
Im trying to to replicate this by going from $(2)$ to $(1)$ but im having trouble managing to do it here is what I've done so far.
Firstly to define things
$$A_{\mu}=(\frac{\phi}{c},-\textbf{A})$$
$d\tau$ is the proper time given as
$$dt=\frac{d\tau}{\sqrt{a-\frac{v^2}{c^2}}}=\gamma d\tau$$
hence the 4-velocity $u^{\nu}$ is given as:
$$u^0=\frac{dct}{d\tau}=\gamma c, \ \ u^i = \frac{dx^i}{d\tau}=\gamma v^i$$
and finally
$$\frac{\partial}{\partial x^{\mu}}=\frac{\partial}{\partial ct}+\nabla$$
Where, $\nabla$, is the 3 dimensional del operator. So starting from the first term in $(2)$ and expanding $u^{\nu}A_{\nu}$ I get
$$u^{\nu}A_{\nu}=u^0\frac{\phi}{c} - u^iA_{i}=\gamma \phi- \gamma v^iA_{i}$$
Applying $\frac{\partial}{\partial x^{\mu}}$, I get,
$$ \begin{align} \frac{\partial(u^{\nu}A_{\nu})}{\partial x^{\mu}} &= \frac{\partial }{\partial ct}(\gamma \phi- \gamma v^iA_{i})+\nabla(\gamma \phi- \gamma v^iA_{i}) 
\\ &=\frac{\partial }{\partial ct}(\gamma \phi) - \frac{\partial }{\partial ct} (\gamma v^iA_{i}) +  \gamma\nabla \phi-\gamma\nabla(v^iA_{i})
\end{align} \tag{3}$$
and for $\frac{dA_{\mu}}{d\tau}$ I get
$$\frac{dA_{\mu}}{d\tau}=\gamma (\mathbf{v} \cdot \nabla)A_{\mu} +\gamma \frac{\partial A_{\mu}}{\partial t} \tag{4}$$
with $(3)$ I know we can write $\nabla(v^iA_{i}) = \nabla(\textbf{v}\cdot\textbf{A}) = \textbf{v}\times (\nabla \times \textbf{A}) +(\textbf{v}\cdot \nabla)\textbf{A}$ since $\textbf{v}$ is only time dependent.
putting $(3)$ and $(4)$ into $(2)$ and using the vector identity, I got
$$\frac{\partial(u^{\nu}A_{\nu})}{\partial x^{\mu}} - \frac{dA_{\mu}}{d\tau} = \frac{\partial }{\partial ct}(\gamma \phi) - \frac{\partial }{\partial ct} (\gamma v^iA_{i}) +  \gamma\nabla \phi-\gamma(\textbf{v} \times(\nabla \times \textbf{A})) -\gamma (\textbf{v} \cdot \nabla)\frac{\phi}{c} -\gamma \frac{1}{c}\frac{\partial \phi}{\partial t}  +\frac{\partial \textbf{A}}{\partial t}$$
Where I've broken up $A_{\mu}$ in the last two terms.
However, I get a bit stuck from this point on, in the first two terms does the partial time derivative skip the $\gamma's$ and the velocity terms?  I'm also unsure what I do with the $\gamma (\textbf{v} \cdot \nabla)\frac{\phi}{c}$ term?
 A: You've got several issues here, the way I see it.

*

*Goldstein's derivation is not very standard today; it's somewhat outdated. The expression you will find in most books on electrodynamics and relativity is:
$$
\frac{dP^{\mu}}{d\tau}=qu^{\nu}\left.F_{\nu}\right.^{\mu}
$$
where $ F_{\nu\mu}:=\frac{\partial A_{\mu}}{\partial x^{\nu}}-\frac{\partial A_{\nu}}{\partial x^{\mu}} $. So you will find it difficult to compare with other bibliographic sources.

*Your Eq.(3) is inconsistent. You've got a 4-vector on the LHS and a scalar on the RHS. You would have to distinguish the 0-term, and then the 3-vector term, which is the one that gives you the Lorentz-force law:
$$
\frac{dP_{k}}{d\tau}=q\left(\frac{\partial(u^{\nu}A_{\nu})}{\partial x^{k}}-\frac{dA_{k}}{d\tau}\right)
$$
The expression I get from your prescription Eq.(2) is,
$$
\frac{d}{d\tau}P^{k}=-q\gamma\frac{\partial\phi}{\partial x^{k}}+q\gamma\left(\boldsymbol{v}\times\boldsymbol{\nabla}\times\boldsymbol{A}\right)^{k}+q\gamma\left(\boldsymbol{v}\cdot\boldsymbol{\nabla}\right)A^{k}-q\gamma\left(\boldsymbol{v}\cdot\boldsymbol{\nabla}+\frac{\partial}{\partial t}\right)A^{k}
$$
Where I have substituted what you seem to have (correctly) implied. Namely: that the total derivative and the partial derivative are related by,
$$
\frac{d}{dt}=\boldsymbol{v}\cdot\boldsymbol{\nabla}+\frac{\partial}{\partial t}
$$
although you seem to be in doubt about whether this is correct: It is. This is sometimes called a "material derivative", and it's a consequence of the chain rule in calculus. When you have,
$$
f\left(t,x\left(t\right)\right)
$$
the correct way to obtain how much $f$ changes with $t$ requires you to add a term depending on the drift:
$$
\frac{d}{dt}f=\frac{\partial}{\partial t}f+\dot{x}\cdot\nabla f
$$
Your vector identity is also correct. As it's the fact that the $\gamma$'s are unaffected by the spatial differential operators.
I think you won't have problems filling in the details. You drop the $\gamma$'s on both sides and cancel the $\boldsymbol{v}\cdot\boldsymbol{\nabla}$'s, and you're there.

I hope that helped. If something wasn't clear, feel free to tell me.
