So I would like to show
\begin{equation} K^{\mu} = \frac{q}{c}F^{\mu\nu}u_{\nu} = \frac{q}{c}\left[\partial^{\mu}(A^{\nu}u_{\nu}) - \frac{dA^{\mu}}{d \tau}\right] \end{equation}
and I went about as follows:
I can re-write the electric and magnetic fields of
\begin{equation} \textbf{E}(\textbf{r},t) = -\nabla\Phi(\textbf{r},t) -\frac{1}{c}\frac{\partial \textbf{A}(\textbf{r},t)}{\partial t} \end{equation}
\begin{equation} \textbf{B}(\textbf{r},t) = \nabla \times \textbf{A}(\textbf{r},t) \end{equation}
by employing the four-gradient notation to get
\begin{equation} E^i = -(\partial^0A^i - \partial^iA^0) \end{equation}
\begin{equation} B^i = -(\partial^jA^k - \partial^kA^j) \end{equation}
where the indices $(ijk)$ are cyclic permutations of $(123)$. I can find the contravariant electromagnetic field tensor by putting these expressions for the electric and magnetic fields together in matrix form as
\begin{equation} F^{\mu\nu} = \partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu} = \begin{bmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{bmatrix} \end{equation}
If I multiply this tensor with the four-velocity I get
\begin{equation*} \begin{aligned} F^{\mu\nu}u_{\nu} = \gamma \begin{bmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{bmatrix} \begin{bmatrix} c \\ -\dot{x} \\ -\dot{y} \\ -\dot{z} \end{bmatrix} = \gamma \begin{bmatrix} \textbf{E} \cdot \textbf{r} \\ cE_x + \dot{y}B_z - \dot{z}B_y \\ cE_y + \dot{z}B_x - \dot{x}B_z \\ cE_z + \dot{x}B_y - \dot{y}B_x \end{bmatrix} \end{aligned} \end{equation*}
which gives me
\begin{equation} F^{\mu\nu}u_{\nu} = \begin{bmatrix} \textbf{E} \cdot \dot{\textbf{r}} \\ c\textbf{E} + \dot{\textbf{r}} \times \textbf{B} \end{bmatrix} \end{equation}
I notice that the space component of the right-hand side is similar to the Lorentz force (minus some factors)
\begin{equation} \textbf{F}_L = e\textbf{E} + \frac{e}{c}\dot{\textbf{r}} \times \textbf{B} \end{equation}
I can find from this the form-invariant generalization of Newton’s second law for this case as
\begin{equation} \frac{e}{c}F^{\mu\nu}u_{\nu} = \frac{d(mu^{\mu})}{d\tau} \end{equation}
Now my question is this: I've been told the reasoning in the next step where I say that before I write the Minkowski force given by the contracted field tensor I note that:
\begin{equation*} \begin{aligned} (\partial^{\mu}A^{\nu})u_{\nu} = \partial^{\mu}(A^{\nu}u_{\nu}) - A^{\nu}\underbrace{\partial^{\nu}u_{\nu}}_\text{=0} \end{aligned} \end{equation*}
and
\begin{equation*} \begin{aligned} \frac{dA^{\mu}}{d\tau} = (\partial^{\nu}A^{\mu})\frac{dx_{\nu}}{d\tau} + \underbrace{\frac{\partial A^{\mu}}{\partial \tau}}_\text{=0} \end{aligned} \end{equation*}
My reasoning for this is that the respective partial derivatives vanish because $u_{\nu}$ is not an explicit function of the spacial component $x_{\mu}$ and $A^{\mu}$ is not an explicit function of the time component $\tau$. I have been told however that there is an error in my argument here. From this reasoning I went on to say that I can write the Minkowski force as
\begin{equation*} \begin{aligned} K^{mu} = \frac{e}{c}F^{\mu\nu}u_{\nu} = \frac{e}{c}\left[\partial^{\mu}(A^{\nu}u_{\nu}) - \frac{dA^{\mu}}{d\tau}\right] \end{aligned} \end{equation*}
but that argument would not hold if my reasoning is incorrect.
tl;dr: Apparently my reasoning to get from
\begin{equation} \frac{e}{c}F^{\mu\nu}u_{\nu} = \frac{d(mu^{\mu})}{d\tau} \end{equation}
to
\begin{equation*} \begin{aligned} K^{mu} = \frac{e}{c}F^{\mu\nu}u_{\nu} = \frac{e}{c}\left[\partial^{\mu}(A^{\nu}u_{\nu}) - \frac{dA^{\mu}}{d\tau}\right] \end{aligned} \end{equation*}
is wrong, but I'm not seeing the problem. Does anyone on the stackexchange see my problem?
