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*}

 A: 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
$$
\frac{d}{dt}\left(\gamma m {\bf u}\right) = q\left[- \nabla\phi - \frac{1}{c}\frac{\partial{\bf A}}{\partial t} + \frac{{\bf u}}{c} \times \left(\nabla \times {\bf A}\right) \right].
$$
A: 
Minkowski metric: $\text{diag}(+1,~+1,~+1,~-1)$

How about deriving from Lagrangian using 4-vector?
The Lagrangian is
\begin{eqnarray}
  L = -\frac{1}{\gamma} mc^2 + Q \dot{x}_{\nu} A^\nu,
  \qquad\left( \gamma = \frac{1}{\sqrt{1-(v/c)^2}} = \frac{c}{\sqrt{-\dot{x}_{\mu}\dot{x}^{\mu}}}\right).
\end{eqnarray}
Taking derivative w.r.t $x_\mu$ and $\dot{x}_\mu$
\begin{eqnarray}
  \frac{\partial L}{\partial x_\mu} &=& Q \dot{x}_{\nu} \partial_\mu A^\nu, \\
  \frac{\partial L}{\partial \dot{x}_\mu} &=& 
    -mc^2 \frac{\partial}{\partial \dot{x}_\mu} \frac{1}{\gamma}
    + Q \frac{\partial \dot{x}_\nu}{\partial \dot{x}_\mu} A^\nu \\
  &=& -mc^2 \left( -\gamma \frac{\dot{x}^\mu}{c^2}\right)
    + Q \delta_{.\nu}^{\mu} A^\nu \\
  &=& p^\mu + Q A^\mu, \qquad (p^\mu = \gamma m \dot{x}^\mu:~\text{4-momentum}).
\end{eqnarray}
So the Eular-Lagrange equation is
\begin{eqnarray}
  \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{x}_\mu} - \frac{\partial L}{\partial x_\mu}
  &=& \dot{p}^\mu + Q \dot{A}^\mu - Q \dot{x}_\nu \partial^\mu A^\nu \\
  &=& \dot{p}^\mu + Q \dot{x}_\nu \partial^\nu A^\mu - Q \dot{x}_\nu \partial^\mu A^\nu \\
  &=& \dot{p}^\mu + Q \dot{x}_\nu( \partial^\nu A^\mu - \partial^\mu A^\nu) \\
  &=& \dot{p}^\mu + Q \dot{x}_\nu B^{\nu\mu} = 0, \qquad (B^{\nu\mu}:~\text{electromagnetic tensor}), \\
  \therefore \dot{p}^\mu &=& Q \dot{x}_\nu B^{\mu\nu}.
\end{eqnarray}
The space components of $(\dot{p}^\mu)$ will be
\begin{eqnarray}
  (\dot{p}^i) &=& \frac{\mathrm{d}\gamma\boldsymbol{p}}{\mathrm{d}t}
  = Q \left( \boldsymbol{E} + \boldsymbol{v} \times \boldsymbol{B} \right). \\
  \because (\dot{p}^\mu) &=& Q (B^{\mu\nu}) (\dot{x}_\nu) \\
  &=&
  \begin{bmatrix}
    0 & B^z & -B^y & -E^x/c \\
    -B^z & 0 & B^x & -E^y/c \\
    B^y & -B^x & 0 & -E^z/c \\
    E^x/c & E^y/c & E^z/c & 0 \\
  \end{bmatrix}
  \begin{bmatrix} v^x \\ v^y \\ v^z \\ -c \end{bmatrix} \\
  &=& \begin{bmatrix}
    v^y B^z - v^z B^y + E^x \\
    v^z B^x - v^x B^z + E^y \\
    v^x B^y - v^y B^x + E^z \\
    \boldsymbol{E} \cdot \boldsymbol{v} / c
  \end{bmatrix}.
\end{eqnarray}
