Maxwell's equation in cylindrical coordinate system I am trying to express the Maxwell equation $$ \partial_a F^{ab} = \mu_0 J^b $$ in cylindrical coordinates. Now, if I express this equation for b=0, the a's would get summed up as $$ \partial_0 F^{00} + \partial_1 F^{10} + \partial_2 F^{20} + \partial_3 F^{30} = \mu_0 J^0 $$
Now in cylindrical coordinates 0,1,2,3 corresponds to t, r, $\phi$, z respectively. The first term would be zero since $F^{00}=0$.  
My doubt is whether I can write the other terms simply as partial derivatives w.r.t. r, $\phi$, z respectively as $$\frac{\partial F^{10}}{\partial r} + \frac{\partial F^{20}}{\partial \phi} + \frac{\partial F^{30}}{\partial z}=\mu_0 J^0$$
Usually standard texts only give the information for gradiant, divergence and curl in cylindrical coordinates and not to such partial derivatives.
 A: As you mention
$$
\partial_\alpha F^{\alpha0} = \partial_0 \underbrace{F^{00}}_{=0} + \partial_iF^{i0} = \mu_0 J^0 \tag{1}
$$
However, the latin index $i$ refers to cartesian coordinates $\{x^1,x^2,x^3\}=\{x,y,z\}$. You can transit to cylindrical coordinates by realizing that
\begin{eqnarray}
x^1 &=& x = r \cos\phi \\
x^2 &=& y = r \sin\phi \\
x^3 &=& z \tag{2}
\end{eqnarray}
So that 
$$
\partial_i f =\frac{\partial r}{\partial x^i} \frac{\partial f}{\partial r} + \frac{\partial \phi}{\partial x^i} \frac{\partial f}{\partial \phi} + \frac{\partial z}{\partial x^i} \frac{\partial f}{\partial z} \tag{3}
$$
where the partial derivatives can be obtained from (2):
\begin{eqnarray}
r &=& \sqrt{(x^1)^2 + (x^2)^2} \\
\phi &=& \arctan\left(\frac{x^2}{x^1}\right) \\
z &=& z \tag{4}
\end{eqnarray}
For example,
$$
\frac{\partial r}{\partial x^1} = \frac{x^1}{\sqrt{(x^1)^2 + (x^2)^2}} = \frac{r\cos\phi}{r} = \cos\phi
$$
A: Another way to do it, and a way that works in general coordinates, is to express Maxwell's equations in a covariant form. Basically you can just replace the usual derivative with a covariant derivative. Then you just need to calculate the connection coefficients. This is more effective when using general coordinates because inverting the equations (2) can be difficult and the derivatives messy.
