Where did I make a mistake in simplifications of equations of EM field?

[olgerm]

All tensors here are contravariant.

from maxwell equation in terms of E-field we know that:
$$\rho=\frac{\partial E_1}{\partial x_1}+\frac{\partial E_2}{\partial x_2}+\frac{\partial E_3}{\partial x_3}$$

from maxwell equation in terms of magnetic 4-potential in lorenz gauge we know that
$$-\rho=-\frac{\partial^2 A_0}{\partial x_0^2}+\frac{\partial^2 A_0}{\partial x_1^2}+\frac{\partial^2 A_0}{\partial x_2^2}+\frac{\partial^2 A_0}{\partial x_3^2}$$

we know how electric field is related to (contravariant) EM-tensor:
$$E_1=F_{10}$$
$$E_2=F_{20}$$
$$E_3=F_{30}$$

we know how EM-tensor is related to magnetic 4-potential:
$$F_{10}=\frac{\partial A_0}{\partial x_1}-\frac{\partial A_1}{\partial x_0}$$
$$F_{20}=\frac{\partial A_0}{\partial x_2}-\frac{\partial A_2}{\partial x_0}$$
$$F_{30}=\frac{\partial A_0}{\partial x_3}-\frac{\partial A_3}{\partial x_0}$$
by combining these equations we get:
$$\rho=\frac{\partial E_1}{\partial x_1}+\frac{\partial E_2}{\partial x_2}+\frac{\partial E_3}{\partial x_3}$$
$$\rho=\frac{\partial^2 A_0}{\partial x_0^2}-\frac{\partial^2 A_0}{\partial x_1^2}-\frac{\partial^2 A_0}{\partial x_2^2}-\frac{\partial^2 A_0}{\partial x_3^2}$$
$$E_1=F_{10}$$
$$E_2=F_{20}$$
$$E_3=F_{30}$$
$$F_{10}=\frac{\partial A_1}{\partial x_0}-\frac{\partial A_0}{\partial x_1}$$
$$F_{20}=\frac{\partial A_2}{\partial x_0}-\frac{\partial A_0}{\partial x_2}$$
$$F_{30}=\frac{\partial A_3}{\partial x_0}-\frac{\partial A_0}{\partial x_3}$$

simplifying:
$$\frac{\partial E_1}{\partial x_1}+\frac{\partial E_2}{\partial x_2}+\frac{\partial E_3}{\partial x_3}=\frac{\partial^2 A_0}{\partial x_0^2}-\frac{\partial^2 A_0}{\partial x_1^2}-\frac{\partial^2 A_0}{\partial x_2^2}-\frac{\partial^2 A_0}{\partial x_3^2}$$
$$E_1=\frac{\partial A_1}{\partial x_0}-\frac{\partial A_0}{\partial x_1}$$
$$E_2=\frac{\partial A_2}{\partial x_0}-\frac{\partial A_0}{\partial x_2}$$
$$E_3=\frac{\partial A_3}{\partial x_0}-\frac{\partial A_0}{\partial x_3}$$simplifying:

$$\frac{\partial }{\partial x_1}(\frac{\partial A_1}{\partial x_0}-\frac{\partial A_0}{\partial x_1})+
\frac{\partial}{\partial x_2}(\frac{\partial A_2}{\partial x_0}-\frac{\partial A_0}{\partial x_2})+
\frac{\partial}{\partial x_3}(\frac{\partial A_3}{\partial x_0}-\frac{\partial A_0}{\partial x_3})=
\frac{\partial^2 A_0}{\partial x_0^2}-\frac{\partial^2 A_0}{\partial x_1^2}-\frac{\partial^2 A_0}{\partial x_2^2}-\frac{\partial^2 A_0}{\partial x_3^2}$$

simplifying:

$$\frac{\partial^2 A_1}{\partial x_0*\partial x_1}-\frac{\partial^2 A_0}{\partial x_1^2}+
\frac{\partial^2 A_2}{\partial x_0*\partial x_2}-\frac{\partial^2 A_0}{\partial x_2^2}+
\frac{\partial^2 A_3}{\partial x_0*\partial x_3}-\frac{\partial^2 A_0}{\partial x_3^2}=
\frac{\partial^2 A_0}{\partial x_0^2}-\frac{\partial^2 A_0}{\partial x_1^2}-\frac{\partial^2 A_0}{\partial x_2^2}-\frac{\partial^2 A_0}{\partial x_3^2}$$

simplifying:
$$
-\frac{\partial^2 A_0}{\partial x_0^2}+
\frac{\partial^2 A_1}{\partial x_0*\partial x_1}+
\frac{\partial^2 A_2}{\partial x_0*\partial x_2}+
\frac{\partial^2 A_3}{\partial x_0*\partial x_3}=0
$$

simplifying:
$$\frac{\partial}{\partial x_0}(
-\frac{\partial A_0}{\partial x_0}+
\frac{\partial A_1}{\partial x_1}+
\frac{\partial A_2}{\partial x_2}+
\frac{\partial A_3}{\partial x_3})=0$$

is this result correct?
This seems wrong, but I do not understand what wrong assumptions or derivation mistakes I did.

using lorenz gauge condition
$$\frac{\partial A_1}{\partial x_1}+\frac{\partial A_2}{\partial x_2}+\frac{\partial A_3}{\partial x_3}+\frac{\partial^2 A_0}{\partial x_0}=0$$
we can also derive that
$$\frac{\partial^2 A_0}{\partial x_0^2}=0$$
and
$$\frac{\partial}{\partial x_0}(
\frac{\partial A_1}{\partial x_1}+
\frac{\partial A_2}{\partial x_2}+
\frac{\partial A_3}{\partial x_3})=0$$

and using maxwell equation in terms of magnetic 4-potential in lorenz gauge again also:
$$-\rho=\frac{\partial^2 A_0}{\partial x_1^2}+\frac{\partial^2 A_0}{\partial x_2^2}+\frac{\partial^2 A_0}{\partial x_3^2}$$

 

[vanhees71]

Let's do the derivation. We start with the homogeneous Maxwell equations (working in natural Heaviside-Lorentz units as you obviously do):
$$\vec{\nabla} \times \vec{E} + \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0.$$
From the 2nd equation you know via Helmholtz's theorem that there's a vector potential for $\vec{B}$,
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Using this in the first equation, leads to
$$\vec{\nabla} \times (\vec{E}+\partial_t \vec{A})=0,$$
and again using Helmholtz's theorem this means that there's a scalar potential for the vector field in the brackets:
$$\vec{E} + \partial_t \vec{A}=-\vec{\nabla} \Phi \; \Rightarrow \; \vec{E}=-\partial_t \vec{A}-\vec{\nabla} \Phi.$$
The potentials $\Phi$ and $\vec{A}$ are not uniquely defined but any other choice,
$$\Phi'=\Phi+\partial_t \chi, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi$$
leads to the same fields $\vec{E}$ and $\vec{B}$. This is the celebrated "gauge invariance" of electrodynamics.

To get equations of motion for the potentials, we need the inhomogeneous Maxwell equations,
$$\vec{\nabla} \times \vec{B} - \partial_t \vec{E}=\vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
Plugging in the potentials we get
$$\vec{\nabla} (\vec{\nabla} \cdot \vec{A} + \partial_t \Phi) + \Box \vec{A}=\vec{j}, \quad -\vec{\nabla} \cdot (\partial_t \vec{A}+\vec{\nabla} \Phi)=\rho$$
with $\Box=\partial_t^2-\Delta$).

Now due to the gauge invariance we can impose a constraint on the potentials, which (partially) "fixes the gauge". Very convenient for the general time-dependent case is the Lorenz gauge,
$$\partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0.$$
Then the equations for the components of the vector potential and the scalar potential decouple:
$$\Box \vec{A}=\vec{j}, \quad \Box \Phi=\rho.$$

In the 4D formalism (using the $(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)$ convention) the Maxwell equations read
$$\partial_{\mu} F^{\mu \nu}=j^{\nu}, \quad \partial_{\mu} {^{\dagger}F}^{\mu \nu}=0,$$
where
$${^{\dagger}F}^{\mu \nu}=\epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma}$$
is the Hodge dual of $F_{\mu \nu}$.

From the latter equation you get the existence of the four-potential
$$F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu},$$
and the gauge transformation reads
$$A_{\mu}'=A_{\mu} + \partial_{\mu} \chi.$$
With the potentials the inhomogeneous equations read
$$\partial_{\mu} (\partial^{\mu} A^{\nu}-\partial^{\nu} A^{\mu})=j^{\nu}.$$
Again gauge invariance allows for a gauge-fixing constraint, and the Lorenz gauge turns out to be also a maniffestly covariant constraint, which explains why it simplifies the task so much:
$$\partial_{\mu} A^{\mu}=0.$$
From this you get
$$\partial_{\mu} \partial^{\mu} A^{\nu}=j^{\nu}.$$
In the (1+3) convention this indeed resolves to the same equations as directly derived in this convention. The Lorenz-gauge constraint translates indeed to
$$\partial_{\mu} A^{\mu} = \frac{\partial A^{\mu}}{\partial x^{\mu}} = \partial_t A^0 + \vec{\nabla} \cdot \vec{A}=0$$
and
$$\partial_{\mu} \partial^{\mu}= \eta^{\mu \nu} \partial_{\mu} \partial_{\nu} =\partial_0^2-\vec{\nabla}^2=\Box.$$