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

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The discussion focuses on the simplifications of equations related to the electromagnetic (EM) field and the derivation of the charge density from Maxwell's equations. It highlights the relationship between the electric field, the electromagnetic tensor, and the magnetic four-potential in the Lorenz gauge. The participant expresses confusion about potential mistakes in their derivations, particularly regarding the gauge invariance and the implications of the Lorenz gauge condition. The derivations lead to the conclusion that the second derivative of the scalar potential is zero, raising questions about the correctness of the results. The conversation ultimately emphasizes the importance of understanding gauge conditions and their effects on the equations governing electromagnetic fields.
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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:
$$-\rho=\frac{\partial^2 A_0}{\partial x_1^2}+\frac{\partial^2 A_0}{\partial x_2^2}+\frac{\partial^2 A_0}{\partial x_3^2}$$
 
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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
$$\partial_{\mu} \partial^{\mu}= \eta^{\mu \nu} \partial_{\mu} \partial_{\nu} =\partial_0^2-\vec{\nabla}^2=\Box.$$
 
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