In Maxwell equations, why time derivatives only appear together with Curl? In the four maxwell's equations, the time dependence only appear in curl of $E$ and $B$ but not divergence.
My question was that:


*

*Why time dependence only appear in curl?

*what's the implication?
(I was thinking there was actually a fifth equation, continuum equation, which appear in divergence. Thus divergence was dependence in space while divergence was the symmetric relation in time.)
 A: That's a great question that is ultimately tied to Lorentz symmetry. Maxwell's equations on tensor form read
$$ \sum_{\mu=0}^3\partial_{\mu} F^{\mu\nu}~\propto~ j^{\nu}_e, \qquad \sum_{\mu=0}^3\partial_{\mu} \widetilde{F}^{\mu\nu}~\propto~ j^{\nu}_m, \qquad \nu~\in~\{0,1,2,3\},$$ 
where we could not resist to introduce magnetic monopoles to make E&M duality manifest.
There are two cases:


*

*Index $\nu=0$ is temporal: There are no time-derivatives $\partial_{\mu=0}$ since the $F$-tensor is anti-symmetric.

*Index $\nu=j\in~\{1,2,3\}$ is spatial: Then there are two types of terms
$$\partial_{0} F^{0j} + \sum_{i=1}^3\partial_i F^{ij}~\propto~ j^{j}_e. $$
The first term has a time-derivative. In the second term
the spatial-spatial components of the $F$-tensor are associate with an Levi-Civita epsilon-tensor, so that the spatial derivative becomes a curl. This is the answer to OP's question.
A: Faraday's law $$\nabla\times\mathbf{E}=-\partial_t\mathbf{B}$$ is the differential form of the flux rule explored by Faraday: $\mathcal{E}=\dot{\Phi}$. Where $\mathcal{E}$ is the electromotive force $\mathcal{E}=\oint\mathbf{E}\cdot d\mathbf{l}$ and $\Phi$ is the magnetic flux $\Phi=\iint\mathbf{B}\cdot d\mathbf{A}$. The divergence equation (Gauss's law) is essentially a statement of the law of conservation of energy and can be derived from coulomb's law.
Ampere's law$$\nabla\times\mathbf{B}=\mu_0\mathbf{J}$$ historically did not have a time-derivative. It is derived from the Biot-Savart law but as such does not satisfy the vector identity $\nabla\cdot\nabla\times\mathbf{v}=0$ but rather $\nabla\cdot\nabla\times\mathbf{B}=\mu_0\nabla\cdot\mathbf{J}$. This is resolved by plugging in Gauss's law into the continuity equation:
$$-\partial_t\rho=\nabla\cdot\mathbf{J}$$
$$-\epsilon_0\nabla\cdot\partial_t\mathbf{E}=\nabla\cdot\mathbf{J}$$
Thus the additive inverse is combined with $\mathbf{J}$, giving $$\nabla\times\mathbf{B}=\mu_0\mathbf{J}+\mu_0\epsilon_0\partial_t\mathbf{E}$$
Now, by taking the second curl of those equations (and assuming vacuum conditions), one can show that the electromagnetic field satisfies the wave equation with velocity $\frac{1}{\sqrt{\mu_0\epsilon_0}}=c$, a starting point for Einstein's relativity.
