# What it means for the wave equation of light in vacuum to be a solution of the maxwell equations? [closed]

I know how to get the wave equation of light from Maxwell's equation, but I never understood why it is called the solution of the Maxwell equations. If say we have a positive charge standing still from our frame of reference, it generates an electric field that is solution of the Maxwell equations (Gauss law) but it does not propagate, it remains "fixed". I don't understand the difference between this electric field and the ones that do propagate freely through space, and how can it get "de-attached" from its source?

• If you plug the wave formula for $\vec{E}(\vec{x}, t)$ and $\vec{H}(\vec{x},t)$ in to the left hand sides of Maxwell's equations, you get the expected right hand sides. Jan 1, 2020 at 23:47
• The inhomogeneous wave equation is not a solution of Maxwells equations, it is a consequence of it. Jan 1, 2020 at 23:51
• Maxwell equations have plenty of solutions, the solutions to the wave equation is only one subset of of them. The wave equation can be derived from the ME's but the reverse is not true, so it does not replace the 4 Maxwell equations: you cannot recover the 4 Maxwell equations just from the wave equation.
– user65081
Jan 1, 2020 at 23:59
• I mean that your title does not make sense. An equation is not a solution of an equation. The wave equation is not a solution of Maxwells equation. Can you reformulate it ? What are the Maxwell equations "without vacumm"? Do you have a spelling checker ? Jan 2, 2020 at 0:29
• @my2cts Right. By wave equation I meant the Electric field satisfying this equation, namely E=E_0 cos(wt-kx), and by maxwell equation without vacumm the ones where neither the charge density or density current are 0. I tought all the solutions must have the form of a wave, but now I see they are only a subset of them Jan 2, 2020 at 0:43