Maximum Electric/Magnetic field in vacuum

The photoelectric effect suggests that the energy of electromagnetic field for monochromatic light should be a function of the frequency \(\omega\). Since waves are not generally monochromatic, to generalize this, we look for an expression of \(\omega\) in terms of the electromagnetic field (electric/magnetic) itself: since $$\textbf{E}(z,t)=E_0\sin(kz-\omega t)\hat{\textbf{x}}, \ \ \textbf{B}(z,t)=\frac{1}{c}E_0\cos(kz-\omega t)\hat{\textbf{y}},$$ we find $$\omega=\frac{1}{E_0}\frac{\Arrowvert\frac{\partial\textbf{E}}{\partial t}\Arrowvert}{\sqrt{1-(\frac{E}{E_0})^2}},$$

Angular acceleration of light from Relativity

 $$v^2=v_x^2+v_y^2=c^2$$

So

$$\begin{cases}v_x=c\cos\phi \\ v_y=c\sin\phi\end{cases}$$

Poynting vector as the source of gravity

 $$u=\rho c^2,$$

so $$\nabla\cdot\textbf{g}=-\frac{4\pi G}{c^2}u.$$

Propose

$$\boxed{\nabla\times\textbf{g}=\alpha \textbf{S}}$$

Using

$$\frac{\partial u}{\partial t}=-\nabla\cdot\textbf{S}$$

we get a new source for \(\textbf{S}\):

$$\boxed{\textbf{S}=\frac{c^2}{4\pi G}\frac{\partial \textbf{g}}{\partial t}}$$