From Maxwell,
\(\nabla.E=\cfrac{\rho}{\varepsilon_o}\)
\(\nabla.B=0\)
\(\nabla\times E+\cfrac{\partial B}{\partial t}=0\)
\(\nabla\times B-\cfrac{1}{c^2}\cfrac{\partial E}{\partial t}=\cfrac{1}{c^2}\cfrac{J}{\varepsilon_o}\)
In an analogous way, if we are able to oscillate negative gravity particles in an equivalent conductor, we will generate gravito-electric waves.
\(\nabla.G_W=\cfrac{\rho_g}{\varepsilon_{go}}\)
where \(G_W\) gravitational field, replaces \(E\) the electric field. \(\rho_g\) is the total negative gravity particle enclosed (expressed as mass density). And \(\varepsilon_{go}\) is equivalent to \(\varepsilon_{o}\) in free space.
\(\nabla.E=0\)
An \(E\) field due to the negative gravity particle spin replaces the \(B\) field due to electron spin.
\(\nabla\times G_W+\cfrac{\partial E}{\partial t}=0\)
\(\nabla\times E-\cfrac{1}{c^2}\cfrac{\partial G_W}{\partial t}=\cfrac{1}{c^2}\cfrac{J_g}{\varepsilon_{go}}\)
where \(J_g\) is the negative gravity particle flow density.
If we compare with Newton's expression for gravity, per unit mass,
\(F_{/m}=G\cfrac{m}{r^2}=4\pi G\cfrac{m}{4\pi r^2} \)
keeping in mind,
\(E=\cfrac{1}{\varepsilon_{o}}\cfrac{q}{4\pi r^2}\)
We can let,
\(\rho_g=\rho_m=\cfrac{m}{Volume\,\,enclosed}\)
\(4\pi G=\cfrac{1}{\varepsilon_{go}}\)
This is using the unmodified gravitational constant \(G\) for gravity.
Such gravito-electric waves can be detected by their varying electric component, just as EMW can also be detected by their varying magnetic field.
Have a nice day.
