Pengwuino said:
Gauss' Law is actually a general mathematical tool - it is not unique to either gravity or electromagnetism. Also, magnetic fields are not relativistic effects. They can clearly be seen at non-relativistic energies.
Your first equation actually does exist. It is Newton's Law of Gravity! You typically see it as \nabla^2 \phi = 4\pi G \rho. I don't think there is any reason to believe there is an analogue to the magnetic field in Newtonian gravity, however.
Why not?
The classical electromagnetic effect is perfectly consistent with the lone electrostatic effect but with special relativity taken into consideration.
The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of \lambda \ and some non-zero mass per unit length of \rho \ separated by some distance R \. If the lineal mass density is small enough that gravitational forces can be neglected in comparison to the electrostatic forces, the static non-relativistic repulsive (outward) acceleration (at the instance of time that the lines of charge are separated by distance R \) for each infinite parallel line of charge would be:
a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho}
If the lines of charge are moving together past the observer at some velocity, v \, the non-relativistic electrostatic force would appear to be unchanged and that
would be the acceleration an observer traveling along with the lines of charge would observe.
Now, if special relativity is considered, the in-motion observer's clock would be ticking at a relative
rate (ticks per unit time or 1/time) of \sqrt{1 - v^2/c^2} from the point-of-view of the stationary observer because of time dilation. Since acceleration is proportional to (1/time)
2, the at-rest observer would observe an acceleration scaled by the square of that rate, or by {1 - v^2/c^2} \, compared to what the moving observer sees. Then the observed outward acceleration of the two infinite lines as viewed by the stationary observer would be:
a = \left(1 - v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho}
or
a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} - \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e - F_m }{\rho}
The first term in the numerator, F_e \, is the electrostatic force (per unit length) outward and is reduced by the second term, F_m \, which with a little manipulation, can be shown to be the classical magnetic force between two lines of charge (or conductors).
The electric current, i_0 \, in each conductor is
i_0 = v \lambda \
and \frac{1}{\epsilon_0 c^2} is the magnetic permeability
\mu_0 = \frac{1}{\epsilon_0 c^2}
because c^2 = \frac{1}{ \mu_0 \epsilon_0 }
So you get for the 2
nd force term:
F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R}
which is precisely what the classical E&M textbooks say is the magnetic force (per unit length) between two parallel conductors, separated by R \, with identical current i_0 \.
Why cannot the same thought experiment be used to obtain a gravitomagnetic effect for two parallel lines of mass and the inverse-square gravitational effect?
