Gaussian to MKS Four-vector Conversion

[Born2bwire]

Is there a straightforward way to determine a given four-vector in MKS units? Our system of units is MKS but I need to use some results from special relativity and of course all the sources use Gaussian units. Jackson has a simple chart in his appendix on converting between the two systems, but it seems to require knowledge of both a left hand and right hand side of the equation. That is,

$$\nabla\cdot\mathbf{E} = 4\pi\rho \quad \rightarrow \quad \nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}$$

The proper conversion here requires you to change units on both sides of the equation to get the correct MKS version. So when I see on Wikipedia and Weisstein's sites that the four-current and four-potential in MKS are:

$$A^\alpha = \left(\frac{\Phi}{c}, \ \mathbf{A}\right)$$
$$J^\alpha = \left(\frac{\rho}{\epsilon_0}, \ \mu_0 \mathbf{J}\right)$$

I can't figure out how you would derive the conversion.

 

[PeterDonis]

I can't figure out how you would derive the conversion

I don't think you can "derive" the conversion without looking at the equations. The different systems of units involve different conventions for how the units are "divided up" between the fields/potentials, the sources, and arbitrary constants. The only way to know what those conventions are is to look at the equations in both systems of units.