#### Born2bwire

Science Advisor

Gold Member

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[tex]\nabla\cdot\mathbf{E} = 4\pi\rho \quad \rightarrow \quad \nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0} [/tex]

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:

[tex] A^\alpha = \left(\frac{\Phi}{c}, \ \mathbf{A}\right) [/tex]

[tex] J^\alpha = \left(\frac{\rho}{\epsilon_0}, \ \mu_0 \mathbf{J}\right) [/tex]

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