Finding a substitution of $\mu_0$ to convert from SI to c.g.s in EM laws

Question

In c.g.s, we set $$k=\frac 1{4\pi\epsilon_0}=1$$

which gives $$\epsilon_0=\frac 1{4\pi}$$

This conversion works, for example, in Gauss' law:

in SI $$\vec{\nabla}\cdot\vec{E}=\frac\rho{\epsilon_0}$$ and in c.g.s $$\vec{\nabla}\cdot\vec{E}=4\pi\rho$$

Now, we know $c^2=\frac 1{\epsilon_0\mu_0}$, so in order to convert $\mu_0$ (that usually appears in the SI form) into c.g.s, I thought the substitution would be $$\mu_0=\frac {4\pi}{c^2}$$

However, this doesn't yield the correct laws in c.g.s.

For example, Amper's law in SI is $$\vec{\nabla}\times\vec{B}=\mu_0\vec{j}$$

But in c.g.s it is $$\vec{\nabla}\times\vec{B}=\frac {4\pi}c \vec{j}$$ And not $$\vec{\nabla}\times\vec{B}=\frac {4\pi}{c^2} \vec{j}$$

as I would have expected. So it seems $\mu_0=\frac{4\pi}c$ is the correct substitution.

How can that be? Where did one $c$ go? What is another assumption I am missing?

Answer 1

You're assuming $c^2\,\mu_0\,\epsilon_0=1$, which is a formula that holds in SI units but not in all unit systems. I find the best "mnemonic" to keep track of all this stuff is the following generalized unit version of Maxwell's equations, detailed in the "Rationalization" section of the Lorentz-Heaviside unit system Wikipedia page:

$$\begin{align} \nabla \cdot \mathbf{D} &= \rho / \beta, \\ \quad \nabla \cdot \mathbf{B} &= 0, \\ \quad \kappa \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t}, \\ \quad \kappa \nabla \times \mathbf{H} &= \frac{\partial \mathbf{D}}{\partial t} + \mathbf{J} / \beta, \end{align}$$

where $\beta$ and $\kappa$ are in general dimensionful units that define the unit system in question. From this set it is clear we must in general use:

$$c^2\,\mu_0\,\epsilon_0=\kappa^2$$

and Gaussian units have $\kappa=c;\,\beta=\frac{1}{4\pi}$. The reason for your "anomaly" should now be clear from the above equation. Of course, often people define their time and length units to be the same, in which case $c=1$ and the problem doesn't arise.

I empathize: this is one of the things I am really bad at - I trip up on units all the time too. If you're like me you really need to keep a few sharp mnemonics like the above handy so that you can get an overall view of what you're doing rather than blindly plugging into conversion formulas.