Magnetic dipole moment of electron I have read an electron has an intrinsic magnetic dipole moment.
Does this mean that because dipole moment can be thought of as a current loop, and a current loop radiates EM Waves due to a changing $dJ/dt$ that electrons must radiate EM  waves when they are stationary?
can someone show me the math of the electrons dipole moment in maxwells equations? i assume the curl of $M$ is in current density
 A: The electron doesn't radiate because the "current" in the Amperian model of dipoles is constant. Alternatively, the dipole can be represented by two magnetic monopoles separated by an infinitesimal distance (Gilbert dipole), which also confirms the static picture.
In reality, the electron's magnetic moment is a quantum-mechanical phenomenon which is related to its intrinsic angular momentum (spin) by
$$\mathbf{m} = -g\mu_B \mathbf{S},$$
where $\mathbf{S}$ is the spin of the electron, $g=2$ classically, and $\mu_B$ is the Bohr magneton.
As you said, the magnetic moment enters the Maxwell's equations via the source term, and is given by
$$\mathbf{J} = -(\mathbf{m}\times \nabla)\delta^{(3)}(\mathbf{r} - \mathbf{r}_0).$$
In media with distributed magnetization, a magnetization density $\mathbf{M}$ is used instead, and the current will need to be integrated over.
Computationally, magnetic contributions from the electron are usually ignored at leading order. This is on one hand due to the massive ratio between the electric and magnetic forces,
$$\bigg(\frac{1}{4\pi\epsilon_0}\bigg)\bigg/\bigg(\frac{\mu_0}{4\pi}\bigg)= \frac{1}{\mu_0\epsilon_0} = c^2.$$
It is on the other hand due to the faster decay of the magnetic field of a dipole, being
$$\mathbf{B} = \frac{\mu_0}{4\pi}\bigg(\frac{3(\mathbf{m}\cdot\mathbf{r})\mathbf{r}}{|\mathbf{r}|^5} - \frac{\mathbf{m}}{|\mathbf{r}|^3}\bigg) \sim \frac{1}{r^3},$$
compared to the $1/r^2$ decay of the electric field.
