Einstein radius

Remarks

By the lens equation for a point mass, $\beta = \theta - \theta_E^2 \theta / \abs{\theta}^2$, a source at $\beta = 0$ is imaged onto the whole circle $\abs{\theta} = \theta_E$. The check: on that circle, $\theta_E^2 \theta / \abs{\theta}^2 = \theta$, so $\beta = 0$. The deflection at the ring is $\hat\alpha = 4 G M / (c^2 D_d \theta_E)$.

Example: a lens of one solar mass halfway to a source at $8$ kpc has $D_d = D_{ds} = 4$ kpc and $D_s = 8$ kpc. With $4 G M_\odot / c^2 \approx 5.9$ km, one gets $\theta_E \approx 4.9 \times 10^{-9}$ rad, about $1$ milliarcsecond.

For an extended lens with circular symmetry the Einstein radius is the radius inside which the mean surface density equals the critical density $\Sigma_{cr} = c^2 D_s / (4 \pi G D_d D_{ds})$. Non-example: an axially symmetric lens whose mean surface density is below $\Sigma_{cr}$ at every radius has no such radius; it forms no Einstein ring.