Goldreich–Julian density

Remarks

Assume that the magnetospheric plasma is a perfect conductor that corotates with the star, with velocity $u = \Omega \times r$, so that $E + u \times B = 0$. By Maxwell's equations, $\rho = \varepsilon_0 \nabla \cdot E$, and $$\nabla \cdot (u \times B) = B \cdot (\nabla \times u) - u \cdot (\nabla \times B) = 2 \Omega \cdot B - u \cdot (\nabla \times B).$$ If the currents are negligible, $\nabla \times B = 0$, and so $\rho = -\varepsilon_0 \nabla \cdot (u \times B) = -2 \varepsilon_0 \Omega \cdot B$. This is valid well inside the light cylinder, where $\abs{\Omega \times r} \ll c$; a fuller relativistic treatment divides by $1 - \abs{\Omega \times r}^2/c^2$. In Gaussian units the density is $-\Omega \cdot B / (2\pi c)$.

Example: for an aligned dipole, with $\Omega$ along the magnetic axis, $\rho_{GJ}$ has one sign near the poles and the opposite sign near the equator. It vanishes on the cones where $\Omega \cdot B = 0$, which for a dipole are at $\cos^2\theta = 1/3$ from the axis.

Non-example: a non-rotating magnetized star, with $\Omega = 0$, has $\rho_{GJ} = 0$; no space charge is needed, since the plasma at rest feels no motional electric field.