Displacement current in astrophysical plasmas Could someone please explain to me why we can easily neglect the displacement current when we consider a plasma in astrophysics?
Thank you
 A: 
Could someone please explain to me why we can easily neglect the displacement current when we consider a plasma in astrophysics?

The simple answer is that electric fields do work to get rid of themselves.  The slightly more complicated answer is that the magnitude of the displacement current under many situations in plasmas is really small.
As an example, suppose you had an oscillation at frequency, $\omega$, that has an amplitude, $E_{o}$, and the oscillation is linear, i.e., given by:
$$
E\left( t \right) = E_{o} \ e^{ \mathbf{k} \cdot \mathbf{x} - \omega \ t } \tag{0}
$$
where $\mathbf{k}$ is the wave vector, $\mathbf{x}$ is the 3-vector of position in the wave field, and $t$ is the time of observation.  Then we can see that $\partial_{t} E\left( t \right) = -i \ \omega \ E\left( t \right)$ from linearizing the displacement current.
For most astrophysical scenarios not in stellar atmospheres, the electron number density, $n_{e}$, is typically small, i.e., $\lesssim$100 cm-3.  The highest frequency fluctuations that directly couple to the plasma are upper hybrid oscillations at frequency, $\omega_{uh}$.  In regions where the magnetic field magnitude is small (i.e., most regions of space), the electron cyclotron frequency, $\Omega_{ce}$, is much smaller than the electron plasma frequency, $\omega_{pe}$.  In other words, most regions of space not near some source (e.g., planetary bodies with intrinsic magnetic fields or stars) the following is satisfied $\omega_{pe} \gg \Omega_{ce}$.  Since $\omega_{uh}^{2} = \omega_{pe}^{2} + \Omega_{ce}^{2}$ and we already determined that $\omega_{pe} \gg \Omega_{ce}$, we can use  the following approximation $\omega_{uh} \approx \omega_{pe}$.
So let's use the upper limit on particle density of $\lesssim$100 cm-3 that we mentioned above then that gives $f_{pe} = \tfrac{ \omega_{pe} }{ 2 \ \pi }$ ~ 89.787 kHz (or ~564146 rad/s).  That is, the largest value of our $\omega$ that would be coupled to plasmas would have a magnitude on the order of ~5 x 105.  The displacement current is normalized by the inverse of the speed of light squared, which has a magnitude ~9 x 1016.  Thus, the magnitude of $\tfrac{ \omega }{ c^{2} }$ < 6 x 10-12 s m-2.
Typical electric fields in the solar wind (not near transient sources like collisionless shock waves) satisfy $E_{o}$ ≤ 10 mV/m [e.g., see https://doi.org/10.3389/fspas.2020.592634] and likely smaller in the interstellar medium (ISM), thus the magnitude of the displacement current goes as:
$$
\frac{ \omega \ E_{o} }{ c^{2} } \lesssim 6 \times 10^{-14} \text{ [T/m]} \tag{1}
$$
Note that the scale length over which this type of calculation is valid for the assumed frequencies requires that $k \ \rho_{ce} \sim 1$, where $\rho_{ce}$ is the electron gyroradius.  If we assume that the electron temperature satisfies ~1–30 eV and the magnetic field magnitude satisfies ~1–20 nT, then $\rho_{ce}$ ~ 0.75–18.5 km.  That puts an upper bound on the equivalent magnetic field generated by such fluctuations from Equation 1 of ~10-12 T or ~10-3 nT.
The magnetic fields in the ISM can be small but over large scales these small-scale contributions from the displacement current would be completely washed out (i.e., they are localized, small-scale, and generally don't contribute to the global average).  On larger scales, the values of $\omega$ would be orders of magnitude smaller and $E_{o}$ not significantly larger (most likely smaller too).  Eventually, the scales become so large and $\omega$ so small that it's no longer really a fluctuating term and the approximation breaks down.  In the absence of a source (e.g., star), large scale electric fields do work to eliminate themselves very quickly in plasmas, thus we do not have sustained or even measurable (often) displacement currents.
