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Almost all papers on plasma mentioned that plasma ions drag on a particle is in the direction is opposite to the velocity of the particle. But no one said anything about the other two components; are the other components zeroes? What about the velocity which is a vector and in any direction will always have three components which would mean three drags in opposite direction. The Plasma drag ( Morfill 1980, Gruen et. al. 1984, Northop 1989,1990,1992) is given by

$F = - 2\sqrt{\pi}b^2 n_i m_i u^2 $

$n_i$ = plasma ions density, $m_i$= masss density, $b$ = radius of the particle and $u$=relative vel0city of the particle w.r.t plasma.

Since the velocity is spherical polar coordinates is:

$\dot{r}= \vec v$ = $\dot r\hat{r}$ + $r\dot{\phi}\hat{\theta}$ + $r\dot{\phi}\sin{\theta}\hat{\phi}, $
$v^2$ = $\dot r^2 + r^2{\dot{\theta}^2}+ r^2\dot{\phi}\sin^2{\theta}$

can the drag force be decomposed into:

$ F_r =- 2\sqrt{\pi}b^2 n_i m_i (\dot r^2) $;

$ F_\theta =- 2\sqrt{\pi}b^2 n_i m_i (r^2{\dot{\theta}^2})$ ;

$ F_\phi =- 2\sqrt{\pi}b^2 n_i m_i (r^2\dot{\phi}\sin^2{\theta}) ? $

If the particle is moving along the azimuthal direction is the total force equivalent to

$ F_\phi =- 2\sqrt{\pi}b^2 n_i m_i (r^2\dot{\phi}\sin^2{\theta}) ? $

Supposing, the plasma is at the origin of a Spherical polar Frame and a particle is passing through it at some radial distance, at ang co-latitude angle (θ) = 45degree and azimulth angle (ϕ) =60degree. How would one find the drag force if one is interested in the effect along the components or let say one is interested in finding the path of the particle?

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    $\begingroup$ I would imagine that the scattering cross sections along the perpendicular are small enough to be negligible. $\endgroup$
    – Kyle Kanos
    Commented Apr 5, 2023 at 15:02
  • $\begingroup$ The scattering angles both for large and small one, Impact parameter, only affect the strength of the force not the nature of the force. $\endgroup$
    – Lunthang Peter
    Commented Apr 5, 2023 at 15:15
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    $\begingroup$ Well that's all well and good after a collision. My comment was referring to the probability of a (perpendicular) collision happening in the first place. $\endgroup$
    – Kyle Kanos
    Commented Apr 5, 2023 at 15:37
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    $\begingroup$ I imagine you'd start with the distribution and use the aforementioned scattering cross section and do some convolution of sorts (can't be more specific because it's not something I've done). Intuitively, I'd say that the scattering cross section for the perpendiculars is small & can be ignored (i.e., it'd be close enough to zero so as to be unimportant). $\endgroup$
    – Kyle Kanos
    Commented Apr 5, 2023 at 20:32
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    $\begingroup$ The Coulomb collision rates at physics.stackexchange.com/a/268594/59023 are basically a probabilistic estimate of 90 degree scattering. In most non-lab plasmas outside of stellar surfaces, the collision rate between charged particles in a plasma is tiny to the point of being negligible for most processes of interest. That is, most interesting things happen much faster than Coulomb collisions, which is why shock waves in space are called collisionless shock waves. $\endgroup$
    – honeste_vivere
    Commented Apr 6, 2023 at 13:11

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