Derivation of "polarization drift" in a plasma

In summary, the derivation of "polarization drift" in a plasma involves analyzing the motion of charged particles under the influence of electric fields. It describes how the asymmetrical distribution of charges leads to a net drift of the plasma as a whole, influenced by factors such as electric field strength and particle density. The concept is crucial for understanding plasma behavior in various physical contexts, including astrophysical phenomena and laboratory experiments.
  • #1
Axel Togawa
6
0
TL;DR Summary
I don't understand the role of "drift velocity" in the derivation of the "polarization drift"
When studying a particle in slowly time varying, uniform electric field E, and in a constant, uniform magnetic field [itex]\textbf{B}[/itex], I found many texts where I can't understand the derivation of the "polarization drift" [itex]\textbf{v}_p[/itex], in particular I quote as reference this book I found online ([1], Pag.93) where they take as expression for the "drift velocity ExB" [itex]\textbf{v}_E = \frac{\textbf{E}\times\textbf{B}}{B^2}[/itex], which is the same formula used in the case where [itex]\textbf{E} = cost[/itex], why? The assumption to find this expression is that [itex]\frac{d\textbf{v}_E}{dt} = 0[/itex] but that is not true in this scenario.
I could consider that the variation in time is small, so [itex]\frac{d\textbf{v}_E}{dt} \approx 0[/itex], but then they find the expression for the polarization drift as [itex]\textbf{v}_p = -\frac{m}{q}\frac{\dot{\textbf{v}}_E\times \textbf{B}}{B^2}[/itex] (where [itex]\dot{\textbf{v}}_E = \frac{d\textbf{v}_E}{dt}[/itex]).

So why is [itex]\textbf{v}_E[/itex] sometimes considered constant and sometimes not?

[1] "Introduction to Plasma Physics C17, Lecture Notes, John Howard": https://people.physics.anu.edu.au/~jnh112/AIIM/c17/chap04.pdf
 
Physics news on Phys.org
  • #2
Ok first off I assume you know that E×b is the cross product between the E and B fields.

Now are you familiar which field diverges and which field does not ?
If not look closely at Maxwell equations, amperes law and Lorentz force. (,right hand rule)
Also make sure your clear on the difference between the magnetic field and the magnetic moment.
Griffiths Introductory to electrodynamics has an excellent section on this.
It would be good to get a handle on how familiar you are with the two fields as per Maxwell.
 
  • #3
I think to be somewhat familiar with these concepts, but I'm not sure how they can explain my doubt; I mean, the central point of my query is based on the derivation of [itex]\textbf{v}_p[/itex] starting from the following differential equation (Lorentz force):
[tex]
m\frac{d(\textbf{v}_{E}+\textbf{v}_{p})}{dt}=q\textbf{E}+q(\textbf{v}_{E}+\textbf{v}_{p})×\textbf{B} \;\;\; (assuming \textbf{E} \perp \textbf{B} \perp \textbf{v}_E)
[/tex]
but I'm also open to other ways to obtain the same result, thanks
 
  • #4
Ok so your somewhat familiar with the RH rule. You recognize the cross product as opposed to the dot product between vectors. Now I'm assuming you recognize the E field is perpendicular to the B field. The magnetic force is perpendicular to the B field. So the magnetic force is maximal when the charged particle is moving parallel to the magnetic field and zero when moving parallel to the magnetic field
Lorentz force
##F=q(E+v\times B##

Magnetic force law
##F=qV\times B##

Now magnetic force perpendicular to V cannot change its magnitude it can only change its direction. This relates to the expression magnetic force does no work. As its magnetic force is perpendicular to V.
Motion of a charged particle under the action of a magnetic field alone is always motion with constant speed.
That should help understand when V_e is constant and when it is not.

Also consider what occurs with the magnetic field when You have a charge current as opposed to a static E field ( hint Helical)

I should mention this isn't the best forum for this question. It really is just classical physics thus far.
 
Last edited:

Similar threads

Back
Top