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Let us assume we have a flow of charged particles in a quasi-neutral state (i.e., a plasma) convecting at some speed, $\mathbf{V}_{sw}$ = V$_{sw}$ $\hat{x}$, and the particles have species-dependent thermal speeds, V$_{Ts}$ $\ll$ V$_{sw}$ (for simplicity, just assume protons and electrons). Assume there is a magnetic field, $\mathbf{B}_{o}$ = B$_{o}$ $\hat{z}$, that is small enough that particles can be said to have effectively balistic trajectories (i.e., ignore the gyration initially and/or V$_{Ts}$ $\ll$ V$_{sw}$). Assume that this flow and magnetic field are entirely laminar and homogeneous everywhere initially and that the magnetic field moves with the flow at V$_{sw}$ (this latter part confuses me since sources move, not really the fields, but just bare with me).

Now what if these particles are incident on a magnetic field gradient, $\nabla$B, which is a vector parallel to $\hat{x}$ and |B| only increases in the $\hat{z}$ component. Assume the gradient is strong enough to reflect a small fraction of the incident particles. For now, let us just focus on protons and assume the reflection is specular. Also, let us work in the reference frame where the magnetic field gradient is at rest.

(1) Since the upstream flow and magnetic field are both homogeneous everywhere upstream of the magnetic field gradient, will any given reflected proton know that the magnetic field is moving?

The purpose of the question is to understand arguments that say once a proton is reflected, it will "see" an electric field, $\mathbf{E}_{conv}$ = -$\mathbf{V}_{sw}$ $\times$ $\mathbf{B}_{o}$. However, in my memory of Lorentz transformations, $\mathbf{v}$ $\equiv$ the speed of a reference frame relative to some other lab frame. The idea of a magnetic field moving in this manner bothers me, but not as much as the expression for $\mathbf{E}_{conv}$.

(2) Should not $\mathbf{E}_{conv}$ be expressed in terms of the reflected proton's instantaneous velocity relative to $\mathbf{B}_{o}$ (i.e., $\mathbf{v}$), where $\mathbf{B}_{o}$ is the magnetic field in the proverbial lab frame?

The arguments to which I refer suggest that particles gain energy from this frame-dependent electric field. This should only work if $\mathbf{v}$ $\parallel$ $\mathbf{E}_{conv}$. However, if $\mathbf{E}_{conv}$ = -$\mathbf{v}$ $\times$ $\mathbf{B}_{o}$ as my intuition suggests, then $\mathbf{v}$ $\cdot$ $\mathbf{E}_{conv}$ = 0, right?

(3) Am I wrong to be bothered by the concept of magnetic fields "moving" in the manner I described above, or am I missing something?

honeste_vivere
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    Somehow this sounds like a Rube Goldberg version of the question why a moving magnet induces a current in a wire. Maybe you could draw a picture for clarification? What, by the way is a "lab frame"? Is that like the picture frame that somebody hung in one of my labs with a drawing of three arrows and a warning that this label is not to be removed by threat of the laws of nature? – CuriousOne Sep 21 '14 at 02:04
  • @CuriousOne: This question is asking some of the basics about Fermi acceleration (often improperly called diffusive shock acceleration, which is really a subset of Fermi acceleration). Perhaps you could tone down your sarcasm? – Kyle Kanos Sep 21 '14 at 02:08
  • @honeste: Have you read Kirk's work on particle acceleration? He's got an excellent Saas-Fee lecture on it, plus some other papers with Duffy. There's also some work by Luke O'C Drury that is probably a good start. – Kyle Kanos Sep 21 '14 at 02:17
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    @KyleKanos: I don't see a question particularly related to Fermi acceleration here. The OP seems to have doubts about how to calculate the proper field components seen by particles. The setup for the moving magnetic field is IMHO identical to a moving magnet scenario. Why not simplify the physics before we get into the question of which frame this has to be analyzed in (which should lead to the conclusion that the physics is the same in all rest frames). – CuriousOne Sep 21 '14 at 02:21
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    @CuriousOne: If you don't see how this is related to Fermi acceleration, than all I can say is you need to do some more reading on it. Magnetized shocks + collisions/reflections + particle acceleration should scream "Fermi acceleration" to everyone. – Kyle Kanos Sep 21 '14 at 02:23
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    @KyleKanos: I can not see how the OPs problem with the scenario is related to Fermi acceleration, it seems more fundamental than that. That's not the same thing as not seeing the physics of the mechanism myself. Let's leave it at that, you are free to explain the lab frame joke to the OP, if you think the he didn't get it. – CuriousOne Sep 21 '14 at 02:31
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    @CuriousOne, my problem is largely about the idea of a magnetic field moving. The problem is specific to collisionless shocks, so yes it does relate to Fermi's original magnetic mirror idea. The issue I have is that if the upstream field is everywhere homogeneous, how can there be $\mathbf{E}$ = -$\mathbf{V}{sw} \times \mathbf{B}{o}$? This may seem like I am missing a trivial point, but if you were sitting at some radial distance away from a bar magnetic (at the center of the bar) in vacuum, could you tell it was rotating? [BTW, yes, I got your joke] – honeste_vivere Sep 21 '14 at 14:48
  • P.S. The answer to my last question is no if the bar magnet has no imperfections and produces an ideal dipole magnetic field. Sources move, not the magnetic fields (which propagate at the speed of light)... So in the case I originally stated, even if it's like the field is moving upstream, its homogeneity makes it impossible for an observer to know this since nothing changes. So how would a particle know to drift in addition to gyration? – honeste_vivere Sep 21 '14 at 14:51
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    The electric & magnetic field is the field of the plasma, so a particle moving at velocity $v$ that suddenly reflects, $v\to-v$, sees the new electric field that is opposite in direction (effectively). Any charged particle in a magnetic field sees an electric field, the fact that $B$ is homogeneous shouldn't change this. – Kyle Kanos Sep 21 '14 at 15:57
  • @KyleKanos what if we had a uniform field and then we insert a charged particle? Would it know the field is moving as well? If so, how? I think I can follow the plasma part, as this would be the source moving. However, if the source is not local, it confuses me how a charged particle could know whether a homogeneous field was moving or not. – honeste_vivere Sep 22 '14 at 13:40
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    As long as the particle is moving at a different speed than the field, it will know the electric field is there. That's basically E&M theory of a charged particle moving in a stationary magnetic field (and probably an okay way of looking at this). Remember that it's the electric field that accelerates particles while the magnetic field changes the direction of the particle. – Kyle Kanos Sep 22 '14 at 14:03
  • @KyleKanos I think you point out the primary issue I have with the picture I outlined in my question. Namely, the idea of a frame-dependent electric field doing work on a particle bothers me. However, I should have realized (since I already know this) that energy is not a Lorentz invariant. Though I still think the electric field observed by the particle will be dependent upon the particle's velocity relative to the background field, not just $\mathbf{V}_{sw}$. I am now thinking that my comment regarding the bar magnet was wrong as well. – honeste_vivere Sep 22 '14 at 16:00
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    @honeste: Often in, shock acceleration, the ambient thermal velocity is smaller than the shock velocity so that you can approximate the velocity of the particle as small & simply use $\mathbf V_{sw}$ instead of $\mathbf V_{rel}$. – Kyle Kanos Sep 22 '14 at 16:51

1 Answers1

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(1) Since the upstream flow and magnetic field are both homogeneous everywhere upstream of the magnetic field gradient, will any given reflected proton know that the magnetic field is moving?

This, unfortunately, displays how easily I can confuse myself about something that should be relatively trivial. As @CuriousOne noted, I managed to over complicate things and this led to my confusion.

The first thing to notice is that for the scenario where $\mathbf{V}_{sw} = V_{sw} \ \hat{\mathbf{x}}$ and $\mathbf{B}_{o} = B_{o} \ \hat{\mathbf{z}}$ are satisfied, there must exist an electric field that allows the plasma to flow across the magnetic field. Let's assume this is the Sun's rest frame. From the Lorentz transformation of the electric field (in the non-relativistic limit) we find that $\mathbf{E}_{conv} = -\mathbf{V}_{sw} \times \mathbf{B}_{o}$ (i.e., parallel to $+\hat{\mathbf{y}}$) in the Sun's rest frame. It is then easy to show that in this frame that $\mathbf{V}_{sw}$ is the ExB-drift speed given by: $$ \mathbf{V}_{sw} = \frac{ \mathbf{E}_{conv} \times \mathbf{B}_{o} }{ B_{o}^{2} } \tag{1} $$

In the plasma rest frame (i.e., where $\mathbf{V}_{sw} \rightarrow 0$), however, there is no electric field.

Thus, part of the answer is not that magnetic fields move, which is clearly wrong, rather that the sources move and the fields respond. When we think specifically about a shock moving through this medium, we need an additional transformation velocity. In the case of, say, Earth's bow shock, the shock rest frame is roughly defined by subtracting $\mathbf{V}_{sw}$ since the Earth is roughly at rest relative to the solar wind.

So a proton initially on a balistic trajectory toward the shock will, if reflected, suddenly "see" an electric field due to the frame transformation. Thus, the second part of the answer to my first question is just that the frame transformation results in the electric field and has nothing to do with moving magnetic fields.

(2) Should not $\mathbf{E}_{conv}$ be expressed in terms of the reflected proton's instantaneous velocity relative to $\mathbf{B}_{o}$ (i.e., \mathbf{v}), where $\mathbf{B}_{o}$ is the magnetic field in the proverbial lab frame?

Again, this shows a clear confusion on my part. The relevant electric field is expressed in the frame of reference where calculations are being performed. The relevant reference frame is the shock rest frame.

The particles can gain energy from $\mathbf{E}_{conv}$ in this frame because upon reflection from the moving magnetic gradient, the magnetic contribution to the Lorentz force will accelerate the particle along the $\hat{\mathbf{y}}$ direction, thus $\mathbf{v} \cdot \mathbf{E}_{conv} \neq 0$ in this frame. The particle can gain energy so long as it continues to "skip" along the upstream edge of the shock ramp.

I wrote a simple Mathematica routine to simulate this with numerous test particles with different initial trajectories (see image below). The particles that undergo multiple reflections gain more energy than those that are transmitted into the downstream after one (or zero) reflection(s), which is evidenced by the larger gyroradii of the particles with increasing distance along $+\hat{\mathbf{y}}$. Example test particle simulation

The above image is an example of something called shock drift acceleration (SDA), where shock-reflected particles gain energy while skipping along the upstream edge of the shock front from the convective electric field. This is not to be confused with diffusive shock acceleration (DSA) (or first order Fermi acceleration) whereby particles gain energy by reflecting off of merging magnetic mirrors. A nice animation of SDA can be found at https://svs.gsfc.nasa.gov/4513.

(3) Am I wrong to be bothered by the concept of magnetic fields "moving" in the manner I described above, or am I missing something?

As I have already stated, I was missing several things. Further, magnetic fields do not "move" (i.e., field lines do not move), sources move or change and the fields respond.

honeste_vivere
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