3
$\begingroup$

In high school physics, we know that a charged particle cannot gain energy from a magnetic field. But, it seems that in the so-called Fermi acceleration, the particle is accelerated by the magnetic field. How does it happen?

Is it because the magnetic field is not stationary?

$\endgroup$
  • $\begingroup$ note: acceleration not always means, there is energy transmitted! Normally the force on a particle in a B-field is perpendicular to its velocity, i.e. does no work. A non-stationary field can do work, on the contrary. $\endgroup$ – Ilja Apr 29 '16 at 7:08
4
$\begingroup$

Note

You should clarify your statement from "...a charged particle cannot gain energy from a magnetic field..." to "...a charged particle cannot gain energy from a static magnetic field..." There is nothing wrong with energy transfer from time-varying magnetic fields.

Background

If the spatial gradient in the magnetic field is slow enough such that the particle can complete multiple gyro orbits through the gradient, we can assume that the magnetic moment of the charged particle's gyro orbit remains constant, or: $$ \gamma \mu = \gamma \frac{ q_{s} \ \Omega_{cs} \ \rho_{cs}^{2} }{ 2 \ c } \sim constant \tag{1} $$ where $c$ is the speed of light, $q_{s}$ is the charge of species $s$, $\Omega_{cs}$ is the cyclotron frequency or gyrofrequency of species $s$, $\rho_{cs}$ is the gyroradius or Larmor radius of species $s$, and $\gamma$ is the relativistic Lorentz factor. The gyrofrequency and radius are given by: $$ \begin{align} \Omega_{cs} & = \frac{ q_{s} \ B_{o} }{ \gamma \ m_{s} \ c } \tag{2a} \\ \rho_{cs} & = \frac{ c \ p_{\perp,s} }{ q_{s} \ B_{o} } \tag{2b} \end{align} $$ where $B_{o}$ is the magnetic field magnitude, $m_{s}$ is the mass of species $s$, and $p_{\perp,s}$ is the momentum orthogonal to $\mathbf{B}_{o}$ of species $s$, where $\mathbf{p}_{s} = \gamma \ m_{s} \ \mathbf{u}$.

The assumption that $\gamma \mu$ ~ constant during an interaction with a magnetic field gradient derives from a form of the WKB approximation, where we say this holds if: $$ \Omega_{cs}^{2} \gg \lvert \frac{ 3 }{ 4 } \left( \frac{ \dot{\Omega}_{cs} }{ \Omega_{cs} } \right)^{2} - \left( \frac{ \ddot{\Omega}_{cs} }{ 2 \ \Omega_{cs} } \right) \rvert \tag{3} $$ where $\dot{Q}$ is the total derivative of quantity $Q$. When Inequality 3 holds, the gradient is said to be slow (to which I eluded earlier).

Adiabatic Invariance

Okay, if we can maintain that $\gamma \mu$ ~ constant during an interaction with a magnetic field gradient and there are no time-varying electromagnetic fields, then the total kinetic energy should be constant during the interaction as well. This gives us the following relationships: $$ \begin{align} \frac{ u_{\perp,f}^{2} }{ B_{o,f} } & = \frac{ u_{\perp,i}^{2} }{ B_{o,i} } \tag{4a} \\ u_{i}^{2} & = u_{\perp,i}^{2} + u_{\parallel,i}^{2} \tag{4b} \\ & = u_{\perp,f}^{2} + u_{\parallel,f}^{2} \tag{4c} \end{align} $$ where the subscript $i(f)$ corresponds to the initial(final) state and $\parallel(\perp)$ corresponds to the direction parallel(perpendicular) to $\mathbf{B}_{o}$. If we assume the gradient is along $\hat{\mathbf{z}}$, we can then express the final parallel velocity as a function of $z$, given by: $$ u_{\parallel,f}^{2} \left( z \right) = u_{i}^{2} - u_{\perp,i}^{2} \frac{ B_{o,f} \left( z \right) }{ B_{o,i} } \tag{5} $$

We can see that the particle will reflect or mirror off of the gradient if $u_{\parallel,f} \rightarrow 0$, which occurs if: $$ \begin{align} \lvert \frac{ u_{\parallel,i} }{ u_{\perp,i} } \rvert & < \sqrt{ \frac{ B_{o,f} }{ B_{o,i} } - 1 } \tag{6a} \\ \frac{ B_{o,f} }{ B_{o,i} } & > 1 + \left( \frac{ u_{\parallel,i} }{ u_{\perp,i} } \right)^{2} \tag{6b} \end{align} $$

How does it happen?

First order Fermi acceleration (also called diffusive shock acceleration) considers two merging (i.e., the distance between decreases with time) regions of enhanced magnetic field, which I will call magnetic clouds. Fermi's original idea was to have two merging magnetic clouds (or scattering centers) moving at high speeds (but still non-relativistic) relative to one another. In the rest frame of a magnetic cloud, the particle will reflect and gain no energy. It will just reverse its momentum parallel to $\nabla B_{o}$, where $p_{\parallel} = p \cos{\theta}$. However, in the "center of momentum" frame of the two clouds, a relativistic particle (non-relativistic cloud speed still) particle will gain energy, $\Delta E$, with each reflection proportional to: $$ \frac{ \Delta E }{ E } \approx \frac{ 3 \ V_{sh} }{ 4 \ c } \cos{\theta} \tag{7} $$ where $V_{sh}$ is the upstream flow speed along a unit normal orthogonal to the cloud surface (i.e., parallel to $\nabla B_{o}$ here). Note that $V_{sh}/c$ is small here and this approximation changes if that ratio becomes a significant fraction of unity. This

There is also a 2nd order Fermi acceleration, whereby multiple magnetic clouds are present that can allow a charged particle to diffuse (in energy and pitch-angle) after multiple interactions with multiple different clouds. On average, the particle will interact more with clouds moving anti-parallel to its motion than parallel, thus it should gain an overall energy. However, for a non-relativistic cloud speed and relativistic particle, this is proportional to $\left( V_{sh}/c \right)^{2}$ (thus why it is called 2nd order), which is very small and slower than first order Fermi acceleration.

Simple Analogy

Imagine you have a perfect ping-pong paddle or tennis racquet that allows for perfectly elastic collisions with the incident ball. If the paddle were stationary in the court rest frame, then the ball would gain no energy upon reflection off of the paddle in that frame. If you were running toward the net when the the incident ball struck the paddle, the ball would be seen to gain energy in the court frame (still no energy gain in the paddle rest frame).

It's a similar idea in Fermi acceleration, where we replace the paddle with the magnetic cloud and the ball with the charged particle.

Caveats

Note that reality is much more complicated than the above picture because magnetic clouds, to which I eluded, often carry tremendous free energy allowing them to generate electromagnetic waves. Thus, there are often time-varying electric and magnetic fields present near these clouds, which can complicate things. Even so, we still observe evidence of diffusive shock acceleration in, for instance, the Earth's foreshock is full of ion velocity distributions called diffuse ions aptly named because they are direct evidence of diffusive shock acceleration.

References

  • J.D. Jackson, Classical Electrodynamics, Third Edition, John Wiley & Sons, Inc., New York, NY, 1999.
$\endgroup$
0
$\begingroup$

In the case of the second order acceleration, two clouds are approaching, therefore the energy the charged particle gains comes from the energy of the clouds.

In the case of the first order acceleration, the charged particle gains energy as it moves repeatedly through the shock front. The region before the shock front (upstream) moves at higher speed than that of the region after the shock front (downstream) in the frame with the shock front at rest, and the gain in energy is provided by the difference in motion of these two regions with magnetic field (we assume that the field is such that it forms magnetic mirrors).

In both cases, the magnetic fields are not stationary.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.