Fermi acceleration is a process by which particle are accelerated to super thermal energies. Usually, the spectrum of thermal distribution (e.g. Maxwell - Boltzmann, Maxwell - Juttner and Planck) decays exponentially for energies in excess of the thermal energy. Fermi acceleration usually adds a power law component to the distribution.

The simplest mathematical model for Fermi acceleration is as follows. Suppose there's some process that can increase a particle's energy by a factor $ q>1 $. Suppose further that the probability for a single boost is $ 1>p>0 $. We are interested in particles at much higher energies than the thermal. For that purpose we can assume that all particles start with an energy $ \varepsilon_0 $. Hence, the number of boosts needed to get to an energy $ E \gg \varepsilon_0 $ is

$ E = \varepsilon_0 q^n \Rightarrow n = \frac{\ln \frac{E}{\varepsilon_0}}{\ln q} $

The probability of having energy $ E $ is equivalent to the probability of experiencing $ n $ boosts

$ P\left(E\right) \propto p^n = p^{\ln\frac{E}{\varepsilon_0}/\ln q} = \left(\frac{E}{\varepsilon_0}\right)^{\frac{\ln p}{\ln q}} $

So the probability decreases as the energy increases.

Diffusive Shock Acceleration Edit

When a shock wave travels through a collision - less plasma, particles can be scatter back and forth across the shock front and thus be Fermi accelerated. The appropriate Boltzmann equation for a collision - less particles in the presence of bulk motion and scattering magnetic fields in planar 1D geometry is

$ \frac{\partial f}{\partial t} + U \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left( \kappa \frac{\partial f}{\partial x} \right) + \frac{1}{3} \frac{\partial U}{\partial x} p \frac{\partial f}{\partial p} $

Where $ U $ is the velocity of the bulk, and $ \kappa $ is the diffusion coefficient (due to interaction with magnetic fields). For simplicity, we will assume that the diffusion coefficient is constant (although the final result remains the same even for variable diffusion coefficient). We are interested in steady state solutions, so we can omit the first term. We assume that we are working in the reference frame of the shock, so it's always at $ x = 0 $. In both the upstream $ x < 0 $ and downstream $ x > 0 $ regions the bulk velocity is uniform and positive, so the last term can also be neglected. Finally, we are left with

$ U \frac{\partial f}{\partial x} = \kappa \frac{\partial^2 f}{\partial x^2} $

where $ U $ is either the upstream or downstream bulk velocity. The general solution is

$ f = a + b \exp\left( \frac{U x}{\kappa} \right) $

where $ a $ and $ b $ are integration constants. Since both $ \kappa $ and $ U $ are positive, $ b = 0 $ in the downstream. The solutions for the upstream is

$ f_u\left(x,p\right) = a_u\left(p \right) + b_u \left(p \right) \exp\left( \frac{U_u x}{\kappa} \right) $

and for the downstream

$ f_d \left(x,p \right) = a_d \left(p \right) $

The partition function must be continuous at $ x = 0 $ so

$ \lim_{x\rightarrow 0^-} f_u = \lim_{x \rightarrow 0^+} f_d $

$ a_u + b_u = a_d $

Although the density of the bulk is not continuous across the shock front, we assume that Fermi accelerated particles do not "collide" with the bulk, so they are oblivious to the density step.

Another condition can be derived by integrating the Boltzmann equation around the shock front $ \lim_{\epsilon\rightarrow 0^+} \int_{-\epsilon}^{\epsilon} dx $

$ \kappa \left( \lim_{x \rightarrow 0^-}\frac{\partial f_u}{\partial x} - \lim_{x \rightarrow 0^+}\frac{\partial f_d}{\partial x} \right) + \frac{1}{3} \left(U_u - U_d \right) p \frac{\partial f}{\partial p} = 0 $

We intentionally left out the subscript in the last term because $ f $ is continuous at $ x = 0 $. Substituting the appropriate expressions yields

$ U_u b_u + \frac{1}{3} \left(U_u - U_d \right) p \left( \frac{d a_u}{d p} + \frac{d b_u}{d p} \right) = 0 $

$ \alpha b_u + p \frac{d b_u}{d p} = - p \frac{d a_u}{d p} $

where $ \alpha = 3 / \left(1 - U_d/U_u \right) $. We continue by multiplying by $ p^{\alpha - 1} $

$ \frac{d}{dp}\left( b_u p^{\alpha} \right) = -p^{\alpha} \frac{d a_u}{d p} $

$ b_u p^{\alpha} = -\int_0^p q^{\alpha} \frac{d a_u}{d q} dq + c $

$ b_u = -p^{-\alpha} \int_0^p q^{\alpha} \frac{d a_u}{d q} dq + c p^{-\alpha} $

The first term is not so interesting because it would have the same features as the incident distribution function. Especially, if the incident partition function $ a_u $ decreases exponentially with energy, so would the first term. The second term is the expected power law addendum. The advantage of this calculation over the previous one is that it yields the index of the power law. From the Rankine Hugoniot rules and the ideal gas law we get

$ U_u / U_d = \frac{\rho_d}{\rho_u} = \frac{\gamma + 1}{\gamma - 1 + M^{-2}} $

where $ \rho $ is the density of the bulk, $ \gamma $ is the adiabatic index and $ M $ is the Mach number of the shock.

Second Order Acceleration Edit

In the previous case we considered Fermi acceleration in a gas with bulk motion. In that case we got that the acceleration is linear in the velocity, which is why it is called first order Fermi acceleration. However, a similar (albeit less efficient) phenomenon can occur in a fluid with zero bulk motion (i.e. just thermal random motion).

Let us consider box filled with two types of particles. The first is some microscopic particle like the electron or the proton, and the other is some macroscopic object much more massive than the first, which we will refer to as cloud. Both are moving at random. To further simplify the analysis, we will consider motion along just one axis, and that the velocity of the atomic particle is much larger than that of the cloud $ u >> v $. If the particle collides with a cloud, its new velocity is

$ u' = 2 v - u $

If both are moving in the same direction, the energy of the particle decreases by a factor $ 1 - 4 \frac{v}{u} $, and if they are moving in opposite direction the energy would increase by a factor $ 1 + 4 \frac{v}{u} $. Since there are equal numbers of clouds moving in both direction, one might think the net contribution of collision would be zero. However, as we will shortly show, there's a slightly higher chance of collision between particle and cloud moving in opposite directions.

For simplicity we would assume that the cross section is energy independent, in which case the mean free path in both directions is the same, and equation to the product of the cross section and the density $ l = \sigma n $. However, the mean free time is shorter for opposite velocity collision

$ \tau_{\pm} = \frac{l}{u \mp v} \approx \frac{l}{u} \left(1 \pm \frac{v}{u} \right) $

We can now write a rate equation for the average particle energy

$ \frac{d e}{d t} = \left( \frac{1}{\tau_+} - \frac{1}{\tau_-} \right) \Delta e \propto v^2 $

Hence the acceleration is proportional to the velocity squared.