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The second order Fermi mechanism describes the interaction of charged particles with magnetic clouds. This model leads to a collision-less acceleration of cosmic rays up to ultra high energies.

A rough computation (classical/extreme cases) of this phenomenon is described in the book of Claus Grupen, "Astroparticle physics" on page 68:

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My question concerns equation 5.13 & 5.14. In my opinion, the expressions for $\Delta E_1$ & $\Delta E_2$ are wrong.

Indeed, in the frame of the cloud the particle's velocity is equal to $v_{in}=(v+u)$. After the collision it becomes $v_{out}=-(v+u)$

Going back to the lab frame: \begin{equation} v^*_{out}=-(v+u)-u=-v-2u \end{equation}

$\Delta E_1$ then yields: \begin{equation} \Delta E_1=\frac{1}{2}m(v+2u)^2-\frac{1}{2}mv^2 \end{equation}

Similarly, for $\Delta E_2$: \begin{equation} \Delta E_2=\frac{1}{2}m(v-2u)^2-\frac{1}{2}mv^2 \end{equation}

Is there really a mistake or am I doing it wrong?

innisfree
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AxelAE
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1 Answers1

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AFAICT, the extra factor of 2 should be there. A particle of energy $E$ in the cloud's reference frame is (with $c=1$ throughout), $$ E'=\gamma v\left(E+up\right) $$ where $p$ is the particle momentum. Transforming back to the observer frame, this is $$ E''=\gamma v\left(E'+up'\right) $$ where $p'=\gamma v\left(p+uE\right)$. The energy gain is then the difference, $$ E''-E=\Delta E\propto\left(2uv-2u^2\right) $$ which carries the factor of 2 you quote.

Two presentations that I am aware of both contain this factor as well:

I don't (any more) have Kirk's Plasma Astrophysics Saas-Fee lecture on particle acceleration (Springer link to book), but I suspect it is also correctly given there.

Kyle Kanos
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