8
$\begingroup$

I've seen many versions of the figure shown below -- the famous Swordy plot. They tend to explicitly point out two features in the CR spectrum, the knee and the ankle. I know that the source of UHECR's above the ankle is currently a mystery. But I'm having a devilish time sussing out the cause of this feature at the knee.

Is it known why the slope changes?

enter image description here

$\endgroup$
7
$\begingroup$

The general consensus is that the knee represents the transition from galactic sources (supernova remnants) and extra-galactic sources (AGN, blazars, etc).

At about $10^{15}$ eV, the gyroradius of a proton is $$ r\sim3\frac{\left(mc^2/GeV\right)\left(v/c\right)}{\left(|q|/e\right)\left(B/T\right)}\simeq10^{16}\,{\rm m} $$ where we assume a background magnetic field of a microgauss ($10^{-10}$ T) and that $v\approx c$. This puts us on the same order of magnitude the size of the galactic arm we live in. Hence, particles with energies less than this should be confined to the galaxy. Particles with energies larger than $10^{15}$ eV cannot be contained within the galaxy, meaning they must be extra-galactic in origin.

$\endgroup$
4
$\begingroup$

The knee is believed to be due to one of the following reasons:

1) the reduced efficiency of the galactic magnetic field to confine the cosmic ray particles with energies above the knee within galaxy,

2) the knee corresponds to the maximum energy that protons can have under diffusive shock acceleration in supernova remnants,

3) the contribution from a nearby source,

4) the contribution from a variety of supernovae,

5) the mass distribution of a progenitor of a cosmic ray source,

5) the cosmic ray propagation effect in the galaxy

6) unusual high energy interaction behaviour (new particle creation) at the knee energy

7) a few other models

$\endgroup$
  • $\begingroup$ So, your answer sounds like a "no", as it, it is not known? Do you have any comment on the weight given to these possibilities in the field? $\endgroup$ – chase Sep 7 '15 at 17:10

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.