Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

The energy spectrum of the cosmic radiation (not CMB) is limited to both sides.

I know about the GZK-cutoff at high energies. Basically the interaction probability for photons of energies above 10^20 eV becomes so high, that all have interacted before they can reach us.

But why is there a limit at lower energies? Earth magnetic field and/or athmosphere, radiation belt? Perhaps someone can explain that to me.

share|cite|improve this question
From where did you here there was a low energy cut-off? Technically, there isn't. Stable particles (alphas, protons, neutrinos etc) can have arbitarily low energies. Of course there comes a point when either; we would tend not to refer to them as cosmic rays (most would be solar wind for example), or when detector technolgy prevents us from seeing them. However, these don't really constitute as a 'cut-off' in the traditional sense of the word. – qftme Feb 23 '12 at 12:58
In a talk a lower limit at about 1GeV was mentiond as a side note. – con-f-use Feb 23 '12 at 13:03
Below 1GeV particles aren't generally considered relativistic and therefore, when they bombard our atmosphere from outer-space, they're just not normally refered to as cosmic rays. Furthmore, sub-GeV particles would very quickly thermalise in our atmosphere (Ie exhibit Brownian motion) and therefore retain no information about from where they came. – qftme Feb 23 '12 at 14:05

2 Answers 2

I am not exactly sure which low energy cutoff you refer to; however, there is a low-energy cutoff for photons that I am aware of. Photons with energies on the order of $H_0\sim10^{-33}\text{eV}$ would be super-horizon modes. That is, their wavelengths would be on the order of the Hubble radius, $H_0^{-1}=14.6~Gly$. Larger than this would mean that the photon's peaks are essentially acausal as they would fall outside the current comoving horizon of every other peak. As such, we also could not measure such signals; because we could not see the full wave or the periodicity, it would look like preferred orientations of the EM fields on those scales.

share|cite|improve this answer

Your question covers a number of different topics.

The GZK process refers to hadrons not photons. Also, it isn't exactly some kind of hard cut-off. The mean free path is on the order of several Mpc (which is comparable to the distance between neighboring galaxies) and even then only 20-50% of the energy is lost at each interaction. So particles with $E>E_{GZK}$ will still traverse quite a distance.

On the lower end, protons can go as slow as they like (I feel like I'm going pretty slow) which brings up the question of relativity. The speed of the proton only depends on some reference frame which, for the GZK effect, is the CMB.

As for other low energy particles there can be some low energy effects depending on the particular conditions, but nothing as general as the GZK against the CMB. See here (wikipedia).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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