Why is the zenith angle distribution of the muons cos2(x) ?

1. Nov 27, 2013

phys_student1

Hello,

Empirically, the flux distribution of cosmic ray muons follow cos^2(θ) where θ is angle of incidence. Looking up the papers, I did not find any clue as to why is this the case. All sources simply consider this an experimental fact.

Is their any real explanation for this?

2. Nov 29, 2013

Staff: Mentor

That is a common question and I never saw a good answer. It is probably just some function which is not so far away from the real distribution, without a deeper physical argument behind it.

3. Nov 30, 2013

atha

First of all consider that the earth's atmosphere is a perfect sphere.
The path length for θ=0 is the minimum possible.
Also note that there no chance to detect a muon for θ=π/2(for this specific angle the muon would have to travel along the earth's crust, it would collide and interact with the crust's atoms with a very high probability).

Those arguments state that we need a cos-like distribution.

The next step is to think that the muon obeys the Bethe-Bloch formula. In simple words the energy loss per unit length, is something like 1/β^2...

This implies that it is very much likely to observe a muon for small θ(where the path length is minimum) rather that large θ, in a non-linear way.

If you also put in mind that the earth's atmosphere isn't a perfect sphere, but it's a "3D ellipsoidal", you are again moving away from linearity.

I believe that Andersson has done some "fitting" on experimental data of cosmic ray flux, which proves the cos^2θ distribution.

So it is an experimental model, as far as I know.

4. Nov 30, 2013