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

by phys_student1
Tags: angle, cos2x, distribution, muons, zenith
phys_student1 is offline
Nov27-13, 08:42 PM
P: 88

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?
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mfb is offline
Nov29-13, 09:06 AM
P: 10,795
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.
atha is offline
Nov30-13, 10:29 AM
P: 7
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.

Vanadium 50
Vanadium 50 is offline
Nov30-13, 11:14 AM
Vanadium 50's Avatar
P: 15,574

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

This is an approximation, and as such, you won't be able to derive it.

If cosmic rays were uniform and isotropic, you'd expect a distribution that goes more like 1/cos(theta), because there's not a lot of area with a small zenith angle compared to a large one. However, as atha points out, we have an atmosphere, so cosmic rays from the horizon are unlikely to make it to us.
phys_student1 is offline
Nov30-13, 02:24 PM
P: 88
Thanks all. Yes I also looked too much in the literature without finding any clue, all state is as empirical formula or experimental fact. Thanks for your responses.

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