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This graph is to be found in various Articles, Papers, and websites.

enter image description here

I am trying to understand, what the "y axis" represent. Are they refering to electromagnetic Flux ? What is this unit being used ? (m^2 , Sr, s Gev)^-1

Furthermore, what is meant by 1 Particle per m^2 sec\year, so on?

Thank you

Mad
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1 Answers1

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This graph shows the distribution of cosmic rays detected on ground (per unit of time [s], surface [m$^2$], energy [GeV] and solid angle ["sr", i.e. steradian]), with respect to their energy [eV].

It is not an electromagnetic flux, but a particle flux. Cosmic rays are not photons, they are high-energy particles traveling through space (hadrons, leptons; for example protons and electrons, etc.)

You can read the graph as follows : on Earth, the incoming flux of cosmic rays with an energy $E= 10^{19}$ eV is approximately $10^{-24}$ m$^{-2}$ sr$^{-1}$ s$^{-1}$ GeV$^{-1}$. It means that on a 1 kilometer surface, it would require (statistically) :

$ 1 ~ / ~ (10^6 \mathrm{[m \rightarrow km~surface]} \times 10^{19} \mathrm{[particle~energy, "x~axis"]} \times 10^{-24} \mathrm{[flux, "y~axis"]} \times 10^{-9} \mathrm{[eV \rightarrow GeV]} \times 2\pi \mathrm{[full~hemisphere~coverage]}) = 10^{8} / 2 \pi~\mathrm{seconds}$

to observe one cosmic ray with that energy, or approximately 6 months.

By the way, this is why the errorbars are larger at the bottom-right end of the graph. These high-energy events are very rare.

Gonstasp
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    Hello, thank you for explaining. Could you please clarify where the calculation is coming from? i understand that the flux is equal to the number you gave when the energy is equal to $10^19$ eV, but how do you conclude that if i divide 1 / the numbers you choose i will get the time required?

    Furthermore, you choose 1 Km as surface Area, why do you choose then 4pi for the steradiant? (full sky coverage) wouldnt that imply, coverage of all earths surface?

     – Mad Dec 13 '22 at 10:44
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  • That is how you work with a flux. Given a flux of particle per second $F$ [$T^{-1}$] (the dimension of $F$ is $time^{-1}$), the time required to observe one particle is $1 / F$ [$T$] (the dimension of $1/F$ is $time^1$). The dimension of this flux is not $T^{-1}$, so I also had to multiply by the surface, solid angle and energy.

  • It's true that using $4 \pi$ does not make sense, except for muons since they can travel through Earth. If we assume our 1 km detector to be sensitive to all directions in a full hemisphere, the correct value would be $2 \pi$ steradian.

    • – Gonstasp Dec 18 '22 at 11:51