# How to check the particle rate of $1\ m^{-2}.s^{-1}$ at an energy of $10^{12}\ TeV$ from the “Swordy plot”? [duplicate]

On this web page, we see the so-called "swordy plot"

https://www.quantamagazine.org/ultrahigh-energy-cosmic-rays-traced-to-hotspot-20150514/

At an energy of $$10^{12}\ eV=1\ TeV$$, we read vertically $$10^{-4}$$.

The radius of earth is 6371 km : that makes $$4\pi R^2=5.1\times 10^{14}\ sr$$.

I deduce that the flux for $$1\ TeV$$ is : $$5.1\times 10^{14} * 10^{-4}=51\times 10^9 m^{-2}.s^{-1}$$.

But wikipedia claims at this energy : $$1 m^{-2}.s^{-1}$$.

Where is my mistake?

I don't know if the "GeV" is important in the vertical axis. But if I multiply by $$10^9$$, it will certain not help to go in the "good" value.

• Also, you calculated surface area, not solid angle. – J. Murray Jan 17 at 14:37
• @J. Murray : thanks a lot for the explanations and reference – Mathieu Krisztian Jan 17 at 14:49