humanino said:
It seems to me confusing in the context of the original question to hide [itex]\hbar[/itex] and c. Vanadium 50's and my result are on the first page of the particle data group booklet or review, so I think we were justified not to give further details (unless requested). I'd like to request a clarification about your formula. It seems to me, energy and length have inverse dimension for instance. Can you please re-establish the proper [itex]\hbar[/itex], c and (probably) G factors ?
Well, I agree that the OP really meant GeV^(-2) and agree with your answers. Now, if you put G = 1, then of course, any power of GeV could be a cross section (because you've made physics dimensionless).
Now, I don't work in particle physics so, I don't have the conversion factors in my head. So, what I always do is use a few well known formulae that contain hbar, c and G to do the conversion.
To convert GeV^2 to a cross section, you can use that in General Relativity, mass and length have the same dimensions (if you put c = G = 1). So, GeV^2 is already a cross section and no additional conversion using hbar needs to be performed.
To restore G and c, we just hijack the formula for gravitational potential energy, so:
m^2 G/r = energy = m c^2
this is a dimensionally correct expression, that doesn't need to make sense. So, we have:
m G/(c^2 r) = dimensionless
Or:
E G/(c^4) = length
where E is an energy. So, we see that:
cross section = E^2 G^2/c^8
If you know the formulas for Planck length, Planck energy etc. etc., you can do the conversion directly. To convert GeV^n to a cross section, you simply divide this by the Planck energy to the power n and multiply by the Planck length squared.