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It would be pretty spectacular.

Each proton would, at the full power of 7 TeV, have 1.12 microjoules of energy. 1.15x10^11 protons per pulse, 2808 pulses per beam and two beams, one of antiprotons, and one of protons; for a total energy of 352,235,520 joules. 87 kilograms of TNT.[1][2]

Your hand would evaporate fairly quickly, then turn into a plasma, then get hot enough to start radiating x-rays. There would be quite a lot of bremsstrahlung from the hyperenergetic protons punching through the cloud of plasma, and producing showers of secondary radiation,[3] which means you would be quite well irradiated by the time the shockwave from the explosion killed you.

1: All this is straight from the wikipedia page, but I double checked the math.

2: I'm not including energy liberated from antimatter annihilation energy, since the total mass of the antiprotons is quite small.

Typing that out, it sounds like a pretty lame excuse. Let's do the math.

3.2292x10^14 protons per beam. Atomic weight of 1, of course, so:

(3.2292x10^14)/(6.0221415x10^23)[4] = 5.362212x10^-10 grams. .536 nanograms of antimatter.

Since annihilating antimatter gets you 9x10^13 joules per gram, you get... 48,259.9089 joules. That's actually larger than I expected, but .01% of the kinetic energy of the beam.

3: Just how much secondary radiation, I don't know, since that depends on the density of the cloud of plasma, which would change over time, be pretty anisotropic, and be a general pain in the ass to model.

4: Avogadro's constant

Buncha edits: forgot HN uses the asterisk to style text.




Can't edit this now, of course, but as fjh pointed out elsewhere, I confused the details of the LHC for the LEP, partially. The LHC collides protons and sometimes lead nuclei; but at no point collides antiprotons.

Posting after midnight: Not always a good idea.


Don't worry, you had me at "bremsstrahlung".


Braking radiation ? What they don't teach that in school anymore ?

Just kidding.




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