How much powerful is an accelerator

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juan avellaneda
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hi all

its has been told in books that a particle accelerator can reproduce the initial conditions in the Universe
But we know that this gadgets have very high voltages but low currents
so the product V*I = Energy is not too high to assert that
I think lighting can carry much more energy than this, although it don't carry much voltage , it carry millions of amperes.
So i think this is a missconception that shoul be reviewed.
 
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Even a falling piece of chalk may have a higher kinetic energy than that achievable in Fermilab's Tevatron.

The difference comes when you consider the energy per particle.
 
Originally posted by juan avellaneda
hi all

its has been told in books that a particle accelerator can reproduce the initial conditions in the Universe
But we know that this gadgets have very high voltages but low currents
so the product V*I = Energy is not too high to assert that
I think lighting can carry much more energy than this, although it don't carry much voltage , it carry millions of amperes.
So i think this is a missconception that shoul be reviewed.

... or that there is a misconception of your understanding. Keep in mind that one doesn't spend billions of dollars to build something that has this kind of, let's face it, elementary misconception.

You need to understand what is meant by "energy per nucleon", energy in a center of mass frame especially when you have two incoming, colliding beams, etc. If you accept that the energy per particle have energies of the order of GeV or TeV (you're welcome to visit Fermilab, RHIC, or CERN to verify this), then there are no "misconceptions" here.

Zz.
 
Units, anyone?

V * I = Watts. Watts = Joules / second, or energy per unit time. The total beam power at Fermilab during collisions at the top of a stack (when beam current is maximum) is typically on the order of 500kW. This is equivalent to one Joule of energy being released in 0.000002 seconds. It is also equivalent to 500kJ in one second. There is a disconnect in the units of the question.
 
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