For think of one stationary particle, and another particle heading towards it at close to the speed of light. Naively, this seems equivalent to two particles heading toward each other at *half* the speed of light. And that’s going to give a crappy, *constant* amount of energy, no matter how close to the speed of light the second particle is going.

I guess the answer has to be that, when you do the Lorentz transform, the situation will *not* be equivalent to two particles heading toward each other at half the speed of light, and will instead be equivalent to two particles heading toward each other at close to the speed of light — but not *as* close as in the collider situation.

But here’s the strange thing: assuming you’re correct that the zero-momentum frames are what we really want to look at, we *still* get a quadratic improvement using a collider over a fixed target — though in this case the improvement is 1/sqrt(ε) versus 1/ε^{1/4}. So I somehow got the high-level answer I wanted to get (quadratic improvement in energy), and from the right source (relativistic addition formulas), but by making a lousy assumption (one of the particles is at rest). Any idea what’s going on here?

*I think there is a typo. (1 – ε^2)/(2-2ε+ε^2) scales as ε.*

Firstly, I meant 1 – ((ε^2)/(2-2ε+ε^2)), not (1 – ε^2)/(2-2ε+ε^2). Secondly, the expression that you wrote scales as 1/2!

]]>let gamma = 1/sqrt(1-v^2)

~ 1/ sqrt(2(1-v))

=1/sqrt(2epsilon)

EXPERIMENT 1:

pa = mgamma[1, v]

~ m gamma [1,1]

pb = m[1, 0]

(pa+pb)_mu(pa+pb)^mu

= m^2 ( ( gamma+1)^2- gamma^2 )

~ m^2 2gamma

so energy in zero-momentum frame

= msqrt(2gamma)

EXPERIMENT 2:

pa = mgamma[1,1]

pb = mgamma[1,-1]

(pa+pb)_mu(pa+pb)^mu

= (2mgamma)^2

so energy in zero-momentum frame

= m2gamma

In conclusion, the energy available for creation of new particles in a collider goes as gamma~1/sqrt(2epsilon),

not as 1/epsilon as you claim.

rrtucci

]]>I mean, you take one particle, accelerate it, that takes energy E,

you take another, that’s 2E.

How on earth do you end up with E^2?

]]>Why isn’t this an immediate consequence of the fact that kinetic energy is proportional to the square of velocity?

]]>