Why I’m not a physicist: reason #4328

There’s a trivial question about particle accelerators that bugged me for a while. Today I finally figured out the answer, and I’m so excited by my doofus “discovery” that I want to tell the world.

In Ye Olde Times, accelerators used to smash particles against a fixed target. But today’s accelerators smash one particle moving at almost the speed of light against another particle moving at almost the speed of light — that’s why they’re called particle colliders (duhhh). Now, you’d think this trick would increase the collision energy by a constant factor, but according to the physicists, it does asymptotically better than that: it squares the energy!

My question was, how could that be? Even if both particles are moving, we can clearly imagine that one of them is stationary, since the particles’ motion with respect to the Earth is irrelevant. So then what’s the physical difference between a particle hitting a fixed target and two moving particles hitting each other, that could possibly produce a quadratic improvement in energy?

[Warning: Spoiler Ahead]

The answer pops out if we consider the rule for adding velocities in special relativity. If in our reference frame, particle 1 is headed left at a v fraction of the speed of light, while particle 2 is headed right at a w fraction of the speed of light, then in particle 1’s reference frame, particle 2 is headed right at a (v+w)/(1+vw) fraction of the speed of light. Here 1+vw is the relativistic correction, “the thing you put in to keep the fraction less than 1.” If v and w are both close to 0, then of course we get v+w, the Newtonian answer.

Now set v=w=1-ε. Then (v+w)/(1+vw) = 1 – ε2/(2-2ε+ε2), which scales like 1-ε2. Aha!

To finish the argument, remember that relativistic energy increases with speed like 1/sqrt(1-v2). If we plug in v=1-ε, then we get 1/sqrt(2ε-ε2), while if we plug in v=1-ε2, then we get 1/sqrt(2ε24). So in the case of a fixed target the energy scales like 1/sqrt(ε), while in the case of two colliding particles it scales like 1/ε.

In summary, nothing’s going on here except relativistic addition of velocities. As with Grover’s algorithm, as with the quantum Zeno effect, it’s our intuition about linear versus quadratic that once again leads us astray.

10 Responses to “Why I’m not a physicist: reason #4328”

  1. Anonymous Says:

    Warning: Dumb question follows.

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

  2. Scott Says:

    If that were all there was to it, then by colliding two particles (and thereby doubling the effective velocity), you’d expect to multiply the kinetic energy by 4 (as happens in head-on car crashes). You certainly wouldn’t expect to square it.

  3. Perseph0ne Says:

    Huh. Is there a good way to see how this doesn’t, you know, defy conservation of energy, or is my brain just glitching?

    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?

  4. Anonymous Says:

    I think there is a typo. (1 – ε^2)/(2-2ε+ε^2) scales as ε. The true expression seems to be (2 – 2ε)/(2-2ε+ε^2), which scales as ε^2.

  5. Anonymous Says:

    Scott, I think you are wrong. You compare the total energies for both experiments in a frame in which one particle is at rest. That is not what you want to do. You want to compare the energies in the zero-momentum frames, because in such frames you use all the available energy to produce the fattest (most massive) possible particle at rest.

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

    pa = mgamma[1, v]
    ~ m gamma [1,1]
    pb = m[1, 0]
    = m^2 ( ( gamma+1)^2- gamma^2 )
    ~ m^2 2gamma
    so energy in zero-momentum frame
    = msqrt(2gamma)

    pa = mgamma[1,1]
    pb = mgamma[1,-1]
    = (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.


  6. Anonymous Says:

    Now Scott, that is not your area. If this turns out to be wrong (going by the previous comment), then you have got reason #4329 (for not being a physicist).

  7. Scott Says:


    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!

  8. Scott Says:

    rrtucci: Thanks! Now that I’ve had a chance to think about your comment, it does indeed look like reason #4329 for my not being a physicist.

    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?

  9. Scott Says:

    perseph0ne: Yeah. See my next post.

  10. Scott Says:

    Here’s a further thought about zero-momentum frames. Instead of asking “How can we possibly get a quadratic improvement in energy?,” we might just as well have asked, “Why do we only get a quadratic improvement, and not a much larger one?”

    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.