Why collide a moving particle with a particle at rest, rather than two moving particles?

Question

I was just reading some lecture notes about relativistic and quantum mechanics, and in the later part of this page the author demonstrates that any relativistic particle collision in the "lab" reference frame (where one particle is at rest and the other is accelerated) requires significantly more initial kinetic energy than it would in the "center of mass" reference frame (where both particles are accelerated equally). So I was wondering:

Why don't we use the center of mass frame, i.e. accelerate both particles in a collision?

I would guess the answer is something simple like "aiming a 0.5c proton directly at another 0.5c proton is much harder than aiming one 0.9c proton into a big block of stationary protons", but I don't know how accurate accelerators are these days or how many particles are actually involved on each side of a typical high-energy collision.

Answer 1

We do. The LHC accelerates two protons, each with 3.5 TeV of energy, giving a total of 7 TeV in the CoM frame (The energies are from the initial phase of the previous LHC run. Later in the run this was increased to 8 TeV and the combination of the two dataset was what discovered the Higgs boson. The energies are roughly doubling now for Run II, to 13 TeV).

The main reason for this is as you mentioned, the energy involved. In any frame, we have the following invariant quantity, $s = (p_1 + p_2)^2$ which its square root, $\sqrt{s}$, gives the Centre of Mass (CoM) energy for the experiment, and here $p_i$ represents the momentum four-vector for each particle i. In a collision where two particles are moving in opposite directions with equal energy we have the following:

$$ s \equiv (p_1 + p_2) \cdot (p_1 + p_2) = (E + E, \mathbf{p}_1 − \mathbf{p}_2 ) \cdot (E + E, \mathbf{p}_1 − \mathbf{p}_2 ) = (2E , 0 ) \cdot (2E, 0) = 4E^2$$

and now the CoM energy is given by the square root of this quantity, $$\to E_{CoM} = \sqrt{s} = 2E$$

In a an experiment where one of the particles is at rest (has mass $m_t$) and the other is travelling with momentum $\mathbf{p}$ (and has mass $m_b$) we have the following:

$$ s \equiv (p_1 + p_2) \cdot (p_1 + p_2) = (E_b + m_t, \mathbf{p}_b) \cdot (E_b + m_t, \mathbf{p}_b) = E_b^2 + m_t^2 + 2E_b m_t − p_b^2 = m_t^2 + m_b^2 + 2E_b m_t $$

Assuming the masses are negligible, we have the fixed target (FT) CoM energy,

$$\to E^{\text{FT}}_{CoM} = \sqrt{s} = \sqrt{2E_b m_t}$$

Thus we would need much more energy input in a fixed target experiment to achieve the same energies as in the case with two co-moving particles.

EDIT: Regarding a comment below which I think arises from confusion of what the CoM frame is. $\sqrt{s}$ gives the CoM energy in both cases. This is useful because we can now compare between a fixed target experiment and an experiment where both particles are accelerated at the same speed but in different directions.

So, say my collider has the capability to produce a magnetic field which at its maximum, can accelerate a charged particle so as to have energy of 3.5 TeV. Now in the case that we have two particles with the same energy going in opposite directions, we will give a total CoM energy of 7 Tev, following the result above. In the second case though, theres only one accelerating particle, hence $\sqrt{s} = \sqrt{2 \times 3.5 \times m_t}$ and since E $\ll$ m, this is always less than in the first case.

So be careful, because both experiments can be transformed into a CoM frame. In the CoM frame $|\mathbf{p_1}| = -|\mathbf{p_2}|$. Note this is true in both experiments, even in the second case where one of the particles is stationary. Well, the whole point is that we can use the above formulas so we can skip transforming to the CoM frame; we can compute this quantity directly.

Answer 2

Many modern particle accelerators do accelerate both particles towards each other. LEP accelerated electrons and positrons in opposite directions in the same chamber, and the Tevatron did the same for protons and antiprotons. The LHC is a proton-proton collider, and so it has two stacked rings that accelerate protons in different directions. For the BaBar experiment, SLAC accelerated electron and positron beams toward each other (though with a slight difference in energy so that the B mesons had drift, making them easier to separate).

There are also still accelerators that use fixed targets. Some of these are earlier stages for the above accelerators. This is one way to make anti-protons, for example. Usually the experiments being done are different.

Answer 3

Why collide a moving particle with a particle at rest, rather than two moving particles?

The inverse question ("What's advantageous about colliding beams?") has been conclusively addressed already in the answer by Constandinos Damalas.

Why don't we [... always] accelerate both particles in a collision?

We also like to study collisions involving neutral particles, of course, which by themselves would be quite difficult to accelerate. But of course we can for instance accelerate neutrons as constituents of nuclei or (heavy) ions; and we can produce for instance neutral pions or neutrinos at fairly large speeds wrt. our labs based on first having accelerated certain suitable charged particles.

I would guess the answer is something simple like "aiming a 0.5c proton directly at another 0.5c proton is much harder than aiming one 0.9c proton into a big block of stationary protons"

Rather than aiming one individual proton at another, at the Large Hadron Collider (LHC), there are beams of protons with cross-section $\approx 1~\text{mm}^2$ being aimed at each other quite accurately.

These beams are structured in so-called bunches (of $\approx 100~\text{mm}$ "instantaneous lab length"), where each bunch contains about $10^{11}$ protons; cmp. https://lhc-data-exchange.web.cern.ch/lhc-data-exchange/ruggiero.pdf

This gives a proton density (with respect to the lab) of $10^9~\text{p}^{+} / \text{mm}^3$.

By comparison, the proton density in water, which contains 10 protons and 8 neutrons per water molecule is roughly

$$\frac{10~\text{p}^{+}}{18}~\left(\frac{6 \times 10^{23}/\text{mol}}{18~\text{g}/\text{mol}}\right) ~10^{-3}~\text{g}/\text{mm}^3 \approx 2 \times 10^{19}~\text{p}^{+} / \text{mm}^3,$$ where $10^{-3}~\text{g}/\text{mm}^3 = 1~\text{g}/\text{cm}^3$ is of course the density of water, and $18~\text{g}/\text{mol}$ is its approximate molar mass.

Now, there are some experimental tasks where having a target of such high density (comparable to water), sitting in the lab, are obviously much more sensible than creating a comparatively sparse beam of targets; especially neutrino observatories, or detectors searching for some proposed types of dark matter.

Answer 4

Concerning the title of this question, I think luminosity needs to be mentioned. The number of protons per bunch mentioned by @user12262 is related to that.

In essence if you collide your beam with a proton at rest into a macroscopic target there are so many protons in the macroscopic target that you get all the beam particles to collider, whereas if you collide beam on beam most of the particles do not collide.