Ok, this might sound like a silly question, but I was wondering, when particles (e.g. protons) are smashed together in the LHC, why do they break up into dozens of other particles, as opposed to just bouncing off of each other elastically?

I'm guessing the full explanation is probably going to involve some fairly in-depth quantum analysis of particle interactions, but can anyone explain it in a fairly straightforward way that someone who isn't an expert in QM can understand?

Presumably, there will be some threshold energy level below which this doesn't happen? For example, I assume protons in a lower-energy hydrogen plasma will be bouncing off each other all the time?

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    $\begingroup$ A the right impact parameter they do, but they don't end up in the detector. $\endgroup$ Commented Apr 30, 2015 at 16:26
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    $\begingroup$ This answer might be helpful. $\endgroup$ Commented Apr 30, 2015 at 16:38
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    $\begingroup$ As an analogy - Collide two marbles together. If the collision energy is small, then they simply bounce off each other. If the energy is large enough, they smash and disintegrate into their constituent pieces. The same is happening with protons at the LHC. $\endgroup$
    – Prahar
    Commented Apr 30, 2015 at 17:21
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    $\begingroup$ BTW - A proton is NOT an elementary particle. It is comprised of quarks. $\endgroup$
    – Prahar
    Commented Apr 30, 2015 at 17:22
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    $\begingroup$ I had a short assignment to work on a design proposal for a detector for LHC that was meant to measure these soft collisions in the late 1990s. The problem was that the number of these scattered particles was so large that it was heating up large copper blocks quite considerably. No solid state detector material can withstand this radiation for more than a couple of days, if I remember correctly. The proposal was therefor ti use an old style gas ionization chamber that could, of course, not be damaged by the radiation. I don't know what happened to the design in the end. $\endgroup$
    – CuriousOne
    Commented May 1, 2015 at 3:18

3 Answers 3


Elastic collisions do happen at the LHC. The TOTEM experiment measures the differential cross section (rate as a function of angle) for proton-proton elastic scattering at the LHC. Here is their latest result. They don't publish an estimate of the elastic cross section, but according to their data it must be at least 25 mb (millibarns) (my first version of this post had a mistake--the headline 100 mb number shown in the abstract is a measure of the total pp cross-section which includes both elastic and inelastic contributions). Compare this to the production cross-section of the Higgs boson at the same collision energy, which is about 20 pb (picobarns). This means that when two protons collide at 8 TeV, they are over a billion times more likely to bounce off each other than they are to produce a Higgs boson.

As others have pointed out, the general-purpose detectors like CMS and ATLAS are not designed to detect the elastic collisions. The elastic collisions occur mostly at forward angles, meaning the protons are just barely deflected from their original trajectory (think of a glancing collision between two billiard balls rather than a head-on collision), while the more exotic physics tends to produce particles that go more perpendicular to the beam direction.

  • $\begingroup$ Great answer, thanks! However, do all the collisions that occur 'head-on' cause the protons to smash and produce other particles? If so, why couldn't elastic scattering happen in that case? $\endgroup$
    – Time4Tea
    Commented Apr 30, 2015 at 21:24
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    $\begingroup$ @Time4Tea A greater deflection angle means that more momentum is transferred from one proton to the other which means that smaller distance scales are probed (smaller $\lambda$), which (see James Cowley's answer) means that you are more likely to interact with one of the constituents of the proton rather than with the proton as a whole. In the TOTEM paper you can see that the elastic differential cross-section decreases exponentially with increasing momentum transfer. Is there an angle at which the elastic DCS becomes so vanishingly small that we could say it 'never' happens? I'm not sure. $\endgroup$ Commented Apr 30, 2015 at 23:03
  • $\begingroup$ @Will The "never happens" value can be defined for an experiment just fine, even though it never goes away in theory. When the expectation for the number of such events observed drops far below one over the entire course of the experiment you are in "never happens" territory. $\endgroup$ Commented May 1, 2015 at 17:16

Anything that is not forbidden must happen.

That's an important statement to keep in mind when approaching quantum physics. It doesn't mean that anything that can happen always happens, but it must happen at some time or another just like someone eventually has to win the lottery.

That said, some protons do go through the LHC, ram into each other and bounce off elastically. Nothing forbids this; the energy and momentum is conserved as is charge, particle number, etc. Therefore it must happen. But this is a boring result, so we don't care to have detectors specifically to measure this occurrence (tell you the truth, someone is probably interested in that result because that's not forbidden either. So maybe they have a detector). Additionally, at the high energies these protons are colliding (it was $4TeV$ per proton beam but as of this year it's $6.5TeV$), quantum physics starts allowing for other exciting things to happen.

I'll spare you the in-depth quantum details, but when that much energy is contained in the space the protons are colliding, there exists the possibility for many things to happen.

  • The protons can break apart into other particles, such as mesons if the collision kicks one or more quarks out of the a proton.

  • The high amount of energy can allow for production of particle-antiparticle pairs (so long as this production conserves all the necessary quantities).

  • The quarks and gluons that make up each proton can combine and interact to produce larger particles, heavier quarks, or they can interact and produce heavier bosons or other rare particles, like the Higgs boson.

The high energy is what allows for all of this to happen (that is, it dramatically increases the probability of some things and "unforbids" other things). There certainly is an energy scale where you can't get heavier particles produced and where the protons don't break apart. You need a minimum energy to overcome the forces holding the proton together. And for particle production, you need at least as much energy as the mass of the produced particles holds.

Because there are around 600 million[citation needed] collisions per second, this really is a case of that all important statement above. Quantum physics allows for all these fabulous other particles (and the resulting decays and interactions of them) during the collisions, therefore they must happen. It also allows for elastic scattering of the protons and that does happen, but who wants to see two protons bounce off each other at high speed? Not I. The detectors and energy of collisions are designed in order to maximize (within our technological capability) the probability of interesting things happening.

  • $\begingroup$ Thanks for your answer @ACuriousJim. That description of quantum mechanics reminds me in some ways of Murphy's Law (although I'm not saying protons smashing apart = 'going wrong') :-) $\endgroup$
    – Time4Tea
    Commented Apr 30, 2015 at 21:30

elementary particles (e.g. protons)

Protons aren't elementary particles, they're made of partons (quarks and gluons) in "soup".

Below, $\lambda$ is the wavelength corresponding to the energy of the interaction via the usual de Broglie relation and $r_p$ is the radius of the proton.

At low energy with $\lambda >> r_p$ the interactions are just like you describe, the protons are point-like and postively charged and repel each other.

At mid-range energy with $\lambda \approx r_p$ the proton is no longer point-like but behaves as a uniform charged body.

At high energy with $\lambda < r_p$ the spatial resolution is precise enough to involve interactions between individual quarks and the proton behaves as a bundle of 3 quarks (uud).

At very high energy with $\lambda << r_p$ such as those currently reached by the LHC the 'true' nature of the proton is revealed as containing an ever changing, bubbling soup of quarks and gluons popping in and out of existence (even other quarks besides the u and d that are "normally" present, which is how we can produce B-mesons in the LHCb detector even though at first glance you'd think there are no b quarks available: a certain fraction of the time a b quark from the quark-gluon soup within one proton interacts with a quark or gluon from the other proton via an exchanged gluon and then goes flying off before hadronising into a jet of hadrons, including B-mesons).

So basically, at low energies protons do behave just like positively charged point-like particles and bounce off one another, but at high enough energies the wavelengths of the exchanged bosons become small enough to single out the individual constituent quarks and gluons within the protons and they therefore interact individually.

why do they break up into dozens of other particles

This is the process I mentioned above, "hadronisation". It's a consequence of QCD confinement. Wikipedia will give you a more detailed explanation if you want one, but a qualitative understanding can be gained by knowing that coloured particles (i.e. quarks and gluons) cannot exist on their own and have to pair-up (red-antired, blue-antiblue or green-antigreen), or trio-up (red-green-blue or antired-antigreen-antiblue) to become "colourless" mesons or baryons, respectively. Mesons and baryons are each types of hadrons.

The coloured partons produced by high energy proton collisions emit gluons which pair-produce quark-antiquark pairs by the hundreds or even thousands. These can all then cluster together and become colourless hadrons. You can imagine how complicated and chaotic this process is, which is one reason it's very hard to model.

  • $\begingroup$ Thanks for your answer! So, is $r_p$ the radius of a proton? $\endgroup$
    – Time4Tea
    Commented Apr 30, 2015 at 21:34
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    $\begingroup$ Yes, I should have stated that explicitly - I'll update that now. $\endgroup$
    – jamcowl
    Commented May 1, 2015 at 12:04
  • $\begingroup$ Of course, you can scatter protons coherently at high momentum-transfer it just makes up a very small fraction of the overall cross-section (falling as $1/t^2$). in my dissertation work we used values of $|t|$ up to $8.1 \,\mathrm{GeV}^2$, which is quite small by LHC standards but still represents a significantly smaller size than the naive proton. $\endgroup$ Commented May 1, 2015 at 17:16
  • $\begingroup$ Yes, the energy scale regions I skimmed over above were slightly hand-wavey and qualitative, really just to describe the overall behaviour but yes, as you say, it's not quite as discrete as all that. $\endgroup$
    – jamcowl
    Commented May 1, 2015 at 22:30
  • $\begingroup$ Could you say a few words about the "soup" ? Are the gluons and quarks "in" a broth? I'm not a physicist. Is there something that holds the particles in suspension? Gravity gravy? $\endgroup$
    – TRomano
    Commented Jul 13, 2017 at 13:24

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