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After some reading about the Large Hadron Collider and it's very impressive instruments to detect and investigate the collision results, there is a remaining question.

What would happen if the scientists would use leptons instead of hadrons?

Especially: What would happen if they would collide electrons?

Isn't it intrinsic that all particles consist of smaller particles? With current technology, could we detect them?

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@EduardoGuerrasValera: that's a very strange idea. I don't know any working physicist who believes that leptons are composite. –  Vibert Feb 10 '13 at 22:33
I'm with Kostya in that answer not mentioning LEP are missing a important point, though I don't think "Do you know about LEP?" is enough to answer the question. –  dmckee Feb 10 '13 at 22:53
@Vibert, as long as it is not needed for explaining something or at least simplifying the existing models, there is no need for that but... It seems sort of fishy that the electron carries exactly three times the electric charge of a quark. –  Eduardo Guerras Valera Feb 11 '13 at 0:27
Hi Mare Infinitus, and welcome to Physics Stack Exchange! Could you be more specific about what aspect of electron collisions you're curious about? For example, "If the LHC could collide electrons, would we be able to detect their constituents, if they have any?" is much better than "What would happen if...?" –  David Z Feb 11 '13 at 2:52
@EduardoGuerrasValera Well, at least within the context of grand unified theories the charges of the leptons and quarks are fixed by group theory without invoking any composite structure of leptons. Similarly you can argue for charge relationships within the standard model by anomaly cancellation, and anomaly matching conditions place very strong constraints on any theory of lepton or quark compositeness. Nevertheless such models have been proposed and model-independent searches performed but so far nothing as turned up. –  Michael Brown Feb 11 '13 at 8:54

6 Answers 6

up vote 18 down vote accepted

First of all -- it wouldn't be called "the Large Hadron Collider", right?
Looks like one would rather call it something like "Large Electron-Positron Collider".
In that case one definitely would need another abbreviation for it. Something like "LEP" instead of "LHC"...

Now, guess what was there in the same tunnel before?

Edit: since my shenanigan got popular, I'll elaborate.

  • Yes, they actually was colliding electrons and positrons, not electrons-electrons. Mainly because of the richer physics of such collisions. (But for my theoristish point of view: positron is just an electron going back in time.)

  • Why the same tunnel? Perhaps surprisingly, tunnel is taking a substantial part of the cost of an accelerator. Digging a new one for LHC would have definitely burnt a large hole in CERN's pocket.

  • Given a fixed circular tunnel (it's radius) you actually have a bound on energy you can have for your particles. Due to synchrotron radiation -- see @emarti answer for more.

  • 27 kilometers seems to be a reasonable limit on the size of a circular tunnel. (Actually people think about 233 km, but that sounds crazy to me.) So the next accelerator most probably will be linear and it will be electron-positron.

P.S. Have you heard of a Photon Collider?

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An electron-positron collision is rather different from an electron-electron collision –  Martin Beckett Feb 10 '13 at 22:21
@MartinBeckett: it seems reasonable to assume that OP meant e+e- and not e-e- collisions. Just try and draw some tree level diagrams for the latter - it's a fun exercise, but not so nice in terms of EW physics. –  Vibert Feb 10 '13 at 22:37
@Kostya, nice link!(+1) You fooled me, I already was about to downvote. Why did the dismantled the LEP for building the LHC? Is it not a waste? Couldn't they simply build the LHC in another, nearby place? –  Eduardo Guerras Valera Feb 11 '13 at 1:48
@EduardoGuerrasValera - there weren't many spare 27km long tunnels available! Colliders have a natural life, once they have generated a statistical number of events they are no further use, the data remains to be further analysed but new colliders will target new energies and new physics. –  Martin Beckett Feb 11 '13 at 3:12
@MartinBeckett: Nice answer! –  Mare Infinitus Feb 11 '13 at 17:33

There are two points in answering this question:

  1. Design: The design of the collider would have to be different. Electrons/positrons in a cyclotron radiate synchrotron radiation when they are accelerated (which itself is a useful device). To get above a few GeV, researchers use linear accelerators, such as SLAC. The proposed International Linear Collider is a design intended to reach TeV energies, close to what the LHC already achieves with protons.

  2. Science: Yes, electron-electron or electron-positron collisions are very useful in studying particle physics. The signal is 'cleaner', since electrons are are not composite particles, and it's easier to calculate the cross-sections. By contrast, it is very challenging to calculate how two colliding protons, with six quarks, will decay, to say the least. A classic story of electron-electron and proton-proton colliders complement each other was the discovery of the J/psi meson. The general idea I've heard is that proton-proton colliders can reach higher energies, but electron-positron colliders tend to have better energy resolution and cleaner signals.

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Good answer; to improve it, you should mention LEP! –  Vibert Feb 10 '13 at 22:33
Note that there is currently a lot of interest in electron–ion collisions, though mostly on the intensity frontier rather than the energy frontier (to use the current vernacular). –  dmckee Feb 10 '13 at 22:56
I don't get point one. The use protons and ions now, which are also charged particles. So they also radiate synchroton radiation, what's the difference beside the mass? –  Noldig Feb 11 '13 at 8:46
@Noldig The mass is the key difference. The radiated power is proportional to $m^{-4}$ (see the wiki page), so electrons radiate $10^{13}$ times more than protons! –  Michael Brown Feb 11 '13 at 8:59
That's ok, but for me point one suggests that there is no synchrotron radiation at all. Maybe one could include the m^-4 behaviour into the awnser and say, that they would radiate much more synchrotron radiation than protrons –  Noldig Feb 11 '13 at 9:32

The very basic reason LEP stopped going to higher energies ( it reached over 200GeV center of mass, at the last stage, LEPII) and the tunnel was used for LHC is synchrotron radiation .

Note that radiated power is proportional to 1/m^4

It is not possible to feed a circular beam of electrons the energy needed to raise it to higher energies at the radius of LEP, it is a loosing game, The energy would go into feeding synchrotron radiation. The reason the same radius can be used for much higher energy protons is the ratio of the masses of electron to proton.

Synchrotron radiation is not present in linear colliders and that is why the next electron positron accelerator will be a linear collider the ILC.


What would happen if the scientists would use leptons instead of hadrons?

It has happened at LEP and LEPII with electrons on positrons. If the scattering is not elastic a lot of hadrons appear, as well as leptons and Z bosons. The data from LEP confirmed the calculations of the standard model for elementary particles to great accuracy.

Especially: What would happen if they would collide electrons?

from the previous paragraph, the standard model predicts what would happen if electrons were scattered on electrons : all the variants of the feynman diagram possibilities would appear also.

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Especially: What would happen if they would collide electrons?

In case you really meant electron on electron (and not electron on positron): for a 'discovery' machine it is useful to have a initial state which is 'neutral' in all respects: no net charge (electrons and positrons have opposite charge; protons not only contain quarks but also a significant amount of anti-quarks and gluons), no net lepton flavour etc.

So far, lepton flavour is almost entirely conserved, so an e-e- collider would predominantly produce final states with an electron flavour number of two which would be quite unnatural for a new fundamental particle.

What would happen if the scientists would use leptons instead of hadrons?

Others have pointed out that this happened until the year 2000 in the LHC tunnel (LEP) and is limited by the synchrotron radiation losses (while LHC is limited by the achievable fields in the bending magnets).

There is also the concept of a muon collider which would have features similar to an e+ e- collider (known initial state four momentum etc.) but this is technologically very challenging, mainly due to the liftetime of the muon of only two microseconds. It would however for example allow to measure the mass of the Higgs particle to keV precision if I remember well (through scanning of the beam energy, similar to the determination of the mass and decay width of the Z particle at LEP).

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What would happen if the scientists would use leptons instead of hadrons?

They would go in the opposite direction - electrons are the opposite charge to protons. The LHC doesn't use electrons because protons are 2000x heavier so you get a lot more energy in the collision.

Especially: What would happen if they would collide electrons?

Not a lot - they would bounce off each other. Electrons (as far as we know) don't break apart (and not at these low energies)

Isn't it intrinsic that all particles consist of smaller particles?

Probably not. We don't know of any components of an electron, but we also don't know why it should have the same charge as a proton when it's not made of the same stuff - so I wouldn't take any bets.

With current technology, could we detect them?

Then logically those would be the smallest particles we could detect ... and so on ....

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Thank you for your fast answer! –  Mare Infinitus Feb 10 '13 at 21:23
@MareInfinitus - somebody who is more familiar with the LHC will be along to provide a better answer soon, but it's late on sunday night at CERN –  Martin Beckett Feb 10 '13 at 21:27
"Electrons (as far as we know) don't break apart (and not at these low energies)": I'm scared to imaging what "high" energies are! E.g.: slac.stanford.edu/cgi-wrap/getdoc/slac-pub-7436.pdf –  emarti Feb 10 '13 at 21:55
If the electron has constituents it must break apart at "some" energy! And if it doesn't - you need a good explanation of why it has the same charge as the quarks –  Martin Beckett Feb 10 '13 at 22:20

This is an interesting question, for which some interesting answers have been given from various points of view.

I am intrested in what will happen in the case of the electron-electron collisions at these high energies. It is true that at low energies the two electrons will bounce of each other, not much happening. At very high energies, however, I thought that the following outcome might be possible (it is possible to draw a Feynman diagram, to calculate the probability amplitude for such an event.) Sorry I have not learnt how to bring in drawings, so I will just describe it:

As the $e^-$ and $e^-$ approach each other they could interact via the weak force. Electron 1 would emit a $W^-$ and turn into a $\nu_e$ which would emerge as a free particle. The $W^-$ could interact electromagnetically with electron 2 and both particles would scatter at some other directions. The $W^-$ would then decay into $e^-+\bar{\nu_e}$. Therefore, the $e^--e^-$ collision could serve as a $\nu_e-\bar{\nu_e}$ generating machine?

$e^-+e^- -->e^-+e^- + \nu_e+\bar{\nu_e}$.

Unfortunately, we don't know of any other mechanism to forcast entirely new physics.

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I nearly upvoted this for "Unfortunately, we don't know of any other mechanism to forcast entirely new physics." :) –  Michael Kjörling Feb 11 '13 at 8:43
At LHC energies many other processes would be possible. The Ws could decay into many other particles, or interact and produce a Higgs, or a photon mediated exchange could produce quark/antiquark pairs.... many many possibilities, just scratching the surface here. I'm sure someone has done detailed calculates to see which processes would dominate, but I don't know off hand. –  Michael Brown Feb 11 '13 at 9:06
@John: Please leave a link if you find something. –  Mare Infinitus Feb 11 '13 at 17:31
@John Very interesting, thanks! The left-right model is on my list to consider carefully for my Ph.D project, although I'm not working so much on the collider phenomenology side of things. Very interesting read nevertheless! –  Michael Brown Feb 12 '13 at 13:12
@John Not at all. Even the standard model breaks lepton number non-perturbatively. In fact, it's a good thing for leptogenesis. :) Of course you have to make sure the rate below the weak scale is low enough to evade detection... –  Michael Brown Feb 13 '13 at 0:57

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