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Before closing it as a dupe of this. Please go through the question once .

Is it theoretically possible that quarks make up an electron ( like you may get a particle with the same electronic charge $(-e)$ with three down quarks however the binding energy for that down quarks triplet should be a great number since Dr jh pointed out that the mass of even a single down quark is greater than that of an electron) ?

According to This link, the mass of a down quark is approximately $4.8\; MeV$ . So after converting it into $kg's$ and multiplying by $3$ (since I considered three down quarks) , I got roughly $(256 × 10^{-31})kg$ . So the difference in the mass of an electron and three down quarks can be calculated (which is roughly $28 \; Times\; of \;mass\; of\; electron)$ and this serves as our binding energy . So , $E = (247) (9 × 10^{16}) J$.

Have the particle accelerators reached this energy level (since I have heard that the maximum number they reached is $7 \; TeV$) ? I don't know.

And Can this huge energy requirement be the reasons why we can't break down an electron ? Or am I misinterpreting something here ?

Ankit
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Yes! The electron definitely doesn't have to be fundamental. In fact the LHC does searches that rule out electron compositeness up to a certain energy scale.

If you're trying to make up the electron out of Standard Model (SM) quarks, you are going to run into problems:

  1. Why is the electron being bound together at such a higher scale than the typical strong force (or QCD) confinement scale? This suggests that the force holding the electron together is an exotic force. That means whatever quarks are living inside the electron need to be charged under this exotic force (we are now building a BSM model).

  2. If the exotic force confined to form the electron, when the exotic force confined, how do we know we didn't trigger QCD breaking? Worse, in your example, how do you know we didn't trigger electroweak symmetry breaking (EWSB) a la technicolor models? The electron compositeness scale has been ruled out up to scales far above the EWSB scale.

I'm not 100% sure you can't find a clever way to address these two points, but it's hard for me to see a fruitful model that manages to get around these constraints.

The easier way to build a model of a composite electron is to do it with truly exotic fermions that aren't charged under the SM QCD group. You can think of dark quarks charged only under a dark QCD that bind to form the electron.

Another question for further reading: how is the electron so light if its compositeness scale is so high? What happened to the binding energy? Baryons tend to be living at the scale of QCD confinement in the SM.

Well...
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An electron cannot be composed of quarks because quarks are affected by the strong nuclear force whereas an electron is not.

If you combine three down quarks so that they have the same negative charge as an electron, what you have is a particle called a "delta minus". We know that is not the same thing as an electron because it is more than $2000$ times as massive as an electron, and quickly decays into a pion and a neutron.

We believe (very strongly) that the electron is a fundamental particle because in all the millions (billions ?) of particle collisions observed at the LHC and other particle colliders, we have never seen an electron split apart into other particles or show any sign of internal structure.

gandalf61
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  • Your first sentence is an unambiguous answer. The mass argument is not unequivocal. – my2cts Oct 06 '20 at 14:06
  • One can always make a particle from heavier particles. That's what binding energy is for. It always decreases the mass. The fact that 3 quarks don't make an electron is purely observational. – fraxinus Oct 06 '20 at 16:13
  • @fraxinus Surely it is the other way round ? $99%$ of a proton’s mass is binding energy. The individual quarks have very little mass. But when you combine quarks the binding energy due to the strong force gives you a total mass that is very much greater than an electron. – gandalf61 Oct 06 '20 at 16:35
  • Binding energy is negative. That's why it is "binding" in the first place. A hydrogen atom is ~14eV lighter than free proton and free electron combined. – fraxinus Oct 06 '20 at 16:38
  • Your first sentence is incorrect. QCD forms color neutral states when it makes bound states. If the binding energy is quite high, we wouldn't see the interactions with QCD because they would require probing that energy scale. – Well... Oct 06 '20 at 16:54
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    @fraxinus it depends on the force and the charge configuration. How do you explain the difference between the mass of the proton and the mass of its valence quarks? – Well... Oct 06 '20 at 16:58
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Electrons and the quarks are fundamental in that (as far as we know) they are not comprised of other particles. And you cannot form an electron from three down quarks (even though the total charge will be -1) because even one down quark is much more massive than an electron. And all protons are comprised of 2 up and 1 down quark meaning they all have the same charge to mass ratio. It is possible that quarks and even electrons are not elementary, but there is no evidence to suggest this possibility (the standard model would suggest the opposite).

Nevertheless still there is the possibility that the elementary particles may in fact not be elementary, but there is nothing thus far to show this to be true.

joseph h
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  • one down quark is massive but can't they lose their mass in the form of binding energy ? Also I asked for other properties of fundamental particles as the one in the link can not satisfy me . – Ankit Oct 06 '20 at 07:28
  • Hi Ankit. It is true that quarks combine to form hadrons. And it is true that the gluons bind together these quarks. Although gluons themselves are massless (I think this is what you are getting at), their energy is many times more greater than that for the mass/energy of the electron. – joseph h Oct 06 '20 at 07:37
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Sorry for this stupid question . I found the thing which I misinterpreted and where I made the mistake. I couldn't delete this question . So I am writing it as an answer.

Actually , the difference in mass is $(247 × 10^{-31} \; kg)$. So, the binding energy in this case would be

$ E = (247×10^{-31})(9×10^{16}) = 2223 × 10^{-15}$

And this is very minute when it comes to the energy of LHC's. This is actually a mathematical mistake which I did in my question .

Am I right ?

Ankit
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  • It's not a stupid question. It has attracted some interesting answers which add to the information given at the duplicate target. BTW, in particle physics, it's generally not useful to give masses in kg, it's better to use the equivalent energy like MeV etc. Sometimes people will use MeV/c² when they want to make it clear that they're talking about mass. – PM 2Ring Oct 07 '20 at 09:34
  • @PM 2Ring is my answer correct ? – Ankit Oct 07 '20 at 09:38
  • Well, the bare mass of a down quark is roughly 4.8 MeV, as you mentioned in the question. And the mass of an electron is 0.511 MeV. So we need a minimum of about 13.89 MeV of binding energy. That's actually quite small, compared to the energies the LHC can probe. However, as user Well... mentioned above, to actually bind 3 quarks so tightly that we cannot detect any structure or residual strong force would require a much higher binding energy, greater than what the LHC can currently manage. – PM 2Ring Oct 07 '20 at 10:00
  • You may enjoy this article on The Origins of Mass by Nobel laureate Frank Wilczek. It was published before LHC found the Higgs boson, but it's still very good. There's another version of that article, plus other goodies, on Frank's site. – PM 2Ring Oct 07 '20 at 10:17
  • @PM 2Ring thanks for the links :) , – Ankit Oct 07 '20 at 10:33