Is it a coincidence that quarks have exactly -1/3 or 2/3 the electron's charge?

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I have read these questions:



Why do quarks have a fractional charge?



Is there an explanation for the 3:2:1 ratio between the electron, up and down quark electric charges?



Hypercharge for $U(1)$ in $SU(2)times U(1)$ model



Is there any idea why the electric charges of electron and muon are equal?



My question is different. We live in a world, where quarks can have a real fraction of the elementary charge (-1/3 or 2/3). I do understand that the experimental data fits the models and that Baryons are made up of three quarks, and that those quarks can have -1/3 or 2/3 the elementary charge. This way, the quarks can combine so that the Baryon will have an integer of the elementary charge. This way, the nucleus and the electron can be in a stable atomic state, where their electric charges cancel (attract) exactly. Any other way, the atom would not be stable.



So electron says: hey quarks, let's team up, let's make an atom.



Quarks: Hey, great idea, how much electric charge do you have? Let's call it e. OK so we will team up of three of us, so we will just take each of us -1/3 or 2/3 of your charge.



Electron says: Great!, I feel it, this way we can have a stable atom.



Quarks: Great!



I mean come on! I do understand an respect that the experimental data tells us that electrons and quarks are both elementary particles.



It seems that the quarks teamed up exactly so three of them in Baryons so they can cancel out (attract) exactly the electron's electric charge. Now I do understand that there could be Baryons made up of four quarks, and they could have then -1/4 and 3/4 charge of the electron's elementary charge. This would work too, and the neutron and proton would have the same way an integer of the electron's charge. So the atom would be stable.



We could do this with any integer number of quarks. But, come on, my question is, what if the quarks' charge could only be -1/sqrt(3) and 2/sqrt(3)? How many quarks would we need then to make up the nucleus of to make it match the electron's charge? Is it mathematically possible to have any kind of electric charge for the quark? We could simply put as many valence quarks in a Baryon to make a stable atom?



Question:



  1. Is it a coincidence that quarks have -1/3 and 2/3 the electron's charge and there are three quarks in a Baryon? Could we have quarks with any kind of electric charge, like Sqrt(3)*elementary charge? Could we make a stable atom this way too? How many quarks would we put in this case into a Baryon to make atoms stable?


  2. I accept and respect that currently elecrons are elementary particles. Is it impossible as of today's view that both electrons and quarks are made up of the same something smaller (strings)?










share|cite|improve this question























  • This is an unanswerable question. Physics doesn't tell us "why" the laws of physics are what they are. It just is, it has always been, and it works out. Something smaller (the generic term is "preons") is possible, but there is no evidence of preons to date. The binding energy of any preons would have to be very high given experimental constraints.
    – ohwilleke
    2 hours ago






  • 1




    There are some constraints like anomaly cancellations. Those tell you about some of the discrete parameters like this -1/3 etc.
    – AHusain
    2 hours ago










  • Related: physics.stackexchange.com/q/21753/2451 and links theren.
    – Qmechanic♦
    14 mins ago






  • 1




    Possible duplicate of Why do electron and proton have the same but opposite electric charge?
    – AccidentalFourierTransform
    11 mins ago














up vote
4
down vote

favorite












I have read these questions:



Why do quarks have a fractional charge?



Is there an explanation for the 3:2:1 ratio between the electron, up and down quark electric charges?



Hypercharge for $U(1)$ in $SU(2)times U(1)$ model



Is there any idea why the electric charges of electron and muon are equal?



My question is different. We live in a world, where quarks can have a real fraction of the elementary charge (-1/3 or 2/3). I do understand that the experimental data fits the models and that Baryons are made up of three quarks, and that those quarks can have -1/3 or 2/3 the elementary charge. This way, the quarks can combine so that the Baryon will have an integer of the elementary charge. This way, the nucleus and the electron can be in a stable atomic state, where their electric charges cancel (attract) exactly. Any other way, the atom would not be stable.



So electron says: hey quarks, let's team up, let's make an atom.



Quarks: Hey, great idea, how much electric charge do you have? Let's call it e. OK so we will team up of three of us, so we will just take each of us -1/3 or 2/3 of your charge.



Electron says: Great!, I feel it, this way we can have a stable atom.



Quarks: Great!



I mean come on! I do understand an respect that the experimental data tells us that electrons and quarks are both elementary particles.



It seems that the quarks teamed up exactly so three of them in Baryons so they can cancel out (attract) exactly the electron's electric charge. Now I do understand that there could be Baryons made up of four quarks, and they could have then -1/4 and 3/4 charge of the electron's elementary charge. This would work too, and the neutron and proton would have the same way an integer of the electron's charge. So the atom would be stable.



We could do this with any integer number of quarks. But, come on, my question is, what if the quarks' charge could only be -1/sqrt(3) and 2/sqrt(3)? How many quarks would we need then to make up the nucleus of to make it match the electron's charge? Is it mathematically possible to have any kind of electric charge for the quark? We could simply put as many valence quarks in a Baryon to make a stable atom?



Question:



  1. Is it a coincidence that quarks have -1/3 and 2/3 the electron's charge and there are three quarks in a Baryon? Could we have quarks with any kind of electric charge, like Sqrt(3)*elementary charge? Could we make a stable atom this way too? How many quarks would we put in this case into a Baryon to make atoms stable?


  2. I accept and respect that currently elecrons are elementary particles. Is it impossible as of today's view that both electrons and quarks are made up of the same something smaller (strings)?










share|cite|improve this question























  • This is an unanswerable question. Physics doesn't tell us "why" the laws of physics are what they are. It just is, it has always been, and it works out. Something smaller (the generic term is "preons") is possible, but there is no evidence of preons to date. The binding energy of any preons would have to be very high given experimental constraints.
    – ohwilleke
    2 hours ago






  • 1




    There are some constraints like anomaly cancellations. Those tell you about some of the discrete parameters like this -1/3 etc.
    – AHusain
    2 hours ago










  • Related: physics.stackexchange.com/q/21753/2451 and links theren.
    – Qmechanic♦
    14 mins ago






  • 1




    Possible duplicate of Why do electron and proton have the same but opposite electric charge?
    – AccidentalFourierTransform
    11 mins ago












up vote
4
down vote

favorite









up vote
4
down vote

favorite











I have read these questions:



Why do quarks have a fractional charge?



Is there an explanation for the 3:2:1 ratio between the electron, up and down quark electric charges?



Hypercharge for $U(1)$ in $SU(2)times U(1)$ model



Is there any idea why the electric charges of electron and muon are equal?



My question is different. We live in a world, where quarks can have a real fraction of the elementary charge (-1/3 or 2/3). I do understand that the experimental data fits the models and that Baryons are made up of three quarks, and that those quarks can have -1/3 or 2/3 the elementary charge. This way, the quarks can combine so that the Baryon will have an integer of the elementary charge. This way, the nucleus and the electron can be in a stable atomic state, where their electric charges cancel (attract) exactly. Any other way, the atom would not be stable.



So electron says: hey quarks, let's team up, let's make an atom.



Quarks: Hey, great idea, how much electric charge do you have? Let's call it e. OK so we will team up of three of us, so we will just take each of us -1/3 or 2/3 of your charge.



Electron says: Great!, I feel it, this way we can have a stable atom.



Quarks: Great!



I mean come on! I do understand an respect that the experimental data tells us that electrons and quarks are both elementary particles.



It seems that the quarks teamed up exactly so three of them in Baryons so they can cancel out (attract) exactly the electron's electric charge. Now I do understand that there could be Baryons made up of four quarks, and they could have then -1/4 and 3/4 charge of the electron's elementary charge. This would work too, and the neutron and proton would have the same way an integer of the electron's charge. So the atom would be stable.



We could do this with any integer number of quarks. But, come on, my question is, what if the quarks' charge could only be -1/sqrt(3) and 2/sqrt(3)? How many quarks would we need then to make up the nucleus of to make it match the electron's charge? Is it mathematically possible to have any kind of electric charge for the quark? We could simply put as many valence quarks in a Baryon to make a stable atom?



Question:



  1. Is it a coincidence that quarks have -1/3 and 2/3 the electron's charge and there are three quarks in a Baryon? Could we have quarks with any kind of electric charge, like Sqrt(3)*elementary charge? Could we make a stable atom this way too? How many quarks would we put in this case into a Baryon to make atoms stable?


  2. I accept and respect that currently elecrons are elementary particles. Is it impossible as of today's view that both electrons and quarks are made up of the same something smaller (strings)?










share|cite|improve this question















I have read these questions:



Why do quarks have a fractional charge?



Is there an explanation for the 3:2:1 ratio between the electron, up and down quark electric charges?



Hypercharge for $U(1)$ in $SU(2)times U(1)$ model



Is there any idea why the electric charges of electron and muon are equal?



My question is different. We live in a world, where quarks can have a real fraction of the elementary charge (-1/3 or 2/3). I do understand that the experimental data fits the models and that Baryons are made up of three quarks, and that those quarks can have -1/3 or 2/3 the elementary charge. This way, the quarks can combine so that the Baryon will have an integer of the elementary charge. This way, the nucleus and the electron can be in a stable atomic state, where their electric charges cancel (attract) exactly. Any other way, the atom would not be stable.



So electron says: hey quarks, let's team up, let's make an atom.



Quarks: Hey, great idea, how much electric charge do you have? Let's call it e. OK so we will team up of three of us, so we will just take each of us -1/3 or 2/3 of your charge.



Electron says: Great!, I feel it, this way we can have a stable atom.



Quarks: Great!



I mean come on! I do understand an respect that the experimental data tells us that electrons and quarks are both elementary particles.



It seems that the quarks teamed up exactly so three of them in Baryons so they can cancel out (attract) exactly the electron's electric charge. Now I do understand that there could be Baryons made up of four quarks, and they could have then -1/4 and 3/4 charge of the electron's elementary charge. This would work too, and the neutron and proton would have the same way an integer of the electron's charge. So the atom would be stable.



We could do this with any integer number of quarks. But, come on, my question is, what if the quarks' charge could only be -1/sqrt(3) and 2/sqrt(3)? How many quarks would we need then to make up the nucleus of to make it match the electron's charge? Is it mathematically possible to have any kind of electric charge for the quark? We could simply put as many valence quarks in a Baryon to make a stable atom?



Question:



  1. Is it a coincidence that quarks have -1/3 and 2/3 the electron's charge and there are three quarks in a Baryon? Could we have quarks with any kind of electric charge, like Sqrt(3)*elementary charge? Could we make a stable atom this way too? How many quarks would we put in this case into a Baryon to make atoms stable?


  2. I accept and respect that currently elecrons are elementary particles. Is it impossible as of today's view that both electrons and quarks are made up of the same something smaller (strings)?







charge standard-model quarks elementary-particles baryons






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 25 mins ago









Qmechanic♦

98.2k121721066




98.2k121721066










asked 2 hours ago









Árpád Szendrei

3,4331521




3,4331521











  • This is an unanswerable question. Physics doesn't tell us "why" the laws of physics are what they are. It just is, it has always been, and it works out. Something smaller (the generic term is "preons") is possible, but there is no evidence of preons to date. The binding energy of any preons would have to be very high given experimental constraints.
    – ohwilleke
    2 hours ago






  • 1




    There are some constraints like anomaly cancellations. Those tell you about some of the discrete parameters like this -1/3 etc.
    – AHusain
    2 hours ago










  • Related: physics.stackexchange.com/q/21753/2451 and links theren.
    – Qmechanic♦
    14 mins ago






  • 1




    Possible duplicate of Why do electron and proton have the same but opposite electric charge?
    – AccidentalFourierTransform
    11 mins ago
















  • This is an unanswerable question. Physics doesn't tell us "why" the laws of physics are what they are. It just is, it has always been, and it works out. Something smaller (the generic term is "preons") is possible, but there is no evidence of preons to date. The binding energy of any preons would have to be very high given experimental constraints.
    – ohwilleke
    2 hours ago






  • 1




    There are some constraints like anomaly cancellations. Those tell you about some of the discrete parameters like this -1/3 etc.
    – AHusain
    2 hours ago










  • Related: physics.stackexchange.com/q/21753/2451 and links theren.
    – Qmechanic♦
    14 mins ago






  • 1




    Possible duplicate of Why do electron and proton have the same but opposite electric charge?
    – AccidentalFourierTransform
    11 mins ago















This is an unanswerable question. Physics doesn't tell us "why" the laws of physics are what they are. It just is, it has always been, and it works out. Something smaller (the generic term is "preons") is possible, but there is no evidence of preons to date. The binding energy of any preons would have to be very high given experimental constraints.
– ohwilleke
2 hours ago




This is an unanswerable question. Physics doesn't tell us "why" the laws of physics are what they are. It just is, it has always been, and it works out. Something smaller (the generic term is "preons") is possible, but there is no evidence of preons to date. The binding energy of any preons would have to be very high given experimental constraints.
– ohwilleke
2 hours ago




1




1




There are some constraints like anomaly cancellations. Those tell you about some of the discrete parameters like this -1/3 etc.
– AHusain
2 hours ago




There are some constraints like anomaly cancellations. Those tell you about some of the discrete parameters like this -1/3 etc.
– AHusain
2 hours ago












Related: physics.stackexchange.com/q/21753/2451 and links theren.
– Qmechanic♦
14 mins ago




Related: physics.stackexchange.com/q/21753/2451 and links theren.
– Qmechanic♦
14 mins ago




1




1




Possible duplicate of Why do electron and proton have the same but opposite electric charge?
– AccidentalFourierTransform
11 mins ago




Possible duplicate of Why do electron and proton have the same but opposite electric charge?
– AccidentalFourierTransform
11 mins ago










1 Answer
1






active

oldest

votes

















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2
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The charges of the quarks must be simple fractions of the electron charge $e$, because otherwise there would be a breakdown of charge conservation in quantum corrections. The fractions do not need to be $-frac23$ and $frac13$ specifically. In simple models with $2n+1$ quarks making up the proton (the number of quarks must be odd, so that the nucleons are still fermions with spin-$frac12$), the quarks naturally carry charges $-fracn+12n+1e$ and $fracn2n+1e$. And more elaborate composite nucleon models can have constituent partons with other rational multiples of $e$.



However, not all all fractions of $e$ are going to be allowed as charges of the constituent quarks. Moreover, irrational multiples of the electron charge are generally not possible. The reason is tied to the structure of the full electroweak interaction, of which electromagnetism is only one part. In relativistic quantum field theory, there are are quantum corrections that involve the interactions of three separate gauge bosons (photons, $Z^0$, and $W^pm$) with virtual fermion-antifermion pairs. The fermions involved can be quarks, electrons, muons, neutrinos, etc, and the sizes of the quantum corrections are determined (in part) by the charges of those fermion species. If the quark and electron charges are not in the correct rational ratios, certain quantum corrections will be nonzero.



These quantum anomaly terms lead to nonconservation of the classically conserved currents in the theory, and this destroys several necessary properties of the theory—stability, unitarity, and renormalizability. Equivalently, if we want to have a well-defined quantum theory, these anomaly terms need to be exactly zero. This is known as "anomaly cancelation," and this places the stringent constraints on how the values of the various fermion species can be related.






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    The charges of the quarks must be simple fractions of the electron charge $e$, because otherwise there would be a breakdown of charge conservation in quantum corrections. The fractions do not need to be $-frac23$ and $frac13$ specifically. In simple models with $2n+1$ quarks making up the proton (the number of quarks must be odd, so that the nucleons are still fermions with spin-$frac12$), the quarks naturally carry charges $-fracn+12n+1e$ and $fracn2n+1e$. And more elaborate composite nucleon models can have constituent partons with other rational multiples of $e$.



    However, not all all fractions of $e$ are going to be allowed as charges of the constituent quarks. Moreover, irrational multiples of the electron charge are generally not possible. The reason is tied to the structure of the full electroweak interaction, of which electromagnetism is only one part. In relativistic quantum field theory, there are are quantum corrections that involve the interactions of three separate gauge bosons (photons, $Z^0$, and $W^pm$) with virtual fermion-antifermion pairs. The fermions involved can be quarks, electrons, muons, neutrinos, etc, and the sizes of the quantum corrections are determined (in part) by the charges of those fermion species. If the quark and electron charges are not in the correct rational ratios, certain quantum corrections will be nonzero.



    These quantum anomaly terms lead to nonconservation of the classically conserved currents in the theory, and this destroys several necessary properties of the theory—stability, unitarity, and renormalizability. Equivalently, if we want to have a well-defined quantum theory, these anomaly terms need to be exactly zero. This is known as "anomaly cancelation," and this places the stringent constraints on how the values of the various fermion species can be related.






    share|cite|improve this answer
























      up vote
      2
      down vote













      The charges of the quarks must be simple fractions of the electron charge $e$, because otherwise there would be a breakdown of charge conservation in quantum corrections. The fractions do not need to be $-frac23$ and $frac13$ specifically. In simple models with $2n+1$ quarks making up the proton (the number of quarks must be odd, so that the nucleons are still fermions with spin-$frac12$), the quarks naturally carry charges $-fracn+12n+1e$ and $fracn2n+1e$. And more elaborate composite nucleon models can have constituent partons with other rational multiples of $e$.



      However, not all all fractions of $e$ are going to be allowed as charges of the constituent quarks. Moreover, irrational multiples of the electron charge are generally not possible. The reason is tied to the structure of the full electroweak interaction, of which electromagnetism is only one part. In relativistic quantum field theory, there are are quantum corrections that involve the interactions of three separate gauge bosons (photons, $Z^0$, and $W^pm$) with virtual fermion-antifermion pairs. The fermions involved can be quarks, electrons, muons, neutrinos, etc, and the sizes of the quantum corrections are determined (in part) by the charges of those fermion species. If the quark and electron charges are not in the correct rational ratios, certain quantum corrections will be nonzero.



      These quantum anomaly terms lead to nonconservation of the classically conserved currents in the theory, and this destroys several necessary properties of the theory—stability, unitarity, and renormalizability. Equivalently, if we want to have a well-defined quantum theory, these anomaly terms need to be exactly zero. This is known as "anomaly cancelation," and this places the stringent constraints on how the values of the various fermion species can be related.






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        The charges of the quarks must be simple fractions of the electron charge $e$, because otherwise there would be a breakdown of charge conservation in quantum corrections. The fractions do not need to be $-frac23$ and $frac13$ specifically. In simple models with $2n+1$ quarks making up the proton (the number of quarks must be odd, so that the nucleons are still fermions with spin-$frac12$), the quarks naturally carry charges $-fracn+12n+1e$ and $fracn2n+1e$. And more elaborate composite nucleon models can have constituent partons with other rational multiples of $e$.



        However, not all all fractions of $e$ are going to be allowed as charges of the constituent quarks. Moreover, irrational multiples of the electron charge are generally not possible. The reason is tied to the structure of the full electroweak interaction, of which electromagnetism is only one part. In relativistic quantum field theory, there are are quantum corrections that involve the interactions of three separate gauge bosons (photons, $Z^0$, and $W^pm$) with virtual fermion-antifermion pairs. The fermions involved can be quarks, electrons, muons, neutrinos, etc, and the sizes of the quantum corrections are determined (in part) by the charges of those fermion species. If the quark and electron charges are not in the correct rational ratios, certain quantum corrections will be nonzero.



        These quantum anomaly terms lead to nonconservation of the classically conserved currents in the theory, and this destroys several necessary properties of the theory—stability, unitarity, and renormalizability. Equivalently, if we want to have a well-defined quantum theory, these anomaly terms need to be exactly zero. This is known as "anomaly cancelation," and this places the stringent constraints on how the values of the various fermion species can be related.






        share|cite|improve this answer












        The charges of the quarks must be simple fractions of the electron charge $e$, because otherwise there would be a breakdown of charge conservation in quantum corrections. The fractions do not need to be $-frac23$ and $frac13$ specifically. In simple models with $2n+1$ quarks making up the proton (the number of quarks must be odd, so that the nucleons are still fermions with spin-$frac12$), the quarks naturally carry charges $-fracn+12n+1e$ and $fracn2n+1e$. And more elaborate composite nucleon models can have constituent partons with other rational multiples of $e$.



        However, not all all fractions of $e$ are going to be allowed as charges of the constituent quarks. Moreover, irrational multiples of the electron charge are generally not possible. The reason is tied to the structure of the full electroweak interaction, of which electromagnetism is only one part. In relativistic quantum field theory, there are are quantum corrections that involve the interactions of three separate gauge bosons (photons, $Z^0$, and $W^pm$) with virtual fermion-antifermion pairs. The fermions involved can be quarks, electrons, muons, neutrinos, etc, and the sizes of the quantum corrections are determined (in part) by the charges of those fermion species. If the quark and electron charges are not in the correct rational ratios, certain quantum corrections will be nonzero.



        These quantum anomaly terms lead to nonconservation of the classically conserved currents in the theory, and this destroys several necessary properties of the theory—stability, unitarity, and renormalizability. Equivalently, if we want to have a well-defined quantum theory, these anomaly terms need to be exactly zero. This is known as "anomaly cancelation," and this places the stringent constraints on how the values of the various fermion species can be related.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        Buzz

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