Is the Lagrangian in the Standard Model exact or approximate?

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We say that the Standard Model has $SU(3)_C otimes SU(2)_L otimes U(1)_Y$ symmetries. However, the $SU(2)_L$ symmetry of the doublet ($u, d$) is not exact because $u$ and $d$ have different masses. Does this imply that the Standard Model Lagrangian is not precise because it treats ($u, d$) as though it has an exact $SU(2)$ symmetry? Furthermore, if treating ($u ,d$) as an $SU(2)$ doublet makes the Lagrangian imprecise, can we treat $u$ and $d$ separately so as to develop a more precise Lagrangian?










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    We say that the Standard Model has $SU(3)_C otimes SU(2)_L otimes U(1)_Y$ symmetries. However, the $SU(2)_L$ symmetry of the doublet ($u, d$) is not exact because $u$ and $d$ have different masses. Does this imply that the Standard Model Lagrangian is not precise because it treats ($u, d$) as though it has an exact $SU(2)$ symmetry? Furthermore, if treating ($u ,d$) as an $SU(2)$ doublet makes the Lagrangian imprecise, can we treat $u$ and $d$ separately so as to develop a more precise Lagrangian?










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      We say that the Standard Model has $SU(3)_C otimes SU(2)_L otimes U(1)_Y$ symmetries. However, the $SU(2)_L$ symmetry of the doublet ($u, d$) is not exact because $u$ and $d$ have different masses. Does this imply that the Standard Model Lagrangian is not precise because it treats ($u, d$) as though it has an exact $SU(2)$ symmetry? Furthermore, if treating ($u ,d$) as an $SU(2)$ doublet makes the Lagrangian imprecise, can we treat $u$ and $d$ separately so as to develop a more precise Lagrangian?










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      We say that the Standard Model has $SU(3)_C otimes SU(2)_L otimes U(1)_Y$ symmetries. However, the $SU(2)_L$ symmetry of the doublet ($u, d$) is not exact because $u$ and $d$ have different masses. Does this imply that the Standard Model Lagrangian is not precise because it treats ($u, d$) as though it has an exact $SU(2)$ symmetry? Furthermore, if treating ($u ,d$) as an $SU(2)$ doublet makes the Lagrangian imprecise, can we treat $u$ and $d$ separately so as to develop a more precise Lagrangian?







      particle-physics lagrangian-formalism standard-model quarks gauge-invariance






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      edited Aug 13 at 15:44

























      asked Aug 13 at 12:21









      Shen

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          You seem to be mixing up a few different things. The Standard Model does not say the up and down quark must have the same mass.



          • The up and down quark form a doublet of the flavor symmetry $SU(2)_F$. This is an approximate symmetry that is broken explicitly by the up and down quark mass difference.

          • The left-handed up and down quarks form a doublet $Q_L$ of weak isospin $SU(2)_L$. This is a factor of the Standard Model gauge group.

          • The right-handed up and down quarks form two singlets $U_R$ and $D_R$ of weak isospin $SU(2)_L$.

          • The gauge symmetry of the Standard Model does place restrictions on the masses of the quarks; namely it requires them to all be zero. The quarks instead acquire masses through the Higgs mechanism, which breaks electroweak symmetry.

          • There is no requirement for the up quark and down quark to get the same mass, because their masses come from independent Yukawa couplings, namely $H Q_L U_R$ and $H Q_L D_R$.

          You've asked almost ten questions about $SU(2)_F$ and $SU(2)_L$, and I really recommend just picking up any book on the Standard Model and reading through it.






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            The standard model Lagrangian does not model quantum gravity nor dark matter (as far as we know), hence it is certainly not an 'exact' description of the universe anyway.



            However, your question is with regards to the SU(2) symmetry in the standard model Lagrangian. In this respect, there is no loss of precision as far as I know when one treats the SU(2) doublets on the same footing because it is a spontaneously broken symmetry and spontaneous symmetry breaking need not be put into the Lagrangian by hand.



            As a simple example, in the Ising spin model of Ferromagnetism, a spin can either point up or it can point down and the up and down spin configurations are treated on an equal footing and this is manifest as a symmetry of the system, even though in the low-temperature regime the system will spontaneously choose a preferred spin direction and form a Ferromagnet.






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              2 Answers
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              2 Answers
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              up vote
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              down vote













              You seem to be mixing up a few different things. The Standard Model does not say the up and down quark must have the same mass.



              • The up and down quark form a doublet of the flavor symmetry $SU(2)_F$. This is an approximate symmetry that is broken explicitly by the up and down quark mass difference.

              • The left-handed up and down quarks form a doublet $Q_L$ of weak isospin $SU(2)_L$. This is a factor of the Standard Model gauge group.

              • The right-handed up and down quarks form two singlets $U_R$ and $D_R$ of weak isospin $SU(2)_L$.

              • The gauge symmetry of the Standard Model does place restrictions on the masses of the quarks; namely it requires them to all be zero. The quarks instead acquire masses through the Higgs mechanism, which breaks electroweak symmetry.

              • There is no requirement for the up quark and down quark to get the same mass, because their masses come from independent Yukawa couplings, namely $H Q_L U_R$ and $H Q_L D_R$.

              You've asked almost ten questions about $SU(2)_F$ and $SU(2)_L$, and I really recommend just picking up any book on the Standard Model and reading through it.






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                up vote
                11
                down vote













                You seem to be mixing up a few different things. The Standard Model does not say the up and down quark must have the same mass.



                • The up and down quark form a doublet of the flavor symmetry $SU(2)_F$. This is an approximate symmetry that is broken explicitly by the up and down quark mass difference.

                • The left-handed up and down quarks form a doublet $Q_L$ of weak isospin $SU(2)_L$. This is a factor of the Standard Model gauge group.

                • The right-handed up and down quarks form two singlets $U_R$ and $D_R$ of weak isospin $SU(2)_L$.

                • The gauge symmetry of the Standard Model does place restrictions on the masses of the quarks; namely it requires them to all be zero. The quarks instead acquire masses through the Higgs mechanism, which breaks electroweak symmetry.

                • There is no requirement for the up quark and down quark to get the same mass, because their masses come from independent Yukawa couplings, namely $H Q_L U_R$ and $H Q_L D_R$.

                You've asked almost ten questions about $SU(2)_F$ and $SU(2)_L$, and I really recommend just picking up any book on the Standard Model and reading through it.






                share|cite|improve this answer






















                  up vote
                  11
                  down vote










                  up vote
                  11
                  down vote









                  You seem to be mixing up a few different things. The Standard Model does not say the up and down quark must have the same mass.



                  • The up and down quark form a doublet of the flavor symmetry $SU(2)_F$. This is an approximate symmetry that is broken explicitly by the up and down quark mass difference.

                  • The left-handed up and down quarks form a doublet $Q_L$ of weak isospin $SU(2)_L$. This is a factor of the Standard Model gauge group.

                  • The right-handed up and down quarks form two singlets $U_R$ and $D_R$ of weak isospin $SU(2)_L$.

                  • The gauge symmetry of the Standard Model does place restrictions on the masses of the quarks; namely it requires them to all be zero. The quarks instead acquire masses through the Higgs mechanism, which breaks electroweak symmetry.

                  • There is no requirement for the up quark and down quark to get the same mass, because their masses come from independent Yukawa couplings, namely $H Q_L U_R$ and $H Q_L D_R$.

                  You've asked almost ten questions about $SU(2)_F$ and $SU(2)_L$, and I really recommend just picking up any book on the Standard Model and reading through it.






                  share|cite|improve this answer












                  You seem to be mixing up a few different things. The Standard Model does not say the up and down quark must have the same mass.



                  • The up and down quark form a doublet of the flavor symmetry $SU(2)_F$. This is an approximate symmetry that is broken explicitly by the up and down quark mass difference.

                  • The left-handed up and down quarks form a doublet $Q_L$ of weak isospin $SU(2)_L$. This is a factor of the Standard Model gauge group.

                  • The right-handed up and down quarks form two singlets $U_R$ and $D_R$ of weak isospin $SU(2)_L$.

                  • The gauge symmetry of the Standard Model does place restrictions on the masses of the quarks; namely it requires them to all be zero. The quarks instead acquire masses through the Higgs mechanism, which breaks electroweak symmetry.

                  • There is no requirement for the up quark and down quark to get the same mass, because their masses come from independent Yukawa couplings, namely $H Q_L U_R$ and $H Q_L D_R$.

                  You've asked almost ten questions about $SU(2)_F$ and $SU(2)_L$, and I really recommend just picking up any book on the Standard Model and reading through it.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 13 at 12:50









                  knzhou

                  33.9k897170




                  33.9k897170




















                      up vote
                      2
                      down vote













                      The standard model Lagrangian does not model quantum gravity nor dark matter (as far as we know), hence it is certainly not an 'exact' description of the universe anyway.



                      However, your question is with regards to the SU(2) symmetry in the standard model Lagrangian. In this respect, there is no loss of precision as far as I know when one treats the SU(2) doublets on the same footing because it is a spontaneously broken symmetry and spontaneous symmetry breaking need not be put into the Lagrangian by hand.



                      As a simple example, in the Ising spin model of Ferromagnetism, a spin can either point up or it can point down and the up and down spin configurations are treated on an equal footing and this is manifest as a symmetry of the system, even though in the low-temperature regime the system will spontaneously choose a preferred spin direction and form a Ferromagnet.






                      share|cite|improve this answer
























                        up vote
                        2
                        down vote













                        The standard model Lagrangian does not model quantum gravity nor dark matter (as far as we know), hence it is certainly not an 'exact' description of the universe anyway.



                        However, your question is with regards to the SU(2) symmetry in the standard model Lagrangian. In this respect, there is no loss of precision as far as I know when one treats the SU(2) doublets on the same footing because it is a spontaneously broken symmetry and spontaneous symmetry breaking need not be put into the Lagrangian by hand.



                        As a simple example, in the Ising spin model of Ferromagnetism, a spin can either point up or it can point down and the up and down spin configurations are treated on an equal footing and this is manifest as a symmetry of the system, even though in the low-temperature regime the system will spontaneously choose a preferred spin direction and form a Ferromagnet.






                        share|cite|improve this answer






















                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          The standard model Lagrangian does not model quantum gravity nor dark matter (as far as we know), hence it is certainly not an 'exact' description of the universe anyway.



                          However, your question is with regards to the SU(2) symmetry in the standard model Lagrangian. In this respect, there is no loss of precision as far as I know when one treats the SU(2) doublets on the same footing because it is a spontaneously broken symmetry and spontaneous symmetry breaking need not be put into the Lagrangian by hand.



                          As a simple example, in the Ising spin model of Ferromagnetism, a spin can either point up or it can point down and the up and down spin configurations are treated on an equal footing and this is manifest as a symmetry of the system, even though in the low-temperature regime the system will spontaneously choose a preferred spin direction and form a Ferromagnet.






                          share|cite|improve this answer












                          The standard model Lagrangian does not model quantum gravity nor dark matter (as far as we know), hence it is certainly not an 'exact' description of the universe anyway.



                          However, your question is with regards to the SU(2) symmetry in the standard model Lagrangian. In this respect, there is no loss of precision as far as I know when one treats the SU(2) doublets on the same footing because it is a spontaneously broken symmetry and spontaneous symmetry breaking need not be put into the Lagrangian by hand.



                          As a simple example, in the Ising spin model of Ferromagnetism, a spin can either point up or it can point down and the up and down spin configurations are treated on an equal footing and this is manifest as a symmetry of the system, even though in the low-temperature regime the system will spontaneously choose a preferred spin direction and form a Ferromagnet.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 13 at 12:39









                          Mason

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