Why Is Abelian Gauge Theory So Special?What is a “free” non-Abelian Yang-Mill's theory?(coordinates) Invariance/Covariance of Chern-Simons theory and Yang-Mills theoryEquations of motion for the Yang-Mills $SU(2)$ theoryEinstein-Yang-Mills ConnectionsLattice gauge theory + Chern-Simons TheoryHodge star operator in Yang-Mills theory and derivation of the YM equationsInterpretation of the field strength tensor in Yang-Mills TheoryMistake or Rewriting of Yang-Mills in NakaharaClassical Yang-Mills equation of motion with both electric and magnetic sources?Global Part of Non-Abelian Gauge Transformation
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Why Is Abelian Gauge Theory So Special?
What is a “free” non-Abelian Yang-Mill's theory?(coordinates) Invariance/Covariance of Chern-Simons theory and Yang-Mills theoryEquations of motion for the Yang-Mills $SU(2)$ theoryEinstein-Yang-Mills ConnectionsLattice gauge theory + Chern-Simons TheoryHodge star operator in Yang-Mills theory and derivation of the YM equationsInterpretation of the field strength tensor in Yang-Mills TheoryMistake or Rewriting of Yang-Mills in NakaharaClassical Yang-Mills equation of motion with both electric and magnetic sources?Global Part of Non-Abelian Gauge Transformation
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I have a perhaps stupid question about Maxwell equations.
Let $G$ be a generic Lie group. We consider a $G$-gauge theory. Let $A$ be the associated connection $1$-form, and $F=dA+Awedge A$ be the corresponding curvature tensor. The Yang-Mills equations can be written as
$$Dstar F=star J tag1.1$$
$$DF=0 tag1.2$$
Equation (1.1) follows from variation of the Lagrangian, and equation (1.2) is the Bianchi identity, which holds universally.
If the gauge group $G$ is replaced by $U(1)$, then naively one would expect the Yang-Mills equations to stay the same except $F=dA$.
However, as we all know, the Maxwell equations are
$$dstar F=star J$$
$$dF=ddA=0$$
Is there any deep geometrical reasons why $U(1)$ gauge theory is so special that one has to replace the covariant differential operator $D$ by the de-Rham differential $d$?
differential-geometry gauge-theory maxwell-equations yang-mills
asked 9 hours ago
The Last Knight of Silk RoadThe Last Knight of Silk Road
7242 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
I have a perhaps stupid question about Maxwell equations.
Let $G$ be a generic Lie group. We consider a $G$-gauge theory. Let $A$ be the associated connection $1$-form, and $F=dA+Awedge A$ be the corresponding curvature tensor. The Yang-Mills equations can be written as
$$Dstar F=star J tag1.1$$
$$DF=0 tag1.2$$
Equation (1.1) follows from variation of the Lagrangian, and equation (1.2) is the Bianchi identity, which holds universally.
If the gauge group $G$ is replaced by $U(1)$, then naively one would expect the Yang-Mills equations to stay the same except $F=dA$.
However, as we all know, the Maxwell equations are
$$dstar F=star J$$
$$dF=ddA=0$$
Is there any deep geometrical reasons why $U(1)$ gauge theory is so special that one has to replace the covariant differential operator $D$ by the de-Rham differential $d$?
differential-geometry gauge-theory maxwell-equations yang-mills
asked 9 hours ago
The Last Knight of Silk RoadThe Last Knight of Silk Road
7242 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
I have a perhaps stupid question about Maxwell equations.
Let $G$ be a generic Lie group. We consider a $G$-gauge theory. Let $A$ be the associated connection $1$-form, and $F=dA+Awedge A$ be the corresponding curvature tensor. The Yang-Mills equations can be written as
$$Dstar F=star J tag1.1$$
$$DF=0 tag1.2$$
Equation (1.1) follows from variation of the Lagrangian, and equation (1.2) is the Bianchi identity, which holds universally.
If the gauge group $G$ is replaced by $U(1)$, then naively one would expect the Yang-Mills equations to stay the same except $F=dA$.
However, as we all know, the Maxwell equations are
$$dstar F=star J$$
$$dF=ddA=0$$
Is there any deep geometrical reasons why $U(1)$ gauge theory is so special that one has to replace the covariant differential operator $D$ by the de-Rham differential $d$?
differential-geometry gauge-theory maxwell-equations yang-mills
asked 9 hours ago
The Last Knight of Silk RoadThe Last Knight of Silk Road
7242 silver badges13 bronze badges
$endgroup$
I have a perhaps stupid question about Maxwell equations.
Let $G$ be a generic Lie group. We consider a $G$-gauge theory. Let $A$ be the associated connection $1$-form, and $F=dA+Awedge A$ be the corresponding curvature tensor. The Yang-Mills equations can be written as
$$Dstar F=star J tag1.1$$
$$DF=0 tag1.2$$
Equation (1.1) follows from variation of the Lagrangian, and equation (1.2) is the Bianchi identity, which holds universally.
If the gauge group $G$ is replaced by $U(1)$, then naively one would expect the Yang-Mills equations to stay the same except $F=dA$.
However, as we all know, the Maxwell equations are
$$dstar F=star J$$
$$dF=ddA=0$$
Is there any deep geometrical reasons why $U(1)$ gauge theory is so special that one has to replace the covariant differential operator $D$ by the de-Rham differential $d$?
differential-geometry gauge-theory maxwell-equations yang-mills
differential-geometry gauge-theory maxwell-equations yang-mills
asked 9 hours ago
The Last Knight of Silk RoadThe Last Knight of Silk Road
7242 silver badges13 bronze badges
asked 9 hours ago
The Last Knight of Silk RoadThe Last Knight of Silk Road
7242 silver badges13 bronze badges
asked 9 hours ago
The Last Knight of Silk RoadThe Last Knight of Silk Road
7242 silver badges13 bronze badges
asked 9 hours ago
The Last Knight of Silk RoadThe Last Knight of Silk Road
7242 silver badges13 bronze badges
asked 9 hours ago
The Last Knight of Silk RoadThe Last Knight of Silk Road
7242 silver badges13 bronze badges
7242 silver badges13 bronze badges
add a comment |
add a comment |
2 Answers
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Normally the $D$ would be the covariant derivative in the adjoint representation (because $F$ and $star F$ transform in the adjoint of the gauge group).
For abelian gauge theories the adjoint representation is the trivial one (the representation with charge zero). You can either see this because $f^abc = 0$ or because the photon is electrically neutral. Hence $D$ in the adjoint representation for an abelian gauge theory is just $d$.
answered 8 hours ago
MannyCMannyC
1,5641 gold badge4 silver badges15 bronze badges
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We don't "replace" the gauge covariant derivative by the ordinary differential, it is simply that case that $DF = mathrmdF$, since the adjoint representation of $mathrmU(1)$ is trivial.
answered 8 hours ago
ACuriousMind♦ACuriousMind
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2 Answers
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2 Answers
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$begingroup$
Normally the $D$ would be the covariant derivative in the adjoint representation (because $F$ and $star F$ transform in the adjoint of the gauge group).
For abelian gauge theories the adjoint representation is the trivial one (the representation with charge zero). You can either see this because $f^abc = 0$ or because the photon is electrically neutral. Hence $D$ in the adjoint representation for an abelian gauge theory is just $d$.
answered 8 hours ago
MannyCMannyC
1,5641 gold badge4 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
Normally the $D$ would be the covariant derivative in the adjoint representation (because $F$ and $star F$ transform in the adjoint of the gauge group).
For abelian gauge theories the adjoint representation is the trivial one (the representation with charge zero). You can either see this because $f^abc = 0$ or because the photon is electrically neutral. Hence $D$ in the adjoint representation for an abelian gauge theory is just $d$.
answered 8 hours ago
MannyCMannyC
1,5641 gold badge4 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
Normally the $D$ would be the covariant derivative in the adjoint representation (because $F$ and $star F$ transform in the adjoint of the gauge group).
For abelian gauge theories the adjoint representation is the trivial one (the representation with charge zero). You can either see this because $f^abc = 0$ or because the photon is electrically neutral. Hence $D$ in the adjoint representation for an abelian gauge theory is just $d$.
answered 8 hours ago
MannyCMannyC
1,5641 gold badge4 silver badges15 bronze badges
$endgroup$
Normally the $D$ would be the covariant derivative in the adjoint representation (because $F$ and $star F$ transform in the adjoint of the gauge group).
For abelian gauge theories the adjoint representation is the trivial one (the representation with charge zero). You can either see this because $f^abc = 0$ or because the photon is electrically neutral. Hence $D$ in the adjoint representation for an abelian gauge theory is just $d$.
answered 8 hours ago
MannyCMannyC
1,5641 gold badge4 silver badges15 bronze badges
answered 8 hours ago
MannyCMannyC
1,5641 gold badge4 silver badges15 bronze badges
answered 8 hours ago
MannyCMannyC
1,5641 gold badge4 silver badges15 bronze badges
answered 8 hours ago
MannyCMannyC
1,5641 gold badge4 silver badges15 bronze badges
1,5641 gold badge4 silver badges15 bronze badges
add a comment |
add a comment |
$begingroup$
We don't "replace" the gauge covariant derivative by the ordinary differential, it is simply that case that $DF = mathrmdF$, since the adjoint representation of $mathrmU(1)$ is trivial.
answered 8 hours ago
ACuriousMind♦ACuriousMind
74.5k18 gold badges137 silver badges340 bronze badges
$endgroup$
add a comment |
$begingroup$
We don't "replace" the gauge covariant derivative by the ordinary differential, it is simply that case that $DF = mathrmdF$, since the adjoint representation of $mathrmU(1)$ is trivial.
answered 8 hours ago
ACuriousMind♦ACuriousMind
74.5k18 gold badges137 silver badges340 bronze badges
$endgroup$
add a comment |
$begingroup$
We don't "replace" the gauge covariant derivative by the ordinary differential, it is simply that case that $DF = mathrmdF$, since the adjoint representation of $mathrmU(1)$ is trivial.
answered 8 hours ago
ACuriousMind♦ACuriousMind
74.5k18 gold badges137 silver badges340 bronze badges
$endgroup$
We don't "replace" the gauge covariant derivative by the ordinary differential, it is simply that case that $DF = mathrmdF$, since the adjoint representation of $mathrmU(1)$ is trivial.
answered 8 hours ago
ACuriousMind♦ACuriousMind
74.5k18 gold badges137 silver badges340 bronze badges
answered 8 hours ago
ACuriousMind♦ACuriousMind
74.5k18 gold badges137 silver badges340 bronze badges
answered 8 hours ago
ACuriousMind♦ACuriousMind
74.5k18 gold badges137 silver badges340 bronze badges
answered 8 hours ago
ACuriousMind♦ACuriousMind
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74.5k18 gold badges137 silver badges340 bronze badges
add a comment |
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