- #1
tomdodd4598
- 138
- 13
Hi there,
I have just read that the gauge field term Fμν is proportional to the commutator of covariant derivatives [Dμ,Dν]. However, when I try to calculate this commatator, taking the symmetry group to be U(1), I get the following:
[tex]\left[ { D }_{ \mu },{ D }_{ \nu } \right] =\left( { \partial }_{ \mu }-iq{ A }_{ \mu } \right) \left( { \partial }_{ \nu }-iq{ A }_{ \nu } \right) -\left( { \partial }_{ \nu }-iq{ A }_{ \nu } \right) \left( { \partial }_{ \mu }-iq{ A }_{ \mu } \right) ={ \partial }_{ \mu }{ \partial }_{ \nu }-{ q }^{ 2 }{ A }_{ \mu }{ A }_{ \nu }-iq\left( { \partial }_{ \mu }{ A }_{ \mu }+{ \partial }_{ \nu }{ A }_{ \mu } \right) -{ \partial }_{ \nu }{ \partial }_{ \mu }+{ q }^{ 2 }{ A }_{ \nu }{ A }_{ \mu }+iq\left( { \partial }_{ \nu }{ A }_{ \mu }+{ \partial }_{ \mu }{ A }_{ \mu } \right) =0[/tex]
So it seems that the commutator is zero, which doesn't seem right... where have I gone wrong?
I have just read that the gauge field term Fμν is proportional to the commutator of covariant derivatives [Dμ,Dν]. However, when I try to calculate this commatator, taking the symmetry group to be U(1), I get the following:
[tex]\left[ { D }_{ \mu },{ D }_{ \nu } \right] =\left( { \partial }_{ \mu }-iq{ A }_{ \mu } \right) \left( { \partial }_{ \nu }-iq{ A }_{ \nu } \right) -\left( { \partial }_{ \nu }-iq{ A }_{ \nu } \right) \left( { \partial }_{ \mu }-iq{ A }_{ \mu } \right) ={ \partial }_{ \mu }{ \partial }_{ \nu }-{ q }^{ 2 }{ A }_{ \mu }{ A }_{ \nu }-iq\left( { \partial }_{ \mu }{ A }_{ \mu }+{ \partial }_{ \nu }{ A }_{ \mu } \right) -{ \partial }_{ \nu }{ \partial }_{ \mu }+{ q }^{ 2 }{ A }_{ \nu }{ A }_{ \mu }+iq\left( { \partial }_{ \nu }{ A }_{ \mu }+{ \partial }_{ \mu }{ A }_{ \mu } \right) =0[/tex]
So it seems that the commutator is zero, which doesn't seem right... where have I gone wrong?