Why do the extra terms cancel in the derivation of the EM field strength tensor?

Click For Summary
SUMMARY

The discussion centers on the derivation of the electromagnetic field strength tensor (Fuv) using covariant derivatives defined by Du = ∂u - iqAu. The field strength tensor is expressed as Fuv = ∂uAv - ∂vAu. The user initially encounters extra terms, iq(Av∂u - Au∂v), during the commutator expansion but later realizes that these terms cancel due to the properties of the operator commutator. The clarification provided emphasizes the importance of understanding the operator's action on functions.

PREREQUISITES
  • Understanding of covariant derivatives in differential geometry
  • Familiarity with electromagnetic theory and field strength tensors
  • Knowledge of operator algebra in quantum mechanics
  • Basic proficiency in tensor calculus
NEXT STEPS
  • Study the properties of covariant derivatives in the context of gauge theories
  • Learn about the derivation and applications of the electromagnetic field strength tensor
  • Explore operator commutation relations in quantum mechanics
  • Investigate the role of gauge invariance in electromagnetism
USEFUL FOR

This discussion is beneficial for physicists, particularly those specializing in theoretical physics, quantum mechanics, and electromagnetism, as well as students seeking to deepen their understanding of field theory and tensor calculus.

dyn
Messages
774
Reaction score
63
Hi
I am trying to follow the derivation in some notes I have for the field strength tensor using covariant derivatives defined by Du = ∂u - iqAu . The field strength is the defined by [ Du , Dv ] = -iqFuv
The given answer is Fuv = ∂uAv - ∂vAu .When I expand the commutator I get this answer but I have 2 terms left over which I presume should cancel but I don't understand why . I have the extra terms iq(Av∂u - Au∂v ) . Do these terms cancel and if so , why ?
Thanks
 
Physics news on Phys.org
You are forgetting that this commutator is an operator that should act on something and it should be seen as the operator commutator. The ##\partial A## terms represent first multiplying by A then differentiating. Therefore ##\partial A f = (\partial A) f + A \partial f##. Add the corresponding indices.
 
  • Like
Likes   Reactions: dyn
Thanks for your reply. It all works out now.
 

Similar threads

Contravariant derivative of a tensor field in terms of generalized coordinates?
  • · Replies 2 ·
Replies
2
Views
2K
Graduate Weyl tensor and coordinate acceleration
  • · Replies 12 ·
Replies
12
Views
3K
Undergrad Deriving Maxwell's Equations from Field Tensor (Griffith 4ed)
  • · Replies 6 ·
Replies
6
Views
3K
Four Tensor Derivatives -- EM Field Lagrangian Density
  • · Replies 16 ·
Replies
16
Views
2K
Undergrad Riemann curvature tensor derivation
  • · Replies 4 ·
Replies
4
Views
2K
Graduate Valence Tensor of "DVu/Du" Acting on Vector: Analyzing 1 to 1
  • · Replies 3 ·
Replies
3
Views
1K
Graduate Understanding E-Field Transformation in Feynman II_26
  • · Replies 12 ·
Replies
12
Views
2K
Undergrad Using the derivative of a tangent vector to define a geodesic
  • · Replies 7 ·
Replies
7
Views
5K
Undergrad Why Must Pure Force Depend on 4-Velocity?
  • · Replies 47 ·
2
Replies
47
Views
5K
Graduate Why Do These Riemann Tensor Terms Cancel Each Other Out?
  • · Replies 3 ·
Replies
3
Views
2K