Undergrad Interaction Term in EM Lagrangian - Explained

Click For Summary
The discussion centers on the role of the interaction term -Aμjμ in the Lagrangian for electrodynamics, which includes the field energy density -FμνFμν/4. It highlights that while field energy can explain forces between charged particles, the interaction term is essential for generating a non-zero electromagnetic field from charges. Without this term, the equations of motion derived from the Euler-Lagrange equation would not accurately describe the dynamics of charged particles. The conversation also notes that calculations often consider the static limit, which can lead to misunderstandings about the necessity of the interaction term. Overall, the interaction term is crucial for a complete understanding of electromagnetic interactions.
Anypodetos
Messages
17
Reaction score
1
The (classical, relativistic) Lagrangian for electrodynamics contains the field energy density -FμνFμν/4 and the interaction term -Aμjμ. I understand the maths of that - for one thing, the equations of motion turn out right if you plug this into the Euler Lagrange equantion.
Now I recall having learned that you can explain the forces between charged particles solely with the field energy: pushing 2 electrons together increases field energy because it goes with the square of the field strength, and pushing an electron and a positron together decreases field energy. If this is true, why do we need the interaction term at all? What am I missing?
 
Physics news on Phys.org
If you do not have an interaction term, the charges do not generate a non-zero electromagnetic field.

Furthermore, it is worth noting that you are considering the static limit when you are doing that type of calculation.
 
  • Like
Likes Anypodetos
Perfect answer, thanks! Makes me feel a bit stupid, though.
 
In this video I can see a person walking around lines of curvature on a sphere with an arrow strapped to his waist. His task is to keep the arrow pointed in the same direction How does he do this ? Does he use a reference point like the stars? (that only move very slowly) If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in? So ,although one refers to intrinsic curvature...

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K