1
$\begingroup$

It is known that electromagnetic (EM) fields action on particles is limited to the Lorentz force action. In terms of spinors and currents, the EM field:

(i) rotates the Dirac current around the direction of the field $B$ to the angle proportional to the magnitude of $B$, and

(ii) "accelerates" the Dirac current in the direction of $E$ to the extent proportional to the magnitude of $E$.

Action of Lorentz force exhausts all possible rotations and boosts.

In fact, nothing else could "happen" to the Dirac current (which is a Minkowski space 4-vector) except (a) rotation and boost described above, and (b) change in magnitude of the Dirac current vector prohibited by conservation of charge and/or mass.

Unless the action of weak and strong forces is always limited to change of mass/charge of the particle, is it possible to consider the effect of these forces on particles as a kind of "special" electromagnetic action?

Improve this question
$\endgroup$
1
  • $\begingroup$ This is not true--- you are thinking of the classical effects. There are also spin interactions in the Dirac equation. Your question would be better (and answered!) if you asked "what is the classical force from the SU(2) and SU(3) gauge field on a heavy source" This is physical, and similar to the Lorentz force. $\endgroup$
    – Ron Maimon
    Commented Apr 15, 2012 at 7:32

1 Answer 1

Reset to default
2
$\begingroup$

No.

The weak force acts on uncharged particles, has the wrong range behaviors, violates symmetries that the electromagnetic force is known to respect, and is much too, well, weak.

The strong force has the wrong range behavior and is much to, well, strong.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ well, if we go into the su(2)xsu(3)xU(1) standard model we may is say that it is inspired by the gauge invariance of the electromagnetic U(1) extended to higher dimensional and different symmetries? $\endgroup$
    – anna v
    Commented Mar 24, 2012 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.