- #1
Abrain
- 8
- 0
Hi everybody!
I'd like to understand the physical meaning of the Feynman's vector potential definition:
[tex]$ A_{m}^{(b)}(x) = e_b \int \delta (xb_{\mu}xb^{\mu})db_m(b), \qquad m=0,1,2,3 $[/tex]
(component m of the vector potential of the particle b at the point x)
Here
- the integration is done over the whole world line of the b particle, parametrized by the invariant parameter b
- [tex]\delta[/tex] is the Dirac delta function
- [tex]$ xb_{\mu}xb^{\mu} $[/tex] is the 4-norm of [tex]x - b[/tex]
- So the [tex]$\delta$[/tex] function is non-zero [tex]$\iff x$[/tex] is in the boundary of the light cone of b.
Can you give me any clues?
Thanks!
I'd like to understand the physical meaning of the Feynman's vector potential definition:
[tex]$ A_{m}^{(b)}(x) = e_b \int \delta (xb_{\mu}xb^{\mu})db_m(b), \qquad m=0,1,2,3 $[/tex]
(component m of the vector potential of the particle b at the point x)
Here
- the integration is done over the whole world line of the b particle, parametrized by the invariant parameter b
- [tex]\delta[/tex] is the Dirac delta function
- [tex]$ xb_{\mu}xb^{\mu} $[/tex] is the 4-norm of [tex]x - b[/tex]
- So the [tex]$\delta$[/tex] function is non-zero [tex]$\iff x$[/tex] is in the boundary of the light cone of b.
Can you give me any clues?
Thanks!