Conversation started Aug 26, 2020 at 7:34.
bolbteppa
bolbteppa
Aug 26, 2020 07:34
The Lorentz force law can be written as $\frac{d p^{\mu}}{d \tau} = e F^{\mu \nu} U_{\nu}$, where $F^{\mu \nu} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}$, you can then view this as the extremum of a Lagrangian and integrate it against $\delta x^{\mu} d \tau$ and use integration by parts to end up with the potential
Lorentz force
In physics (specifically in electromagnetism) the Lorentz force (or electromagnetic force) is the combination of electric and magnetic force on a point charge due to electromagnetic fields. A particle of charge q moving with a velocity v in an electric field E and a magnetic field B experiences a force of F = q E + q v × B {\displaystyle \mathbf {F} =q\mathbf {E} +q\mathbf {v} \times \mathbf {B} } (in SI units)...
For one way to remember the potential, just remember $S = \int [\frac{1}{2} m v^2 - V(x)] d t$, this is the non-relativistic Lagrangian of a particle moving in a scalar potential $V(x)$. What should the potential term look like if it is not a scalar potential $V(x)$ but is instead a four-vector potential $A_{\mu}(t,x,y,z) = (V,\vec{A})$, added in a relativistically-invariant way, such that when $\vec{A} = 0$ we obtain the usual scalar potential.
Clearly $S = \int [\frac{1}{2} m v^2 - \frac{e}{c} A_{\mu} \frac{dx^{\mu}}{d t}] dt = \int [\frac{1}{2} m v^2 - e [V(x) \frac{ dt}{d t} + \frac{1}{c} A_i \frac{dx^i}{dt} ] d t$ will reduce to the right form
 
Conversation ended Aug 26, 2020 at 7:43.