The given action $S$ on the above picture is said to give the Lorentz force law and electromagnetic field tensor. And $A_a$ is the covector which is the 4-vector potential. I am curious why the integral of the potential term has plus sign in front of it. $q$ is just the charge. For mechanics, lagrangian is $L=T-U$. Why does electromagnetism seem different? Could anyone please explain?
Why does the Lagrangian of the electromagnetic vector potential has plus sign on the potential term?
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3$\begingroup$ There is a convention of the action where it could get an overall minus sign that is originated from the signature of the metric. Please check what the metric signature is for that lecture note. $\endgroup$– Oktay DoğangünCommented Apr 27, 2018 at 18:07
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$\begingroup$ "L=T-U, U is potential energy" is valid only when U is position dependent only. Here magnetic vector energy is also speed dependent so different. See this question physics.stackexchange.com/questions/420155/… and comments there. I agree that Lagrangian is just something that when differentiated yields the correct motion equation. It is not more fundamental. $\endgroup$– verdeliteCommented Feb 19, 2019 at 15:52
1 Answer
Conventionally, the velocity-dependent potential energy is $$U~=~q(\phi \dot{x}^0 -\vec{A}\cdot \dot{\vec{x}})~=~\mp q A_{\mu}\dot{x}^{\mu} ,$$
where
$$A^{\mu} ~=~(\phi,\vec{A} ), \qquad A_{\mu} ~=~(\mp \phi,\pm \vec{A} ), $$
in Minkowski signature $(\mp,\pm,\pm,\pm)$, respectively. (Here we work in units where $c=1$.)
Note in particular that the electric potential $\phi$ always comes with a plus sign in the velocity-dependent potential energy $U$ (when the charge $q$ is positive).
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$\begingroup$ Is $U$ the potential energy of a particle of charge $q$ obtained from the 4-vector potential $A_u$? $\endgroup$– KeithCommented Apr 27, 2018 at 18:23