This is a trivial question probably, i have a doubt when considering the singssigns for the Lagrangian for a charge particulecharged particle in an electromagnetic field.
Considering that the Lagrangian for a charge particulecharged particle in an electromagnetic field is given as $$L = \frac{m v^2}{2} + q \vec{v} \cdot \vec{A} - q \phi \ \ \ \ or \ \ \ \ L = \frac{m v^2}{2} - q \vec{v} \cdot \vec{A} + q \phi$$$$L = \frac{m v^2}{2} + q \vec{v} \cdot \vec{A} - q \phi \ \ \ \ \text{or} \ \ \ \ L = \frac{m v^2}{2} - q \vec{v} \cdot \vec{A} + q \phi$$ And that the equation of motion for this Lagrangian is associated with the Lorentz force as $$m \dot{v} = q(E + v \times B) \ \ \ \ or \ \ \ \ m \dot{v} = -q(E + v \times B)$$ My doubt is:
If iI have the Lagrangian as $$L = \frac{m v^2}{2} - q \vec{v} \cdot \vec{A} - q \phi$$ that gives the equation of motion as $$m \dot{v} = q(-E + v \times B)$$
Does this equation of motion still represents the Lorentz force even with the difference with singsin signs, or E$E$ and B$B$ must have the same sign?
Is it correct to interpret this Lagrangian as a charge particulecharged particle in an electromagnetic field?
