- #1

Malamala

- 299

- 26

$$L = \frac{1}{2}mv^2+fv$$

where ##v = \dot{x}##, where x is my coordinate and f is a function of v only (no explicit dependence on t or x). What I get by solving the Euler-Lagrange equations is:

$$\frac{d}{dt}(mv+f+\frac{\partial f}{\partial v} v) = 0$$

$$m\ddot{x} + \frac{\partial f}{\partial v}\ddot{x} + \frac{\partial f}{\partial v}\ddot{x} + \frac{\partial^2 f}{\partial v^2}\ddot{x} = 0$$

$$(m+2\frac{\partial f}{\partial v}+ \frac{\partial^2 f}{\partial v^2})\ddot{x} = 0$$

Is this correct? Can I think of this system as a particle of effective mass ##M = m+2\frac{\partial f}{\partial v}+ \frac{\partial^2 f}{\partial v^2}## moving without any force acting on it? Thank you!