Effective mass from the Lagrangian

[Malamala]

Hello! I have the following Lagrangian:

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

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!

 

[Demystifier]

You made a trivial error, $\partial^2f/\partial v^2$ is multiplied with $\dot{x}$, not $\ddot{x}$. Hence, in addition to an effective mass multiplying $\ddot{x}$, there is also a $v$-dependent force. Otherwise, the idea of effective $v$-dependent mass seems OK to me. After all, an effective $v$-dependent mass appears also in old formulations of special relativity. In fact, with a right choice of $f$, you can reproduce the special relativistic $v$-dependent mass exactly. What remains to be seen is whether the $v$-dependent force could be interpreted as the magnetic force, I leave it as a research/exercise problem for the others.

 

[Malamala]

You made a trivial error, $\partial^2f/\partial v^2$ is multiplied with $\dot{x}$, not $\ddot{x}$. Hence, in addition to an effective mass multiplying $\ddot{x}$, there is also a $v$-dependent force. Otherwise, the idea of effective $v$-dependent mass seems OK to me. After all, an effective $v$-dependent mass appears also in old formulations of special relativity. In fact, with a right choice of $f$, you can reproduce the special relativistic $v$-dependent mass exactly. What remains to be seen is whether the $v$-dependent force could be interpreted as the magnetic force, I leave it as a research/exercise problem for the others.

Thank you! For the $\partial^2f/\partial v^2$ term, don't we have $\frac{d}{dt}(\partial f/\partial v)v = \partial^2f/\partial v^2 \frac{dv}{dt}v = \partial^2f/\partial v^2 \ddot{x}\dot{x}$? So indeed I did a mistake, but that term would still contribute as an effective mass by $\partial^2f/\partial v^2 \dot{x}$ (I missed the $\dot{x}$ term before), no? Or am I doing my derivatives wrong?

 

[Demystifier]

Thank you! For the $\partial^2f/\partial v^2$ term, don't we have $\frac{d}{dt}(\partial f/\partial v)v = \partial^2f/\partial v^2 \frac{dv}{dt}v = \partial^2f/\partial v^2 \ddot{x}\dot{x}$? So indeed I did a mistake, but that term would still contribute as an effective mass by $\partial^2f/\partial v^2 \dot{x}$ (I missed the $\dot{x}$ term before), no? Or am I doing my derivatives wrong?

You are right. There is no "force", everything can be put into the effective mass. I made an error too.