# Effective mass from the Lagrangian

• I
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!

Gold Member
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.

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• vanhees71
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?

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