I Effective mass from the Lagrangian

AI Thread Summary
The discussion revolves around the derivation of effective mass from a given Lagrangian, specifically addressing the terms involving the function f dependent on velocity v. The Euler-Lagrange equations lead to the conclusion that the effective mass can be expressed as M = m + 2(∂f/∂v) + (∂²f/∂v²). A correction was made regarding the multiplication of the second derivative term, which should involve v rather than acceleration. The participants agree that the effective mass concept is valid and can relate to special relativity, while also considering the implications of a velocity-dependent force. The conversation concludes with a consensus on the nature of the effective mass and its derivation.
Malamala
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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!
 
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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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Demystifier said:
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?
 
Malamala said:
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.
 
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