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
Messages
345
Reaction score
28
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!
 
Physics news on Phys.org
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.
 
Last edited:
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.
 
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
Back
Top