Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Physics
Classical Physics
Mechanics
Effective mass from the Lagrangian
Reply to thread
Message
[QUOTE="Malamala, post: 6828917, member: 649307"] 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! [/QUOTE]
Insert quotes…
Post reply
Forums
Physics
Classical Physics
Mechanics
Effective mass from the Lagrangian
Back
Top