Classical Mechanics of NonConservative Systems

[Slightly]

There is an article I am reading,

http://authors.library.caltech.edu/38643/1/PhysRevLett.110.174301.pdf

I don't quite understand a lot of where this guy is coming from. I do not have much background in Lagrangian and Hamiltonian formulations, but it is vital that I can formulate some sort of explanation of what this article is talking about.

Is there a picture that anyone can make to describe this? or maybe help describe his usage of q+ and q-?

 

[Joey21]

I'll have a read and see what I can get out of it. Thanks for the link by the way.

 

[Maurice7510]

I think this might actually help explain a problem I had come across a month or so ago; I was having trouble coming up with equations of motion via Lagrangian mechanics for an object with air resistance. I don't know how much you know about Lagrangian/Hamiltonian formalisms, but both involve a linear combination of kinetic (T) and potential (V) energies; L=T-V and H=T+V, resp. But neither kinetic nor potential energy is affected by the presence of air resistance, so the equations of motion come out the same as though there were no air resistance at all. The abstract on this paper appears to address this issue, and I, too, will look through it and see what I can make of it. Either way, thanks for a good read

 

[Maurice7510]

Ok, I've gone through it and, while it's quite in depth, it seems reasonable. As I mentioned above, it does indeed address the issue I was having, and my understanding is that you don't quite get what the paper is trying to do. I was a little brief above so I'll try and be a little more thorough without inundating you.
Hamiltonian and Lagrangian mechanics essentially use kinetic and potential energies to find equations of motion. An example is an object falling through the air: the kinetic energy is

T=1/2mv^2 (v should be xdot, I'm unsure how to actually do that. Likewise, a will represent xdoubledot)

and the potential is

V=mgx

The Lagrangian would then be

L=T-V=1/2mv^2 - mgx

The equations of motion come out of

∂L/∂q = d/dt(∂L/∂qdot) (and yes, those should be partial derivatives)

where q denotes an arbitrary position variable (in our case, this is x). Going through with this,

∂L/∂x = -mg

and

d/dt(∂L/∂xdot) = d/dt(∂L/∂v) = d/dt(mv) = ma

We now equate these and cancel the m to find a=-g. Bet you didn't see that coming.
Now this was a problem where only conservative forces were concerned; namely, gravity. What if there was a dissipative, nonconservative force? In the same example, that would be in the form of air resistance. The paper actually goes through the example, albeit rather conservatively, and finds the equation of motion to be

ma=-bv|v|^n-1

where I've substituted b for his alpha, a for xdoubledot and v for xdot. Hopefully you're familiar with objects falling in fluids enough to recognize this equation, the absolute value being necessary to ensure the force is in the opposite direction of the motion.

Hopefully I didn't completely misunderstand where you were having trouble and was of some help

 

[Slightly]

thank you so much.

You helped so much.

I was really confused with the x+ and the x- stuff top

 

[Maurice7510]

No problem. The x± was just a change of variables for the sake of convenience, he mentions going through it without the change too. It's like when a problem has distinctly spherical characteristics; it's most convenient, though not always necessary, to convert to spherical/polar coordinates