One quick question about a Lagrangian

atomqwerty

Let be

L=(x^2 + y^2)x* +2xyy*

where x* = dx/dt and y* = dy/dt. Which physical system is referred it to? why?

Thanks

gabbagabbahey

See here

w4k4b4lool4

$$
L= (x^2+y^2)\dot{x}+ 2xy\dot{y}
$$
Your Lagrangian is a total derivative, and so I don't see any dynamics at all ...
$$
L = \frac{d}{dt}\Big(\frac{1}{3}x^3+xy^2\Big)
$$

gabbagabbahey

You should also read the link in my previous post. We do not provide solutions to students' homework problems here. It teaches the student very little (if anything) and is unethical. Please stick to giving only hints to problems, and only after the student has shown some effort.

w4k4b4lool4

Yes, thanks for that,
I just read the link you posted.
Apologies,
Wakabaloola

atomqwerty

Yes, sorry about that, I get it (and thanks for the reply). The thing is this is from an exam problem, and after thinking about it, I couldn't reach to any plausible answer. I showed that it fits with the Euler-Lagrange equations, and then I tried writing it in polar coordinates (r, theta), maybe I could see that way what kind of system was it. I know the Lagrangian for several systems, but none of them fit with this one, so I didn't know even how to make a hypothesis. Thanks.

gabbagabbahey

No problem. In future, please include descriptions of what you've tried (like the above) in the "attempt at solution" part of your posts.

The fact that the Lagrangian "fits with the Euler-Lagrange equations" for all [itex]x[/itex], [itex]y[/itex], [itex]\dot{x}[/itex], and [itex]\dot{y}[/itex] tells you there are no meaningful dynamics in this system (i.e. no external forces).

atomqwerty

I see... And what about the fact that L doesn't have a term with x and y only? Can it be used to end up with the same conclusion? I mean, can I say that L is pure kinetic energy, and V=0, and so that F=0? Thanks

gabbagabbahey

I don't see how. As a counterexample, consider [itex]L=xy\dot{x}\dot{y}[/itex].

w4k4b4lool4

A particle with only kinetic energy and no potential energy DOES have dynamics ...

Classical evolution minimises the action,
$$
S = \int_{t_i}^{t_f} dt L.
$$
In your case,
$$
S = \frac{1}{3}x^3(t_f) + x(t_f)y(t_f)^2 - \frac{1}{3}x^3(t_i) + x(t_i)y(t_i)^2,
$$
which is constant and independent of what x(t) and y(t) do when t_i<t<t_f. That is, any x(t) or y(t) you write down with these endpoints satisfy the equations of motion,
$$
\delta S=0.
$$
All paths are allowed. Even discontinuous paths! Such a particle would be very strange!

Food for thought: Is it possible for such a particle to violate energy conservation, travel faster than light, violate momentum conservation ... provided the endpoints are fixed as above, which are arbitrary anyway?

As far as I understand, all this is saying is that there are no physical degrees of freedom! Much like gravity in 2 or 3 spacetime dimensions: any gravitational ripple in a 2D universe can be gauge away (i.e. one can find a coordinate system where the ripple is absent), so that all metrics are equivalent (assuming trivial topology).

atomqwerty

That was very helpful, and thought-feeding as well! Thank you for your help, I'll follow the forum rules in the future.

Carlos