Understanding the Dissipative Lagrangian for Coupled Oscillators

  • Thread starter muzialis
  • Start date
  • Tags
    Lagrangian
In summary, the conversation discusses a Lagrangian and its application to two coupled oscillators with opposite friction signs. The Lagrangian is described as the physical Lagrangian for the oscillators, but the physical interpretation is unclear to the speakers. The Lagrangian is shown to conserve energy, but the concept of separate energies for each oscillator is deemed artificial. The speakers struggle to understand the physical picture and are looking for a useful application of the Lagrangian.
  • #1
muzialis
166
1
Hello,

I run across the following Lagrangian, $$\mathcal{L} = m \dot{x}\dot{y} + \frac{1}{2} \gamma (x \dot{y} - \dot{x} y) $$

I can see how a variation with respect to $$ x, y $$ yields the (viscous) equations of motions

$$ \ddot{x} + \dot{x} = 0 \quad, \quad \ddot{y} - \dot{y} = 0 $$.

In the paper I attach this Lagrangian is described as being the physical Lagrangian for two coupled oscillators, one with negative and the other with positive friction (hence one is exponentially stable, while the other is not, as the equation of motion stress).

I do not understand how the lagrangian $$ \mathcal{L}$$ can correspond to two such oscillators.
Moreover, the whole idea of the formalism seems to conserve energy, in the sense that the (non-dampening) oscillator will absorb all the energy dissipated by the viscous oscillator.
But then the two speeds should be in magnitude equal, and not like $$ e^{t}$$ and $$e^{-t}$$.

I struggle to understand the physical picture, I wounder if anybody could help.

Thanks you very much
 

Attachments

  • Riewe Mechanics with Fractional Derivatives.pdf
    138.5 KB · Views: 505
Physics news on Phys.org
  • #2
I do not understand how the lagrangian
L
can correspond to two such oscillators.
You wrote two equations above, one is the equation of h.o. with friction, the other one has friction term with opposite sign, which leads to run-away.

The energy here is defined as
[tex]
E = p_x \dot x + p_y \dot y - L = m \dot x\dot y.
[/tex]
and since the Lagrangian does not depend on time, it should be constant in time.

You can verify this by multiplying the solutions for [itex]\dot x, \dot y[/itex] - the exponentials will cancel out and the result does not depend on time.

However, all this seems very artificial - I would like to see some useful application of it.
 
  • #3
Jano,

many thanks for your response.
I am still confused though.
1) Should not the Lagrangian for two harmonic oscillators (with equal mass m) be something like (I have the lingering feeling I am not seeing something obvious)
$$ \frac{1}{2}m (e^{-2t}+e^{2t}) $$
2) As you underline, oner h.o. will dampen exponentially, the other one will run away.
How is it possible then the energy dissipated in one equals the energy input in the other, as the article states?

Thank you ever so much
 
  • #4
No problem muzialis.

1) Should not the Lagrangian for two harmonic oscillators (with equal mass m) be something like (I have the lingering feeling I am not seeing something obvious)
[tex]
1/2m(e^{−2t}+e^{2t})
[/tex]

No, the Lagrangian is usually not a function of time, but a function of coordinates and velocities (their derivatives). The first formula for L you wrote is mathematically valid Lagrangian, but the function in the quote is not, because there are no coordinates/velocities.

2) As you underline, oner h.o. will dampen exponentially, the other one will run away.
How is it possible then the energy dissipated in one equals the energy input in the other, as the article states?

This seems like very imprecise way to say much simpler thing, that the total energy [itex]m\dot x\dot y[/itex] is conserved. I do not see any simple way to define separate energy of the first and second oscillator; the Lagrangian describes the system as whole and the energy derived from it corresponds to the whole system as well.

If the energy came out as a sum of independent terms, then we could ascribe the terms to subsystems, but in this case, there is just one term, so it makes no sense to speak about the energy of one oscillator.
 
  • #5
for your question. The dissipative Lagrangian for coupled oscillators is a mathematical representation of a system where energy is dissipated through friction or other forms of dissipation. In this case, the Lagrangian includes a term for the kinetic energy of the system (m dot{x}dot{y}) and a term for the dissipation of energy through friction (1/2 gamma (x dot{y} - dot{x}y)).

The equations of motion that you have derived from this Lagrangian are correct, and they show that the two oscillators are coupled together through this dissipative term. The presence of positive and negative friction terms indicates that one oscillator is being damped while the other is being driven by an external force. This results in the exponential behavior of the two oscillators, with one decaying and the other growing.

The physical picture of this system is that the energy from the undamped oscillator is being dissipated by the damped oscillator. This is why the two speeds are not equal in magnitude - the damped oscillator is absorbing the energy from the undamped one, resulting in a decrease in the energy of the system over time.

I hope this helps to clarify the physical interpretation of the dissipative Lagrangian for coupled oscillators. If you have any further questions, please don't hesitate to ask. Thank you for your interest in this topic.
 

1. What is a dissipative Lagrangian?

A dissipative Lagrangian is a mathematical function that describes the dynamics of a physical system by taking into account energy dissipation. It is used in the branch of physics known as classical mechanics to model systems that lose energy due to friction, heat transfer, or other dissipative forces.

2. How is a dissipative Lagrangian different from a traditional Lagrangian?

The traditional Lagrangian, also known as the conservative Lagrangian, only takes into account conservative forces that do not dissipate energy. In contrast, a dissipative Lagrangian considers both conservative and dissipative forces, providing a more accurate description of the system's behavior.

3. What are some examples of systems that can be described using a dissipative Lagrangian?

Some examples include a pendulum with air resistance, a car driving on a rough surface, and a damped harmonic oscillator. Essentially, any physical system that experiences energy dissipation can be modeled using a dissipative Lagrangian.

4. How is a dissipative Lagrangian used in practice?

In practice, a dissipative Lagrangian is used to derive the equations of motion for a system, which can then be solved to predict the system's behavior over time. It is also used in numerical simulations and mathematical models to study the effects of dissipative forces on a system.

5. What are the limitations of using a dissipative Lagrangian?

One limitation is that it assumes the dissipative forces are linear, meaning they have a constant relationship with the system's velocity. It also does not take into account any external forces acting on the system. Additionally, it may not provide an accurate description of the system if the dissipative forces are large compared to the conservative forces.

Similar threads

Replies
19
Views
1K
  • Classical Physics
Replies
3
Views
646
Replies
2
Views
1K
Replies
5
Views
1K
Replies
4
Views
1K
  • Classical Physics
Replies
1
Views
546
Replies
4
Views
967
Replies
2
Views
128
  • Classical Physics
Replies
1
Views
929
Replies
6
Views
930
Back
Top