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muzialis Dec14-12 06:34 AM

Dissipative Lagrangian
 
1 Attachment(s)
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

Jano L. Dec14-12 08:25 AM

Re: Dissipative Lagrangian
 
Quote:

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.

muzialis Dec14-12 08:50 AM

Re: Dissipative Lagrangian
 
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

Jano L. Dec14-12 02:55 PM

Re: Dissipative Lagrangian
 
No problem muzialis.

Quote:

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.

Quote:

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.


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