What is the physical interpretation of the Lagrangian condition b2-ac ≠ 0?

[humanist rho]

Homework Statement

A Lagrangian for a particular physical system can be written as,

$L^{\prime }=\frac{m}{2}(a\dot{x}^{2}+2b\dot{x}\dot{y}+c\dot{y}^{2})-\frac{K% }{2}(ax^{2}+2bxy+cy^{2})$

where a and b are arbitrary constants but subject to the condition that b²
-ac≠0.What are the equations of motion?Examine particularly two cases a=0=c and b=0,c=a.What is the physical system described by above lagrangian.? What is the significance for the condition b²-ac?

2. The attempt at a solution

I've done the mathematics.But donno the physics!

Equations of motion are,
$ma\ddot{x}+mb\dot{y}+Kax+Kby=0$

$ma\ddot{y}+mb\dot{x}+Kcy+Kbx=0$

I think these equations represent coupled 2D harmonic oscillator.(i'm not sure)

when a=c=0,
$mb\dot{y}+Kby=0$
$mb\dot{x}+Kbx=0$

when b=0,c=-a,

$ma\ddot{x}+Kax=0$
$ma\ddot{y}+Kay=0$

Thanks.

 

[francesco85]

Hello! In general, the system described by the lagrangian represents two "independent" harmonic oscillators, in a sense I'm going to explain: you can collect the coordinates in a vector $\vec{v}=(x,y)^t$ and the coefficient a,b,c in a matrix M such that $M_{11}=a$, $M_{12}=M_{21}=b$, $M_{22}=c$; the lagrangian takes the following form:

$L=\frac{m}{2}\dot{\vec{v}^t}M\dot{\vec{v}}-\frac{K}{2}\vec{v}^t M \vec{v}$.

Since M is symmetric, we diagonalize it through an orthogonal matrix O:
$M=O^t M^{\text{diag}} O$.

We can now define two new coordinates x' and y' that can be incorporated in a vector $\vec{v'}=(x',y')^t$ which is equal by definition to

$\vec{v'}=O\vec{v}$.

In this case the lagrangian has manifestly the form of two decoupled harmonc oscillators (if b^2-ac different from zero).
The significance of b^2-ac different from zero means that the two eigenvalues of M are different from zero and, so there are two modes which oscillate.

As far as I know (and if I don't forget any hypothesis), this is a quite general feature of lagrangian which are at most quadratic in the coordinates. I hope this is right and the answer you need,
Francesco