Re: [Phys-L] gravitational waves

[David Bowman <David_Bowman@georgetowncollege.edu>]

Regarding bc's remembrance:

> This reminds me of the counter intuitive energy loss of an
> oscillator with quadratic dissipation for each cycle is
> proportional to the cube of the initial amplitude for each
> cycle.  
>
> This assumes a rather high Q and a short period oscillator.
>
> bc

This is all correct to leading order in a weakly quadratically velocity dissipative oscillator.  But I don't quite know why you claim the result is counter-intuitive.  Why would you expect it to be different, and what would be the difference you would expect?

FWIW, in case anyone is possibly interested in the particular value of the proportionality constant in the proportionality mentioned above, the energy loss |DE| per *full* cycle in the weak dissipation limit is given by

|DE| = (8/3)*(k*b/m)*|x_0|^3

where x_0 is the displacement amplitude, k is the spring stiffness, m is the mass, and b is the quadratic velocity damping force coefficient, i.e. |F_damp| = b*(dx/dt)^2.  The half-cycle energy loss is half of this amount per swing.

One really cool (at least cool IMO) thing about such an oscillator is that the solution of its dynamics can be calculated from a *dimensionless* D.E. which has no internal parameters whatsoever, and thus has a universal behavior in terms of its motion regardless of the specific values of k, m, and b.  (No distinction or switching between under-damped and over-damped motions here as the given parameters change).  Changing the values of these constants merely changes the scale of the dimensionless time unit, or of the dimensionless displacement unit, or both.  The dimensionless D.E. is:

d^2u/dT^2 + u + (1/2)*sgn(du/dT)*(du/dT)^2 = 0

where the dimensionless time T == sqrt(k/m)*t, and the dimensionless displacement u == (2*b/m)*x.  Not only that, but the exact amount of amplitude damping per cycle (or, better yet per half-cycle) can be solved for exactly without needing to integrate any nonlinear D.E., but rather only solving a nonlinear *algebraic* equation.  And the algebraic equation is mild enough that it is fairly easy to find the leading order small amplitude damping by making a straightforward Taylor expansion for the nonlinear algebraic function involved in powers of the amplitude (and dividing out the trivial u = 0 fixed point equilibrium solution).  Suppose we define the real-valued function g(z) == (1 - z)*exp(z).  Then if |u_0| is the dimensionless amplitude magnitude , i.e. absolute value of the maximum magnitude displacement at the beginning of a half-cycle swing of the oscillator when it is momentarily at rest, and |u_1| is the slightly reduced dimensionless amplitude on the other side of the equilibrium position at the other end of the oscillator's (half-cycle) swing, then u_1 is related to u_0 by the simple equation:

g(-|u_1|) = g(|u_0|), or equivalently:

|u_1| + ln(1-|u_1|) = ln(1+|u_0|) - |u_0|.

The fixed point found by infinite composition of the recursion over many sequential half-cycles is simply the trivial equilibrium solution u = 0.  For small amplitude swings the recursion relation has the approximate solution:

1/|u_1| = 1/|u_0| + 2/3,

and therefore for many swings

1/|u_n| = 1/|u_0| + 2*n/3 .

This solution is good to higher order in small |u| than it looks because of the fortuitous vanishing of the next higher order correction.

The reason this nonlinear system can be solved so easily (at least compared to more typical nonlinear systems) is that the system has an easily findable first integral and even has a Lagrangian formulation for each separate half-cycle swing (whereupon the damping term switches sign and the Lagrangian has a discontinuous change), and the corresponding 'Hamiltonian' for each separate swing is conserved up until a discontinuity occurs at the ends of each of the swings.  The dimensionless Lagrangian for the problem has the form:

L(u, du/dT) = exp(s*u)*[(1/2)*(du/dT)^2 + 1 - s*u]

where s is the velocity signum i.e. s == sgn(du/dT) = (du/dT)/|du/dT| = +/-1.

BTW, this problem is special case of a class of single-degree-of-freedom classical dynamical problems that have quadratic velocity damping.  It ends up that if such a problem has a simple Lagrangian formulation without the dissipation, then putting the dissipation into the problem will not destroy the existence of a Lagrangian formulation, and it is fairly simple procedure to find the Lagrangian with the damping included from the one without it.  Of course the price paid is that the Lagrangian with the quadratic damping is *not* T - V, and the corresponding 'Hamiltonian' is *not* the total energy E = T + V.

David Bowman