A Does continuous mass distribution implies finite propagation

[mma]

speed?

This question emerged in my mind while studying a discrete and continuous mathematical model of a falling slinky.

In the discrete model, we suppose an instantaneous interaction between mass points at a distance, so the action propagates through the chain of mass points with infinite speed, in the sense that any after any \varepsilon>0 time after releasing the upper end of the hanging slinky, the displacement of the lower end of the slinky is greater than 0.

In contrast, in the continuous limit, there is a \varepsilon > 0 time interval so, that after releasing the upper end of the hanging slinky, the lower end remains exactly motionless during this interval.

However, this is only a special case, assuming a special (linear) force law between the masspoints, I conjecture that for a broader class of force laws (or any force law) this feature holds. Is it true, or I'm wrong, and this effect is due to just the linear force law?

 

[wrobel]

The wave equation implies the finite propagation of waves indeed. It is a some kind of weak solutions to the wave equation. The sound waves for example.
In the discrete model we consider a finite dimensional system of analytic differential equation. It can not provide finite propagation since the solution to such a system is an analytic function. If an analytic function is equal to zero on some interval then it is equal to zero identically.

 

[vanhees71]

This is of course not what in this very nice paper is done. Rather it starts with the very useful model of a discrete number of mass points connected by massless springs, and of course then you don't get an effective action at a distance but nearest-neighborhood bounds, i.e., the Lagrangian reads
$$L=\sum_{j=1}^N \frac{m}{2} \dot{x}_j^2 + \frac{k}{2} \sum_{j<k} (x_j-x_k)^2-mg \sum_{j=1}^N x_j,$$
with $x_0=x_{N+1}=0$. This leads to the equation of motion (2.1) of the manuscript, and of course local perturbations from equilibrium in this chain leads to wave-like propagation of this behavior also for the discrete model of the slinky.

 

[mma]

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In the discrete model we consider a finite dimensional system of analytic differential equation. It can not provide finite propagation since the solution to such a system is an analytic function. If an analytic function is equal to zero on some interval then it is equal to zero identically.

I like this nice argument that proves the statement, that continuous mass distribution is a necessary condition for the finite speed of propagation. However the sufficiency is still not quite clear to me.

 

[vanhees71]

No, the statement is wrong, as explained in my posting. I strongly recommend to read and understand the excellent paper in the OP.

 

[mma]

No, the statement is wrong, as explained in my posting. I strongly recommend to read and understand the excellent paper in the OP.

I think, you mean something else of "finite propagation speed" than I. As I wrote in the OP, I define here the finite propagation speed as

there is a \varepsilon&gt;0 time interval so, that after releasing the upper end of the hanging slinky, the lower end remains exactly motionless during this interval.

As far as I see, Wrobel's proof is good if we insist to this definition.

 

[vanhees71]

What I mean is that a local distortion at one place of the chain from the equilibrium state of course leads to a finite speed of propagation of this perturbation to other parts of the chain. The very funny behavior when releasing the hanging slinky demonstrated in the manuscript in the OP and confirmed by experiment can be described with the model of discrete mass points with nearest-neighborhood harmonic interactions, and this is a demonstration of the finite speed of propagation of perturbations along the chain.

 

[wrobel]

It seems I must clarify my post. I wrote that the wave equation implies the finite propagation of waves. It is better to say that the wave equation admits the finite propagation of waves. There also may be waves that perturb the whole space simultaneously . For example $u_{tt}=u_{xx},\quad u(t,x)=\frac{1}{1+(t-x)^2}$

 

[vanhees71]

Yes, but the "wave propagation" is also of finite speed for the discrete model!

 

[wrobel]

I have already explained why it is wrong. By the way, by the same reason there is no solutions with finite speed of propagation in the heat equation: typically parabolic problems have analytic solutions