I Dropping an extended Slinky -- Why does the bottom of the Slinky not fall?

[DaTario]

Discretization involves its own set of approximations. The claim that it is “a more fundamental certification to this result” is not valid, regardless of the open question.

You seem to be just posing and axiom here.
I may agree with you in that both the wave and the many subsystem approaches lead to imperfect descriptions.

The tension does not change instantaneously. Therefore there must be some time where the one discrete element is not in equilibrium but the next is.

where is this 'dead time' (or 'waiting time') defined? I think the tension is a function of the position of the masses (i.e., the length of the spring). Once the equilibrium configuration is changed the movement starts -- no waiting time.

Your arguments in post 17 are simply wrong:

Okay, maybe so. I'm not here trying to defend a closed theory that I think is already free from flaws. But how do you refute my arguments from post #17?

 

[PeroK]

Okay, maybe so. I'm not here trying to defend a closed theory that I think is already free from flaws. But how do you refute my arguments from post #17?

It's a common mistake to use $dt$ as a differential and then to assume it's a finite time interval and thereby reach a contradiction (such as the impossibility of circular motion).

Or, in this case, if I understand your argument, the impossibility of a wave pulse.

 

[DaTario]

Yes. This is the time that it takes for the wave to cross the subsystem’s length.

you are referring to a time interval that is a global parameter of the dynamics. But what I would like to know is how those micro intervals of time, related to elementary occurrences of waiting, are distributed on the temporal axis.

 

[DaTario]

It's a common mistake to use $dt$ as a differential and then to assume it's a finite time interval and thereby reach a contradiction (such as the impossibility of circular motion).

By dividing a differential time interval by two one is not necessarily commiting a mistake, is he?

Or, in this case, if I understand your argument, the impossibility of a wave pulse.

I believe in the possibility of a wave pulse, but I also believe in a probable underlying dynamic that is hidden in formalism due to approximations. Consider the video I sent in post #15 as inspiration.

 

[Dale]

where is this 'dead time' (or 'waiting time') defined?

I already answered this:

This is the time that it takes for the wave to cross the subsystem’s length

But how do you refute my arguments from post #17?

I already answered this too:

If that is the case then your discretization time step is far too coarse and your solution will be grossly inaccurate.

Your whole reasoning is based on the premise that

if the highest mass element dm falls dy in a time dt, I think it is reasonable to propose that the equilibrium condition for the second mass element is no longer guaranteed

That premise is wrong. In any case where it holds it indicates that your time step is too coarse. This will result in a pathologically wrong solution, one that reflects the incorrect numerical method used to solve it rather than one that is faithful to the system being modeled.

 

[DaTario]

I already answered this:

I very respectfully desagree. You are pointing me a global time parameter. How does this finite time value form from elementary contributions? When do they occur? If $\Delta t = \int dt$, at what times do these $dt$ occur?

 

[PeroK]

By dividing a differential time interval by two one is not necessarily commiting a mistake, is he?

It depends what you are doing. There is no equivalent of $\frac 1 2 \Delta t$, where $\Delta t$ is a small, finite time interval. You can't divide a differential because it's not a number.

 

[DaTario]

It depends what you are doing. There is no equivalent of $\frac 1 2 \Delta t$, where $\Delta t$ is a small, finite time interval. You can't divide a differential because it's not a number.

I agree that we must be careful with these operations.
But if $y(t)$ is the coordinate of the highest loop and its initial value is zero. If the fall occurs from time zero I believe there is no error in the inference:

If $y(dt) = dy \ge 0$, then $y(dt/2) = \alpha dy \ge 0$, with $\alpha \in I\!\!R^*_+$

given that we are dealing with the usual concepts of gravitational and elastic forces within Classical Physics framework.

 

[Dale]

I very respectfully desagree. You are pointing me a global time parameter.

I am pointing you to your time step that you used in your reasoning. If your $dt$ that you used in your reasoning is large enough that one element being out of equilibrium implies that the next element is out of equilibrium then it is too coarse. I am not pointing you to anything new, I am pointing out that your reasoning fails in the very beginning.

 

[PeroK]

If $y(dt) = dy \ge 0$, then $y(dt/2) = \alpha dy \ge 0$, with $\alpha \in I\!\!R^*_+$

Note that $\frac{dy}{dt}dt = dy$. What you have is mathematically meaningless.

 

[Dale]

You can't divide a differential because it's not a number.

It is not a real number.

There are number systems, like the hyperreal numbers, with infinitesimal numbers that could be divided. That is a very nit picky point that has nothing to do with the present discussion. I just like hyperreal numbers.

 

[DaTario]

I am pointing you to your time step that you used in your reasoning. If your $dt$ that you used in your reasoning is large enough that one element being out of equilibrium implies that the next element is out of equilibrium then it is too coarse. I am not pointing you to anything new, I am pointing out that your reasoning fails in the very beginning.

Please do not interpret my $dt$ as finite. I'm thinking that $dt$ is any infinitely small, non-zero interval.

 

[PeroK]

It is not a real number.

There are number systems, like the hyperreal numbers, with infinitesimal numbers that could be divided. That is a very nit picky point that has nothing to do with the present discussion. I just like hyperreal numbers.

Then you have to be clear about that. But, simply waving your arms and saying these are hyperreals without understanding the implications is perhaps where this thread has gone all wrong.

 

[Dale]

Please do not interpret my $dt$ as finite. I'm thinking that $dt$ is any infinitely small, non-zero interval.

If your $dt$ is not finite then your reasoning is not even plausible. There is no infinitesimal value for which your premise holds.

By the way, the standard term for “infinitely small” is “infinitesimal”.

 

[DaTario]

Note that $\frac{dy}{dt}dt = dy$. What you have is mathematically meaningless.

Okay, I won't go any further into this dangerous forest of infinitesimals. But I don't think you are presenting a reason for the non-existence of some immediate response from the last loop.

In principle, when it comes to classical physics, we can elaborate on finite $\Delta t$ intervals that are worth less than $10^{-43} s$. In these terms we are probing the formal machine of classical physics and my argument from post #17 would not be dethroned.

 

[DaTario]

the standard term for “infinitely small” is “infinitesimal”.

Thank you.

 

[Frabjous]

Okay, I won't go any further into this dangerous forest of infinitesimals. But I don't think you are presenting a reason for the non-existence of some immediate response from the last loop.

In principle, when it comes to classical physics, we can elaborate on finite $\Delta t$ intervals that are worth less than $10^{-43} s$. In these terms we are probing the formal machine of classical physics and my argument from post #17 would not be dethroned.

Another way to think about it. You are essentially discretizing things which means that you you be thinking about stability criterion. For hyperbolic equations, one should use the Courant-Friedrich-Lewy condition.

 

[Dale]

Then you have to be clear about that. But, simply waving your arms and saying these are hyperreals without understanding the implications is perhaps where this thread has gone all wrong.

I 100% agree. Hyperreals have to be treated correctly and are not a panacea for sloppy calculus

 

[DaTario]

I just like hyperreal numbers.

Do you think this concept is applicable in the present discussion? I don't know what they are.

 

[DaTario]

What do you think of the attitude of saying that this problem in its most fundamental treatment is a many-body problem and therefore we only have access to smart solutions, which make use of approximations and simplifications which are capable of eliminating some interesting details on the micro scale?

 

[DaTario]

Another way to think about it. You are essentially discretizing things which means that you you be thinking about stability criterion. For hyperbolic equations, one should use the Courant-Friedrich-Lewy condition.

Hi, Frabjous. I don't believe we are tempted here to proceed to a numerical solution of the entire dynamics of the spring. The problem seems to me to be more where to allocate small waiting intervals on the micro temporal scale that ultimately lead to a finite and 'macro' rest time, maintained by the last turns.

 

[PeroK]

Do you think this concept is applicable in the present discussion? I don't know what they are.

That may be an interesting question. In the case of a slinky, the classical model is of a large but finite number of discrete particles. This, in itself, is an argument against infinite rigidity - even without the theory of relativity. And, the standard calculus of a continuous mass distibution is an approximation to the finite reality. In that sense, infinitesimals take us further from the reality. They are, IMO, even more unphysical than the real numbers.

 

[DaTario]

I would like to send a fraternal greeting to everyone I debated here. I'll see if I learn introductory notions of hyperreals. As my last comment to this thread:

Perhaps a reasonably acceptable way to momentarily pacify this discussion is to say that the wave model offers a basic and efficient explanation for what can be measured and that a more detailed and in-depth understanding of the dynamics of this system, involving the possibility of new effects arising in the micro-scales of space and time, depends on a more fundamental theory (for example with quantum or relativistic ingredients).

I confess that I am already somewhat happy with the early manifestation of the torsion effects.

 

[Frabjous]

Hi, Frabjous. I don't believe we are tempted here to proceed to a numerical solution of the entire dynamics of the spring. The problem seems to me to be more where to allocate small waiting intervals on the micro temporal scale that ultimately lead to a finite and 'macro' rest time, maintained by the last turns.

You missed my point. Looking at numerical stability will give you insights on your “small waiting intervals.” Both involve discretization.

 

[Dale]

Do you think this concept is applicable in the present discussion?

Not really. It is just that some of the arguments that you have proposed are not even well formulated statements in the real numbers. @PeroK correctly objected to those statements, and I just nitpicked that he said “number” where he should have said “real number”.

I'll see if I learn introductory notions of hyperreals.

Here is a good free online introductory calculus textbook based on hyperreals:

https://people.math.wisc.edu/~hkeisler/calc.html

I confess that I am already somewhat happy with the early manifestation of the torsion effects.

Even those happen with a measurable delay, just small enough that an ordinary camera won’t see it. There simply is no way around it. Regardless of whether you are talking about the torsion wave or the longitudinal wave, the bottom does not move until the wave from the top reaches it.

In principle, when it comes to classical physics, we can elaborate on finite $\Delta t$ intervals that are worth less than $10^{-43} s$. In these terms we are probing the formal machine of classical physics and my argument from post #17 would not be dethroned.

On the contrary, such a small time interval dethrones your argument at the beginning, as I have said multiple times. With “$\Delta t$ intervals that are worth less than $10^{-43} s$” it is not “reasonable to propose that the equilibrium condition for the second mass element is no longer guaranteed”. The argument is dead at the start.

 

[DaTario]

Thank you, Dale.

Best wishes.

Thank you all for the contributions.

 

[DaTario]

You missed my point. Looking at numerical stability will give you insights on your “small waiting intervals.” Both involve discretization.

You may well have a point, Frabjous. Perhaps the stability analysis may yield an interesting result associated with the present discussion. But I still don't see cristal clear how does this waiting interval build up from the many subsystems approach.

 

[Dale]

I still don't see cristal clear how does this waiting interval build up from the many subsystems approach.

Then you should actually go through the effort to calculate it. The way to build intuition is through experience. You don’t have to artificially add in the waiting interval. It will come out naturally.

 

[DaTario]

I think you are right. The way most of the time is to do the math. Thank you for the contributions.

 

[DaTario]

Hi All,

I think I found the missing piece to the many-body argument to allow the conclusion that the last loop is at rest. When the first loop falls, the movement away from the initial position undoes the equilibrium condition on the one hand, but the acceleration of its mass produces a compensating force that maintains the equilibrium. In short, what the Slinky does is that, when falling, it exchanges elastic tension for forces associated with inertia, so that throughout the fall the elastic forces gradually weaken and disappear, but accelerated masses appear that provide the necessary support for the balance to be maintained. It's a case of perfect (or almost perfect) compensation.

An example that may help to see the point is the following. Imagine a man holding a simple 5 meter wooden ladder. He is initially positioned at the base of the ladder, i.e., at its lowest point (see fig). Let's transport the man and the ladder to a height of 30 meters, with the help of a helicopter. From there let's let the system fall. If at the very beginning of the fall the man begins to climb the ladder quickly (in an accelerated manner), he will be able to obtain from the ladder a force capable of keeping him static in space as long as there is a ladder to 'climb'.