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

[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'.

 

[Orodruin]

Microscopically, approximating a slinky with a set of mass-and-springs subsystems is just as wrong as using a continuum approximation. Neither are good descriptions if you go to the fundamental level. The question becomes: is either a better approximation? Assuming a sufficient number of sub-masses and springs, the predictions of the models will be indistinguishable. The ”better” model becomes the one where making the predictions is easier, which is arguably the continuum model. (I at least prefer solving a rather simple PDE over diagonalizing a ##10^{20}\times 10^{20}## matrix - no matter how sparse …)

 

[DaTario]

Microscopically, approximating a slinky with a set of mass-and-springs subsystems is just as wrong as using a continuum approximation. Neither are good descriptions if you go to the fundamental level. The question becomes: is either a better approximation? Assuming a sufficient number of sub-masses and springs, the predictions of the models will be indistinguishable. The ”better” model becomes the one where making the predictions is easier, which is arguably the continuum model. (I at least prefer solving a rather simple PDE over diagonalizing a ##10^{20}\times 10^{20}## matrix - no matter how sparse …)

Hi, Orodruin. I have never seen the PDE associated with this problem. It seems that it must be somewhat hard to solve, as relevant properties of the medium, including its size, are changing in time and differently in each location. The matrix I would leave to quantum computers,

 

[Orodruin]

Hi, Orodruin. I have never seen the PDE associated with this problem. It seems that it must be somewhat hard to solve, as relevant properties of the medium, including its size, are changing in time and differently in each location. The matrix I would leave to quantum computers,

The PDE is just the wave equation (written in the correct variables), but the pure wave equation is not really describing the case completely as it results in a solution where the top falls past the other parts of the slinky before the bottom moves. In a real slinky, compressions are not actually possible and therefore there will be collisions of the slinky windings. These could be modelled separately, but even the wave equation (that is sufficient to describe the lower part where collisions still have not occured) is sufficient to conclude the bottom does not move until some time has passed. The wave equation solution is relatively easy and outlined and illustrated here in a thread from yesterday:
https://www.physicsforums.com/threa...inearly-elastic-string-under-gravity.1061144/

 

[DaTario]

The PDE is just the wave equation (written in the correct variables), but the pure wave equation is not really describing the case completely as it results in a solution where the top falls past the other parts of the slinky before the bottom moves. In a real slinky, compressions are not actually possible and therefore there will be collisions of the slinky windings. These could be modelled separately, but even the wave equation (that is sufficient to describe the lower part where collisions still have not occured) is sufficient to conclude the bottom does not move until some time has passed. The wave equation solution is relatively easy and outlined and illustrated here in a thread from yesterday:
https://www.physicsforums.com/threa...inearly-elastic-string-under-gravity.1061144/

It seems to be a nice work, in deed. Considered as one spring, the Slinky has mass and its natural length in the horizontal is such that it does not admit compression. But one consequence of its having mass is that when it is put in the vertical the new natural length, accompanied by a pattern of distensions that are not homogeneously distributed, is such that it admits a certain degree of compression.

But my point is basically that the wave argument is somewhat opaque for those who study this problem in general don't know how to associate the concept of information with the fundamental interactions which are present in this context. The tentative explanation (or the outline of an explanation) I provided in post #60 has a bit more transparency (IMO), showing dinamical reasons for the observed rest of the loop which is closest to the ground.

 

[Orodruin]

But my point is basically that the wave argument is somewhat opaque.

I disagree with this. The wave argument is very transparent. A disturbance to any shape of the slinky cannot travel faster than the wave speed as long as the wave equation holds (which it admittedly does not once you start getting hit by the upper part of the slinky, but that too takes time). As long as the shape at the bottom is not disturbed, the bottom won't move because it is subject to exactly the same forces it was subjected to when the full slinky was hanging still.

 

[DaTario]

I disagree with this. The wave argument is very transparent. A disturbance to any shape of the slinky cannot travel faster than the wave speed as long as the wave equation holds (which it admittedly does not once you start getting hit by the upper part of the slinky, but that too takes time). As long as the shape at the bottom is not disturbed, the bottom won't move because it is subject to exactly the same forces it was subjected to when the full slinky was hanging still.

Ok, but what about the more newtonian argument that the rest is caused by the fact that the decrease of the elastic force in time is compensated by the appearance of accelerated masses in the system? Do you think it makes some sense?

 

[Orodruin]

Ok, but what about the more newtonian argument that the rest is caused by the fact that the decrease of the elastic force in time is compensated by the appearance of accelerated masses in the system? Do you think it makes some sense?

On the level of the end of the slinky, no. The end of the slinky is subject to exactly the same forces as before. If you consider the full slinky, obviously the average acceleration is g downwards. As the bottom is not accelerating, upper parts must be accelerating faster than g. The distribution of the acceleration is determined by the internal tension in the slinky.

 

[DaTario]

Is the example of the man climbing the ladder well stated in your opinion?

 

[DaTario]

On the level of the end of the slinky, no. The end of the slinky is subject to exactly the same forces as before. If you consider the full slinky, obviously the average acceleration is g downwards. As the bottom is not accelerating, upper parts must be accelerating faster than g. The distribution of the acceleration is determined by the internal tension in the slinky.

the last turn is subject to a stack of subsystems. As time passes, the superior subsystems lose the elastic forces that kept them in balance and in return, masses begin to accelerate. The logic of this exchange seems to allow the maintenance of balance in this stack of subsystems.

 

[A.T.]

Thank you A.T., for your contribution, but regarding your comment on relativity, this is a Classical Physics' discussion, isn't it?

Yes, and in Newtonian physics, when appropriate, we use the approximation of extended bodies as perfectly rigid objects (infinite propagation speed of mechanical disturbances). But this approximation definitely not appropriate for a slinky.