Lagrangian density of a linearly elastic string under gravity

[pines-demon]

TL;DR Summary

Trying to write down the Lagrangian density of an elastic spring, with relation to the slinky problem

This was inspired by this:Dropping an extended Slinky -- Why does the bottom of the Slinky not fall?. There is that famous demonstration of dropping a slinky, and the bottom of the slinky does not move until the center of mass reaches the bottom. I was trying to figure out how hard are the equations? This attempt: Modelling a Falling Slinky w/ Lagrangian only considers masses at top and bottom.

I was wondering if it would suffice to just write the Lagrangian density of some (linear elastic) slinky-stringy kind of system (in 1D):
$$\mathcal{L}=\frac12 \rho \dot{u}^2-\frac12 \rho c^2 \left(\frac{\partial u}{\partial z}\right)^2 -\rho g u $$
where $c$ is the longitudinal speed of sound, $\rho$ the density, $u(z,t)$ the displacement field along $z$, the $g$ gravitational acceleration.

Which leads to the equation
$$\ddot{u}-c^2 \frac{\partial^2 u}{\partial z^2} + g=0$$

So the question is (1) how do I impose that the string is finite and falling along? (2) how to solve such an equation? (3) would this be enough to simulate the situation of the slinky (bottom at rest until the center of mass reaches the bottom)?

Edit: I might be missing some length constant in the gravitational potential energy

Edit2: Another possibility would be to add the center of mass (CM) Lagrangian $L=\frac12 m \dot{z}^2- mgz$ but then I would not know how to couple the CM Lagrangian and the field density.

 

[Orodruin]

So the question is (1) how do I impose that the string is finite and falling along?

Start by imposing an initial condition. That the initial condition is the slinky hanging at rest under its own weight is crucial for the problem.

(2) how to solve such an equation?

Find the static solution for the case when the slinky hangs under its own weight with one of the ends supported and the other end is hanging freely. (Edit: This is why using a model with weights at two ends of the string is bound to fail in describing the slinky.) This gives a Dirichlet boundary condition for the first end and the requirement that the spatial derivative at the other is zero. Together this is sufficient to fix the stationary solution, call the stationary solution $\tilde u(z)$. Now introduce $w(t,z) = u(t,z) - \tilde u(z)$. You now have a homogeneous wave equation for $w$, but be careful with the boundary conditions(!) - both ends are free for $t > 0$ and so you will have an inhomogeneous boundary condition for $w$.

Now find the solution for $w$ by using $w(t,z) = f(z-ct) + h(z+ct)$ and knowing that $h = 0$ because there will be no wave travelling upwards from the bottom. Solve for $f$.

The solution will look like this:

I have here plotted the positions of 11 points along the slinky (each interval between them representing 1/10 of the slinky) at consecutive time points. Note how the lower point remains in place for the full duration (units are scaled so the time for the wave to reach the bottom is normalised to 1, as is the slinky length at $t = 0$).

The main observation however is that this is not really what we observe from a real slinky. The top point passes all other points on the way down in this solution. This does not typically happen for a real slinky if you watch the video. Instead, the top collides rather inelastically with the rest of the slinky and this is not captured by the model.

(3) would this be enough to simulate the situation of the slinky (bottom at rest until the center of mass reaches the bottom)?

You don't need to simulate it. There is an analytic solution. See above.

Edit: This solution focuses on the time until the initial wave from letting the supported end go reaches the bottom. If you want to extend the analysis you can do so, but it basically just requires some fiddling with $f$ and $h$.

Edit 2: This problem had me stumped for quite some time back in the day because I was looking for a particular form for the solution where the upper part was essentially falling freely under gravity. There is no such solution.

 

[pines-demon]

The solution will look like this:

Excellent solution. I will try it out. It might not reflect reality but at least it shows that the bottom remains fixed.

Together this is sufficient to fix the stationary solution, call the stationary solution u~(z).

Does this stationarity means the solution before letting go?

Edit: This solution focuses on the time until the initial wave from letting the supported end go reaches the bottom. If you want to extend the analysis you can do so, but it basically just requires some fiddling with f and h.

Is it too much fiddling, would such extension look anything like reality?

Instead, the top collides rather inelastically with the rest of the slinky and this is not captured by the model.

What is the source of the inelasticity? Internal heating?

 

[Orodruin]

Excellent solution. I will try it out. It might not reflect reality but at least it shows that the bottom remains fixed.

It is not that far away from reality. You can kind of imagine how it would go if you added some inelastic collisions into the mix.

Does this stationarity means the solution before letting go?

Yes. The slinky is stationary in this situation and does not move.

Is it too much fiddling, would such extension look anything like reality?

To get the functional form, you would need to add quite some - essentially an infinite train of terms in both. However, you can kind of understand what the solution would look like from the shape at the end above - which is the same as we started with (just reversed) and now moving at some speed this immediately gives you the solution.

No, it will not make the solution more realistic in the initial fall. It is the same as before in the initial phase.

 

[Orodruin]

Ok, I deleted a couple of posts here earlier as I realised they might wander off into the direction original research that is disallowed on PF. I have since scoured the literature and found a paper by William Unruh (yes, the Unruh of the Unruh effect) that does essentially what I had considered:

https://arxiv.org/abs/1110.4368​

(NB: Not published mind you, but the entire subject is kind of an application of mechanics ... I am going to assume Unruh did not consider it worth submitting anywhere but rather just wanted it out there)

Here follows my take, which closely resembles that of Unruh:

The thing is, you can make a rather easy argument for the collapse of the spring. Using $z(t,s)$ as the extension of the spring in the down direction a fraction $s$ of spring, the initial condition is
$$
z(0,s) = 2L_0 s (1 - s/2)
$$
where $L_0$ is the total extension of the spring. We know that the bottom part cannot move until it receives the impulse from the top so we assume that at time $t$, everything below $s = \sigma(t)$ will remain in the initial state. It is easy to see that this part will satisfy the wave equation. If we now assume that the top part of the spring inelastically collides with the parts of the spring it is moving past, the part of the spring with $s < \sigma(t)$ will be moving downwards at some velocity $v(t)$. This part will carry momentum $m\sigma(t)v$ as its mass is $m\sigma(t)$. But - and this is a beautiful argument - since the rest of the spring is not moving and the only external force on the full spring is $mg$, this momentum must be equal to $mgt$, leading to:
$$
\sigma(t) v = gt.
$$
In order to use this equation, we must find an expression for the velocity $v$, preferably in terms of $\sigma(t)$. Luckily there is a good argument for this too. Since the lower part of the spring is stationary, the position of the moving upper part is always given by $z(0,\sigma(t))$ and the velocity $v$ is the time derivative of this. In other words,
$$
v = \frac{dz(0,\sigma(t))}{dt} = \frac{dz(0,\sigma)}{d\sigma} \frac{d\sigma}{dt} = 2L_0(1-\sigma) \dot\sigma.
$$
Inserted into the momentum equation, we now have
$$
2 L_0 \sigma(1-\sigma) \dot \sigma = gt
$$
which is a separable ODE. Assuming that $\sigma(0) = 0$ it easily integrates to
$$
\sigma^2\left(\frac{1}{2} - \frac{\sigma}{3}\right) = \frac{gt^2}{4L_0},
$$
which is an implicit equation for $\sigma(t)$. In particular, we can see that the moving part of the spring will meet the bottom when $\sigma(t) = 1$, i.e., when
$$
t = \sqrt{\frac{4L_0}{g} \left(\frac{1}{2}-\frac{1}{3}\right)}
= \sqrt{\frac{4L_0}{g} \frac{1}{6}} = \sqrt{\frac{2L_0}{3g}}
$$

One thing to note in particular is that, if you compute $\dot\sigma$, you will find that it is larger than the wave speed. This is somewhat contrary to what seems to be the prevailing popular belief that the bottom starts falling when a longitudinal wave from the top travelling at the wave speed reaches the bottom. Instead, we have a signal propagating faster than the longitudinal wave speed and is accompanied by an abrupt change in the density of the coils before and after the signal has passed. This is characteristic not of a wave propagating at the wave speed in its medium, but of a shock wave. Indeed, it would take far too long for the bottom to start moving if it had to wait for a wave to propagate down, as seen in the previous posts. If you look at a slow motion capture of an actual slinky drop, it should be clear that the slinky actually curls up almost completely, meaning its center of mass passes the bottom when $gt^2/2$ is equal to the original distance of the center of mass from the bottom - which happens to be $L_0/3$. Again, we find that the time for this to occur should be $\sqrt{2L_0/3g}$, just as predicted by the inelastic model above.

I also played around a bit with my Python script for the illustrations. Here is the model of the slinky drop in the inelastic model (compare to the slow motion capture of any slinky drop):

(I added a non-zero rest length to the slinky for visual effect.)
The red and blue dots represent the top and bottom of the slinky, respectively. I used it to easier see how they moved in the original model with only a propagating wave:

 

[pines-demon]

View attachment 342341​

Again amazing graphics and nice paper!

One question though, Unruh approach is a bit different to mine. I added a displacement field $u$ and derive with respect to the real coordinate of space $z$. While Unruh does the inverse he writes the real coordinate $x$ derived by the coordinate in the slinky. Is one approach better than the other? I would have used mine if I was dealing with vibrations in a solid for example...

If we now assume that the top part of the spring inelastically collides with the parts of the spring it is moving past,

Why do you assume the real collision is ineslastic? Where does the energy go? Heat? I though it just became torsional or something else.

 

[Orodruin]

Is one approach better than the other?

Yes. If you use the real spatial coordinate, tension and density wont be constant and you will not obtain a simple wave equation. You will only obtain a wave equation if you do it with the slinky fraction as the variable.

Why do you assume the real collision is ineslastic?

Because it certainly looks that way in all of the slinky drop videos and it gives an excellent approximation in the end. Experiment is the ultimate judge and guide in physics.

Where does the energy go? Heat? I though it just became torsional or something else.

For the purposes we have discussed it is not that relevant. I never even considered energy, but Unruh did discuss it. Exactly where it goes depends, but in the ideal slinky drop that looks exactly like the image, heat. In a real slinky drop you will often see the slinky hitting somewhat unevenly at the end so the approximation is then not as good.