Monday, February 13, 2017

Brian Greene's spring trick and his weird explanation based on locality

Update: A blog post basically arguing that Greene is right was published hours later.

Esquire has argued that one third of the U.S. employees have been less productive since Trump's triumph because they were distracted by political posts on the Internet. I guess that Hillary's supporters were more affected than Trump's fans. In particular, unless he is joking and unless I misunderstood something, Brian Greene has forgotten the lectures of classical mechanics that he took at the elementary school.

His daughter is dropping a spring. A camera records what's going on and the (slowed down) recording shows that the bottom of the spring remains at a fixed place – before the top of the spring arrives and the whole spring starts to fall down. At least that's what it looks like.

So far so good. It's not quite trivial to notice that something like that is going on and record it.

However, Brian Greene also provides us with an "explanation": the bottom of the spring isn't moving because it doesn't know for a while that Greene's daughter released the top of the spring. It takes some time for the bottom of the spring to learn, Greene effectively says, because the information is only moving at some finite speed – he probably means the speed of sound. And that's why the ignorant bottom of the spring hovers in mid-air. This bottom hasn't learned yet that it should obey Newton's gravitational law which is why this bottom ignores this law. ;-)

In this explanation, the situation is analogous to the explanation why vibrations (sound) move at a finite speed through a rigid rod. For a while, the opposite side of the rod doesn't know that a melody entered our end of the rod from a speaker, and that's why it doesn't oscillate. The oscillations only spread through the rod by the speed of sound.

My reading of Greene's tweet is that the situations are analogous and the bottom of the spring hovers in mid-air basically because of locality or the finite speed of propagation of influences. Do you understand the tweet in the same way?

This explanation may sound convincing to other folks who want to be high-tech but who missed some lectures at the basic school. But those who still include the basic school among their foundational knowledge know that this explanation isn't quite right.

The Earth's gravitational attraction acts at all times, it can't be turned off, and both the upper portions of the spring as well as the bottom of the spring are exposed by this downward gravitational force.

The acceleration \(a = -d^2 z / dt^2\) of the bottom parts of the spring (let's think about the lowermost circle) is determined by Newton's\[

F = ma

\] where \(m\) is its mass and \(F\) is the total force acting on that. The total force – which I consider downward if the value is positive – is the sum of the gravitational \(F_g\) piece and other pieces \(-F_o\):\[

F = F_g - F_o.

\] Our conventions are that \(F_g\) and \(F_o\) are positive but the other forces mainly include the force exerted by the upper parts of the spring, i.e. the force trying to shrink the spring.

The bottom of the spring may only hover in mid-air if these two forces are equal (with the opposite sign). It doesn't happen automatically. The stiffness of the spring has to be fine-tuned so that it's true. If the spring constant is too high, the shrinking of the spring will beat gravity and the bottom will go up. If the spring constant is too low, gravity wins.

What will be the motion of the top of the spring and the bottom of the spring? We may easily take the gravitational force into account by describing the system from the viewpoint of a freely falling frame. Let's pick the initial speed as \(v=0\) at the moment when the top of the spring is released. The center of mass (com) – let us assume that the spring is uniform so it's the same thing as the middle of the spring – will therefore fall as\[

z_{com}(t) = -\frac L2 - \frac{gt^2}{2}

\] where \(L\) is the initial length of the spring, \(g\) is the gravitational acceleration, and \(t\) is time. Also, \(z=0\) is the initial height of the top of the spring and \(z\lt 0\) for \(t \gt 0\) indicates the lower attitudes than the initial location of the top of the spring.

Great. The center of the spring moves along the usual parabola with the right acceleration. What about the top of the spring and the bottom of the spring? They are moving relatively to the center of mass. So their location will be \(z_{com}\) plus some extra functions of time that we could derive in empty space without the gravitational field.

If you've learned the quantum harmonic oscillator really thoroughly :-), you also know its classical limit. The spring wants to oscillate as \(\sin(\omega t)\) where \(\omega\) is some angular frequency that you may know from \(E=\hbar\omega\). I am sort of joking because I still expect a greater number of people to understand the classical harmonic oscillator than the quantum harmonic oscillator.

Note that \(\omega\sim \sqrt{k/m}\) where \(k\) is the spring constant, the proportionality coefficient between the separation and the force, and \(m\) is the mass. One would have to carefully study the question whether the relevant \(m\) in this problem is the mass of the whole spring or its one-half or its one-quarter etc. But this only affects the constant, not the qualitative functional dependence.

OK, so the length of the spring will behave as\[

\ell(t) = L \cdot \cos(\omega t)

\] with some angular frequency. The normalization \(L\) is the same original length of the spring. Great, \(\pm \ell/2\) is clearly the offset from the center of mass that we need to calculate the position of the top and bottom of the spring, respectively. In particular, the bottom of the spring has the coordinate\[

z_{bot}(t) &= z_{com}(t) - \frac{\ell(t)}{2}\\
&=-\frac L2 \zav{ 1+\cos (\omega t) } -\frac{gt^2}{2}

\] You may see that these two terms can't precisely cancel for a whole interval of \(t\). However, you may Taylor-expand the cosine and pick the leading and subleading term only to get\[

z_{bot}\approx -L \zav{ 1- \frac{\omega^2 t^2}{4} } - \frac{gt^2}{2}

\] You may see that as long as the cosine may be approximated well by the quadratic function, the two terms proportional to \(t^2\) have the potential to cancel. They will cancel if\[

\frac{gt^2}{2} = L\frac{\omega^2 t^2}{4}\quad {\rm i.e.} \quad 2g = L\omega^2

\] When this is true, the bottom of the spring will hover in mid-air in the initial part of the process. However, at some moment, the quadratic function won't be quite an accurate approximation of the cosine. And because the terms in the Taylor expansion of the cosine have alternating signs, the quadratic function will overstate how much the cosine is capable of shrinking the spring, and gravity starts to win at some time.

One may reduce the errors by making the spring a little bit stiffer, i.e. \[

L\omega^2 = 2g + \epsilon

\] which will have the effect that before gravity starts to win (because cosine isn't quite the quadratic function), the spring will be winning and the bottom of the spring will accelerate upwards for a while. If you look at the spring carefully, you may see that this is indeed the case here, too. The spring slowly accelerates in the direction up for a while, because \(\epsilon\gt 0\), but when the spring is already short, the difference between the cosine and the quadratic function starts to play some role, gravity will start to dominate, and the bottom of the spring returns approximately to the original place.

This continues up to the moment when the cosine (i.e. the length of the spring) drops to zero. At that moment, the cosine no longer describes the length well. The length is set to \(\ell(t)=0\) and the spring falls down along a quadratic trajectory – which coincides with the trajectory that the center of mass of the spring followed from the beginning. Check e.g. this graph to see how very accurately the cosine and the quadratic function may cancel for \(0\leq x\leq \pi /2\).

So sorry, Brian and his followers, but this unusual if not miraculous hovering of the bottom of the spring in mid-air has nothing to do with the finite speed of propagation of signals. ;-)

Bonus I: While the bottom of the spring can't exactly hover in the air in any interval because the cosine and the quadratic function aren't equal, you might wonder whether the "rather good fine-tuning" given by the condition \(L\omega^2=2g+\epsilon\) is the result of some very good luck or fine-tuning or whether it is unavoidable for any piece of a slinky. Well, it's the latter: All slinky will have the same math. Why?

Because at the beginning, the slinky is hanging which means that the gravitational force matches the maximum spring force. The spring force acting on the bottom is some \(kL\). If this is equal to the force of gravity, we have \(kL=mg\) where \(m\) is the mass of the spring, up to a universal numerical coefficient of order one. But recall that \(k/m=\omega^2\) so that \(kL=mg\) is equivalent to \(L\omega^2=g\). The numerical coefficient could be hard to get but again, it's universal. Regardless of the material, length, gravitational field etc., the experiment, if the gravitational field is enough to stretch the slinky at all, will look like in Brian Greene's video.

That also establishes that there cannot be any "extreme case", as Michael incorrectly says in the comment. For all choices of the parameters, the situation is the same.

Bonus II: There is a non-uniformity in the initial compression of the spring. The top is more stretched because a greater mass attracted by the Earth is underneath it, the bottom is less stretched. To get the numerical coefficients right, one needs to solve the partial differential equations for locations that depend both on time \(t\) and the vertical coordinate \(z\). The relevant partial differential equation will contain second derivative and be a cousin of the wave equation. It will indeed locally have some causality that arises in all these wave-like situations. But it's still an emergent situation and you can't rely on it because signals and compression may still propagate through the slinky by the much higher speed of sound. It seems plausible to me after some time that the speed-of-sound signals won't have an observable effect on the slinky because they are longitudinal waves that act by twisting the slinky around the vertical axis and the changes of the length are negligible. So in some rather good approximation, it could be true that there is an emergent causal description of these signals propagating through the waves. But because fundamentally the information generically does propagate by the much higher speed of sound, I think it's misleading to say that the bottom "cannot know".

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