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Let's say you had a perfect pencil, with a point which was just that one point (see this question). The pencil's mass was perfectly distributed, and there are no flaws in the craftsmanship. Let's say you had it oriented vertically (like you were going to balance it), and it was exactly straight up and down i.e. the pencil's center of mass is directly above the point where the pencil is touching the ground. In the universe where this pencil is, there are no outside forces which can affect the pencil, other than gravity.

Which way will the pencil fall, after you let go?

Is it random?

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    $\begingroup$ It's not random. The question is simply flawed. In classical mechanics you would have to look at the trajectories as a function of perturbations of the initial conditions, which is what modern treatments of classical mechanics do. In reality, of course, such a pencil can not exist, so you are really analyzing the properties of Hamiltonian equations that approximate certain physical systems. $\endgroup$ – CuriousOne May 1 '15 at 22:04
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    $\begingroup$ Down. It will fall down. Not random at all. $\endgroup$ – rob May 1 '15 at 23:12
  • $\begingroup$ In the context of classical mechanics, this is a valid (but fictitious) question, and the answer is that it doesn't fall. In the context of quantum mechanics, this is a nonsense question. There is no pencil whose point is a perfect point in QM. Even worse, there is no way to position the pencil so it is stationary and exactly vertical. That would violate the uncertainty principle. The uncertainty-principle tag should be deleted (or perhaps the entire question should deleted). $\endgroup$ – David Hammen May 2 '15 at 14:29
  • $\begingroup$ No outside forces other than gravity? That's wrong; the pencil is subject to the normal force (otherwise it would sink into the ground). $\endgroup$ – David Hammen May 2 '15 at 14:40
  • $\begingroup$ there's a deeper question, here, though: can the groundstate break rotational invariance? $\endgroup$ – innisfree May 2 '15 at 15:42
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Which way will the pencil fall, after you let go?

Facetious answer: it would fall in the direction to which it was leaning when you let go.

Okay, now to justify that: you cannot balance the pencil before letting go. For an object resting on a surface to be balanced, its centre of mass (a single point) must be directly above a point within the area of contact between it and the surface beneath.

In your scenario the area of contact is a single point (i.e. the surface area is zero), so it is impossible to find a point within this area above which to position the pencil's centre of mass and balance it. There will always be a line you can draw between the point of contact and the point on the surface below the pencil's centre of mass - the pencil will fall along this line.

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  • $\begingroup$ Since the center of mass is one point and the pencil is touching the "ground" at one point, can't they line up? I'm not saying you have to balance it, you just have to hold it so that the center of mass is directly above the point on which the pencil is touching the ground, then let the pencil fall. $\endgroup$ – Tdonut May 2 '15 at 0:24
  • $\begingroup$ Having the centre of mass above the region where the pencil is touching the ground means it's balanced by definition, s you are saying you have to balance it. If that region has zero area it's impossible to align the two exactly and it's impossible to balance the pencil to begin with. $\endgroup$ – jamcowl May 2 '15 at 15:28
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In the universe where this pencil is, there are no outside forces which can affect the pencil, other than gravity.

But there are forces afoot inside the pencil.

Unless you also chill the pencil to absolute zero and thus stop all molecular activity, the trillions of atoms in the pencil are vibrating in random directions. It won't take long for the random force vectors to push the pencil's center of mass outside the column of stability, which will result in a tiny amount of torque, which will quickly cascade until the pencil is lying on it's side.

[control of random molecular movement] were often used to break the ice at parties by making all the molecules in the hostess's undergarments leap simultaneously one foot to the left, in accordance with the theory of indeterminacy.

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What you've fabricated is of course unrealistic in the physical world as CuriousOne stated, but not so in the virtual world of simulation. All the conditions you ask for can be arranged in a simulated universe. If perfectly balanced as its initial condition, the virtual pencil will not fall in this virtual world. It is unstable, but it will not fall until a perturbing force, however small knocks it off balance. It will then fall in the direction of that perturbing force.

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    $\begingroup$ This world would ignore quantum mechanics. $\endgroup$ – Jimmy360 May 2 '15 at 0:20
  • $\begingroup$ @Jimmy360 Only if you wanted it to. Tdonut said no forces other than gravity. $\endgroup$ – docscience May 2 '15 at 3:09
  • $\begingroup$ +1, this is the right answer for this fictitious question. $\endgroup$ – David Hammen May 2 '15 at 14:26
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I'll look at the question from multiple aspects.

Classical mechanics, exact measurements, no thermodynamics, no perturbing forces

In this fictitious universe it is possible to stand our perfectly balanced pencil exactly vertically and perfectly stationary. This is an unstable equilibrium position.

With no perturbing forces, no thermodynamics, no quantum mechanics, the pencil won't fall.


Classical mechanics, exact measurements, no thermodynamics, no perturbing forces except tides

You can balance the pencil vertically for an instant. A moment later, the tidal forces from the Moon and the Sun will have shifted the direction of local vertical. (Aside: tides are gravitational.)

The pencil will fall away from the direction in which local vertical is moving.


Classical mechanics, exact measurements, classical thermodynamics

The pencil will fall in some random direction.


Classical mechanics, realistic measurements

In theory, one can measure to infinite precision in classical mechanics. In practice, we cannot do that. In the world of pencils and bridges, the engineering limitations on ability to measure precisely swamp the tiny quantum mechanical errors inherent in measuring conjugate variables.

The pencil will fall in some random direction.

Comment: This is an inverted pendulum. They fall over -- unless something acts to move the inverted pendulum back toward the unstable equilibrium. Thousands of mechanical engineering students face this problem every year. They have to construct a robot that keeps an inverted pendulum inverted.


Quantum mechanics

The question is tagged . The only way to have the pencil be in this unstable equilibrium state is to have perfect simultaneous measurements of its orientation and angular momentum (or equivalently, the position of its center of mass position and its linear momentum). These are conjugate variables. The uncertainty principle says the product of the uncertainty of a pair of conjugate variables is at least $\hbar/2$.

This question doesn't make sense in a quantum mechanics context; it is akin to asking what the laws of physics says will happen given a violation of the laws of physics.

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  • $\begingroup$ in QM, would be it possible to find a spherically symmetric preparation of the pencil that satisfied the HUP? (the pencil wouldn't be known to be exactly upright and still). then I can ask, does it fall and in which direction? $\endgroup$ – innisfree May 2 '15 at 15:44
  • $\begingroup$ @innisfree - A spherical pencil doesn't fall. It rolls. $\endgroup$ – David Hammen May 2 '15 at 15:51
  • $\begingroup$ sorry, i meant a rotationally symmetric preparation, not spherically. $\endgroup$ – innisfree May 2 '15 at 16:07
  • $\begingroup$ that's basically the question really, if I have many degenerate, rotationally symmetric vacua, which one will nature pick? $\endgroup$ – innisfree May 2 '15 at 16:14
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Suppose that we align a perfectly cylindrical pencil with the z-axis. If the initial conditions are rotationally symmetric about that axis, then because the laws of physics are rotationally symmetric, the final state must also be symmetric under rotations about the z-axis. This means that the wavefunction of the universe will have to evolve into a superposition of the pencil falling in every possible direction with the rest of the universe being entangled with these possible outcomes.

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