Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip?

Question

I was asked by an undergrad student about this question. I think if we were to take away air molecules around the pencil and cool it to absolute zero, that pencil would theoretically balance.

Am I correct?

Veritasium/Minutephysics video on Youtube.

Answer 1

TL;DR: there are many factors that prevent a pencil remaining perfectly balanced. The most important of these is the uncertainty principle that will make the pencil fall over in less than four seconds. For details, read on...

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Short answer: NO. The first photon of light that hits it would disturb your perfect equilibrium. The moon's tidal forces (which are not always pointing in the same direction) would disturb it. The sun's tidal forces would disturb it. I could go on.

The equation of motion of a pencil tells us that as soon as you are off center by the smallest amount, the motion will build up. It is not a stable equilibrium.

And graphite cannot sustain the weight of a pencil on a mono-atomically sharp tip... According to this supplier of high quality graphite the compressive strength is about 25 ksi (~170 MPa - figure 5-2 from the reference). The smallest tip that can support the weight of 0.05 N would be a circle with a radius of 0.01 mm. That's a pretty sharp tip, for a pencil. It's not nearly "atomic".

Finally, even at absolute zero, the uncertainty principle requires that the position of the center of mass not be known perfectly. The (quantum mechanically required) fluctuations of the position of the center of mass should be sufficient to cause the pencil to fall over eventually.

UPDATE - the impact of a single photon

It is instructive to calculate how long it takes a pencil to drop for a given deviation from equilibrium (assuming for a moment a perfect pivot at the bottom - i.e. the only torque applied is due to gravity). I am showing it here for a pencil that was hit by a single green photon - and the result is a surprisingly short time.

Modeling the pencil as a uniform rod with mass $m$, length $\ell$, moment of inertia $I=\frac{1}{3}m\ell^2$, the torque $\Gamma$ when it is at an angle $\theta$ to the vertical is

$$\Gamma = \frac12 m g \ell \sin \theta$$

For small deflections, $\sin\theta = \theta$ and we will use that assumption below. Then the equation of motion becomes

$$I\ddot\theta = \frac12 m g \ell \theta$$ $$\frac13 m \ell^2 \ddot\theta = \frac12 m g \ell \theta$$ $$\ddot\theta = \frac{3 g}{2\ell}\theta$$

This looks a lot like the equation for a simple harmonic oscillator, but with the wrong sign. We do indeed get a very similar solution, but with hyperbolic functions.

Putting $\frac{3 g}{2 \ell} = \alpha^2$, we can integrate this twice to get a

$$\theta = C_1 e^{\alpha t} + C_2 e^{-\alpha t}$$

If the pencil starts at equilibrium, we can set $\theta=0$ at $t=0$, which reduces the above to

$$\theta = 2C_1 \sinh{\alpha t}$$

Given an initial velocity $v_0$, we see that

$$v_0 = \frac{\ell}{2}\dot\theta = \ell C_1 \alpha \cosh{\alpha t}$$ so

$$C_1 = \frac{v_0}{\ell \alpha} = \frac{v_0}{\ell\sqrt{3 g /2 \ell}} = \frac{v_0}{\sqrt{\frac32 g \ell}}$$

Now we have an expression for $\theta$, we can solve with a given initial velocity:

$$t = \frac{1}{\alpha}\sinh^{-1}\frac{\theta}{C_1} $$

$$= \sqrt{\frac{2\ell}{3 g}}\sinh^{-1}\left(\frac{\theta \sqrt{\frac32 g \ell}}{v_0}\right)$$

Now comes the fun part. Let's assume we hit the perfectly balanced pencil with a single photon of green light. The momentum of such a photon is approximately

$$p = \frac{E}{c} = \frac{h}{\lambda} \approx 10^{-27} Ns$$

Let's assume the pencil is black so the photon is not reflected. The mass of a pencil is about 0.005 kg, length 20 cm. The velocity of the pencil after the collision with the photon is roughly (yes, there are some factors to account for offset impact etc. I am ignoring all those - it doesn't change the basic answer):

$$v_0 = \frac{p}{m} = 2 \cdot 10^{-25} m/s$$

Let's assuming that "definitely falling" corresponds to an angle of 0.5 degrees, or about 0.01 rad. We can put the values in the above equation, and find t $\approx$ 6 s.

One photon. Six seconds. That is a shockingly short time... but it's about to get worse:

UPDATE 2 - the importance of the uncertainty principle

It is also interesting to see how long it would take for a pencil to fall given an initial deflection from the vertical - because we can then get rid of the one photon and use the uncertainty principle to estimate the maximum time the pencil will balance.

If the center of mass is off by $\Delta x$, then the angle is $\Delta \theta = \frac{2\Delta x}{\ell}$.

Using the same equations as before, we find $C_1 + C_2 = \Delta \theta$. Let's assume the initial velocity is zero - because "on average" it will be, given that the direction of initial velocity is equally likely to point back towards equilibrium and away from it - so we get $C_1 = C_2$, and the solution is a $\cosh$ function:

$$\theta = 2 C_1 \cosh(\alpha t)$$

Where $C_1 = \frac{\Delta\theta}{2} = \frac{\Delta x}{\ell}$.

We now have the time to fall (time to reach a certain $\theta$) as

$$t = \frac{1}{\alpha}\cosh^{-1}\left(\frac{\theta\ell}{2\Delta x}\right)$$

The equation we derived earlier for the time taken with a given initial velocity can be rewritten as

$$t = \frac{1}{\alpha}\sinh^{-1}\left(\frac{\theta\ell\alpha}{\Delta v}\right)$$

and we know that

$$\Delta x \Delta p = \hbar$$

Obviously the longest time to balance will be reached when the two times are the same - otherwise, one will be longer and the other shorter, and it's the shorter time that will dominate. To solve, we substitute for $\Delta x = \frac{\hbar}{m\Delta v}$ and get

$$\cosh^{-1}\left(\frac{\theta\ell m \Delta v}{\hbar}\right) = \sinh^{-1}\left(\frac{\theta\ell\alpha}{\Delta v}\right)$$

If the term in the parentheses is sufficiently large, then we can (for the purpose of estimating) set

$$\frac{\theta\ell m \Delta v}{\hbar} = \frac{\theta\ell\alpha}{\Delta v}$$

From which it follows that

$$\Delta v = \sqrt{\frac{\theta \ell \alpha \hbar}{\theta \ell m }} = \sqrt\frac{{\alpha \hbar}}{m}$$

Substituting the values for the pencil, we find

$$\Delta v = 4\cdot 10^{-16}m/s$$

which is many orders of magnitude larger than the velocity due to the pencil being hit by a photon. The time to fall is then

$$t = \frac{1}{\alpha}\sinh^{-1}\left(\frac{\theta\ell\alpha}{\Delta v}\right)\approx 3.7 s$$

So a perfectly balanced, "theoretical" pencil that stands on its mono-atomic tip, will fall on average in a handful of seconds because of the uncertainty principle.

AFTERWORD my daughter just pointed me to an interesting post that calculates the same thing - and comes up with a very similar answer. The author signs himself as "Alemi". There is a contributor on this site with the same handle. I think I recognize the style of the thought process, so I am going to give a belated tip of the hat. Incidentally, that post comes up with a value of about 3.6 seconds. Which is astonishingly similar to the value I got.

Answer 2

No. To balance perfectly, the pencil would have to be perfectly upright and perfectly still. The uncertainty principle limits how well you can do both at the same time.

Momentum and position form a conjugate pair. $$\Delta x \Delta p \geq \frac{\hbar}{2}$$.

Angular momentum and angular position form one too. $$\Delta L \Delta \Theta \geq \frac{\hbar}{2}$$

This doesn't guarantee that angular momentum and angular position will be non-zero. It is an uncertainty - The actual values can be anything, including $0$.

But it does prevent you from arranging them both so the pencil stays upright. Furthermore, if you ask what the probability of finding both values very close to $0$, you find that it is very small. In the limit, infinitely improbable.

If it turns out that $L = \Theta = \sqrt{\hbar}$, and you plug in reasonable values for the mass and length of the pencil, you will find it falls over in a few seconds.

Belated update

I was waiting until the weekend to add an update. By the time it got here, Floris had left very little to add. And he did a better job than I would have. Good answers.

A number of users felt that an ideal pencil sharpened to an atomic tip was not realistic. The pencil should have a flat on the bottom.

My own thought is that the pencil should be mounted on one of those massless, frictionless pulleys that seem to be so common in high school physics classrooms.

Never the less, a pencil with a flat can be treated semi classically. Because of the uncertainty principal, the pencil has an initial momentum, and therefore an initial energy. This will cause the pencil to tip. Which in turn will cause the pencil to rotate about an edge of the flat. The center of mass will rise until it is directly over the edge of the flat. If the initial "uncertainty" energy is larger than the energy needed to raise the center of mass, the pencil will tip over.

A quantum mechanical treatment would treat the region where the center of mass is over the interior of the flat as a potential well. There is a probability of tunneling out.

Both of these scenarios are treated in full detail (with diagrams in case my description is unclear) here. I found this link by following Floris' "interesting post that calculates the same thing." That post had some comments at the bottom. The very last comment contains the link.

Answer 3

No. The weight of the pencil is roughly 1 Newton, and the area is about 500 square picometer (5 * 10^-22) which means the pressure on the tip is around 2 ZettaPascal. That's quite a bit more than what graphite (or diamond) can withstand (that's measured in GigaPascal)

Answer 4

The question is so ambiguous that it allows a resounding yes. This is because "balance" is not defined, neither are the dimensions and the material used for the pencil, nor the location of where the "balancing" is to happen.
The material and the shape of the surface to balance the pencil on, are not specified, nor the length of time it should stay balanced. So, if you use a titanium tipped "pencil," on a planet/moon with a gravitational force 1/10,000 of earth with no atmosphere, on a surface made of a "complementary" material such that the titanium molecule will "fit" in a "hole" created by the surrounding complementary molecules, then "balancing" the "pencil" will be a snap.

Answer 5

No. Firstly, the point of the pencil is generally not sharp enough to have just one single atom. People attempt to make that kind of tip in STM's. Even if you somehow did manage to get it sharp enough, graphite is so soft the the weight of the pencil will crush the tip. It won't stay a single atom wide. So there is no way to balance the pencil on a single atom because there is no single atom tip.

Secondly, even if you do get a tip a few atoms wide, it is impossible to make a perfectly symmetrical pencil. There are several ways this asymmetry can be introduced. When you sharpen the pencil, there is a point at which the wood forms a step-like structure(where the blade was when you stopped sharpening), making the structure asymmetric. If you used a CNC machine to sharpen it, then you will have to think about whether the paint was applied perfectly, whether the graphite core has a uniform density and more importantly whether the wood has absolutely uniform. Usually you can tell the two kinds of wood in a pencil just by looking at the color difference.

In short, 'theoretically' in this case is not attainable just by cooling everything to zero and creating a vacuum. The level of finishing that the pencil itself will require is too high to even call it a pencil anymore.

Answer 6

GIVE IT A SPIN, with the max possible rpm for such massive object. Do the experiment at the ISS, well above Earth surface.
Unconstrain the problem from the 'surface' issue. Remove the 'surface' below the atom tip -- where is bellow without gravity? -- or approximate the surface as much as you can but without contact (the atom's electronic cloud prevent this, so the definition of 'contact' is sensible, i.e. substitute with momentum transfer).
The Earth is not a perfect uniform sphere and the experiment is sensible to that changes. So move the experiment to a geostationary orbit.
I can use a set of controlled pulsed lasers to put the pencil in rotation, at the same time I chose to align the axis of rotation with Earth's center.
I can Use magnetic levitation (Meissner effect - video) to counteract and control automatically any change of the orientation.
I can not make an actual computation of the time the setup stays in rotation -- it depends on max rpm -- but I expect a very long time, maybe several years, as long as the liquid He stays confined (see Herschel space observatory fate). The other answers are talking on seconds ?.

I've translated the OP theoretically to: the best physical favorable configuration.

Answer 7

Let me be the firs to answer 'Yes' (more or less).

As the saying goes:

In theory there is no difference between theory and practice. In practice there is.

What I'm getting at is that there will always be differences between theory and practice, and that it is up to the physicist to decide which assumptions/simplications are suitable and which ones are not.

Therefore, if you are theoretically describing your pencil in a sufficiently simple manner, e.g. classical mechanics, a vacuum, no external influences,..., then yes, in that theory you will be able to balance a pencil on its tip.

However, regarding that you are already considering the atomic nature of matter as an important element to the theory, it is likely that the other effects that my colleagues have put forth, and which will make the pencil drop, will not be negligible either. Hence, in those theories, it is impossible to balance a pencil on its tip.

Edit: I think the real point might be that "theoretically" is at best poorly defined, and at worst completely nonsensical. I see two interpretations:

1. Does there exist a theory in which...? This was my understanding of it, in which case the answer is 'Yes'
2. For the state-of-the-art theory of reality, ...? This is the understanding of most, with answer 'No'. Though a reasonable interpretation as well, I'd like to add that this is a poor definition as
   • What is the state-of-the-art? Commonly accepted? A rather vague definition.
   • There is not single state-of-the-art theory describing all of physics (e.g. standard model and gravity)
   • The state-of-the-art changes with time

Answer 8

If the pencil were at absolute zero it would necessarily assume its lowest energy state, which is not the vertical state.

If the pencil were modeled as a quantum rotor with an infinite potential barrier covering half its solid angle space (i.e. the table) then there are certainly excited but stable states where the pencil remains in a more or less vertical position.

Answer 9

If it were not for the word theoretically, the question could be answered simply that it is practically impossible to balance such a pencil. The tip would crumble under the pressure. It is inherently unstable, and would topple within a few seconds, because of air currents, vibrations, Brownian Motion, the impact of a single photon. Etc etc.

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When you say theoretically, what theory do you have in mind? The original context of the question, as posted, was Newtonian Mechanics. Tags for QM and the Uncertainty Principle were added later by others, without editing the question itself to remove the ambiguities.

As stated, the question is contradictory and confusing. It asks theoretically if the pencil can balance on a one-atom tip. But then it mentions practical precautions to avoid real-world disturbances such as air currents and (presumably) the thermal vibrations of the pencil. Atoms and thermal vibrations are part of Classical Physics but not part of Newtonian Mechanics, so what does a one-atom tip mean in this context?

Another difficulty is that there is no definition of what it means for the pencil to balance. It would be highly unstable even if the tip were a blunted cone one atom in diameter. However, in theory we can position the pencil with sufficient accuracy such that it would take an arbitrarily large time (eg the age of the universe) to topple. (The required accuracy in position would be orders of magnitude smaller than the Planck Length, but that is not an objection which is inherent within the theoretical model of Newtonian Mechanics.) Conversely, if balanced means that the pencil has not started to topple, then it is never balanced. In theory it is always toppling.

So I agree with Guill that theoretically, and with a suitable definition of balanced, the pencil can be balanced.

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I also agree with Daniel Sank that the use of the Uncertainty Principle to calculate a maximum time of about 3s for which the pencil could remain "balanced" is misleading. Because this is close to times achieved in practice, it gives the false impression that the pencil falls because of quantum mechanics, as in this answer to Explain why quantum behavior is not observed in daily life. As Floris explains, any small deviation from perfect balance will give a similar result.

It is shown by Peter Lynch (University College Dublin) in Balancing a Pencil on its Point, May 2014, that the pencil falls in 3.72s if the Uncertainty Principle is used to decide the initial uncertainty in its position, and in 2.51s if the uncertainty in position is arbitrarily chosen to be the width 1 atom. He concludes that "The toppling is not a quantum effect." Any reasonably small value for the uncertainty - whether it is 0.1mm or 0.1nm or 0.1fm - results in a time of about 2-3s. The Uncertainty Principle sets an upper limit for the time, but it is not the practical cause of toppling.

In The quantum mechanical tipping pencil - a caution for physics teachers, Don Easton of Stony Brook University uses the Uncertainty Principle to estimate the maximum time for which the COM of the pencil could remain within its atom-sized base to be around 10$^{12}$s - ie about 1 million years. In another calculation he estimates the same value for the time it takes the pencil to tunnel out of its potential well. It is only when the inverted pendulum model of the pencil is used - with its exponential instability - that the answer comes out as about 3s.