In this post I'd like to address a fun physics problem.

How long can you balance a pencil on its tip? I mean in a perfect world, how long?

No really. Think about it a second. Try and come up with an answer before your proceed.

What this question will become by the end of this post is something like the following:

Given that Quantum Mechanics exists, what is the longest time you could conceivably balance a pencil, even in principle?

I will walk you through my approach to answering this question. I think it is a good problem to illustrate how to solve non-trivial physics problems.

I will try and go into some detail about how I arrived at my solution. For most of you this will probably be quite boring, so feel free to skip ahead to the last section for some numbers and plots.

### Finding an Equation of Motion

The first thing we need to do is find an equation of motion to describe our system. Lets consider the angle theta that the pencil makes with respect to the vertical. Lets treat this as a torque problem. Dealing with rotating systems is almost identical to dealing with free particles in Newtonian mechanics. Instead of Netwon's first law, relating forces to acceleration\[ F = m \ddot x \]

we just replace it with the rotational analogue of force - torque, the rotational analogue of acceleration - rotational acceleration, and the rotational analogue of mass - the moment of inertia.

\[ T = I \ddot \theta \]

(I've taken the usual physics notation here, dots represent time derivatives) We need to determine the torque and moment of inertia of our pencil.

At this point I need to

*model*the system. I need to break up the real world, rather complicated idea of a pencil, and turn it into an approximation that retains all of the important bits but enables me to actually proceed.

So, I will model a pencil as a rod, a uniform rod with a constant mass density. In doing so, I can proceed. The moment of inertia of a rod about its end is rather easy to calculate. If you are not familiar with the result I recommend you try the integral yourself.

\[ I = \int r^2 \ dm = \frac{1}{3} m l^2 \]\

where m is the total mass of my pencil and l is its length. I will take a pencil's mass to be 5 g and its length to be 10 cm.

Now the torque. The only force the pencil feels is the force due to gravity, which acts from the center of mass, which for my model of a pencil occurs at half its length. I additionally wish to express the force in terms of the parameter I decided would be useful, namely theta, the angle the pencil makes with the vertical. I obtain

\[ T = r \times F = \frac{1}{2} m g l \sin \theta \]

Great, putting the pieces together we obtain an equation of motion for our pencil

\[ \frac{1}{2} m g l \sin \theta = \frac{1}{3} m l^2 \ddot \theta \]

rearranging I get this into a nicer form

\[ \ddot \theta - \frac{3}{2} \frac{g}{l} \sin \theta = 0 \]

in fact, I'll utilize another time honored physics trick of the trade and simplify my expression further by making up a new symbol. Since I've done these kinds of problems before I can make a rather intelligent replacement

\[ \omega^2 = \frac{3}{2} \frac{g}{l} \]

obtaining finally

\[ \ddot \theta - \omega^2 \sin \theta = 0 \]

And we've done it.

### Looking at the equation of motion

Now that we've found the equation of motion, lets look at it a bit.

First off, what does an equation of motion tell us? Well, it tells us all of the physics of our system of interest. That little equation contains all of the information about how our little model pencil can move. (Notice that while I haven't yet been explicit about it, in my model of the pencil, I also don't allow the tip to move at all, the pencil is only able to pivot about its tip).

Great. A useful thing to do when confronting a new equation of motion is to try and find its fixed points. I.e. try and find states in which your system can be which do not evolve in time. How can I do that? Sounds complicated. In fact, I'll sort of work backwards. I want to know the values that do not evolve in time, meaning of course that if I were to find such a solution, all of the terms that depend on time would be zero. So, if such a solution exists, for that solution the derivative term will vanish. So the solutions have to be solutions to the much simpler equation

\[ \sin \theta = 0 \]

Which we know the solutions. In fact, lets be a little smart about things and only worry about theta = 0 and theta = pi. Thinking back to our model this suggest a pencil being straight up (theta = 0) and straight down (theta = pi).

These are the stable points of the equation. The second one you are familiar with. If you instead of balancing a pencil 'up', try to balance it 'down', you know that if you start with the pencil pointing straight down it stays that way and doesn't do anything interesting.

But what about that first solution theta =0? That indicates that if you could start this model pencil exactly straight up, it would stay that way forever, and also not do anything interesting.

Oh no you cry. It seems as though we've already answered the question. How long can you balance a pencil? It looks like you could do it forever if you did it perfectly.

But you and I both know that is impossible. You can't ever balance a pencil forever. I've never done it, and tonight I've spent a lot of time trying.

So what went wrong?

### When your approximations fail

So what went wrong again? It seems like I've gotten an answer, namely in my model you could, at least in principle balance your pencil forever. But you and I both know you can't. Something is amiss.

Hopefully, the first thought that occurs to you is something along the lines of the following.

Of course you dummy! You couldin principlebalance a pencil forever, but in the real world, you can't set the pencil up standing perfectly straight. Even if its tilted just a little bit, its going to fall. This is exactly the problem with your physicists, you don't live in the real world!

Whoa, lets not be so harsh there. I made some rather crude approximations in order to get such a simple equation. You are allowed to make approximations provided (1) they are still right to as many digits as you care about, and (2) you keep in mind the approximations you made, and think a bit about how they could go wrong.

So, before we do anything too drastic, lets do with your gut. I agree, it seems like if the pencil would be at any small angle, it ought to fall. Lets double check that our equation does that.

So for the moment imagine theta being some small number. In fact, I will use the usual notation and call it epsilon. What does our equation say then?

\[ \ddot \theta = \omega^2 \sin \epsilon \]

lets make another approximation (I know I know, we've already run into trouble, but bear with me). If epsilon is going to be a really small number, then we can simplify this equation even more. That sine being in there is really bugging me. Sines are hard. So lets fix it. Can we say something about how sine behaves when the angles are super small?? In fact we can. Such an approach is super common in physics.

### A short side comment on Taylor Series

Imagine a function. Any function. Picture the graph of the function. I.e. imagine a line on a graph. No matter what function you imagine, if you zoom in far enough, at any point that function ought to look like a line. Seriously. Zoom out a little bit and it will look like a line plus a parabola. Zoom out a little more and it will look like a cubic polynomial.You can make these statements precise, and thats the Taylor Expansion. But the idea isn't much more complicated than what I've described. Taylor expanding the sine, we obtain

\[ \ddot \theta \approx \omega^2 \left( \epsilon - \frac{\epsilon^3}{3!} + \cdots \right) \]

So if you are at

*really*small angles, sin(x) looks just like x. Whats really small? Well as long as x^3 is too small for you to care about. For me, for the rest of the problem that will be for angles that are less than about 0.1 radians, for which that second term is about 0.00017 radians or 0.01 degrees, which is too small for me to care.

### Coming back to the approximation bit

Anywho, for really small angles, our equation of motion is approximately\[ \ddot \theta = \omega^2 \theta \]

So, notice for a second that if theta is positive, since omega^2 has to be positive, then our angular acceleration is going to be positive.

So your intuition was right. If your pencil ever gets to any positive angle, even the smallest of angles, then our angular acceleration is positive and our pencil will start to fall down.

So. The next question becomes. How can we capture this bit of reality. It looks like my model has this unphysical solution. How can I make it more real worldy?

Ah, this is the real fun of physics. You could go in any number of directions. Perhaps you could try and estimate how good you can actually prepare the angle of the pencil, perhaps you could ask whether air bouncing into the pencil would make it fall, perhaps you could wonder whether adding more realism to the moment of inertia would make the pencil fall easier, maybe you could wonder whether the thermal motion of the pencil would make it fall? Maybe you could consider the pencil as an elastic object and consider it vibrating as well as pivoting. Maybe you could model the tip as being able to move? Maybe you could introduce the gravitational pull of the sun? or the moon? or you? or the nearest mountain?

The sky's the limit.

So what am I going to do? Quantum Mechanics. Seriously, bear with me a bit.

### A little preliminaries

Before proceeding any further, lets actually solve the equation of motion we just got for the smallest angles. To remind you, the equation I got for my model of a pencil in the limit of the smallest angles was\[ \ddot \theta = \omega^2 \theta \]

This is an equation I can solve. Its a very common differential equation, and one that we use and abuse in physics so I know the solution by heart. So lets write it down.

First just to let know you, this sort of equation, second derivative of thing being linearly proportional to thing gives you solutions that are always pairs, usually, depending on the sign of the constant, they are written as sines and cosines, or decaying and growing exponentials.

Naturally of course in order to solve a second order differential equation, we need to specify two initial conditions. In this case I will call them theta_0 and dot theta_0, representing the initial position and initial angular velocity.

In this form the solution can actually most easily be written in terms of the exponential pair associated with sine and cosine, sinh and cosh (you can read more about them here, they are really neat functions).

The solution is

\[ \theta(t) = \theta_0 \cosh \omega t + \dot \theta_0 /\omega \sinh \omega t \]

which as you could probably convince yourself, exponentially grows for any positive theta_0 or dot theta_0.

### Abandoning Realism

At this point of considering the question, I turned down a different route. I don't really care about balancing pencils on my desk. You see, I know a curious fact. I know that in quantum mechanics there is an uncertainty principle which says that you cannot precisely know both the position and momentum of an object.This of course means that

*even in principle*, since our world is dominated by quantum mechanics, I could never actually balance even my model pencil forever, because I could never prepare it with perfect initial conditions.

The uncertainly principle tells us that the best possible resolution I could have in the position and momentum of an object are set by Planck's constant:

\[ \Delta x \Delta p \geq \frac{\hbar}{2} \]

This has to be true for our pencil as well. In fact I can translate the uncertainty principle into its angular form

\[ \Delta \theta \Delta J \geq \frac{\hbar }{2} \]

where theta is our theta and J is the angular momentum, which for our pencil we know is

\[ J = I \dot \theta = \frac{1}{3} m l^2 \]

So the uncertainty principle for our pencil is,

\[ \Delta \theta \Delta \dot \theta \geq \frac{3 \hbar }{2 m l^2 } \approx 3.2 \times 10^{-30} \text{ Hz} \]

So what?

So, I'm going to approximate the effects the uncertainty relation has on our pencil problem by saying that when I start off the classical mechanical pencil, I'm going to require that my initial conditions satisfy the uncertainty relation:

\[ \theta_0 \dot\theta_0 = \frac{3 \hbar }{2 ml^2} \]

which we decided is going to mean that our pencil has to fall.

The real question is? How long will it take this pseudo-quantum mechanical pencil to fall?

In order words the question I am really trying to answer is:

Assuming a completly rigid pencil which you place in a vaccum and cool down to a few millikelvin so that it is in its ground state. Roughly how long will it take this pencil to fall?

### Do it to it

So lets do it. This is going to be a bit quick, mostly because its getting late and I want to go to bed.

But the procedure is kind of straight forward now. I need to choose initial conditions subject to the above constraint, figure out how long a pencil with those initial conditions takes to get to theta = pi/2 (i.e. fall over), and then do it over and over again for different values of the initial conditions.

So, the first thing to do is figure out how to pick initial conditions that satisfy the constraint. I'll do this systematically by parameterizing the problem in terms of the ratio of the initial conditions. I.e. lets define

\[ \log_{10} \frac{\theta_0}{\dot \theta_0} = R \]

where I've taken the log for convenience. Now, figuring how long the pencil takes to fall in principle is just numerically integrating forward the full equation of motion

\[ \ddot \theta = \omega^2 \sin \theta \]

where I need to do it numerically because the sine makes this equation too hard to solve analytically. In order to do the numerical integration I implemented a Runge-Kutta algorithm in python.

The only problem is that I am dealing with really small numbers and my algorithm can't play well with those in any reasonable amount of time. But, I can solve analytically for the equations of motion in the small angle limit, so I actually use the solution

\[ \theta(t) = \theta_0 \cosh \omega t + \dot \theta_0 / \omega \sinh \omega t \]

to evolve the system up to an angle of 0.1 radians, and then let the nonlinear equation and runge kutta algorithm take over.

The full python code for my problem is available here (if you want to run it, remove the .txt extension, I did that so that it would be previewable).

And what do I obtain? First looking over 20 orders of magnitude in differential initial conditions:

And second, zooming into the interesting region.

So, what is the best time you could balance a quantum mechanical pencil, i.e. what is the absolute longest time you could hope to balance a pencil in our universe? About 3.5 seconds. Seriously.

Think about that for a second. Usually you hear about the uncertainty principle, and it seems like a neat parlor trick, but something that couldn't influence your day to day lives, and here is a remarkable problem where even in the best case, the uncertainty principle puts a hard limit on championship pencil balancing which seems tantalizingly close.

And there you have a graduate student working through a somewhat non trivial problem. I probably went into way too much details with the basics, but we are still trying to feel out who our audience is. Please leave comments and let me know if I either could have left things out, or should have went into more details at parts.

### EDIT

As per request, here is how the max fall time scales with the length of the pencil assuming a pencil with uniform density.Plotted on a log-log plot, that is a pretty darn good line. The power law dependence is

\[ t \sim l^{0.514} \]

Neat.

Strangely enough, if I trust my numbers here, the longest you could hope to balance a 'pencil' 1 km long would be about 6 minutes. Thats a very strange mental picture.

Awesome to see the uncertainty principle at work on a more tangible scale.

ReplyDeleteAs for your audience, perhaps some links to explain equations and how you manipulate them for those of us not in either math or physics fields. Nevertheless, very well written and an intriguing read!

sin(pi/2) = 1

ReplyDeleteThank you Forbes, I goofed. Fixed

ReplyDeleteI like how much detail you went into, I like to see the limits of certain models and how there are more accurate (but usually more complicated) models that can be used to sharpen the prediction.

ReplyDeleteI'd like to see how the time scales with the length of the pencil.

Interestingly, I remember this exact problem from the sophomore year quantum mechanics class required of all students at Caltech.

ReplyDeleteThanks, I enjoyed that a lot! I'd love to see a followup from a statistical mechanics viewpoint showing how long it would take at room temp.

ReplyDeleteRegarding audience, if you put even one single equation, no matter how simple, you will lose nearly all laypeople, even if they could understand it in principle. They just won't stick around for the explanation. Since you really do need the equations for a post of this type, you have to assume an audience of interested freshmen as a baseline. In that case you probably don't have to explain Taylor series because if they've stuck around up to that point, they probably already know them.

There are a couple of typos. One is in the solution for theta(t) -- your cos should be a cosh. You have it right in the second appearance of the equation. The other thing is that I don't understand the theta=pi fixed point. Wouldn't that put the pencil inside the table? The horizontal pencil would be theta=pi/2.

Finally, the equation of motion is the same as for a physical pendulum, so shouldn't it have a solution in terms of elliptic functions?

particle_person:

ReplyDeletefixed the typo. Thanks for the input on audience.

Now, about the other stable point at pi, you're right in that this would be inside the table, but my equation of motion has no knowledge of the table. As you pointed out, this is the equation of motion for a pendulum, which if were suspended above the table could swing back and forth about theta=pi.

Finally, saying that we can write the exact solution to the differential equation in terms of elliptic integrals (while true), is kind of cheating seeing as the elliptic integrals are special functions, the elliptic integrals of the first kind are more or less defined to be the function that solves this equation.

ReplyDeletemy equation of motion has no knowledge of the tableYeah, that's why I was confused by the following when I read it the first time:

If you instead of balancing a pencil 'up', try to balance it 'down', you know that if you start with the pencil pointing straight down it stays that way and doesn't do anything interesting.Like you added, the equation doesn't know about the table so it's just not a physical solution of the original problem. It think it's less confusing to say it's unphysical than to switch the mental picture (without telling the audience) to one of someone holding the pencil by the tip, which is the only way I can make sense of balancing it down.

Anyway, that's my suggestion for improving clarity — don't switch the mental image in mid-stream without telling the audience.

Habit Melon: As per request, how fall time scales with length for a constant density pencil has been appended to post.

ReplyDeleteI was able to integrate the equations to find a simple analytic expression.

ReplyDeleteTime to Fall = 1/k ln(4(sqrt(2)-1) / (Theta0)min)

where

k = sqrt(3 g / 2 L )

(Theta0)min = sqrt((hbar k)/ M L g))

For realistic input, I get about 4 sec.

How would balance time be affected if you add resistance by adding weight and width to the pencil, i.e. x4 adjacent 45 degree angled triangular "wings" (like if you draw an 'x' on a square page and cut out the 'x' to produce 4 equal triangle shaped cut outs), perhaps placed at the top end of the balancing pencil, like a dart or arrow, with a width of about 3mm?

ReplyDeleteNice read. But half way thru its i realized that one could

ReplyDeleteOnly answer this by actually trying to do it for real. Would be nice to have the time and equipment to create a vacuum and a mechanical device to actually balance the pencil 100%. Taking in account the earth's rotation ( trying to counter it some how) and in my mind I think that one could balance the pencil on its tip as long as you can counter the earths rotation. Any thoughts any one?

Noel: Earth's rotation doesn't figure into it, except to very slightly reduce the effect of gravity by generating a centrifugal force. (Yes, centrifugal forces "exist" in rotational frames.) This tiny correction is less significant than the variation of g at different parts of the earth.

DeleteAlthough small, I say you probably are better off trying this somewhere on the equator at noon on the spring or fall equinox when the sun's gravity, earth's gravity, and earth's centripetal force are all colinear in your frame.

Deletehttp://users.mccammon.ucsd.edu/~jcsung/qmp.pdf

ReplyDeleteComments?

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ReplyDeleteI think if we were to take away air molecules around the pencil and cool it to absolute zero, that pencil would theoretically balance.This info was used from essay writers world

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