避风塘里抽风: Puzzle from Terrence Tao

I was recently at an international airport, trying to get from one end of a very long terminal to another. It inspired in me the following simple maths puzzle, which I thought I would share here:

Suppose you are trying to get from one end A of a terminal to the other end B. (For simplicity, assume the terminal is a one-dimensional line segment.) Some portions of the terminal have moving walkways (in both directions); other portions do not. Your walking speed is a constant $v$ , but while on a walkway, it is boosted by the speed $u$ of the walkway for a net speed of $v+u$ . (Obviously, given a choice, one would only take those walkways that are going in the direction one wishes to travel in.) Your objective is to get from A to B in the shortest time possible.

Suppose you need to pause for some period of time, say to tie your shoe. Is it more efficient to do so while on a walkway, or off the walkway? Assume the period of time required is the same in both cases.
Suppose you have a limited amount of energy available to run and increase your speed to a higher quantity $v'$ (or $v'+u$ , if you are on a walkway). Is it more efficient to run while on a walkway, or off the walkway? Assume that the energy expenditure is the same in both cases.
Do the answers to the above questions change if one takes into account the various effects of special relativity? (This is of course an academic question rather than a practical one. But presumably it should be the time in the airport frame that one wants to minimise, not time in one’s personal frame.)

It is not too difficult to answer these questions on both a rigorous mathematical level and a physically intuitive level, but ideally one should be able to come up with a satisfying mathematical explanation that also corresponds well with one’s intuition.

I have read several blogs talking about this puzzle from Terrence Tao (here and here). I have no idea how I can solve the third question. But maybe one of comments from his blog provided the whole strategies. (It is a pity that Terrence Tao's blog has been blocked.)

I’ll be honest. It’s getting late, and I’m not feeling up to reworking all my algebra for a blog comment. But I think there are some subtleties to the special relativity case. For example, there is not a single answer to question one if you analyze if from your own point of view (get there are fast as possible, by the time on your wristwatch.) Instead, the answer of where you should tie your shoe actually depends on $u$ and $v$

The problem states,

“Your objective is to get from A to B in the shortest time possible.”

But it does not state in the shortest time possible according to whom. In SR, you will perceive the journey to take a shorter time than the airport clocks indicate, because you are moving. So do you want to minimize the time on your wristwatch, due to impatience, or minimize the time on the airport clocks, due to being late?

Let all velocities be proportions of the speed of light. i.e. c=1. You have to cover a distance $d_n$ of non-moving and $d_m$ of moving floor.

Q1, SR) airport time
Your speed while walking is still $v$ , but your speed while walking on the walkway is

$\frac{v+u}{1+vu}$ .

If you tie your shoe for a time $t$ while on the non-moving part, your total time is the time spent walking the non-moving distance, the time spent tying your shoe, and the time spent walking the moving distance:

$t_n = \frac{d_n}{v} + t + \frac{d_m (1+uv)}{u+v}$

If you tie it while on the moving part, your total time will different for two reasons:
1) you perceive yourself to spend $t$ seconds tying your shoe, but from the airport’s point of view you are experiencing time dilation. They think you take longer to tie your shoe.
2) the distance you walk on the moving platform is reduced because you are getting closer to the end of it while you tie your shoe.

Note that these effects work in opposite directions. Effect 1) means you take longer than you did before, while effect 2) means you take less time to get there. The relativistic solution must converge to the nonrelativistic one in the limit where the velocities are much smaller than one, but we might look for some transition velocity at which it is no longer beneficial to wait for the moving walkway to tie your shoe. The equation for the case where you tie on the moving part is

$t_m = \frac{d_n}{v} + \frac{t}{\sqrt{1-u^2}} + \frac{(d_m - \frac{t}{\sqrt{1-u^2}}u)(1+uv)}{u+v}$

Subtract to get the delta. The first term is the same, but the other terms don’t cancel exactly. Here are the leftovers:

$\Delta t = t \left(1 - \frac{1}{\sqrt{1-u^2}} + \frac{tu(1+uv)}{u+v} \right)$

Limiting cases: u->0, we get zero. This means that if the sidewalk isn’t moving, it doesn’t matter where you tie your shoe.
u->1, we get infinity. This gives the result that if the sidewalk is moving near the speed of light, tying your shoe on it is a terrible idea. You’re already going close to the speed of light, so the time dilation factor is huge, and you will be tying your shoe in super-slow motion. For any given $v$ , there is eventually some speed of the moving sidewalk above which you are better off tying on the non-moving portion. However, this equation only makes sense as long as $dm - \frac{tu}{\sqrt{1-u^2}}$ is positive, because that’s the distance you walk on the walkway. Eventually, for very high $u$ , you’ll spend the entire walkway tying your shoe and still not be done yet. At that point the formula breaks down.

There is a 1-D family of values of $(u,v)$ such that it doesn’t matter whether you tie your shoe on the walkway or off it. If I did the algebra right, $v_{crit}$ , the critical transition velocity, turns out to be quadratic in $u$ . The solution I got was:

$v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

with

$a = u^2(1-u^2-u^4)$

$b = 2u(1-2u^4)$

$c = u^2(3-4u^2)$

I might have preferred to solve for $u(v)$ , but that was a sixth degree polynomial.

At this point, you’d want to check where the determinant of that solution was positive and where you get unphysical solutions, and throw those out. But the interesting conclusion was that this transition velocity does exist.

Q1 SR) wristwatch time
The time that passes for you, $dt_y$ , analyzed from the airport frame as a function of airport time $dt_a$ is

$dt_y = \frac{dt_a}{\sqrt{1-v_r^2}}$

where $v_r$ is the relative speed between you and the airport floor. So the total time that passes on your wristwatch if you tie you shoe on the non-moving portion is

$t_n = \frac{d_n \sqrt{1-v^2}}{v} + t + \frac{d_m (1+uv)\sqrt{1-\left(\frac{u+v}{1+uv}\right)^2}}{(u+v)}$

I got this by taking the answer for Q1 and multiplying by the time-dilation factor in each term. For the case where you tie your shoe in the moving portion, the total time elapsed on your wristwatch is

$t_m = \frac{d_n \sqrt{1-v^2}}{v} + t + \frac{(d_m - \frac{tu}{\sqrt{1-u^2}})(1+uv)\sqrt{1-\left(\frac{u+v}{1+uv}\right)^2}}{u+v}$

Again perform the subtraction. The first two terms are identical, which is good news since the last term is getting ridiculous.

$\Delta t = \frac{tu(1+uv)\sqrt{1-\frac{u+v}{1+uv}^2}}{(u+v)\sqrt{1-u^2}}$

This quantity is always positive, so from the standpoint of listening to the fewest number of songs on your ipod, it’s always best to tie your shoe on the walkway. That makes good sense because the effect of drawing out the time it takes to tie your shoe is gone in this frame. Three seconds from your point of view is always three seconds.

It’s interesting to note that for certain values of $u$ and $v$ , it’s actually possible for one strategy to be best in terms of minimizing your personal time, while a different strategy can be best for minimizing how late you are to the flight.

For the special relativistic case of question 2, I’ll leave the details to somebody else. From the standpoint of airport time, running on the walkway is not going to help you as much as running on the nonmoving portions, due to velocity addition. So like the nonrelativistic case, you should always save your running for the nonmoving portions (or run those entire things if you have the energy).

From the standpoint of your wristwatch, time dilation is much more effective if you run while on the walkway, if the walkway is pretty fast. So it feels like there should again be transition velocities at which you best strategy switches.

Okay, please tell me about the copious mistakes I made, since it’ll actually be interesting to know the real answer.