**at sunrise at the bottom of a mountain path. He winds his way to the top — taking several rest breaks along the way — arriving at the summit around sunset.**

*A monk begins his day*He meditates throughout the night.

At sunrise the next morning he begins his journey home on the same pathway. (Indeed, there is only one way up, and one way down the mountain, unless one jumps and subsequently dies. The monk chooses not to jump that day.) He arrives an hour or so before dusk — since it is far less laborious going down the hill than up it. Besides, he has just become enlightened.

Now to the question:

**Is there a specific place on the path where the monk was at the exact same location at the exact same time of day on both days? Please explain your answer.****Ok, two days and 20 posts later, here's the first installment**

**of my two part answer : The Feynman diagram:**

## 22 comments:

OK Bill, I'll give this a shot. let H(x) = distance from the bottom of the mountain with respect to time on day one (ranging from 0 at t=0 to H at t=sunset) assuming the monk hasn't learned to teleport or astrally project or something.

let G(x) = distance from the top of the mountian with respect to time on day two (ranging from H at t=0 to 0 at t=sunset). also assuming the monk has learned to teleport, activate some of his later neural circuits and whatnot.

Consider the function H(x) - G(x). This is a composition of continuous functions, along the same domain, and thus is also continuous. Now, it ranges from -H to +H and must achieve every value in between at some point along the way. But the moment when [H(x) - G(x)] achieves a height of 0 is exactly the moment when the monk was equidistant with respect to time to the base camp on subsequent days.

Note: that the monk must not teleport, and the time interval must be the same so as to not allow for the possibility that he could roll down the mountain really quick after sunset.

OR: You could hike the mountain yourself, and imagine yourself walking down the mountain as you go up the mountain, and the point where you imagine yourself running into yourself, is your answer.

ORR: The concept of time as we know it is a human manifestation of our conciousness, therefore the monk exists at all points along the mountain at the same time.

ORRR....

(Too much coffee, sorry)

Nope: Earth moved 16 million miles from yesterday's position on its path around the sun. (Better number crunchers can also factor in movement of solar system and galaxy.) As Heraclites said, we never step in the same river twice (he wasn't thinking of orbital dynamics, but you get the idea.)

Good point Cory. Heraclites also said "the path up and down is one and the same"

Man! Remind me not to take Heraclites hiking.

Ignoring messy cosmic mechanics, assuming two nice 15-hour days around the solstice, with sunrise at 06:00, and steady, nonstop walking, the monk will pass the same geolocation at about 13:14 local time.

Just what I needed, Bill F.: algebra to distract me from getting some paid work done! ;-)

Cory: You also assumed that he was walking the same speed both up and down. And of course, when hiking, you tend to go faster down, then up! (unless you decide to take a break, have something to eat, look, there is an interesting plant to look at, I think I'll look at that, etc..)

Great discussion so far! We've got yes, no and maybe all kinds of ways. Very cool.

My initial thought coincided with Cory's. Since the world moved for an entire day since his last trip, no, he dosn't reach the same spot.

Au contraire, mon ami Bill K!My model assumes ascent speed of distance/(15 hours), descent speed of distance/(14 hours). If I had assumed the same speed up and down, I would have come up with a magic time of 13:30, exactly at the middle of the day (and clearly, the monk is in an area on Daylight Savings Time).corey and bill k:

both of your explanations assume a constant speed. this is not necessarily the case. perhaps the mountain only takes 45 minutes to climb, when climbing at full speed? in that case, he may have not even been on ANY POINT on the trail at the same time.

in addition, it is given how long the breaks are.

Hey Bill (and others), this is probably off topic, but it's related enough that I feel compelled to share. I just discovered it today...

An Atheist Talks to God on a Train

http://www.fullmoon.nu/articles/art.php?id=tal

You people are awesome. Best discussion I've ever heard on this. And Neal, no problem being off-topic. I'll check it out, and if so moved, perhaps elevate it to it's own topic.

Ok, as for the puzzle. Let's nail some things down a little. I think it's safe to say that a mere 100 years ago, very few people on our planet would have given the answers some of you have given here, especially as they pertain to Relativity and Quantum Mechanics, which some of your answers do.

Of course, there were Zen Monks back then, and perhaps they may have given approximately the same answers as some of you have here as well, (or even more esoteric) and so I salute you all for your thoroughly modern, new-paradigmatic thought capabilities. Bravo. You've all, each in your own way broken the Cartesian/Newtonian juggernaut.

But (and this pains me to say this) let's just look at the problem from a "Classical" physics perspective first and formost.

That should allow you all to be more confident in giving a definite "Yes" or "No" answer, and a convincing accompanying explanation.

Then, once we've agreed on that, then, we'll blow things back up again, ok?

(Again, this is along the lines of Frank Wilzecks idea that "the opposite of a profound truth is another profound truth." Of which, I must say, that all of you here are hot on the trail. I'm just surprized Sibby hasn't checked in on this one yet. His being "In Search of the Truth" and all.)

Actually Aaron, Nowhere did I make the assumption that the time taken was at a constant speed. I just assumed that the function was continuous.

Corey: Thanks for the correction, I guess if I actually would have done the math, I would have figured that out!

And I thought of another solution, the moment that the monk decides that his journey "up" is complete, will be at the exact moment that the journey "down" begins, hence at that exact moment, he will be in the same spot on the journey up and down.

That said, I must confess that I solved the problem "correctly" and graphically using a Feynman diagram, much to the surprise of the person who first posed the puzzle to me.

Further, suffice it to say that neither of us used any formal mathematics (algebra, arithmetic, calculus, etc.) to derive what is considered by most to be the most convincing solution.

All-rightie then, those are the best two clues I have to offer you at this time. Please carry on, good gentle readers.

Indeed, Aaron, inconstant speed and breaks foul up my calculations (so knock it off! ;-) ).

If you want an answer from a classical physics perspective, I think that answer is no, there is no place on the path where he is in the same place at the same time both coming and going. Or at least there is no such moment that can be identified based on the facts that have been presented.

This is primarly because he went at different rates of speed up and down the mountain. Also, the question indicates that he took "several rest breaks" on the way up, but apparently none on the way down. We don't know the length or quantity of the breaks. Thus we would be unable to conclude that he was ever in the same place at the same time.

In other words, what Aaron said.

Now, if Bill used some fancy graphs and charts that I've never heard of to figure out the answer, that tells me we're both wrong, Aaron.

Fair enough, Neal.

I've now provided a "Feynman diagram" to the bottom of the original post. It's a pretty simple thing really, should be self-explanatory.

"Feynman diagrams" are really just very simple space/time diagrams. (The more you get to know me the more you'll understand that I am so stupid, I HAVE to keep things really, really simple.)

The vertical line represents space, in this case altitude, from the bottom of the mountain to the top, and the horizontal line represents the passage of time, in this case from sunrise to sunset.

Now notice that I'm suggesting you use two different colored lines, one for each day you're going to plot on the graph.

The first day-line starts at the bottom left (bottom of the mountain at sunrise) and ends at the upper right. Go ahead and draw it in your mind's eye, or make your own chart. Draw the line any way you think kind of fits the story.

The second day-line, of course starts at the top left (top of the mountain at sunrise) and proceeds toward the bottom right, ending a little before sunset at the baseline.

So, what do you think? Draw as many Feynman diagrams as you want to, and let me know what you find out.

BTW, Bill K's 2nd answer is very close to the explanation that was given to me after I worked out my Feynman diagram thingy. I'll go over it in more detail tomorrow. G'night all.

So the correct answer is:

It depends?

Actually, in terms of everyday, run of the mill experience — thinking only on the scale of the monk going up and down the hill, Aaron, I'm thinking the answer is always "Yes."

Here's another way to look at it if you don't like the Feynman diagrams. Instead of just one monk on that first day, now there are two.

One monk starts at the bottom of the hill at sunrise and goes up, exactly as described in the story. The other monk starts at the top of the hill and comes down, again exactly as described in the story.

Since the first monk is on the pathway all day, from sunup to sun down, and the second monk arrives at the bottom of the hill before sundown, THEY WOULD HAVE TO MEET EACH OTHER SOMEWHERE ON THE PATH THAT DAY at a certain specific time in a certain specific place.

So the answer is Yes.

The fact that you don't have enough information to determine exactly WHERE on the path they would cross or WHEN is pretty much irrelevant, as is the fact that the monk coming down the hill is recently enlightened. The fact remains, there IS such a space/time. It happened when and where it happened.

Are you followin' me camera guy?

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