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October 11

In a city, how do you see more people, walking around or standing?

In a city, how do you see more people, walking around or standing on a concrete place? Suppose the same number of people come and go from any side. And all move at the same speed as you.--Jubilujj 2015 (talk) 19:01, 11 October 2015 (UTC)[reply]

I would say more walking. Under the conditions you give, assume the city is also uniformly dense of people and infinite, or at least large enough to make no difference. Then you will have the same density of people at any time. But the rate at which you meet them will depend on their speed. The faster they are going, relative to you, on average, the more you will meet.

If you are stationary they all have the same speed relative to you. If you are walking then some appear faster some slower but it is not symmetric. Those travelling in the same direction as you seem slower, even stationary. Those travelling the opposite direction seem faster, up to twice as fast. But those moving at right angles seem faster, about √2 as fast. More seem faster than slower so you meet more of them, in theory.--JohnBlackburne^words_deeds 19:59, 11 October 2015 (UTC)[reply]

In general you should have people seen when walking / people seen when stationary =

{\frac {\int _{||v||=1}||v-{\hat {x}}||\,dv}{\int _{||v||=1}\,dv}}

=

{\frac {\int _{0}^{\pi }(\sin \theta )^{d-2}{\sqrt {2+2\cos \theta }}\,d\theta }{\int _{0}^{\pi }(\sin \theta )^{d-2}\,d\theta }}

. In 1, 2, 3, 4 dimensions I get 1, 4/π, 4/3, 64/15π ≈ 1, 1.27, 1.33, 1.36. It looks like it converges to √2 in the d → ∞ limit. -- BenRG (talk) 22:12, 13 October 2015 (UTC)[reply]

Your 2-dimensional result only applies to the squares in the city, with the added assumption that people walk randomly over any point and with any direction. Most of the walking in a city is done along 1-dimensional routes, such as sidewalks. Therefore, the actual ratio should be close to 1. In other words, it's almost a wash. — Sebastian 23:48, 13 October 2015 (UTC)[reply]

It looks like you've assumed that all the people are converging to a point, assuming I understand what the "v" means. That doesn't seem like a very realistic matter model governing "people". Presumably "people" are not infinitely compressible. I think a better matter model would be that people are totally incompressible. In which case, a local solution is fairly obvious: just go against the flow of traffic. Or have I missed something? Sławomir
Biały 00:57, 14 October 2015 (UTC)[reply]

v is the velocity of other people relative to the ground. Yours is either

{\hat {x}}

or 0. I assumed that the people are noninteracting (pass through each other and you, or are so sparse that they may as well be treated that way), their velocities are uniformly distributed over the (d−1)-sphere, and people with a given velocity are more or less uniformly distributed through space, so the rate at which you see new people with that velocity is proportional to their speed relative to you. Obviously this is unrealistic; I thought the problem was intended that way. I think JohnBlackburne made the same assumptions. -- BenRG (talk) 03:06, 14 October 2015 (UTC)[reply]

I see, it's really a sparse limit. Sławomir
Biały 23:28, 14 October 2015 (UTC)[reply]

Oh, and we're of course interpreting the condition "And all move at the same speed as you" as not applying to the v=0 case, or else you would not meet anybody new. — Sebastian 23:53, 13 October 2015 (UTC)[reply]