To save from heat and rain, the open air market workers use tents. It was curious to ponder: how much do they protect from Sun? What if the twice as fat coverage is used? The black body model is the simplest to analyze: every surface absorbs all the light and emits it equally in both directions:

The net result is that every screen with these properties passes only half of heat and another half is reflected back. Now, what if we use two such screens?

This time we have got an infinite series: 1/4 + 1/16 + 1/64 = ∑(1/4ⁿ) = 1/3. Two thirds is back. What about three screens?

1/8+1/16+1/32+... = 1/4(1/2 + 1/4 + 1/8 + ...) = 1/4 * ∑(1/2ⁿ) = 1/4 * 1 = 1/4. The series is not so obvious for the four and more screens but you see the rule: n screens reflect n/n+1 of light and 1/n+1 is passed.

The flow evolution can be described by (a Markov) matrix. For 1 screen,

For three screens, the matrix would be

When multiplied by matrix, every vector component selects a column. The column has zero at its row (main diagonal), which is surrounded by 1/2. This means that all the energy is moved from the surface to its immediate neighbors on the next round. For instance, multiplying the matrix by initial vector always results in the half reflected to the Sun and half proceeded to surface_2 on the next step: [0 1 0 0 0 ...] -> [1/2 0 1/2 0 0 0 ...]. The elements a

Computer simulation of large matrices with this structure converges to 1/n+1 light passed indeed. That page derives a nice formula to compute the winning probability for light originating at any surface in the middle for a fixed number of surfaces using characteristic equation.

It turns out that the infinite number of surfaces problem corresponds to the random walk. According to Mosteller, the probability of ending up on the Sun is greater than 1 if particle cannot escape from it (as in our case).

In case the random walk starts in the middle on infinite series of surfaces, we'll get Pascal's triangle. In this case probability to diverge

To conclude, when the light falls onto the first black surface and there is infinite sequence of them, all light will be reflected back. That is, absolutely black matter becomes white. Isn't it paradoxical that to get absolutely white body you need to take more absolutely black bodies? Why one large enaught lump of black matter doesn't look white? How big must it be to start shining? Well, it seems that there is no paradox because absolutely black actually means that the body is only half black and the secret of turning it white is vacuum separation between the slices. What you need is not (only) black matter but some vacuum in between its slices to prevent molecular heat conduction.

The net result is that every screen with these properties passes only half of heat and another half is reflected back. Now, what if we use two such screens?

This time we have got an infinite series: 1/4 + 1/16 + 1/64 = ∑(1/4ⁿ) = 1/3. Two thirds is back. What about three screens?

1/8+1/16+1/32+... = 1/4(1/2 + 1/4 + 1/8 + ...) = 1/4 * ∑(1/2ⁿ) = 1/4 * 1 = 1/4. The series is not so obvious for the four and more screens but you see the rule: n screens reflect n/n+1 of light and 1/n+1 is passed.

The flow evolution can be described by (a Markov) matrix. For 1 screen,

1 .5 0 0 0 0 0 .5 1with initial state [0 1 0]. Here the first and last rows accumulate the reflected and passed light correspondingly (that is why the ones are in the corners). For two screens, the matrix is

1 .5 0 0 0 0 .5 0 0 .5 0 0 0 0 .5 1

For three screens, the matrix would be

1 1/2 0 0 0 0 0 1/2 0 0 0 1/2 0 1/2 0 0 0 1/2 0 0 0 0 0 1/2 1You see the structure of 1/2 halves. Every component of state vector is a unit of light at a corresponding surface. The border surfaces, the Sun and heat consumer, are virtual - they accumulate the light and do not give it away (the ones are in the corners). The initial vector is always [0 1 0 0 0 ...], which gives a unit pack of photons to the Sun-adjacent surface. At every step, this pack is distributed between adjacent surfaces by the matrix.

When multiplied by matrix, every vector component selects a column. The column has zero at its row (main diagonal), which is surrounded by 1/2. This means that all the energy is moved from the surface to its immediate neighbors on the next round. For instance, multiplying the matrix by initial vector always results in the half reflected to the Sun and half proceeded to surface_2 on the next step: [0 1 0 0 0 ...] -> [1/2 0 1/2 0 0 0 ...]. The elements a

_{12}and a_{n,n-1}, next to the corner 1s, are the amounts of light that go into reflected and passed trough accumulators correspondingly. The large powers of matrix will show the steady state -- the total amount of light reflected and passed through.Computer simulation of large matrices with this structure converges to 1/n+1 light passed indeed. That page derives a nice formula to compute the winning probability for light originating at any surface in the middle for a fixed number of surfaces using characteristic equation.

It turns out that the infinite number of surfaces problem corresponds to the random walk. According to Mosteller, the probability of ending up on the Sun is greater than 1 if particle cannot escape from it (as in our case).

In case the random walk starts in the middle on infinite series of surfaces, we'll get Pascal's triangle. In this case probability to diverge

*l*steps away from 0 after*n*steps, P(l,n) = C^{l+n}/2_n/2^n (*n*must be even).To conclude, when the light falls onto the first black surface and there is infinite sequence of them, all light will be reflected back. That is, absolutely black matter becomes white. Isn't it paradoxical that to get absolutely white body you need to take more absolutely black bodies? Why one large enaught lump of black matter doesn't look white? How big must it be to start shining? Well, it seems that there is no paradox because absolutely black actually means that the body is only half black and the secret of turning it white is vacuum separation between the slices. What you need is not (only) black matter but some vacuum in between its slices to prevent molecular heat conduction.

Nov 2016, Ross textbook:The argument used in circular table also shows that a gambler who is equally likely to either win or lose one unit on each gamble will be down n before being up 1 with probability 1/(n + 1): P{gambler is up 1 before being down n} = n / (n + 1).P{gambler is up k before being down n} = n / (n + k)

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