# Forest for the trees

## Contents

### Problem 1: the Olbers' paradox

If the Universe was infinitely old and infinitely extended, and stars could shine eternally, then in whatever direction you look the line of your sight should cross the surface of a star, and therefore all the sky should be as bright as the Sun's surface. This notion is known under the name of Olbers' paradox. Formulate the Olbers' paradox quantitatively

L. Anchordoqui, arXiv:physics.ed-ph/0706.1988.

Let $n$ be the mean density of stars in the Universe and $L$ be the mean luminosity of stars. The observed luminosity (energy flow's surface density) on the Earth from a star with luminosity $L$ at distance $r$ is \[F(r) = {L \over {4\pi r^2 }}.\] Consider a spherical shell of radius $r$ and thickness $dr$ centered at Earth. The radiation intensity of stars inside this shell (the power which reaches a unit surface from one steradian) is $$ dP(r) = F(r) \cdot n \cdot r^2 dr = {{nL} \over {4\pi }}dr. $$ It is important to note that total intensity of radiation from the shell is independent on the distance to it. Thus, the total intensity of radiation from all stars in the Universe \[P = \int_0^\infty {dP = {{nL} \over {4\pi }}\int_0^\infty {dr} }\] diverges in the case of infinite stationary Universe. Olbers' paradox is an example of the so-called "law of incorrect naming", which states that no law is called after a person who in fact discovered it. This paradox had been known 150 years before Olbers' formulated it (Diggers (1576)).

### Problem 2: a down to earth setting

There are $n$ trees per hectare in a forest. Diameter of each one is $D$. Starting from what distance will we be unable to see the forest for the trees? How is this question connected to the Olbers' paradox$^*$?

$^*$ Ryden, Introduction to cosmology, ADDISON-Wesley.

Lets draw a circle of radius $x$ around an observer. Part of the observer's field of view $s(x),\;0 \le s(x) \le 1$ is hidden by trees. Next, expand the observation region by $dx$, passing to a circle of radius $x+dx$. There are $n\times 2\pi xdx$ trees in the ring between the circles. They overlap the portion $nDdx$ of the observed circle. Since portion $s(x)$ was already overlaped, the additional overlap is $\left[1 - s(x)\right]nDdx$. Hence \[s(x + dx) - s(x) = \left[ {1 - s(x)} \right]nDdx\] or \[{{ds} \over {dx}} = \left[ {1 - s(x)} \right]nD.\] This equation could be easily integrated \[\int_0^s {{ds} \over {1 - s}} = \int_0^x {nDdx},\] giving the solution \[s(x) = 1 - e^{ - nDx}.\] This is the probability that when looking at random direction, the line of sight is blocked by a tree at distance no greater than $x$. Hence, the mean "free sight" length is \[\lambda = \int_0^\infty x {{ds(x)} \over {dx}}dx = {1 \over {nD}}.\] If there are $1000$ trees per hectare with the diameter of the trunk $20 \mbox{cm}$, the mean length of sight is $50 \mbox{m}$. This result can be generalized to the 3-dimensional case. There are $n$ stars per unit volume and each star has diameter $D$. If the surface perpedicular to the line of sight is $A=\pi D^2/4$, then \[s(x) = 1 - e^{ - nAx}\] and \[\lambda = {1 \over {nA}}.\] If the Universe was infinitely old and infinitely extended, then the line of sight would reach a surface of a star in finite time (and at finite distance). This is the meaning of Olbers' paradox.

### Problem 3: a Robin Hood setting

In a more romantic formulation this problem looks as follows. Suppose that in the Sherwood Forest the average radius of a tree is $30cm$ and the average number of trees per unit area is $n=0.005m^{-2}$. If Robin Hood shoots an arrow in a random direction, how far, on average, will it travel before it strikes a tree?

$\displaystyle \lambda = {1 \over {nD}} \approx 333\mbox{m}. $

### Problem 4: the cosmological setting

The same problem in the cosmological formulation looks this way. Suppose we are in a infinitely large infinitely old Universe, in which the average density of stars is $n_{st}=10^11 Mpc^{-3}$ and the average stellar radius is equal to the Sun's radius: $R_{st}=R_{\bigodot}=7\cdot10^8m$. How far, on average, could we see in any direction before your line of sight hits a star?

$\displaystyle \lambda = {1 \over {n_{st} \pi R_ \odot ^2 }} \approx 6 \cdot 10^{15} Mpc. $

### Problem 5: stars collisions

Demonstrate that stars in galaxies can be considered a collisionless medium.

Lets make some simplifying assumptions. First, assume that the total energy of a pair of stars is conserved when they are attracted due to gravitational interaction. Second, the masses of stars and their velocities are equal at the start of the process (formally, at infinitely large distance). These assumptions are sufficient for estimate. Energy conservation in the center of momentum frame reads \[mV^2 - G{m^2\over r}=mV_0^2,\] where $V$ is the current velocity of stars (the velocities are equal due to symmetry of the system), $r$ is the distance between them and $V_0$ is velocity at infinite distance. Considering the collision to happen at the moment when kinetic energy of the stars is doubled, we obtain \[G{m\over r}=V_0^2,\] and, hence, the collision cross-section is \[\sigma = {\pi G^2m^2\over 4V_0^2}.\] Characteristic time before the first collision is \[t_{st} = (nV_0\sigma)^{-1}.\] For the Milky Way we have $n \simeq 3\cdot 10^{-56}\mbox{cm}^{-3}$, $V_0\simeq 20\mbox{km/s}$, so that $\tau_{st} \simeq 10^{21}\mbox{s}$. This time is three orders larger than the age of the Universe. This estimate allows one to consider stars to comprise a collisionless medium.