Difference between revisions of "Forest for the trees"
(→Problem 1: the Olbers' paradox) |
(→Problem 5: stars collisions) |
||
| Line 67: | Line 67: | ||
=== Problem 5: stars collisions === | === Problem 5: stars collisions === | ||
Demonstrate, that stars in galaxies can be considered a collisionless medium. | Demonstrate, that stars in galaxies can be considered a collisionless medium. | ||
| − | + | <div class="NavFrame collapsed"> | |
<div class="NavHead">solution</div> | <div class="NavHead">solution</div> | ||
<div style="width:100%;" class="NavContent"> | <div style="width:100%;" class="NavContent"> | ||
| − | <p style="text-align: left;"></p> | + | <p style="text-align: left;">Let's make some simplifying assumptions. First, assume that total energy of a pair of stars conserves when they converge due to gravitational interaction. Second, the masses of stars and their velocities are equal at the start of convergence (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 current velocity of stars (which velocities are equal due to symmetry of the system), $r$ is the distance between them and $V_0$ is velocity at infinite distance. Considering collision as a moment, when kinetic energy of stars doubled, we obtain | ||
| + | $$ | ||
| + | G{m\over r}=V_0^2, | ||
| + | $$ | ||
| + | and, hence, collision cross--section is | ||
| + | $$ | ||
| + | \sigma = {\pi G^2m^2\over 4V_0^2}. | ||
| + | $$ | ||
| + | Characteristic time before first \underline{(sic!)} 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 to consider stars as collisionless medium.</p> | ||
</div> | </div> | ||
| − | </div | + | </div></div> |
Revision as of 18:11, 5 November 2012
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.
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.
Problem 3: a Robin Hood setting
In a more romantic formulation this problem looks as following. Suppose that in the Sherwood Forest, the average radius of a tree is $30cm$ and 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?
Problem 4: the cosmological setting
The same problem in the cosmological formulation looks this way. Suppose you 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 you see in any direction before your line of sight strucks a star?
Problem 5: stars collisions
Demonstrate, that stars in galaxies can be considered a collisionless medium.
Let's make some simplifying assumptions. First, assume that total energy of a pair of stars conserves when they converge due to gravitational interaction. Second, the masses of stars and their velocities are equal at the start of convergence (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 current velocity of stars (which velocities are equal due to symmetry of the system), $r$ is the distance between them and $V_0$ is velocity at infinite distance. Considering collision as a moment, when kinetic energy of stars doubled, we obtain $$ G{m\over r}=V_0^2, $$ and, hence, collision cross--section is $$ \sigma = {\pi G^2m^2\over 4V_0^2}. $$ Characteristic time before first \underline{(sic!)} 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 to consider stars as collisionless medium.
