Forest for the trees - Revision history

Igor: /* Problem 5: stars collisions */

2012-11-10T17:34:11Z

‎Problem 5: stars collisions

Igor: /* Problem 4: the cosmological setting */

2012-11-10T17:29:15Z

‎Problem 4: the cosmological setting

Igor: /* Problem 3: a Robin Hood setting */

2012-11-10T17:28:20Z

‎Problem 3: a Robin Hood setting

Igor: /* Problem 2: a down to earth setting */

2012-11-10T17:27:48Z

‎Problem 2: a down to earth setting

Igor: /* Problem 1: the Olbers' paradox */

2012-11-10T17:21:09Z

‎Problem 1: the Olbers' paradox

Cosmo All at 18:19, 5 November 2012

2012-11-05T18:19:43Z

Cosmo All: /* Problem 5: stars collisions */

2012-11-05T18:11:07Z

‎Problem 5: stars collisions

Igor: /* Problem 1: the Olbers' paradox */

2012-10-11T14:52:11Z

‎Problem 1: the Olbers' paradox

Igor: /* Problem 1: the Olbers' paradox */

2012-10-11T14:51:58Z

‎Problem 1: the Olbers' paradox

Igor at 14:50, 11 October 2012

2012-10-11T14:50:32Z

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 === 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.
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-     <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
+     <p style="text-align: left;">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
-$$
+\[mV^2 - G{m^2\over r}=mV_0^2,\]
-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
-$$
+\[G{m\over r}=V_0^2,\]
-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
+and, hence, the collision cross-section is
-$$
+\[\sigma = {\pi G^2m^2\over 4V_0^2}.\]
-G{m\over r}=V_0^2,
+Characteristic time before the first collision is
-$$
+\[t_{st} = (nV_0\sigma)^{-1}.\]
-and, hence, collision cross--section is
+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.</p>
-$$
 \sigma = {\pi G^2m^2\over 4V_0^2}.
-$$
-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>
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 === 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?
+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?
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 === Problem 5: stars collisions ===
 Demonstrate, that stars in galaxies can be considered a collisionless medium.

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 === 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?
+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?
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 === 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?

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-     <p style="text-align: left;">Let's draw a circle of radius $x$ around an observer. Part of an observers' viewfield $s(x),\;0 \le s(x) \le 1$ 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 a ring between the circles. They overlap the portion $nDdx$ of observed circle. Since portion $s(x)$ was already overlaped, the additional overlap is $\left[ {1 - s(x)} \right]nDdx$. Hence,
+     <p style="text-align: left;">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
-$$
+\[s(x + dx) - s(x) = \left[ {1 - s(x)} \right]nDdx\]
-s(x + dx) - s(x) = \left[ {1 - s(x)} \right]nDdx
+or
-$$ or
+\[{{ds} \over {dx}} = \left[ {1 - s(x)} \right]nD.\]
-$$
+This equation could be easily integrated
-{{ds} \over {dx}} = \left[ {1 - s(x)} \right]nD.
+\[\int_0^s {{ds} \over {1 - s}}  = \int_0^x {nDdx},\]
-$$
+giving the solution
-This equation could be easily integrated as
+\[s(x) = 1 - e^{ - nDx}.\]
-$$
+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
-\int_0^s {{ds} \over {1 - s}}  = \int_0^x {nDdx},
+\[\lambda  = \int_0^\infty  x {{ds(x)} \over {dx}}dx = {1 \over {nD}}.\]
-$$
+If there are $1000$  trees per hectare with the diameter of the trunk $20 \mbox{cm}$, the mean length of sight is $50 \mbox{m}$.
-thus obtaining a solution
-$$
 s(x) = 1 - e^{ - nDx}.
-$$
-This value is a probability that looking at random direction, sight could reach no more than $x$, i.e. it reach a tree at distance $x$. Hence, the mean length of "free sight" is
-$$
 \lambda  = \int_0^\infty  x {{ds(x)} \over {dx}}dx = {1 \over {nD}}.
-$$
-If there are $1000$  trees on one hectare and diameter of trunk is $20 \mbox{cm}$, the mean length of sight is $50 \mbox{m}$.
-This result could be generalized to 3-dimensional case. There are $n$ stars in unit volume and each star has diameter $D$. If the surface, perpedicular to the line of sight is $A=\pi D^2/4$, then
+This result can be generalized to the 3-dimensional case. There are $n$ stars per unit volume and each star has diameter $D$. If the surface perpedicular to the line of sight is $A=\pi D^2/4$, then
-$$
+\[s(x) = 1 - e^{ - nAx}\]
 s(x) = 1 - e^{ - nAx}
-$$
 and
-$$
+\[\lambda  = {1 \over {nA}}.\]
-\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.</p>
-$$
 If the Universe was infinitely old and infinitely extended, then line of sight would reach a surface of a star in finite time (and at finite distance). This is the meaning of Olbers' paradox.</p>
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 === 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?

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-     <p style="text-align: left;">Let $n$ be the mean density of stars in the Universe and $L$ be the mean luminosity of stars. The observed luminosity (surface density of light flux) on the Earth from a star with luminosity $L$ at distance $r$ is
+     <p style="text-align: left;">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
-$$
+\[F(r) = {L \over {4\pi r^2 }}.\]
-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
-$$
-Consider a spherical shell of radius $r$ and thickness $dr$ with center in the Earth. The radiation intensity of stars inside this shell (the power which reaches the square unit from one steradian) is
 $$
 dP(r) = F(r) \cdot n \cdot r^2 dr = {{nL} \over {4\pi }}dr.
 $$
-It is important to note, that total intensity of radiation from the shell is independent on the distance to it. Thus, the total intensity of radiation from all stars in the Universe
+It is important to note that total intensity of radiation from the shell is independent on the distance to it. Thus, the total intensity of radiation from all stars in the Universe
-$$
+\[P = \int_0^\infty  {dP = {{nL} \over {4\pi }}\int_0^\infty  {dr} }\]
-P = \int_0^\infty  {dP = {{nL} \over {4\pi }}\int_0^\infty  {dr} }
+diverges in the case of infinite stationary Universe.
-$$
 diverges in case of infinite stationary Universe.
-Olbers' paradox is an example of a so-called "law of incorrect naming", which states, that neither law is called after a person, who in fact discovered it. Considered paradox was known 150 years before Olbers' had formulated it (Diggers (1576)).</p>
+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)).</p>
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 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, [http://arxiv.org/abs/0706.1988 , arXiv:physics.ed-ph/0706.1988].
+L. Anchordoqui, [http://arxiv.org/abs/0706.1988 arXiv:physics.ed-ph/0706.1988].
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 === 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 [http://arxiv.org/abs/0706.1988 L. Anchordoqui, arXiv:0706.1988, physics.ed-ph].
+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
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 === 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$^*$?