Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Olbers Paradox and The Sherwood Forest

  1. Jul 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose that in sherwood forest, the average radius of a tree is equal to R= 1m and the average number of trees per unit area sigma = 0.005 m^-2 . If Robin Hood shoots an arrow in a random direction , how far , on average, will it travel before it strikes a tree?


    2. Relevant equations
    Since in the problem robin is shooting in a random direction, im assuming that the area in question is circular. and the area of a circle is A = pi X r^2.


    3. The attempt at a solution
    Im not sure how to tie in the area with the distance is question. Ive tried integrating over a circle ( sin, cos) but I dont know if I should solve for the radius some how. Or if i should find a max distance and a min distance somehow and average them. Also the radius of the trees thing is throwing me off. ......... Any insight and help would be great
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 6, 2010 #2
    Hrm, the first issue that strikes me is that the manner in which a random shot can be made is more than a circle; you must take elevation of the **** into account, excepting those angles which would make drawing the bow impossible. There is a chance of NOT hitting a tree after all, especially if the arrow is fired vertically.
     
  4. Jul 6, 2010 #3
    The case of a "vertically" fired arrow shouldn´t be treated in this problem - one assumes horisontal shots. This is so to speak a 2-dim problem.

    The 3-dimensions, vertical and all directions arises naturally when we consider another type but still a similar problem e.g. the average length of the line of sight to galaxies - that is, the average distance to galaxies as seen from earth. In this particullar problem we need an estimate of the radius and the average density of galaxies. This would correspond to the radius and the average density of trees. This would then be a 3-dim problem and vertical and all directions would have to be considered.
     
  5. Jul 13, 2010 #4
    Hi,

    I've been lurking this forum for months but this is my first post. Also, excuse my english. I'm brazilian and I've never studied math or physics in english, so I might end up using wrong technical terms.

    Well, here's an approach for this problem. I'm sure it's not 100% correct but it's a starting point.

    Robin Wood chooses a randon direction and shoots his arrow. Since we can't privilege any direction, if a given distance d is overcome by Robin Wood's arrow, we better take into account the area pi*d^2 when it comes to determining how many trees it may encounter. So, the number of trees inside the area pi*d^2 equals sigma*pi*d^2.

    Now to the other part of the problem. Consider that there's only one tree in the whole forest, and that the distance of the center of the tree to the origin is r. If r is much bigger than 2R (where R is the radius of a tree), we can approximate and say that:

    2R = fi*r, where fi is the angle between the two segments that start on the origin and are tangent to the tree. So, fi = 2R/r

    Robin can choose any direction on the plan to shoot, but if he chooses any direction over the fi angle, his arrow encounters the tree. All of this was done to say: we can swap a tree of radius R for a angle fi that equals 2R/r, where r is the distance of the tree's center to the origin.

    Back to our equation of how many trees there are inside a given circle or radius d:

    Quantity of trees = sigma*pi*d^2

    If we imagine that we can swap each tree inside for a angle fi = 2R/d. The angle that we would end up with is: theta = sigma*pi*d^2*2R/d = sigma*pi*d*2R

    Since we want the average "d", then theta must be equal to pi, half the plan. So:

    pi = sigma*pi*d*2R, which gives: d = 1/(sigma*2R)

    Since sigma = 0.005 trees/m^2 and R = 1m,
    d = 100m

    The problems with this resolution are evident, but I'd like to point out a particular one. It was assumed that all trees inside the area pi*d^2 occupied the periphery of the circle, which obviously is false. But as I said, it's a starting point.

    Hope I have helped.
     
  6. Jul 13, 2010 #5

    Ich

    User Avatar
    Science Advisor

    Welcome to PF!
    Your English is excellent. Normally you'd call an angle phi, not fi, but that's all.

    Acili, why didn't you post in the homework section?
    Anyway, what you're supposed to do is IMHO the following:
    Figure out how many trees there are in a thin ring with diameter r and thickness dr.
    What's the chance "p*dr" of hitting a tree if the arrow goes through the ring (distance dr)?
    So the number of arrows lost within dr is n*p*dr, if n was the number of arrows entering. So you have dn/dr=-n*p.
    The solution is an exponential function in r. The average shooting distance is where the function has decreased e-fold.
     
  7. Jul 13, 2010 #6
    Ich and harroxelas. Thank you very much for your replies - they were certainly usefull for me... and I will study your remarks further. I think harroxelas is on the right track, the exponential solution seems/feels not true at first glance imo... But I´ll look into it.

    This was also my first post - so I thought this would be the right forum.

    Harroxelas, can you explain further why you equal theta to pi, half the plan, when averaging "d"? Your english is excellent by the way - I agree with ich.

    Regards
    Acili
     
  8. Jul 13, 2010 #7
    Well, thinking of it now, it was more of a intuitive thing. I thought about using 2*pi but I'd be "forcing" the arrows to find a tree, it didn't seem right. It would be like finding a upper limit for d, not an avarage, and since d = theta/(pi*sigma*2R), d increases linearly with theta and all the other variables are just parameters, they don't change. The mean value for d is given by the mean value of theta, which is pi, since 0 <= theta < 2*pi. I know my approach is not 100% correct but it was the best I could think of.

    This problem is quite interesting. Ich, do you care to explain your idea further? I didn't get it yet... Solving that diff. equation you'd get a exponential function kinda like n = k*exp(-r*p) + C, correct? But how do you account for the number of arrows fired?
     
  9. Jul 13, 2010 #8

    Ich

    User Avatar
    Science Advisor

    Correct.
    That's irrelevant, I put it in for illustration only. You only need percentages. You may set n=1. But k=n, obviously, as all arrows are still there at r=0.

    That's more or less a standard calculation for absorption. Try http://en.wikipedia.org/wiki/Beer-Lambert_law#Derivation" in Wikipedia. It's a fairly simple concept with a real application.
     
    Last edited by a moderator: Apr 25, 2017
  10. Jan 3, 2011 #9
    Sorry to resurrect an old post but I was considering this problem recently and came here to try and verify my answer. Answering this post seemed easier than typing it all over again. This problem comes from Barbara Ryden's Introduction to Cosmology. The text is an upper level undergraduate text (according to it's introduction) so I suppose the question rightly belongs in the homework section. After thinking about the question for a while, it dawned on me that this question seemed related to mean free path concept from statistical mechanics. Search mean free path at Wikipedia for a derivation that is very similar to ones that are already posted but from a different viewpoint. My answer comes from the tree density being the "particle" density, n, and the cross section being the tree width as seen from the archer, or 2R = 2 meters. This gives a mean free path of 100 m. The wiki article also shows why the exponential factor is defined as the "mean" from a statistics standpoint.
     
  11. Feb 13, 2011 #10
    i cant solve this problem....i cant understand how chrisas relate mean free path and this problem....would you mind solve this question completely?
    thank you very much
     
  12. Feb 13, 2011 #11
    Mean free path
    <http://en.wikipedia.org/wiki/Mean_free_path> [Broken]

    In physics, the mean free path is the average distance covered by a moving particle (such as an atom, a molecule, a photon) between successive impacts (collisions) [1] which modify its direction or energy or other particle properties.

    Translated to problem terms:

    Mean free path:
    In Sherwood Forest, the mean (or average) distance covered by a moving arrow between successive impacts (first being shot by Robin, second hitting a tree) which modifies the arrows properties (direction: deflected, energy: arrow sticks in tree, arrow properties: arrow shatters).

    The derivation follows the wiki page (except Robin in the forest is a 2-D problem instead of 3-D. Replace Volume with area, and area with line lengths):

    Imagine a beam of particles (bunch of arrows) being shot through a target (forest), and consider an infinitesimally thin slab of the target (Thin circular ring). The atoms (trees) that might stop a beam particle (arrow) are shown in red. The magnitude of mean free path depends on the characteristics of the system (forest) the particle (arrow) is in:

    (see wiki: We consider many arrows being shot since this is a statistics problem and we don't know which way Robin is facing.)

    Where \ell is the mean free path, n is the number of target particles (trees) per unit volume (area), and σ is the effective cross sectional area (width of tree) for collision.

    The area (line width) of the slab is 2*pi*r and its volume (area) is 2*pi*r*dr. The typical number of stopping atoms (trees) in the slab is the concentration n (=trees/area =sigma) times the volume (area), i.e., 2*pi*r*sigma*dr. The probability that a beam particle (arrow) will be stopped in that slab is the net area (line length) of the stopping atoms (trees) divided by the total area (line length) of the slab.

    (see wiki: but note that sigma in the wiki is tree cross section not the trees/area given in the problem. The sigma value given in the problem related to tree density, n)

    where σ is the area (or, more formally, the "scattering cross-section") of one atom (tree).

    The drop in beam intensity (many arrows) equals the incoming beam intensity (arrows) multiplied by the probability of being stopped within the slab

    dI = − Inσdx

    This is an ordinary differential equation

    (see wiki)

    whose solution is known as Beer-Lambert law and has form I = I_{0} e^{-x/\ell}, where x is the distance traveled by the beam (arrows) through the target (forest) and I0 is the beam intensity (number of arrows) before it entered the target; ℓ is called the mean free path because it equals the mean distance traveled by a beam particle (single arrow) before being stopped.

    So mean free path = 1/(tree density * tree cross section)

    tree density = 0.005 trees/m^2
    tree cross section = 2*r = 2 m

    mean free path = 100 m

    The wiki continues on to explain why the parameter /ell in the exponent is the mean free path.
     
    Last edited by a moderator: May 5, 2017
  13. Feb 14, 2011 #12

    Chronos

    User Avatar
    Science Advisor
    Gold Member

    Olber's paradox is not solved. It still rules out spatial and temporal infinity. We already have good evidence our universe is temporally finite, so that pretty much resolves the matter.
     
  14. Feb 14, 2011 #13
    hmmm,now i get it.thank you chrisas...i have been thinking on this problem for 3days !!!!
    wish U all the best
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Olbers Paradox and The Sherwood Forest
  1. Olber's paradox (Replies: 36)

  2. Olber's paradox (Replies: 14)

Loading...