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Thermodynamics problem

  1. Jan 26, 2005 #1
    I am reading a book of thermodynamics. I have some problems and I hope you can help. But I am not sure whether these are math problems or physics problems. I am sorry if I post in a wrong place.

    page 1
    Please read (10.19), the definiton of [tex]W(r)[/tex]

    The writer defined [tex]W(r)[/tex] as "probability that particle will travel a distance r without suffering a collision". Does this definition make sense? since distance should be regarded as continuous variable. Why isn't defined as "... a distance [tex]r + dr[/tex]..."?

    Furthermore, since [tex]W(r)[/tex] is probability, we must have
    [tex]
    \int_{0}^{\infty} W(r) dr = 1
    [/tex]

    Page 2

    How come (10.27) is true? How can A be 1?

    Thanks in advance!
     
  2. jcsd
  3. Jan 26, 2005 #2

    Doc Al

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    Staff: Mentor

    Yes, it makes sense. [tex]W(r)[/tex] is a probability, not a probability density.

    No. [tex]W(\infty) = 0[/tex], since the probability of avoiding a collision over an infinite distance is zero. (If you travel far enough you are guaranteed to hit something.)

    10.27 is just the solution to the differential equation 10.25. The key step in the derivation of that equation is given in 10.23.

    A = 1 since [tex]W(0) = 1[/tex]. That just means that the probability of avoiding a collision over zero distance is 1. (You haven't moved so how could you have hit something?)
     
  4. Jan 26, 2005 #3
    Thank you!

    Thank you for your explanation! I understand all except the following:

    When we deal with continuous variables, the "total number of outcomes" is infinite but the "number of outcomes favorable to E" is finite. So

    [tex]\hbox{the probability of an event } E = P(E) = \frac{\hbox{number of outcomes favorable to } E}{\hbox{total number of outcomes}} \rightarrow \infty[/tex]

    so why does it make sense to say "the probability of distance r"?

     
  5. Jan 29, 2005 #4

    Doc Al

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    Staff: Mentor

    I'm struggling to understand your point, but I'm not seeing it. Maybe an analogy will help. Imagine you are walking in a minefield (explosive mines are scattered randomly throughout). Now let W(r) be the probablity that you could walk r meters without hitting a mine. Will you agree that there is a finite probability that for a given distance that you walk you will hit a mine? And that W(r = 1) > W(r = 2), since the farther you walk the greater your chance of stepping on a mine? And that W(r = 0) = 1, since if you stay put, you won't hit anything? And that W(r = infinity) = 0, since sooner or later you will definitely hit a mine?

    If that makes some sense to you, then apply the same thinking to the case at hand. It's really the same idea: a molecule moving a distance r colliding with another randomly moving molecule.
     
  6. Jan 30, 2005 #5
    Thank you for your illustrating example. maybe I should re-state my problem in this way.

    When I read my textbook(Resnick Halliday & Krane, "Physics" 5 ed) and other reference books, I found a sentence like this:

    "The probability that a molecule has a precisely stated speed, such 600.34326759...m/s, is exactly zero. However, the probability that molecules
    whose speeds lie in a narrow range such as 600 m/s to 602 m/s has a definite nonzero value."

    In my last message, I made a typing mistake.
    it should be:
    [tex]\hbox{The probability of an event }E = P(E) = \frac{\hbox{the number of outcomes favorable to} E}{\hbox{total number of outcomes}} \rightarrow 0 [/tex]

    I think it is a mathematical statement of what my textbooks mentioned.

    So, I think we shouldn't say [tex]\hbox{the probablity that you could walk r meters without hitting a mine}[/tex], but [tex]\hbox{the probablity that you could walk a distance between } r + dr \hbox{ meters without hitting a mine}[/tex].

    I hope it's clearer now :biggrin:
     
  7. Jan 30, 2005 #6

    Doc Al

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    Staff: Mentor

    Your statement is clear, but your point is not. :smile: I'm still struggling to understand the problem with the definitions given in your book.

    The probability of not getting hit in walking from 0 to r is very different from the probability of not getting hit in walking from r to r+dr. (Of course, the probability of not getting hit in walking from 0 to r+dr is only incrementally different from the probability of not getting hit in walking from 0 to r.)

    By definition, W(r) is the probability of traveling a distance r (that means traveling from 0 to r) without hitting a mine. Also defined is the probability of hitting a mine in traveling between r and r+dr: [itex]\alpha dr[/itex] (the probability is assumed independent of r). So, using the definitions in your text, the probability of not hitting a mine in traveling between r and r+dr must be [itex]1 - \alpha dr[/itex].

    Note that this means if you travel between r and r+dr, the probability of not hitting a mine is [itex]1 - \alpha dr[/itex]. That "if" is key.

    So, in terms of our definitions, what must W(r+dr) be? In order to not get hit in traveling from 0 to r+dr, you first must not get hit in the distance 0 to r. (Obviously!) So [itex] W(r+dr) = W(r)(1 - \alpha dr)[/itex].

    (I think I'm just repeating stuff better said in your text, so I'll stop here. Let me know if I'm getting closer to your issue.)
     
  8. Jan 30, 2005 #7
    OH~~!
    I see your point~
    Thank you so much for your explanation!
    Your rephrasing of my text helps me a lot ~ thanks
     
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