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1-dimensional random walk

  1. Jan 2, 2016 #1
    Hello!
    I'm struggling with a probably easy physics question concerning random walks. Here I have the slide regarding this:
    Delta is the distance that a particle moves.
    upload_2016-1-2_16-32-9.png

    Can someone explain where the n-1 initially comes from? Does it have to do wtih the concept of the degrees of freedom?
    Than you already! Regards, Vera
     
  2. jcsd
  3. Jan 2, 2016 #2
    Not sure if I understand the question. So your first equation states that after ##n## steps the particle is at the position it was after the previous step (which is the ##(n-1)##th step) plus some ##\delta##. If you put in ##n=1## you get that the position after the first step is the initial position plus some ##\delta##: ##x_1(1) - x_i(0) \pm \delta##. Every next step a new ##\delta## is added, so of course the position after ##n## steps depends on the position after ##(n-1)## steps. Why do you think this has to do anything with degrees of freedom?
     
  4. Jan 2, 2016 #3

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    After ##n-1## steps, particle i is at position ##x_i(n-1)##.
    One step after that, at step ##n##, particle i has moved by ##\pm \delta##: therefore ##x_i(n)=x_i(n-1)\pm \delta##.
     
  5. Jan 2, 2016 #4

    fresh_42

    Staff: Mentor

    The notations ##x_n^{(i)}## and ##<x_n^{(i)} | 1≤i≤N >## might have been less confusing. Or more.
     
  6. Jan 2, 2016 #5
    Oh okay, thank you. After some thinking ad not giving that much emphasis of on this n-1 (which so muchly looks like the n-1 from the df concept) I finally realized what both of you mean ;) I was really confused about why suddenly this n-1 would appear. But its probably quite reasonable. oO Thanks a lot!
     
  7. Jan 2, 2016 #6
    @fresh42: More (for I have no clue what the vertical bar means AND SO ON) ;)
     
  8. Jan 2, 2016 #7
    Another question:
    Equation 1.) is followed by equation 2.). In equation 2.) the delta is treated separately as encircled in red colour. Is this just a process of pulling the expression after the sigma sign in equ. 1 apart?
    But then, why is the second expression (in the circle) SUBTRACTED from the first part of equ. 2.)? I know it doesn't really matter as delta is 0... but just assume that the encircled expression without the minus in the beginning would equal a positive number. Then the outcome would be different compared to when the expression was negative. I know there's a mistake in thinking but I I just can figure out what's behind this. ^^

    1.) upload_2016-1-2_18-17-55.png


    2.) upload_2016-1-2_18-10-26.png


    Just how I'd approach this: The first part of equ. 2.) is the average of all preceding steps (n-1) and the second part is the average of all distances after one step. Then, if I subtract all deltas and assume their average was negative (which just means this whole particle is moving in the opposite direction compared to a positive delta) this would result in a position (x) actually in the opposite direction of the negative delta average for it is subtracted from the first average...

    Is that right or complete bullsh**???? ;) Thanks!
     

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    Last edited: Jan 2, 2016
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