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Path Integral Derivation Question

  1. Nov 13, 2011 #1
    Hello, this will be my first post on the physics forum, so i wanted to make it decent :P

    I've been trying recently to derive for myself a path integral formulation (not quantum mechanical or anything feynman like but for finding the length of a curve on a given interval). Heres my attempt at deriving it for a simple curve, but its a very time consuming tedius process to actually calculate. I know real path integrals involve Reinman sums, but after trying to learn them on my own, I decided to give it a shot on my own, here goes:

    f(x) = x^2, from x1 = 3 to x2 = 4 using a small interval dx = 0.01

    distance between two points on a function is defined as:
    d = √(x2 - x1)^2 + (f(x2) - f(x1))^2

    so breaking the curve down into infentesibly small portions, (or in this case our small interval dx), and adding the distances together, we should get a reasonably good approximation for the distance on the curve... right?

    so in this case, between x1 = 3 and x2 = 4, we have several peices of the curve, ex, the distance between 3 - 3.01, 3.01 - 3.02, 3.02 - 3.03.........3.99 - 4.0. i've worked it out so that the number of these peices is defined by (x2 - x1)*(1/dx). so in this case we have:

    1*100 , so 100 individual peices we have to sum up. now this is where i've run into some problems. i've defined the sum as:

    Ʃ(i = 1 to 100) = √((3 + i*dx) - (3 + (i-1)*dx))^2 + (f(3 + i*dx) - f(3 + (i-1)*dx))^2

    which gives a reasonable approximation. now my problem is, i want to evaluate this sum as dx tends to zero, so we would have an exact answer. so we could write this as:

    lim(dx -> 0) Ʃ(i = 1 to ∞) = √((3 + i*dx) - (3 + (i-1)*dx))^2 + (f(3 + i*dx) - f(3 + (i-1)*dx))^2

    but i am stuck here, i feel like i should take an integral, but i don't know how and that probably isn't correct (or even doable). which leads me to my question, what now?

    thank you very much

    EDIT: using 0.01 i got ~6.94, and continuing to decrease it it seems to converge at ~7.074
     

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    Last edited: Nov 13, 2011
  2. jcsd
  3. Nov 15, 2011 #2

    Stephen Tashi

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    I don't know much about path integrals, but for deriving the formula for arclength by "infinitesimal reasoning" the procedure in calculus is:

    [tex] \sum_n \sqrt{ dx^2 + (f(x+(n+1) dx) - f(x+ n dx))^2} [/tex]

    which is how you began. Then divide inside the radical by [itex] dx^2 [/itex] and multiply on the outside by [itex] dx [/itex]

    [tex] = \sum_n \sqrt{ 1 + (\frac {f(x+(n+1)*dx) - f(x+ n dx)}{dx})^2} dx [/tex]

    Argue that this approximates

    [tex] \sum_n \sqrt{ 1 + f'(x+n dx) } dx [/tex]

    Argue that this is a Riemann sum to approximate the integral

    [tex] \int \sqrt{1 + f'(x)} dx [/tex]
     
  4. Nov 15, 2011 #3
    to get expression 1) you're noticing that [itex]\frac{f(x + (n+1)dx) - f(x + ndx)}{dx}[/itex] is just [itex]f'(x + ndx)[/itex], i feel like that makes sense but i'm not sure why? both these steps make sense to me but I can't seem to reason them out. At any rate i'm going to reveiw integrals maybe that will help clear this up. Thanks again for your time
     
  5. Nov 15, 2011 #4

    lurflurf

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    [tex]dl=\sqrt{dx^2+dy^2}=dx \sqrt{1+\left(\frac{dy}{dx}\right)^2}[/tex]
    [tex]l=\int dl=\int \sqrt{dx^2+dy^2}=\int dx \sqrt{1+\left(\frac{dy}{dx}\right)^2}[/tex]
     
  6. Nov 15, 2011 #5
    AH! Perfect, that makes perfect sense. Thank you very much lurflurf.
     
  7. Nov 15, 2011 #6
    After thinking about it, theres a problem.

    take :
    [tex]y = x[/tex]

    this means:
    [tex]\frac{dy}{dx} = 1 [/tex]

    which means you'll be taking the integral of zero, so you will have no path lenth. also, when taking the definite integral between two points, you can run into imaginary path lengths, theres something wrong
     
  8. Nov 15, 2011 #7

    Mute

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    I'm not sure what you were just looking at that got you confused, but the formula lurflurf posted has 1 + (y')^2 in the square root, not 1 - (y')^2. There's no problem.
     
  9. Nov 15, 2011 #8
    Of course, nevermind. Thanks again everyone.
     
  10. Nov 15, 2011 #9

    lurflurf

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    No, that is a "+" inside the square root not a "-".
     
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