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Path dependent function with a defined path

  1. Mar 6, 2012 #1
    This question is about if I have a path dependent function but a definite path, then can I integrate along the path to form a 2 dimensional field and then take partial derivatives along directions other than the path? And e.g. are points along path A connected to path B? (Note: all paths start at the origin and x and z can only be positive)

    If I consider a non-conservative vector field
    [itex]\vec{dt}= \frac{\vec{ds}}{V(z)}[/itex]

    where

    [itex]ds = \sqrt{dx^{2}+dz^{2}}[/itex]
    and
    V(z) is only a function of Z and not of X

    so I specify the path:

    [itex]p = \frac{sin \Theta}{V(z)} = \frac{dx}{V(z)\sqrt{dx^{2}+dz^{2}}}[/itex] (eq. 1)

    and integrate along the path to create the field:

    [itex]t = \int_{0}^{z }\frac{dz}{\sqrt{1-p^{2}V(z)^{2}}}[/itex] (eq. 2)

    and also:

    [itex]x = \int_{0}^{z }\frac{pV(z)dz}{\sqrt{1-p^{2}V(z)^{2}}}[/itex] (eq. 3)

    may I now treat t as path independent? In other words, may I now take the total derivative and evaluate at a constant x like this:


    [itex]dt = \frac{\partial t}{\partial z}|_{p}dz + \frac{\partial t}{\partial p}|_{z}dp[/itex] (eq. 4)

    [itex]dx = \frac{\partial x}{\partial z}|_{p}dz + \frac{\partial x}{\partial p}|_{z}dp[/itex] (eq. 5)

    and then evaluate eq. 5 at a fixed x so that dx = 0

    It seems to me that I can't do this because I think that I have defined a relationship between dx and dz in equation 1. Since t is path dependent, p must be in the definition of t so setting dx = 0 violates the constraint that p has set on the relationship between dx and dz. Is what I was thinking correct? Or am I allowed to take the total derivative of X and set dx = 0? Is the point t(x,z) connected to t(x,z+dz)?
     
  2. jcsd
  3. Mar 13, 2012 #2
    Can any of the math geniuses comment on this? Thanks,
     
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