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Displacement after Time T in a Velocity Field

  1. Mar 29, 2012 #1
    I have a velocity field that is static in time. At every location, x, there is a corresponding velocity vector.

    I'm trying to work out the displacement of a particle after T seconds if I drop it into the velocity field at time t=0 and at location x_0.

    I was thinking something along the lines of ∫v(x_0)dt between the limits of t=0 and t=T but it doesn't seem right because the integrand doesn't seem to depend on t at all (in fact v(x_0) is a constant so it can't be right). Maybe there's a double integral involved...

    Anyone know what a suitable form for this equation would be?
     
  2. jcsd
  3. Mar 29, 2012 #2
    For simplicity, you could consider a one dimensional case:

    [itex]\frac{dx}{dt}=v(x)[/itex]

    This is a separable differential equation, so:

    [itex]\int_{x_0}^{x_f}\frac{dx}{v(x)}=\int_0^T dt[/itex]

    But it gets you a solution for the final time, T.

    [itex]T = \int_{x_0}^{x_f}\frac{dx}{v(x)} [/itex]

    If for a fixed x0, you can suppose that the total time is a function of the ending position.

    [itex]T = f(x_f) [/itex], where this function is given by the right hand side of the equation above.

    So, to get the ending position for a given total time, you would need to take the inverse of that function

    [itex]x_f = f^{-1}(T)[/itex]

    In order to get a closed form for this, you need to plug in the velocity field. I don't think there's a way to get a general expression for an arbitrary velocity function.

    Now, for a two dimensional problem you need to do this for each component of the position vector.
     
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