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I Velocity vector fields

  1. Apr 6, 2016 #1

    dyn

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    Hi. Given a one-parameter family of maps such as
    Φt : ( x , y ) → ( xet + 2et -2 , ye2t ) the velocity vector field at t=0 is given by d(Φt)/dt = (x+2) ∂/∂x + 2y ∂/∂y
    My question is ; how does differentiating a vector function Φt with respect to t result in a scalar function ? Thanks
     
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  3. Apr 6, 2016 #2

    Orodruin

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    It doesn't.
     
  4. Apr 6, 2016 #3

    BiGyElLoWhAt

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    How do you go from ##\frac{d}{dt}[(xe^t + 2e^t - 2)\hat{x} + (ye^{2t})\hat{y}]## to what you have?
     
  5. Apr 6, 2016 #4

    Orodruin

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    The derivative is evaluated at ##t = 0##. The tangent vector basis is ##\partial_x## and ##\partial_y##.
     
  6. Apr 6, 2016 #5

    BiGyElLoWhAt

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    Ahh. The basis was what I was missing. I suppose that makes sense, maybe. I would think that we would be looking at ##\frac{\partial \hat{x}}{\partial t}|_{t=0}## and ##\frac{\partial \hat{y}}{\partial t}|_{t=0}##, however.
    Not trying to hijack the thread.
    **I think I see the error of my ways, now.**
     
    Last edited: Apr 6, 2016
  7. Apr 6, 2016 #6

    dyn

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    The function Φ t takes 2 coordinates ( x , y ) into 2 different coordinates. I would think this makes them vectors while the velocity vector field doesn't look like a vector.
     
    Last edited by a moderator: Apr 6, 2016
  8. Apr 6, 2016 #7

    Orodruin

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    Yes it does. Remember that the tangent vector basis is ##\partial_\mu##.
     
  9. Apr 7, 2016 #8

    dyn

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    So the original function is a vector with 2 components ? And the velocity vector field is a vector with one component ?
     
  10. Apr 7, 2016 #9

    Orodruin

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    No, both have two components. It is just a different way of writing it as a linear combination of the basis instead of a collection of components.
     
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