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    Why can't a streamline be defined using the dot product:
    http://en.wikipedia.org/wiki/Streaml..._and_pathlines

    Why must they be defined using a cross product? My thinking is that a velocity vector is tangential to a streamline, so why can't we express that using a dot product and setting cosine(theta) = 1?
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    (Original post by djpailo)
    Why can't a streamline be defined using the dot product:
    http://en.wikipedia.org/wiki/Streaml..._and_pathlines

    Why must they be defined using a cross product? My thinking is that a velocity vector is tangential to a streamline, so why can't we express that using a dot product and setting cosine(theta) = 1?
    I don't know any of this stuff but I just looked at the wikipedia page and basically, if you define it with the cross product, you can write a single equation whereas with the dot product, you would need two i.e. (using wikipedia's notation)

    \frac{d\mathbf{x}_s}{ds}\cdot \mathbf{u}(\mathbf{x}_s) = \mid\frac{d\mathbf{x}_s}{ds}\mid \mid\mathbf{u}(\mathbf{x}_s) \mid

    or

    \frac{d\mathbf{x}_s}{ds}\cdot \mathbf{u}(\mathbf{x}_s) = -\mid\frac{d\mathbf{x}_s}{ds}\mid \mid\mathbf{u}(\mathbf{x}_s) \mid

    which is kinda dopey and impractical.

    In general, the cross product is convenient for talking about things being parallel and the dot product is more useful for saying when things are perpendicular.
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    Are we necessarily dealing with unit vectors? That would give +/-1?
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    (Original post by ghostwalker)
    Are we necessarily dealing with unit vectors? That would give +/-1?
    Oops - edited previous post to correct.
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    (Original post by Mark85)
    Oops - edited previous post to correct.
    Definitely looks dopey now.
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    (Original post by Mark85)
    I don't know any of this stuff but I just looked at the wikipedia page and basically, if you define it with the cross product, you can write a single equation whereas with the dot product, you would need two i.e. (using wikipedia's notation)

    \frac{d\mathbf{x}_s}{ds}\cdot \mathbf{u}(\mathbf{x}_s) = \mid\frac{d\mathbf{x}_s}{ds}\mid \mid\mathbf{u}(\mathbf{x}_s) \mid

    or

    \frac{d\mathbf{x}_s}{ds}\cdot \mathbf{u}(\mathbf{x}_s) = -\mid\frac{d\mathbf{x}_s}{ds}\mid \mid\mathbf{u}(\mathbf{x}_s) \mid

    which is kinda dopey and impractical.

    In general, the cross product is convenient for talking about things being parallel and the dot product is more useful for saying when things are perpendicular.
    I don't get why you need two equations?
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    (Original post by djpailo)
    I don't get why you need two equations?
    If u,v are three dimensional vectors then

    u \times v = 0 \Leftrightarrow u \mbox{ and } v \mbox{ are parallel } \Leftrightarrow \frac{u \cdot v}{\mid u\mid \mid v\mid} = 1 \mbox{ or } \frac{u \cdot v}{\mid u\mid \mid v\mid} =-1

    In other words, the cross product of two vectors is 0 if they are parallel whereas the dot product of two vectors is the product of the lengths of the vectors with a sign corresponding to whether the vectors point in the same or opposite direction.

    For a streamline, the definition says that a streamline is parallel to the velocity vector but can either have the same or opposite direction. I know nothing about fluid mechanics but I guess the point is that it you want "points in the same direction as the flow" to be independent of the parameterisation of the streamline.
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    (Original post by Mark85)
    If u,v are three dimensional vectors then

    u \times v = 0 \Leftrightarrow u \mbox{ and } v \mbox{ are parallel } \Leftrightarrow \frac{u \cdot v}{\mid u\mid \mid v\mid} = 1 \mbox{ or } \frac{u \cdot v}{\mid u\mid \mid v\mid} =-1

    In other words, the cross product of two vectors is 0 if they are parallel whereas the dot product of two vectors is the product of the lengths of the vectors with a sign corresponding to whether the vectors point in the same or opposite direction.
    Okay, yes I see what you mean. Thanks for your help!
 
 
 
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