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Math Help - Cross Product Problem

  1. #1
    Super Member craig's Avatar
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    Cross Product Problem

    Let \mathbf{u} = (\Omega \hat{e}_3) \times \mathbf{r}, where \Omega \in \mathbb{R} and \hat{e}_3 is the unit vector aligned to the vertical direction. We need to show that (\mathbf{u} \cdot \Delta) \mathbf{u} is proportional to the horizontal part of \mathbf{r}.

    I'm guessing that \hat{e}_3 = (0,0,1), and using \mathbf{r} = (x,y,z) we get \mathbf{u} = (-\Omega y, \Omega x, 0).

    But then when we "dot" u and \Delta we get 0?

    Am I missing something here, this doesn't seem to make sense?

    Thanks in advance for the help.
    Last edited by craig; February 22nd 2011 at 01:42 AM.
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  2. #2
    Super Member craig's Avatar
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    Oh and I'm using the definition that \Delta = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}).
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  3. #3
    A Plied Mathematician
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    A couple of comments:

    1. Is it \mathbf{v}=(\Omega\hat{e}_{3})\times\mathbf{r}, or \mathbf{u}=(\Omega\hat{e}_{3})\times\mathbf{r}?

    2. The expression \mathbf{u}\cdot\Delta is an operator, and with operators, order matters. That is,

    \mathbf{u}\cdot\Delta\not=\dfrac{\partial u_{x}}{\partial x}+\dfrac{\partial u_{y}}{\partial y}+\dfrac{\partial u_{z}}{\partial z}, but

    \mathbf{u}\cdot\Delta=u_{x}\,\dfrac{\partial }{\partial x}+u_{y}\,\dfrac{\partial }{\partial y}+u_{z}\,\dfrac{\partial }{\partial z}.

    You then have

    (\mathbf{u}\cdot\Delta)\mathbf{u}=\left(u_{x}\,\df  rac{\partial }{\partial x}+u_{y}\,\dfrac{\partial }{\partial y}+u_{z}\,\dfrac{\partial }{\partial z}\right)\mathbf{u}.

    Make sense?
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  4. #4
    Super Member craig's Avatar
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    Thanks for the reply!

    Quote Originally Posted by Ackbeet View Post
    A couple of comments:

    1. Is it \mathbf{v}=(\Omega\hat{e}_{3})\times\mathbf{r}, or \mathbf{u}=(\Omega\hat{e}_{3})\times\mathbf{r}?
    Ohh apologies it's \mathbf{u}=(\Omega\hat{e}_{3})\times\mathbf{r}?

    Quote Originally Posted by Ackbeet View Post
    2. The expression \mathbf{u}\cdot\Delta is an operator, and with operators, order matters. That is,

    \mathbf{u}\cdot\Delta\not=\dfrac{\partial u_{x}}{\partial x}+\dfrac{\partial u_{y}}{\partial y}+\dfrac{\partial u_{z}}{\partial z}, but

    \mathbf{u}\cdot\Delta=u_{x}\,\dfrac{\partial }{\partial x}+u_{y}\,\dfrac{\partial }{\partial y}+u_{z}\,\dfrac{\partial }{\partial z}.

    You then have

    (\mathbf{u}\cdot\Delta)\mathbf{u}=\left(u_{x}\,\df  rac{\partial }{\partial x}+u_{y}\,\dfrac{\partial }{\partial y}+u_{z}\,\dfrac{\partial }{\partial z}\right)\mathbf{u}.

    Make sense?
    Yes that does make sense I think. So using your corrected equation do we get:

    (\mathbf{u}\cdot\Delta)\mathbf{u}=\left(u_{x}\,\df  rac{\partial }{\partial x}+u_{y}\,\dfrac{\partial }{\partial y}+u_{z}\,\dfrac{\partial }{\partial z}\right)\mathbf{u}.

    =\left(-\Omega y\,\dfrac{\partial }{\partial x}+\Omega x\,\dfrac{\partial }{\partial y}+0\,\dfrac{\partial }{\partial z}\right)\mathbf{u}.

    But then surly when you combine this with the \mathbf{u} outside of the brackets you get the zero vector?

    Sorry if I'm missing something obvious.
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  5. #5
    A Plied Mathematician
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    Ok. Just double-checking some things here. We've got

    \mathbf{u}=\left|\begin{matrix}\mathbf{i}&\mathbf{  j}&\mathbf{k}\\<br />
0 &0 &\Omega\\<br />
x &y &z\end{matrix}\right|=-\Omega(y\mathbf{i}-x\mathbf{j})=(-\Omega y,\Omega x,0), just like you got. Ok. The thing you're missing when forming your derivative expression is that the entire operator acts on each component. That is, you've got this:

    \left(u_{x}\dfrac{\partial}{\partial x}+u_{y}\dfrac{\partial}{\partial y}+u_{z}\dfrac{\partial}{\partial z}\right)(-\Omega y,\Omega x,0)=\left(-\Omega y\dfrac{\partial}{\partial x}+\Omega x\dfrac{\partial}{\partial y}\right)(-\Omega y,\Omega x,0)=

    <br />
\left(-\left(-\Omega y\dfrac{\partial}{\partial x}+\Omega x\dfrac{\partial}{\partial y}\right)(\Omega y),\left(-\Omega y\dfrac{\partial}{\partial x}+\Omega x\dfrac{\partial}{\partial y}\right)(\Omega x),\left(-\Omega y\dfrac{\partial}{\partial x}+\Omega x\dfrac{\partial}{\partial y}\right)0\right).<br />

    Can you simplify from here?
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  6. #6
    Super Member craig's Avatar
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    Ahh of course because the dot product produces a scalar, which you have to then 'multiply' each term in the vector by!

    So working this out I get - \Omega^2 (x,y,0).

    I think this ok, but how do we know exactly that it's proportional to the horizontal part of \mathbf{r}? Is it because the value for z is zero?

    Thanks again for the reply, really helped.
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  7. #7
    A Plied Mathematician
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    What is the horizontal part of r?
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  8. #8
    Super Member craig's Avatar
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    Quote Originally Posted by Ackbeet View Post
    What is the horizontal part of r?
    That's what I'm trying to visualise lol. Well the vertical part is the z component, so the horizontal part is the (x,y) components.....

    Now why couldn't I come up with that a second ago!!!

    Thankyou again for all your help with this.

    (PS I hope what I've just said is actually correct?)
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  9. #9
    A Plied Mathematician
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    Everything looks correct. The expression with the \mathbf{u}'s is proportional to the horizontal component of \mathbf{r}, with proportionality constant -\Omega^{2}.

    You're very welcome!
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