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Thread: Cross Product Problem

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

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

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

    But then when we "dot" u and $\displaystyle \Delta$ 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; Feb 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 $\displaystyle \Delta = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})$.
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  3. #3
    A Plied Mathematician
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    A couple of comments:

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

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

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

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

    You then have

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

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

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

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

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

    You then have

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

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

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

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

    But then surly when you combine this with the $\displaystyle \mathbf{u}$ 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

    $\displaystyle \mathbf{u}=\left|\begin{matrix}\mathbf{i}&\mathbf{ j}&\mathbf{k}\\
    0 &0 &\Omega\\
    x &y &z\end{matrix}\right|=-\Omega(y\mathbf{i}-x\mathbf{j})=(-\Omega y,\Omega x,0),$ 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:

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

    $\displaystyle
    \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).
    $

    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 $\displaystyle - \Omega^2 (x,y,0)$.

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

    Thanks again for the reply, really helped.
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  7. #7
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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 $\displaystyle z$ component, so the horizontal part is the $\displaystyle (x,y)$ 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
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    Everything looks correct. The expression with the $\displaystyle \mathbf{u}$'s is proportional to the horizontal component of $\displaystyle \mathbf{r}$, with proportionality constant $\displaystyle -\Omega^{2}.$

    You're very welcome!
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