# Cross Product Problem

• Feb 21st 2011, 11:28 AM
craig
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.
• Feb 21st 2011, 11:29 AM
craig
Oh and I'm using the definition that $\Delta = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})$.
• Feb 21st 2011, 12:22 PM
Ackbeet

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?
• Feb 22nd 2011, 01:40 AM
craig

Quote:

Originally Posted by Ackbeet

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
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.
• Feb 22nd 2011, 02:16 AM
Ackbeet
Ok. Just double-checking some things here. We've got

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

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

$
\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?
• Feb 22nd 2011, 08:04 AM
craig
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.
• Feb 22nd 2011, 08:14 AM
Ackbeet
What is the horizontal part of r?
• Feb 22nd 2011, 08:24 AM
craig
Quote:

Originally Posted by Ackbeet
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?)
• Feb 22nd 2011, 08:26 AM
Ackbeet
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!