Cross Product Problem

**craig** · February 21st 2011, 11:28 AM

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.

**craig** · February 21st 2011, 11:29 AM

Oh and I'm using the definition that $\Delta = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})$ .

**Ackbeet** · February 21st 2011, 12:22 PM

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?

**craig** · February 22nd 2011, 01:40 AM

Thanks for the reply!

Originally Posted by Ackbeet

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 $\mathbf{u}=(\Omega\hat{e}_{3})\times\mathbf{r}?$

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.

**Ackbeet** · February 22nd 2011, 02:16 AM

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?

**craig** · February 22nd 2011, 08:04 AM

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.

**Ackbeet** · February 22nd 2011, 08:14 AM

What is the horizontal part of r?

**craig** · February 22nd 2011, 08:24 AM

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?)

**Ackbeet** · February 22nd 2011, 08:26 AM

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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