Vector Triple Product, Alternating Tensor

**craig** · March 8th 2011, 09:49 AM

Consider the field, $\mathbf{u} = \frac{1}{r} \Phi \times \mathbf{r}$ where $\Phi$ is a vector constant. Using suffix notation, show that:

i) $\nabla \cdot \mathbf{u} = 0$
ii) $\nabla \times \mathbf{u} = \frac{1}{r} \Phi + \frac{(\Phi \cdot \mathbf{r})}{r^2} \mathbf{r}$

Using suffix notation, we get $u_i = \frac{1}{r} \epsilon_{ijk} \Phi_j x_k$

And then the dot product gives us $\frac{1}{r} \epsilon_{ijk} \nabla \Phi_j x_k$

Just a little unsure where to go from here?

Thanks in advance

**Ackbeet** · March 8th 2011, 09:58 AM

It should be

$u_{i}=\dfrac{1}{r}\,\epsilon_{ijk}\Phi_{j}x_{k}.$

The dot product would then be

$\partial_{i}u_{i}=\dfrac{1}{r}\,\partial_{i}\epsil on_{ijk}\Phi_{j}x_{k}.$

[EDIT]: This equation is incorrect. The derivative operator cannot go past the 1/r. See below for correction.

Can you see where this is going?

For the cross product, you'll have a double sum, and then you can reduce using your Levi-Civita to Kronecker Delta identity, that should reduce to the RHS. Make sense?

**craig** · March 8th 2011, 10:16 AM

Originally Posted by Ackbeet

It should be

$u_{i}=\dfrac{1}{r}\,\epsilon_{ijk}\Phi_{j}x_{k}.$

The dot product would then be

$\partial_{i}u_{i}=\dfrac{1}{r}\,\partial_{i}\epsil on_{ijk}\Phi_{j}x_{k}.$

Can you see where this is going?

Ahh sorry a little typo in writing out the question. Not too sure how to evaluate the suffix partial differentiation though :\

Originally Posted by Ackbeet

For the cross product, you'll have a double sum, and then you can reduce using your Levi-Civita to Kronecker Delta identity, that should reduce to the RHS. Make sense?

Ahh that does make a little sense, I thought that you would end up using the result from the first part so hadn't really given this much of a go.

Will have a play around with it now, thankyou

**Ackbeet** · March 8th 2011, 10:23 AM

Note that $\Phi$ is a vector constant. Also note that the Levi-Civita symbol is zero if any of its indices are duplicated. Does that give you any direction?

**craig** · March 8th 2011, 12:27 PM

Because it's a constant, when differentiated it just becomes zero, that's the one bit I knew already haha!

I know the fact about the Levi-Civita symbol being zero if any indices appear more than once, but do we 'sum' over the different $x_k$ s ?

Sorry if I'm not getting this, not had much chance at getting my head around the whole suffix notation thing :\

**Ackbeet** · March 8th 2011, 01:15 PM

Hang on a bit, I think I made a mistake earlier. $r$ definitely depends on $x_{i},$ so the partial derivative can't go past that.

Ok, let me make the notation a little clearer:

$\nabla\cdot\mathbf{u}=\dfrac{\partial}{\partial x_{i}}\,u_{i}=\dfrac{\partial}{\partial x_{i}}\,\epsilon_{ijk}\Phi_{j}\dfrac{x_{k}}{r}=\ep silon_{ijk}\Phi_{j}\,\dfrac{\partial}{\partial x_{i}}\,\dfrac{x_{k}}{r}=$ (using the quotient rule)

$\epsilon_{ijk}\Phi_{j}\dfrac{(r)\left(\dfrac{\part ial x_{k}}{\partial x_{i}}\right)-(x_{k})\left(\dfrac{\partial r}{\partial x_{i}}\right)}{r^{2}}.$

Now if $i=k,$ the Levi-Civita symbol cancels out the whole expression. So no need to examine that case. Assume $i\not=k.$ Then how does this expression simplify? Recall that

$r=\sqrt{r_{m}r_{m}}.$

**craig** · March 8th 2011, 03:33 PM

Originally Posted by Ackbeet

Hang on a bit, I think I made a mistake earlier. $r$ definitely depends on $x_{i},$ so the partial derivative can't go past that.

Ok, let me make the notation a little clearer:

$\nabla\cdot\mathbf{u}=\dfrac{\partial}{\partial x_{i}}\,u_{i}=\dfrac{\partial}{\partial x_{i}}\,\epsilon_{ijk}\Phi_{j}\dfrac{x_{k}}{r}=\ep silon_{ijk}\Phi_{j}\,\dfrac{\partial}{\partial x_{i}}\,\dfrac{x_{k}}{r}=$ (using the quotient rule)

$\epsilon_{ijk}\Phi_{j}\dfrac{(r)\left(\dfrac{\part ial x_{k}}{\partial x_{i}}\right)-(x_{k})\left(\dfrac{\partial r}{\partial x_{i}}\right)}{r^{2}}.$

Now if $i=k,$ the Levi-Civita symbol cancels out the whole expression. So no need to examine that case. Assume $i\not=k.$ Then how does this expression simplify? Recall that

$r=\sqrt{r_{m}r_{m}}.$

So on that basis, does that mean the same applies for $i = j$ ?

If so, then we'll just be considering the case $i = i$ ...does this make sense? So:

$\epsilon_{ijk}\Phi_{j}\dfrac{(r)(1) -(x_{k})\left(\dfrac{\partial r}{\partial x_{i}}\right)}{r^{2}}.$

Still not sure how this would become zero? Sorry again for taking up your time haha.

**Ackbeet** · March 8th 2011, 06:53 PM

The reason I'm looking at $i\not=k$ and not $i\not=j,$ is that the expression $\partial x_{k}/\partial x_{i}$ is what's in the numerator of the quotient rule. Note that

$\partial x_{k}/\partial x_{i}=\delta_{ik}.$

Why?

Also, if $i=k,$ what does the Levi-Civita symbol $\epsilon_{ijk}$ do?

**craig** · March 9th 2011, 01:37 AM

Originally Posted by Ackbeet

Also, if $i=k,$ what does the Levi-Civita symbol $\epsilon_{ijk}$ do?

If $i = k$ then the Levi-Civita symbol equals zero, because of the repeated indice right?

Originally Posted by Ackbeet

The reason I'm looking at $i\not=k$ and not $i\not=j,$ is that the expression $\partial x_{k}/\partial x_{i}$ is what's in the numerator of the quotient rule. Note that

$\partial x_{k}/\partial x_{i}=\delta_{ik}.$

Why?

I've never heard that before, is this just basic knowledge that I should know haha?

I'm sorry I'm just not getting it :\

**Ackbeet** · March 9th 2011, 02:46 AM

Originally Posted by craig

If $i = k$ then the Levi-Civita symbol equals zero, because of the repeated indice right?

Right.

I've never heard that before, is this just basic knowledge that I should know haha?

Everyone learns basic things at some point! I would argue, though, that you already know this fact, you just don't recognize that you know it already.

Think about it for a second: I'm taking the partial derivative of one independent variable with respect to another independent variable. If the two variables happen to be the same, then I should get one, right? I mean, you definitely have

$\dfrac{d}{dx}\,x=1,$ and similarly

$\dfrac{\partial}{\partial x}\,x=1.$

That you know from Calculus I and Calculus III (or Differential Calculus and Multivariable Calculus, depending on where you took it).

However, let's suppose that $x$ and $y$ are both independent variables. What should

$\dfrac{\partial}{\partial x}\,y$ be?

Suppose I gave you a function $f=f(x,y,z)=y^{2}+z^{2},$ and asked you to compute

$\dfrac{\partial f}{\partial x}.$

What would you tell me? And then, if I gave you a function $g=g(x,y,z)=y,$ and asked you to compute

$\dfrac{\partial g}{\partial x},$

what would you tell me? But isn't

$\dfrac{\partial g}{\partial x}=\dfrac{\partial y}{\partial x}?$

So what does that tell you?

**craig** · March 9th 2011, 03:05 AM

Originally Posted by Ackbeet

$\dfrac{\partial}{\partial x}\,y$ be?

Right I'm assuming that this is the obvious 0, treating the $y$ as a constant when we take the partial derivative wrt $x$ .

Originally Posted by Ackbeet

Suppose I gave you a function $f=f(x,y,z)=y^{2}+z^{2},$ and asked you to compute

$\dfrac{\partial f}{\partial x}.$

What would you tell me? And then, if I gave you a function $g=g(x,y,z)=y,$ and asked you to compute

$\dfrac{\partial g}{\partial x},$

what would you tell me? But isn't

$\dfrac{\partial g}{\partial x}=\dfrac{\partial y}{\partial x}?$

We know that both $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial g}{\partial x}$ are both zero, by the same principle as above?

Originally Posted by Ackbeet

So what does that tell you?

So does this mean that the two partial derivatives in our previous equations are both zero. Hence the dot product is also zero?

Thanks again for the reply.

**Ackbeet** · March 9th 2011, 04:41 AM

Ok, so it looks like you understand now why

$\dfrac{\partial x_{k}}{\partial x_{i}}=\delta_{ik}.$

Looking back at post # 6, and looking at the expression

$\epsilon_{ijk}\Phi_{j}\dfrac{r\,\dfrac{\partial x_{k}}{\partial x_{i}}-x_{k}\,\dfrac{\partial r}{\partial x_{i}}}{r^{2}},$

I can simplify down to

$\epsilon_{ijk}\Phi_{j}\dfrac{r\,\delta_{ik}-x_{k}\,\dfrac{\partial r}{\partial x_{i}}}{r^{2}}.$

But really, the $\delta_{ik}$ is never going to get a chance to be $1,$ because if $i=k,$ then the Levi-Civita symbol at the beginning of the expression is zero. Thus, we are justified in simply getting rid of that term altogether, and so we now have the expression

$-\epsilon_{ijk}\Phi_{j}\dfrac{x_{k}\,\dfrac{\partia l r}{\partial x_{i}}}{r^{2}}.$

Now, given that $r=\sqrt{x_{m}x_{m}},$ can you tell me what

$\dfrac{\partial r}{\partial x_{i}}$ is?

If all the indices are throwing you off, just think of

$r=\sqrt{x^{2}+y^{2}+z^{2}},$ and compute

$\dfrac{\partial r}{\partial x}.$ What do you get?

**craig** · March 9th 2011, 11:01 AM

Originally Posted by Ackbeet

$-\epsilon_{ijk}\Phi_{j}\dfrac{x_{k}\,\dfrac{\partia l r}{\partial x_{i}}}{r^{2}}.$

Now, given that $r=\sqrt{x_{m}x_{m}},$ can you tell me what

$\dfrac{\partial r}{\partial x_{i}}$ is?

If all the indices are throwing you off, just think of

$r=\sqrt{x^{2}+y^{2}+z^{2}},$ and compute

$\dfrac{\partial r}{\partial x}.$ What do you get?

$\dfrac{\partial r}{\partial x}.$ is just $\frac{x}{r}$ , so

$-\epsilon_{ijk}\Phi_{j}\dfrac{x_{k}\,\dfrac{x}{r}}{ r^{2}}.$

$-\epsilon_{ijk}\Phi_{j}\dfrac{x_{k}\,x}{r}.$

What suffix do I put on the second $x$ ?

**Ackbeet** · March 9th 2011, 11:07 AM

Your partial derivative of $r$ w.r.t. $x$ is correct, but your application of it to the expression at hand is incorrect. You need

$-\epsilon_{ijk}\Phi_{j}\dfrac{x_{k}\dfrac{x_{i}}{r} }{r^{2}}=-\dfrac{1}{r^{3}}\,\epsilon_{ijk}x_{i}\Phi_{j}x_{k} ,$

where I've written the last expression in a (hopefully) suggestive manner. And I think I've answered your question as well.

Do you recognize the expression $\epsilon_{ijk}x_{i}\Phi_{j}x_{k}?$

Think geometrically here.

**craig** · March 9th 2011, 11:48 AM

Originally Posted by Ackbeet

Your partial derivative of $r$ w.r.t. $x$ is correct, but your application of it to the expression at hand is incorrect. You need

$-\epsilon_{ijk}\Phi_{j}\dfrac{x_{k}\dfrac{x_{i}}{r} }{r^{2}}=-\dfrac{1}{r^{3}}\,\epsilon_{ijk}x_{i}\Phi_{j}x_{k} ,$

where I've written the last expression in a (hopefully) suggestive manner. And I think I've answered your question as well.

Ahh of course it is!!

Originally Posted by Ackbeet

Do you recognize the expression $\epsilon_{ijk}x_{i}\Phi_{j}x_{k}?$

Think geometrically here.

Right, I don't recognise it exactly, but according to my lecture notes, $\epsilon_{ijk}x_{j}x_{k} = 0$ .

Because of the cyclic property of the Levi-Cevita symbol, and the fact that $\Phi$ is a vector constant, this means that it's equal to zero?

Either way, could you explain what the geometric interpretation is haha, could be useful...

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