Calculating Curl With Index Notation

1. Sep 30, 2008

poonintoon

Hi, does anyone know a link showing how to calculate curl with a Levi-Civita tensor. I can't figure it out but I am sure if I could see an actual example would be able to work out what is going on.
Thanks.

2. Sep 30, 2008

cristo

Staff Emeritus
Isn't the curl of some vector, A, say $(\nabla \times \vec{A})_i[/tex] just [itex]\epsilon_{ijk}\partial_j A_k$ ?

3. Sep 30, 2008

poonintoon

It is, but I can't find out how to use it.

4. Sep 30, 2008

HallsofIvy

Staff Emeritus
Have you tried just writing it out?

If Ak= <f(x,y,z), g(x,y,z),h(x,y,z)> where x= x1, y= x2, z= x3[/sub], then $\epsilon_{ijk}\partial_j A_k$ is:

We can simplify some of the "writing out" by noting that $\epsilon_{ijk}= 0$ if any of i, j, k are the same, $\epsilon_{123}= \epsilon_{231}= \epsilon{312}= 1$ and $\epsilon_{132}= \epsilon{213}= \epsilon{321}= -1$.

So for B= curl A, we have
$$B_x= B_1= \epsilon_{123}\partial_2\ A_3+ \epsilon{132}\partial_3 A_2$$
$$= \partial_2 A_3- \partial_3 A_2= \frac{\partial h}{\partial z}- \frac{\partial g}{\partial y}$$

$$B_2= \epsilon_{213}\partial_1\ A_3+ \epsilon{231}\partial_3 A_1$$
$$= -\partial_1 A_3+ \partial_3 A_1= \frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}$$

$$B_3= \epsilon_{312}\partial_1\ A_2+ \epsilon{321}\partial_2 A_1$$
$$= \partial_1 A_2- \partial_2 A_1= \frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}$$
which are, of course, the usual formulas for curl A.

5. Sep 30, 2008

nicksauce

So you have
$$\nabla\times\vec{A}=\partial_iA_j\hat{u_k}\epsilon_{ijk}$$.

So to get the x component of the curl, for example, plug in x for k, and then there is an implicit sum for i and j over x,y,z (but all the terms with repeated indices in the Levi-Cevita symbol go to 0)

$$(\nabla\times\vec{A})_x = \partial_yA_z\epsilon_{yzx} + \partial_zA_y\epsilon_{zyx}=\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}$$

6. Sep 30, 2008

poonintoon

Thanks thats what I wanted, I thought once I had seen this I would be able to figure it out, unfortunately it's just not clicking for me.

Can I check I have the right thinking....
$\epsilon_{ijk}\partial_j A_k$

For B1 you set i to 1. then that leaves two combinations for $partial_j A_k$
$partial_2 A_3$ or $partial_3 A_2$

Then you can have any tensor as long as i is 1 i.e $epsilon_{123}\ \epsilon_{112}\ \epsilon_{111}$
but obviously any with two 1's in are zero.

I think this gives the right answer but I should be thinking more in terms of implicit sums than what combination I have left.