# 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.