Calculating Curl With Index Notation

[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.

 

[cristo]

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

 

[poonintoon]

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

$<br /> It is, but I can't find out how to use it.$

 

[HallsofIvy]

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

[itex] <br /> <blockquote data-attributes="" data-quote="poonintoon" data-source="post: 1894763" cite="https://www.physicsforums.com/goto/post?id=1894763" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> poonintoon said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> It is, but I can't find out how to use it. </div> </div> </blockquote> Have you tried just writing it out?<br /> <br /> If A_k= <f(x,y,z), g(x,y,z),h(x,y,z)> where x= x₁, y= x₂, z= x^{3[/sub], then $\epsilon_{ijk}\partial_j A_k$ is:<br /> <br /> 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 $\epsilon_{132}= \epsilon{213}= \epsilon{321}= -1$. <br /> <br /> So for B= curl A, we have <br /> $$B_x= B_1= \epsilon_{123}\partial_2\ A_3+ \epsilon{132}\partial_3 A_2$$<br /> $$= \partial_2 A_3- \partial_3 A_2= \frac{\partial h}{\partial z}- \frac{\partial g}{\partial y}$$<br /> <br /> $$B_2= \epsilon_{213}\partial_1\ A_3+ \epsilon{231}\partial_3 A_1$$<br /> $$= -\partial_1 A_3+ \partial_3 A_1= \frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}$$<br /> <br /> $$B_3= \epsilon_{312}\partial_1\ A_2+ \epsilon{321}\partial_2 A_1$$<br /> $$= \partial_1 A_2- \partial_2 A_1= \frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}$$<br /> which are, of course, the usual formulas for curl A.}[/itex]

 

[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)

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

 

[poonintoon]

Have you tried just writing it out?

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
$$B_x= B_1= \epsilon_{123}\partial_2\ A_3+ \epsilon{132}\partial_3 A_2$$
$$= \partial_2 A_3- \partial_3 A_2= \frac{\partial h}{\partial z}- \frac{\partial g}{\partial y}$$

Thanks that's 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.