I'm having trouble with some concepts of Index Notation. (Einstein notation)

If I take the divergence of curl of a vector, $\nabla \cdot (\nabla \times \vec V)$ first I do the parenthesis:

$\nabla_iV_j\epsilon_{ijk}\hat e_k$ and then I apply the outer $\nabla$...

and get: $\nabla_l(\nabla_iV_j\epsilon_{ijk}\hat e_k)\delta_{lk}$

I am not sure if I applied the outer $\nabla$ correctly. If I did do it correctly, however, what is my next step? I guess I just don't know the rules of index notation well enough. Can I apply the index of $\delta$ to the $\hat e$ inside the parenthesis? Or is that illegal?


1 Answer 1


First some notation

$$\nabla \times \vec B \rightarrow \epsilon_{ijk}\nabla_j B_k$$ $$\nabla \cdot \vec B \rightarrow \nabla_i B_i$$ $$\nabla B \rightarrow \nabla_i B$$

Now, to your problem,

$$\nabla \cdot(\nabla \times \vec V)$$

writing it in index notation

$$\nabla_i (\epsilon_{ijk}\nabla_j V_k)$$

Now, simply compute it, (remember the Levi-Civita is a constant)

$$\epsilon_{ijk} \nabla_i \nabla_j V_k$$

Here we have an interesting thing, the Levi-Civita is completely anti-symmetric on i and j and have another term $\nabla_i \nabla_j$ which is completely symmetric: it turns out to be zero.

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = 0$$

Lets make the last step more clear. We can always say that $a = \frac{a+a}{2}$, so we have

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k + \epsilon_{ijk} \nabla_i \nabla_j V_k \right]$$

Now lets interchange in the second Levi-Civita the index $\epsilon_{ijk} = - \epsilon_{jik}$, so that

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k - \epsilon_{jik} \nabla_i \nabla_j V_k \right]$$

Now we can just rename the index $\epsilon_{jik} \nabla_i \nabla_j V_k = \epsilon_{ijk} \nabla_j \nabla_i V_k$ (no interchange was done here, just renamed).

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k - \epsilon_{ijk} \nabla_j \nabla_i V_k \right]$$

We can than put the Levi-Civita at evidency,

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{\epsilon_{ijk}}{2} \left[ \nabla_i \nabla_j V_k - \nabla_j \nabla_i V_k \right]$$

And, because V_k is a good field, there must be no problem to interchange the derivatives $\nabla_j \nabla_i V_k = \nabla_i \nabla_j V_k$

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{\epsilon_{ijk}}{2} \left[ \nabla_i \nabla_j V_k - \nabla_i \nabla_j V_k \right]$$

And, as you can see, what is between the parentheses is simply zero.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.