1
$\begingroup$

Given a vector $v$, the curl of $v$ is defined as the unique vector field with the property $$(\nabla v - \nabla v^T) a = (\text{curl } v) \times a$$ for every vector $a$. (See pag. 32 of Gurtin's book)

I want to find its cartesian components (pg. 33 of the link above): so I try to compute the l.h.s., since I know that $$(\nabla v)_{ij} = \frac{\partial v_i}{\partial x_j}$$ and $$(\nabla v^T)_{ij} = \frac{\partial v_j}{\partial x_i}$$

Therefore, since $a = a_k e_k$ (I'm writing $a$ w.r.t the basis ${\{ e_k\}}_k$) I obtain

$$\Bigl( \frac{\partial v_i}{\partial x_j} - \frac{\partial v_j}{\partial x_i} \Bigr)(e_i \otimes e_j)a_ke_k = \Bigl( \frac{\partial v_i}{\partial x_j} - \frac{\partial v_j}{\partial x_i} \Bigr)a_je_i $$

I'm using the Einstein notation for repeated indices, but I don't know how to obtain the result. The last term seems to be in the right path, but I can't continue to get the classical formulas for the cartesian components of the curl

$\endgroup$
8
  • 1
    $\begingroup$ Your linked page does not show up, so please actually include this in your MathJax presentation. But I guess what's going on is that (you're off by a factor of $2$, however) you take the skew-symmetric part of the derivative matrix of $v$ and then represent multiplication by a skew-symmetric matrix as the cross product. You also should write $(\nabla v)^\top$. Your $e_i\otimes e_j$ should be $e_i\otimes e_j^*$, and then the result drops out. $\endgroup$
    – Ted Shifrin
    Nov 13, 2020 at 22:22
  • $\begingroup$ @TedShifrin You're right about $(\nabla v)^\top$, but I don't understand your point, I was trying to obtain the usual cartesian components of the curl using tensor calculus $\endgroup$
    – Vefhug
    Nov 13, 2020 at 22:29
  • $\begingroup$ @TedShifrin I updated the link. I'm sorry, but I don't know what you mean by $e_j^{*}$. I am following the book's notation, which is for engineers I think. $\endgroup$
    – Vefhug
    Nov 13, 2020 at 22:34
  • $\begingroup$ Yes, and I said that you need to think of this as a linear map multiplying the vector $a$, not a $2$-tensor. So how do you write the cross-product in terms of coordinates? You need an $\epsilon_{ijk}$ in there. Start by working out a general skew-symmetric matrix, rather than dealing with all these derivatives. By $e_j^*$ I mean the $j$th covector in the dual basis. Linear maps are tensors of type $(1,1)$, not $(2,0)$. $\endgroup$
    – Ted Shifrin
    Nov 13, 2020 at 22:39
  • 1
    $\begingroup$ I'm referring to the fact that $$\begin{bmatrix} 0&-c&b\\c&0&-a\\-b&a&0\end{bmatrix}\begin{bmatrix} x\\y\\z\end{bmatrix} = \begin{bmatrix}a\\b\\c\end{bmatrix}\times \begin{bmatrix} x\\y\\z\end{bmatrix},$$ and so you need to use this identification. $\endgroup$
    – Ted Shifrin
    Nov 13, 2020 at 22:44

1 Answer 1

Reset to default
1
$\begingroup$

Write $w:=\nabla\times v$. Pardon my slight relabelling of indices. Taking components of your equation in blue,$$\epsilon_{ijk}\epsilon_{jlm}\tfrac{\partial v_m}{\partial x_l}a_k=\color{blue}{(\tfrac{\partial v_i}{\partial x_k}-\tfrac{\partial v_k}{\partial x_i})a_k=\epsilon_{ijk}w_ja_k}.$$So $w_j=\epsilon_{jlm}\tfrac{\partial v_m}{\partial x_l}$.

$\endgroup$
5
  • $\begingroup$ Thanks @J.G., are you reading from right to left the definition of $\text{curl}$? $\endgroup$
    – Vefhug
    Nov 13, 2020 at 22:52
  • $\begingroup$ @Vefhug Your definition is the blue part. I just pointed out the left-hand side is expressible in terms of two Levi-Civita symbols. $\endgroup$
    – J.G.
    Nov 13, 2020 at 22:53
  • $\begingroup$ Uh okay, maybe I got it: you started from mine (the blue one, modulo index relabelling) and then you wrote the two Levi-Civita symbols. The point I'm still missing is: how did you see that $\epsilon_{ijk}\epsilon_{jlm}\tfrac{\partial v_m}{\partial x_l}=(\frac{\partial v_j}{\partial x_k}-\frac{\partial v_k}{\partial x_j})$ ? @J.G. $\endgroup$
    – Vefhug
    Nov 13, 2020 at 23:03
  • $\begingroup$ @Vefhug Using $\epsilon_{ijk}\epsilon_{jlm}=\delta_{im}\delta_{kl}-\delta_{il}\delta_{km}$. $\endgroup$
    – J.G.
    Nov 13, 2020 at 23:11
  • $\begingroup$ Thanks @J.G. One last thing: since in $\epsilon_{ijk}\epsilon_{jlm}\tfrac{\partial v_m}{\partial x_l}=(\frac{\partial v_i}{\partial x_k}-\frac{\partial v_k}{\partial x_i})$ the indeces $i,k$ are not repeated, the term $$ (\frac{\partial v_i}{\partial x_k}-\frac{\partial v_k}{\partial x_i}) $$ is a matrix, right? $\endgroup$
    – Vefhug
    Nov 14, 2020 at 0:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .