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$$\nabla \times V= \hat{e_x}\space(\frac{\partial}{\partial{y}} V_z-\frac{\partial}{\partial{z}} V_y)+\hat{e_y}\space(\frac{\partial}{\partial{z}} V_x-\frac{\partial}{\partial{x}} V_z)+\hat{e_z}\space(\frac{\partial}{\partial{x}} V_y-\frac{\partial}{\partial{y}} V_x)$$

In the above provided equation, I have understood "$\nabla \times V$" to be just a notation (which is defined to have above value) and not to indicate any cross product.

Reference: Is $\nabla$ a vector?

The following has been extracted from the book "Mathematical methods for Physicists" by Arfken, Weber, and Harris:

Another possible operation with the vector operator r is to take its cross product with a vector. Using the established formula for the cross product, and being careful to write the derivatives to the left of the vector on which they are to act, we obtain $$\nabla \times V= \hat{e_x}\space(\frac{\partial}{\partial{y}} V_z-\frac{\partial}{\partial{z}} V_y)+\hat{e_y}\space(\frac{\partial}{\partial{z}} V_x-\frac{\partial}{\partial{x}} V_z)+\hat{e_z}\space(\frac{\partial}{\partial{x}} V_y-\frac{\partial}{\partial{y}} V_x)$$

$$= \begin{array}{|ccc|} \hat{e_x} & \hat{e_y} & \hat{e_z} \\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}} \\ V_x & V_y & V_z \end{array} \\\\\space Eq (3.58)$$

This vector operation is called the curl of V. Note that when the determinant in Eq. (3.58) is evaluated, it must be expanded in a way that causes the derivatives in the second row to be applied to the functions in the third row (and not to anything in the top row); we will encounter this situation repeatedly, and will identify the evaluation as being from the top down.

I have trouble in understanding the above passage. I feel the determinant equivalent of the curl of V to be wrong, because we don't get the curl of V when we solve that determinant. Is it right to write in determinant form?

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Sensebe
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  • The determinant form of the curl is just a "formal definition." That means we use it as a heuristic for remembering the formula. Curl is not technically defined that way. In fact, it couldn't be defined that way, because determinants are only defined for ALL scalar components (or ALL vector components, if you want to consider each column to be a vector) but the $\hat e_i$'s are vectors, the $\frac {\partial}{\partial x_i}$'s are operators, and the $V_i$'s are scalars. –  Nov 28 '14 at 14:08
  • BTW, $\nabla \times A= \hat{e_x}\space(\frac{\partial}{\partial{y}} A_z-\frac{\partial}{\partial{z}} A_y)+\hat{e_y}\space(\frac{\partial}{\partial{z}} A_x-\frac{\partial}{\partial{x}} A_z)+\hat{e_z}\space(\frac{\partial}{\partial{x}} A_y-\frac{\partial}{\partial{y}} A_x)$ isn't the technical definition of curl, either. The actual definition is $\nabla \times \vec A = \lim_{V\to 0} \frac {\int_S d\vec \sigma \times \vec A}{\int_V d\tau}$ where $V$ is the volume of a region and $S$ is the surface bounding that volume. –  Nov 28 '14 at 15:31
  • @Bye_World: Thank you very much for the useful comments. Can you say some of the books which contain the info you said? – Sensebe Nov 28 '14 at 15:56
  • Try Div, Grad, Curl and All That by Schey. His first definition of curl is slightly different than mine (he defines it implicitly), but then on page $111$ of the $3$rd edition he gives this as an "alternate definition". His notation is a little different, but it is the exact same definition. –  Nov 29 '14 at 17:15
  • @Godparticle, I've noticed that you ask a lot of questions but rarely accept answers. How do I accept an answer?, Why should we accept answers?. – neptun Feb 02 '15 at 10:32

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The determinant expression is notation that encodes a mnemonic, just like $\nabla \times$ does. Both notations are abuses in the sense applying the usual definition of $\times$ (a product of vectors) or $\det$ (a scalar determined by a matrix) here is nonsense---$\nabla$ isn't a vector, and $\hat{e}_x$ and $\frac{\partial}{\partial x}$ (etc.) are operators, not numbers. In both cases, the notation reminds us of the (slightly complicated) form of the curl.

In the case of the determinant mnemonic, we simply expand across the first row, applying the operators in the second row to the scalar (functions) in the third row, and multiply the results by the appropriate vectors in the first row; NB that the latter is multiplication of a scalar by a vector, not of two scalars, so this too is an abuse despite that we call both operations multiplication.

Travis Willse
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