1
$\begingroup$

Let

$$\mathbf{H} = H_x \mathbf{u}_x + H_y \mathbf{u}_y + H_z \mathbf{u}_z$$

be a vector field whose components are defined with respect to the unit vectors $\mathbf{u}_x$, $\mathbf{u}_y$ and $\mathbf{u}_z$, so in the $(x,y,z)$ system of coordinates.

Question 1: If we computed its curl in a new system of coordinates, that is $(x' = x, y' = y, z' = -z)$, how would we do it?

$$\nabla \times \mathbf{H} = \mathrm{det} \begin{vmatrix} \mathbf{u}_{x'} & \mathbf{u}_{y'} & \mathbf{u}_{z'}\\ H_x? & H_y? & H_z?\\ \displaystyle \frac{\partial}{\partial x'} & \frac{\partial}{\partial y'} & \frac{\partial}{\partial z'} \end{vmatrix} $$

In other words, which quantities should we put in the second line? $H_x$, $H_y$ and $H_z$ are along $\mathbf{u}_x$, $\mathbf{u}_y$ and $\mathbf{u}_z$ and not $\mathbf{u}_{x'}$, $\mathbf{u}_{y'}$ and $\mathbf{u}_{z'}$.

Question 2: And if we would like to keep $H_x$, $H_y$ and $H_z$ in the second line, which would be their meaning in this computation?

These questions are strictly relative (but not equal) to my previous one.

$\endgroup$
4
$\begingroup$

You have two options here:

1) Forget you ever had $x,y,z$ coordinate system and plug $H_{i'}$ into determinant.

2) Compute curl in $x,y,z$ coordinates and see how it looks in $x',y',z'$.

You can easily check that these two give different results. The reason is, curl operator really maps vectors not to vectors but to antisymmetric tensors. Antisymmetric tensors are dual to vectors in 3-dimentional space, but this duality does not commute with reflections.

Now bearing in mind that curl of a vector is actually a pseudo-vector, the first option is the correct one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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