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It is well known that for a vector $\textbf{v}=R\textbf{e}_r+\Theta\textbf{e}_{\theta}+Z\textbf{e}_z$, its 2-norm is $\|\textbf{v}\|_2=\sqrt{R^2+Z^2}$ instead of $\sqrt{R^2+\Theta^2+Z^2}$.

Now, for a 2-tensor, let's take the gradient of a vector field $\textbf{u}=(u_1,u_2,u_3)=(u_r,u_{\theta},u_z)$ as an example:

Under Cartesian coordinates, $\nabla \textbf{u}$ is $$\nabla \textbf{u}=\begin{bmatrix} \partial_{x}u_1 && \partial_{y}u_1 && \partial_{z}u_1 \\ \partial_{x}u_2 && \partial_{y}u_2 && \partial_{z}u_2 \\ \partial_{x}u_3 && \partial_{y}u_3 && \partial_{z}u_3 \end{bmatrix} $$ whose 2-norm (entry-wise Frobenius norm) is simply the square-root of the sum of squares of its entries.

However, under cylindrical coordinates, we have $$\nabla \textbf{u}=\begin{bmatrix} \partial_{r}u_r && \frac{1}{r}(\partial_{\theta}u_r-u_{\theta}) && \partial_{z}u_r \\ \partial_{r}u_{\theta} && \frac{1}{r}(\partial_{\theta}u_{\theta}+u_{r}) && \partial_{z}u_{\theta} \\ \partial_{r}u_z && \frac{1}{r}\partial_{\theta}u_z && \partial_{z}u_z \end{bmatrix} $$ My question is, is the 2-norm (the norm equivalent to the 2-norm under Cartesian coordinates) of it still the square-root of the sum of squares of its entries in the case of cylindrical coordinates? I feel that this is not the case just like the analoge of the 2-norm of a 1D tensor $\textbf{v}$ shown above.

user31899
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1 Answers1

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You need to be careful about the distinction between points and vectors. While a point is described by cylindrical coordinates $(r,\theta,z)$ and has distance $\sqrt{r^2 + z^2}$ from the origin, a vector $v$ based at a point is described by Cartesian coordinates locally aligned to the cylindrical coordinates - i.e. $v = v^r e_r + v^\theta e_\theta + v^z e_z$, where the basis $e_r,e_\theta,e_z$ depends on the base point - it's the orthonormal basis at the point aligned to the corresponding coordinate lines.

Here's a decent picture from this Wikipedia page:

enter image description here

Thus the squared norm of such a vector is indeed defined by the sum of squares of its components.

Anthony Carapetis
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  • Thanks. But we have $|v|=\sqrt{v_r^2+v_z^2}$ instead of $\sqrt{v_r^2+v_{\theta}^2+v_z^2}$, which differs by a squared $\theta$ component. So it is it not always sum of squares of all the components in the case of cylindrical coordinates? – user31899 Jul 15 '16 at 01:35
  • @user31899: that's simply not true. Do you understand the distinction I'm trying to highlight in this answer? You're treating $v$ like a point, not a vector. The distinction is essential when working with non-linear coordinates. – Anthony Carapetis Jul 15 '16 at 01:37
  • Yeah I see the distinction now. But how to express $|\nabla \textbf{u}|$ in terms of $r,\theta,z$ given $\textbf{u}$ being a vector field ? Is it still the sqaure-root of the sum of all the $9$ entries in the matrix representation of $|\nabla \textbf{u}|$ under cylindrical coordinates? – user31899 Jul 15 '16 at 01:43
  • @user31899: yes, that's right. You should be able to verify this by making substitutions using the multivariable chain rule. – Anthony Carapetis Jul 15 '16 at 01:49