0
$\begingroup$

A scale factor in curvilinear coordinates is defined as

$$h_v \equiv \left|\frac{\partial\vec{r}}{\partial v}\right|$$

where $\vec{r}=(x,y,z)^T$ is a position vector. The partial differential can be expressed as a directional derivative:

$$\frac{\partial\vec{r}}{\partial v} = (\vec{v}\cdot\nabla)\vec{r} = v_x\frac{\partial}{\partial x}\vec{r} + v_y\frac{\partial}{\partial y}\vec{r} + v_z\frac{\partial}{\partial z}\vec{r}$$

where $\vec{v}$ is a unit vector pointing in the direction of $v$. The derivatives of $\vec{r}$ are

$$\frac{\partial}{\partial x}\vec{r}=\frac{\partial}{\partial x}\left( \begin{array}{c} x\\ y\\ z\\ \end{array} \right)= \left( \begin{array}{c} 1\\ 0\\ 0\\ \end{array} \right)$$ and similarly for $y$ and $z$. Therefore, I get:

$$\frac{\partial\vec{r}}{\partial v}=v_x\left(\begin{array}{c}1\\0\\0\\ \end{array}\right)+v_y\left(\begin{array}{c}0\\1\\0\\ \end{array}\right)+v_z\left(\begin{array}{c}0\\0\\1\\ \end{array}\right) = \left(\begin{array}{c}v_x\\v_y\\v_z\\ \end{array}\right)=\vec{v}$$

And because of this, the scale factor becomes:

$$h_v = \left|\frac{\partial\vec{r}}{\partial v}\right|=|\vec{v}|=1$$

Obviously, this is wrong. Where is my mistake?

$\endgroup$
  • $\begingroup$ Not entirely sure of your notation, but if $\nabla$ denotes the gradient operator (with respect to a metric, as suggested by the differential-geometry tag), the components of $\nabla r$ involve the metric components, not just partial derivatives. $\endgroup$ – Andrew D. Hwang Apr 30 '16 at 18:35
  • $\begingroup$ I am not sure what you mean by "metric components", but the gradient of a vector is just $\nabla\vec{r} = \frac{\partial r_i}{\partial x_j}$, isn't it? And in this case, $\frac{\partial r_i}{\partial x_j}=0$ for $i\neq j$ and 1 for $i=j$. $\endgroup$ – g3lol Apr 30 '16 at 18:46
  • $\begingroup$ If you're asking about differential geometry, in which the inner products $g(\partial_{i}, \partial_{j}) = g_{ij}$ are generally not the Kronecker symbol $\delta_{ij}$, then $(\nabla r)^{j} = g^{ij}(\partial_{i} r)$, with $[g^{ij}] = [g_{ij}]^{-1}$. $\endgroup$ – Andrew D. Hwang Apr 30 '16 at 18:54
0
$\begingroup$

Partial derivative is to parameter $v$ and $v$ is not a vector so there is no directional derivative and no $v_x$,$v_y$,$v_z$. $$\frac {\partial \vec r}{\partial v}=\frac {\partial \vec r}{\partial x}\frac {\partial x}{\partial v}+\frac {\partial \vec r}{\partial y}\frac {\partial y}{\partial v}+ \frac {\partial \vec r}{\partial z}\frac {\partial z}{\partial v}$$

Therefore $$\frac {\partial \vec r}{\partial v}=\left(\begin{array}{c}\frac {\partial x}{\partial v}\\\frac {\partial y}{\partial v}\\\frac {\partial z}{\partial v}\\\end{array} \right )$$ And $$ h_v=\sqrt{({\frac {\partial x}{\partial v})}^2+{(\frac {\partial y}{\partial v})}^2+{(\frac {\partial z}{\partial v})}^2}$$

$\endgroup$

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.