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?

  • $\begingroup$ 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

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 $$ h_v=\sqrt{({\frac {\partial x}{\partial v})}^2+{(\frac {\partial y}{\partial v})}^2+{(\frac {\partial z}{\partial v})}^2}$$


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