1
$\begingroup$

I am playing around with calculating a line element for cylindrical coordinates. So I tried this in two different ways.

First, I took the position vector to be $$\vec{r} = (x^2+y^2)^{\frac{1}{2}}\hat{r} + tan^{-1}(\frac{y}{x})\hat{\phi} + z\hat{z}.$$

Then, I took the position vector to be $$\vec{r} = rcos\phi \hat{x} + rsin\phi \hat{y} + z\hat{z}.$$

For both of these processes, I used $$e_u = \frac{\partial \vec{r}}{\partial u}$$ to construct the basis vectors where in the first case, $u = x,y,z$ and in the second case, $u=r,\phi ,z$. However, when I go to construct the line element by taking the respective dot products of these basis vectors, I do not get the same result. Since the line element must be invariant, I believe I should get metrics that at least have the same diagonal elements, but in this method, only the position vector $$r = rcos\phi \hat{x} + rsin\phi \hat{y} + z\hat{z}$$ provides the proper diagonal elements on the metric. For the other position vector, $\frac{\partial \vec{r}}{\partial r}$ dotted with itself is not one. What is going wrong here?

$\endgroup$

1 Answer 1

1
$\begingroup$

The error is with the $\phi\hat{\phi}$ term. The position vector in cylindrical/spherical coordinates is NOT $\vec{r}=r\hat{r}+\phi\hat{\phi}+z\hat{z}$! This is obvious once you write down the definition of $\hat{r},\hat{\phi},\hat{z}$ in terms of $\hat{x},\hat{y},\hat{z}$. For example, \begin{align} \hat{r}=\frac{x\hat{x}+y\hat{y}}{r}. \end{align} So, \begin{align} \vec{r}:=x\hat{x}+y\hat{y}+z\hat{z}=r\hat{r}+z\hat{z}. \end{align}


To reiterate: just because $\vec{r}=\sum_{i=1}^3x^ie_i$ is the definition of the position vector in cartesian coordinates, it DOES NOT mean that if you consider some other coordinate system $(\xi^1,\xi^2,\xi^3)$, with corresponding unit vectors $\{e_{\xi,1},e_{\xi,2},e_{\xi,3}\}$ (the precise definition being $e_{\xi,i}$ being the normalized version of the tangent vector field $\frac{\partial}{\partial \xi^i}$), then the position vector is $\vec{r}=\sum_{i=1}^3\xi^i\,e_{\xi,i}$.

$\endgroup$
1
  • $\begingroup$ ah got it, thanks ! $\endgroup$ Commented Jun 24, 2022 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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