differential geometry - What coordinate systems allows the magnitude of the basis vectors to change with position?

I'm familiar with coordinate systems where the direction of the basis vectors changes with position, but I haven't come across any where the relative magnitude of the basis vectors themselves are allowed to change also.

Answer

The only orthonormal coordinate basis is the Cartesian coordinate basis. The basis vectors for the, e.g., polar coordinate basis are orthogonal but not normalized.

That doesn't mean that one can't normalize the polar basic vectors to get the polar unit basis but such a basis isn't a coordinate basis.

For the Cartesian coordinate basis, the basis vectors are orthonormal:

$$\vec e_x \cdot \vec e_x = g_{xx} = 1$$

$$\vec e_y \cdot \vec e_y = g_{yy} =1$$

$$\vec e_x \cdot \vec e_y = g_{xy} = g_{yx}= 0$$

and the line element is

$$dl^2 = g_{xx}dx^2 + g_{yy}dy^2 + 2g_{xy}dxdy = dx^2 + dy^2$$

Now, polar coordinates are defined by

$$r = \sqrt{x^2 + y^2}$$

$$\theta = \tan^{-1}\left(\frac{y}{x}\right)$$

thus

$$x = r \cos \theta$$

$$y = r \sin \theta$$

and the polar coordinate basis vectors are then

$$\vec e_r = \frac{\partial x}{\partial r}\vec e_x + \frac{\partial y}{\partial r}\vec e_y = \cos \theta \; \vec e_x + \sin \theta \; \vec e_y$$

$$\vec e_\theta = \frac{\partial x}{\partial \theta}\vec e_x + \frac{\partial y}{\partial \theta}\vec e_y = -r \sin \theta \; \vec e_x + r \cos \theta \; \vec e_y$$

so

$$\vec e_r \cdot \vec e_r = g_{rr}=1$$

$$\vec e_\theta \cdot \vec e_\theta = g_{\theta \theta}= r^2$$

$$\vec e_r \cdot \vec e_\theta = g_{r\theta} = g_{\theta r} = 0$$

and the line element is

$$ds^2 = g_{rr}dr^2 + g_{\theta \theta}d\theta^2 + 2g_{r\theta}drd\theta = dr^2 + r^2d\theta^2$$

Finally, we ask if coordinates $\{\hat r, \hat \theta\}$ can be found for the unit polar basis such that

$$\vec e_{\hat r} = \vec e_r = \frac{\partial x}{\partial \hat r}\vec e_x + \frac{\partial y}{\partial \hat r}\vec e_y = \cos \theta \; \vec e_x + \sin \theta \; \vec e_y$$

$$\vec e_\hat \theta = \frac{1}{r}\vec e_\theta = \frac{\partial x}{\partial \hat \theta}\vec e_x + \frac{\partial y}{\partial \hat \theta}\vec e_y = -\sin \theta \; \vec e_x + \cos \theta \; \vec e_y$$

If there are coordinates $\{\hat r, \hat \theta\}$, then

$$\frac{\partial^2 x}{\partial \hat r \partial \hat \theta} = \frac{\partial^2 x}{\partial \hat \theta \partial \hat r }$$

$$\frac{\partial^2 y}{\partial \hat r \partial \hat \theta} = \frac{\partial^2 y}{\partial \hat \theta \partial \hat r }$$

but

$$\frac{\partial^2 x}{\partial \hat r \partial \hat \theta} = \frac{\partial}{\partial \hat r} (-\sin \theta) = \frac{\partial}{\partial \hat r}\left(-\frac{y}{r}\right) = \frac{\partial}{\partial r}\left(-\frac{y}{r}\right) = \frac{y}{r^2}$$

$$\frac{\partial^2 x}{\partial \hat \theta \partial \hat r} = \frac{\partial}{\partial \hat \theta} (\cos \theta) \ne \frac{y}{r^2}$$

and thus coordinates ${\hat r, \hat \theta}$ do not exist; the unit polar basis is not a coordinate basis.

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To better see this, consider the level curves of the polar coordinate system:

The concentric circles represent the basis one-form $\tilde dr$ dual to the $r$ basis vector $\vec e_r$. Note that the spacing of the circles is constant which means that magnitude of $\tilde dr$ is constant.

The radial lines represent the basis one-form $\tilde d\theta$ dual to the $\theta$ basis vector $\vec e_\theta$. Note that as the $r$ coordinate increases, the spacing between the radial lines increases or, put another way, the density goes as $\frac{1}{r}$ thus the magnitude of $\tilde d\theta$ is not constant and, in fact, is just $\frac{1}{r}$

$$\tilde d\theta \cdot \tilde d\theta = \frac{1}{r^2}$$

But since

$$\langle\tilde d\theta, \vec e_\theta\rangle = 1$$

it follows that

$$\vec e_\theta \cdot \vec e_\theta = r^2$$

Now it's easy to 'see' why the unit polar basis is not a coordinate basis; if

$$\vec e_\hat \theta \cdot \vec e_\hat \theta = 1$$

then the radial lines (lines of constant $\hat \theta$) must have constant density but radial lines cannot have constant density.