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Alfred Centauri
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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

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

Now, polar coordinates are defined by

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

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

Thus

$$x = r \cos \theta$$

$$y = r \sin \theta$$

The polar coordinate basis vectors are then

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

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

and 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

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

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

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

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.

Alfred Centauri
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