Change of basis in a Euclidean space

Question

I am trying to compute the change in the contravariant components of a vector when the basis is changed from Cartesian (standard basis) to spherical polars.

I understand that a general vector $\mathbf{A}$ can be expressed in either basis:

$$\mathbf{A} = A^x \mathbf{e_x} + A^y \mathbf{e_y} + A^z \mathbf{e_y} = A^r \mathbf{e_r} + A^{\theta} \mathbf{e_{\theta}} +A^{\phi} \mathbf{e_{\phi}}$$

Now, the contravariant vector components transform as

$$ A^{b'} = \frac{\partial x^{b'}}{\partial x^c} A^c$$

and so e.g.

$$ A^r = \frac{\partial r}{\partial x} A^x + \frac{\partial r}{\partial y} A^y + \frac{\partial r}{\partial z} A^z$$

My confusion arises in determining the derivatives. In a curved spacetime, a vector exists in the tangent plane to a point. This point is labeled by some coordinates and so when we calculate e.g. $\frac{\partial r}{\partial x} = \sin \theta \cos \theta$ we can plug in the coordinates $\theta, \phi$ and hey presto. But in a euclidean space, the tangent space is the manifold. It doses not seem to make sense that in order to transform between coordinate systems we need both the 'location' of the vector at the coordinates, and the components of the vector itself?

Any guidance greatly appreciated.

Answer 1

... in a euclidean space, the tangent space is the manifold.

I think this is where your confusion lies. A flat Euclidean manifold $M$ still has a distinct tangent space $T_P$ at each point $P$, and a vector at $P$ lives in the tangent space $T_P$. Each tangent space happens to resemble $M$ iself but this is just coincidence. The vectors in a vector space on $M$ may be informally represented as if they lived in $M$ itself, but this is a simplified and somewhat misleading picture.

Answer 2

I am a bit lost now, why would $\frac{\partial r}{\partial x} = \sin \theta \cos \phi$? Isn't $x = r \sin \theta \cos \phi$? The way I understand it I would rather want $\frac{\partial x}{\partial r} = \sin \theta \cos \phi$. What am I misunderstanding?