Spherical Coordinates and Unit Vectors can you help me on understanding the unit vectors r, φ, θ?
I see in this picture the classical angles φ and θ of the spherical coordinates. Until now it is all clear. 
But I do not understand the vectors with the same names. Obviously r is taken in the direction of r, but φ and θ? How are they defined with respect to the angles (or with respect to x, y, z)?

 A: When I have a fixed coordinate system I may define unit vectors corresponding to this system.  The way we do so is by taking the derivative in the direction of each of these coordinates.  What you get is as every point in the space we are studying is a set of vectors, each of which points in the direction of coordinate increase. So in a Cartesian system for 3 dimension, at every point in space we have a constant set of 3 unit vectors ($\hat{i}, \hat{j}, \hat{k}$) because the direction of the x, y and z increasing is always the same; up for z, and in the positive direction of x and y.  However as you know Cartesian Coordinates are just one of many possible choices.  So in Spherical to see how $\hat{r}, \hat{\theta}, \hat{\phi}$ vary with position just ask yourself in which direction do I have to walk to increase $r,\theta , \phi$ and you get your answer.  The relation between the vectors and derivatives does require some discussion of Differential Geometry so I will hold off on that unless you are curious.  I hope this helps. 
Final remark: the idea of defining unit vectors for a given coordinate system in this way always works, try it with Cylindrical and Polar and maybe Hyperbolic.  Remember coordinate systems are arbitrary and are typically chosen to simplify calculations.
Take Care 
A: These vectors show the direction of infinitesimal displacements when you change one coordinate at a time.
$\hat r$ is in the radial direction; $\hat \theta$ is tangent to a parallel and $\hat \phi$ to a meridian.

You obtain them by differentiating $(x,y,z)$ on one coordinate and normalizing.
E.g. for $\theta$
$$\frac\partial{\partial\theta}(x,y,z)=
\frac\partial{\partial\theta}(r\cos\theta\sin\phi,r\sin\theta\sin\phi,r\cos\phi)=(-r\sin\theta\sin\phi,r\cos\theta\sin\phi,0)$$
and after normalization,
$$(-\sin\theta,\cos\theta,0).$$
A: Let's start with the 2D case. Starting at $(r,\,\theta)$, I can get to $(r+dr,\,\theta)$ by moving (in Cartesian terms) along $dr\left(\cos\theta\hat{x}+\sin\theta\hat{y}\right)$, so we call this $dr$ coefficient $\hat{r}$. You can by the same logic get $\hat{\theta}=-\sin\theta\hat{x}+\cos\theta\hat{y}$, which you'll notice is orthogonal to $\hat{r}$; one is radial, the other tangential. That makes sense, because these are just "complicated" variants on the "obvious" concepts of $\hat{x},\,\hat{y}$, and as in the Cartesian case we want orthogonal unit vectors. The 3D case is just the same principle, so $\hat{r}$, $\hat{\theta}$ and $\hat{\phi}$ end up being mutually orthogonal.
