Why no basis vector in Newtonian gravitational vector field? In my textbook, the gravitational field is given by$$\mathbf{g}\left(\mathbf{r}\right)=-G\frac{M}{\left|\mathbf{r}\right|^{2}}e_{r}$$
  which is a vector field. On the same page, it is also given as a three dimensional gradient$$\mathbf{g}=-\mathbf{\nabla\phi}=-\left(\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y},\frac{\partial\phi}{\partial z}\right)$$
As this second equation is also a vector field, why doesn't it contain a basis vector of some sort and why isn't $\mathbf{g}$ given as a function of something or other? 
Also, how do you actually get from the gravitational potential field $$\phi=\frac{-Gm}{r}$$
to the second equation? I can see that you apply the operator $\nabla$ but how does that give you $\mathbf{g}$? 
Thank you
 A: We want to compute the gradient of
$$ \phi(r) = \phi(<x,y,z>) =\frac{-Gm}{|r|}=\frac{-Gm}{\sqrt{x^{2}+y^{2}+z^{2}}} $$
It is:
$$ \left<\frac{\partial}{\partial x} , \frac{\partial}{\partial y} , \frac{\partial}{\partial z}\right> \phi(<x, y, z>) = $$
$$ = Gm \left< \frac{x}{(x^2+y^2+z^2)^{3/2}} , \frac{y}{(x^2+y^2+z^2)^{3/2}} , \frac{z}{(x^2+y^2+z^2)^{3/2}} \right> = $$
Using $ (x^2+y^2+z^2)^{3/2} = |r|^3 $, we get
$$ = \frac{Gm}{|r|^3} \left< x, y, z \right> = $$
By definition $r=<x,y,z>=|r| e_r$
$$ = \frac{Gm}{|r|^2} e_r = $$
$$ = -g(r) $$
Note that $r$ is a vector, $e_r$ is a vector, $g(r)$ is a vectorfield (maps a vector r to a vector), $\phi(r)$ is a scalarfield (maps a vector r to a scalar).
A: To add to @mtrecseni's derivation, the definition of scalar potential of a field $\mathbf{E}$ is a scalar field $\phi$ such that $\mathbf{E}=-\nabla{\phi}$.
We can easily derive this from the fact that the potential is a path integral (a path-independant path integral actually) $$\phi=-\int\limits_{(x_1,y_1,z_1)}^{(x_2,y_2,z_2)}\mathbf{E}\cdot \vec{dl}$$
We can write the path integral as three integrals, over x, y, z:
$$\phi=-\int\limits_{x_1}^{x_2}E_x(x,y_1,z_1).dx-\int\limits_{y_1}^{y_2}E_y(x_2,y,z_1).dy-\int\limits_{z_1}^{z_2}E_z(x_2,y_2,z_2).dz$$
Basically we have first taken the particle from $x_1\to x_2$, keeing $y,z$ constant, and so on. $E_x$ is the x-component of $\mathbf{E}$.
Taking $\nabla$,since y and z are constant in the first integral and so on, the equation reduces to $\nabla\phi=-(E_x\hat{i}+E_y\hat{j}+E_z\hat{k})\implies \mathbf{E}=-\nabla\phi$
This derivation works for any conservative field. 
