TITLE: Gradient of a Scalar Field is Perpendicular to Equipotential Surface
QUESTION [7 upvotes]: let $\phi = f(x,y,z)$ be a scalar field, is gradient of phi independent of coordinate choice?
Also how to prove that $\nabla \phi$ is perpendicular to scalar potential surface.

REPLY [6 votes]: A simple way to understand this in an arbitrary dimension, is the following (you can make it rigorous if you work a little, but I doubt that there's a way to avoid the implicit function theorem). 
The directional derivative of $\phi$ at the point $\vec r$ and in a direction $\vec u$ is given by $\vec u\cdot \vec \nabla\phi$. It is the derivative of $\phi(\gamma(t))|_{t=0}$ if $\gamma(t)$ is a curve such that $\gamma(0)=\vec r$ and $\dot \gamma(0)=\left.\frac{d\gamma}{dt}\right|_{t=0}=\vec u$. If $\gamma$ is contained in the equipotential surface, then clearly $\dot \phi=0$ identically, and in particular the gradient is perpendicular to all $\vec u$'s which are tangent to the equipotential surface. 
To answer your question about coordinate independence, the gradient is a vector. As such, it is a geometrical entity that does not depend on the choice of coordinates. However, its coordinate representation clearly does depend on the choice of coordinates, but in a simple manner: a rotation of the coordinates will result in a rotation of the vector of components.

REPLY [4 votes]: Here is an intuitive explanation. Let $p$ be any point. We want to show that $\nabla \phi(p)$ is perpendicular to the surface $\phi(x)=\phi(p)$ which passes through $p$. Let $\vec v$ be any unit vector at $p$. By the chain rule, the directional derivative of $\phi$ in the direction $\vec v$ at $p$ is 
$$\vec v \cdot \nabla \phi(p).$$
If $\vec v$ is tangent to the surface $\phi(x)=\phi(p)$, the directional derivative is $0$ because $\varphi$ is constant on the surface $\phi(x)=\phi(p)$. Hence this means $\vec v$ is perpendicular to $\nabla \phi(p)$.