$\nabla$ in curvilinear coordinates

250 Views Asked by At

I am trying to understand how the $\nabla$ operator changes whent the coordinates change (note that I have read other similar questions here and don't really understand what is happening):

In $x,y,z$ coordinates of course $\nabla f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})$

Why then in cylindrical coordinates we have $\nabla f=(\frac{\partial f}{\partial ρ},\frac{1}{ρ}\frac{\partial f}{\partial θ},\frac{\partial f}{\partial z})$?

Ι really can't understand why the $\frac{1}{ρ}$ infront of $\frac{\partial f}{\partial θ}$.

I am familiary with the the chain rule and how to write $\frac{\partial}{\partial x^i}$ in other coordinates but this doesn't seem to help me.

Question: Can you explain how to go from the nabla from one set of coordinates to another? In particular from cartesian to cylindrical or to spherical.

1

There are 1 best solutions below

2
On BEST ANSWER

As this is tagged with differential-geometry I'll follow a more geometric approach, if this doesn't suit, I am happy to tag this answer as community-wiki.

The computation you seek follows from the general definition of $\nabla$ on a manifold, $M$.

Thence \begin{align} \langle\nabla f(p)|v\rangle&=d_pf(v)\\ &=\sum_i\left.\frac{\partial f}{\partial x^i}\right|_pdx^i(v) \end{align} Here, $p$ is a point in some manifold $M$, and $v \in T_p M$.

Summation here is over basis vectors of the tangent space. Expand this in order to find component $i$ $$ (\nabla f)_i=\frac{1}{k_i}\frac{\partial f}{\partial x^i} $$ Where $k_i$ is the modulus of the $i$th tangent vector.