The directional derivative in cylindrical coordinates.

Question

I found the gradient operator in cylindrical coordinates to be

$$\nabla f = \frac{\partial f}{\partial r} \vec{e_r} + \frac{1}{r}\frac{\partial f}{\partial \theta} \vec{e_{\theta}} + \frac{\partial f}{\partial z} \vec{e_z} $$

Is it as easy as defining

$\vec u = u_r\vec{e_r} + u_{\theta}\vec{e_{\theta}} + u_z \vec{e_{z}}$

then taking the dot product and noting that our basis is an orthogonal set to obtain

$$(\vec{u} \cdot \nabla) f = u_r \frac{\partial f}{\partial r} + u_{\theta} \frac{1}{r}\frac{\partial f}{\partial \theta} + u_z \frac{\partial f}{\partial z} $$?

I feel like this is too good to be true. So my question is, Is this the correct expression for the directional derivative of a scalar field $f$?

Answer 1

your final result is correct but the left side is little problematic........you should write ....(grad f) dot product with (u vector)...as del is not a vector itself ...its a vector differential operator...experts will correct me if I am wrong

Answer 2

"...I feel like this is too good to be true". Your question is perfectly legit since changes of coordinates are always tricky for differential operators. But luckily there is a general and well known theory, dubbed "curvilinear coordinates".

The answer is: YES, it is correct, but for a scalar field only. Do not try to extend naively this formula to higher-rank fields. Example: the directional derivative of a vector field in cylindrical coordinates is much more complex than this. Take a look at:

https://link.springer.com/content/pdf/bbm%3A978-3-0348-8579-9%2F1.pdf

The general topic of your question is how "orthogonal curvilinear coordinates" work: this includes the standard cartesian, cylindrical and spherical coordinates.