# The directional derivative in cylindrical coordinates.

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 $f$?

• Why is this too good to be true? And what EXACTLY is your question? Commented Feb 28, 2016 at 16:31
• During my attempts at deriving the gradient operator I made the big mistake of ignoring the $\theta$ dependence of my radial basis basis vector, I thought there would be something I would have overlooked and I tried to google to check it and nothing come up. I thought the question was pretty clear, I have updated to make things clearer.
– user197848
Commented Feb 28, 2016 at 16:35
• Your result is correct. Commented Feb 28, 2016 at 16:40

"...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: