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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 $f$?

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  • $\begingroup$ Why is this too good to be true? And what EXACTLY is your question? $\endgroup$
    – SchrodingersCat
    Commented Feb 28, 2016 at 16:31
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    $\begingroup$ 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. $\endgroup$
    – user197848
    Commented Feb 28, 2016 at 16:35
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    $\begingroup$ Your result is correct. $\endgroup$
    – SchrodingersCat
    Commented Feb 28, 2016 at 16:40

2 Answers 2

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"...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.

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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

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  • $\begingroup$ I have seen it both ways in textbooks, I understand what you are saying though. $\endgroup$
    – user197848
    Commented Feb 28, 2016 at 16:51
  • $\begingroup$ I think it is better to go with safer notation :) ...so no prof will cry :p $\endgroup$
    – Debajyoti Datta
    Commented Feb 28, 2016 at 16:59
  • $\begingroup$ but I have already given an answer and when the author said that he has seen both notations in texts then I told him that it was better to go with safer notation $\endgroup$
    – Debajyoti Datta
    Commented Feb 28, 2016 at 17:29
  • $\begingroup$ @DebajyotiDatta Welcome to the math.SE! The above comment comes from the review portal which suggests fixed statements to certain situations like yours. If you just want to tell someone that his/her answer is correct (also with minor critique) then please do it as a comment - you can easily extend your half answer to a complete one while adding some context and using TeX $\endgroup$
    – user190080
    Commented Feb 28, 2016 at 17:57

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