Is del defined in non-cartesian coordinates

Question

I am trying to understand a list of differential operator identities in different coordinate systems. I already understand how they work in Cartesian coordinates. I will try my best to ask the question systematically:

• In cylindrical coordinates, the gradient of a scalar function $f$ is given by $\nabla f=\frac{\partial f}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial f}{\partial\theta}\hat{\theta}+\frac{\partial f}{\partial z}\hat{z}$
• The above point would seem to imply that I can define an operator called 'del' which is given by $\nabla=\frac{\partial}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial}{\partial\theta}\hat{\theta}+\frac{\partial}{\partial z}\hat{z}$
• If I then want to calculate the divergence of a vector function $\mathbf{F}$, I take the dot product and obtain $\nabla\cdot\mathbf{F}=\frac{\partial F_r}{\partial r}+\frac{1}{r}\frac{\partial F_{\theta}}{\partial\theta}+\frac{\partial F_z}{\partial z}$
• This final point is incorrect, according to the books I have checked. Apparently the divergence is instead given by $\nabla\cdot\mathbf{F}=\frac{1}{r}\frac{\partial (r F_r)}{\partial r}+\frac{1}{r}\frac{\partial F_{\theta}}{\partial\theta}+\frac{\partial F_z}{\partial z}$

My question is: why does this discrepancy arise? If somebody said "the form you gave in your fourth point is indeed the divergence, but in non-Cartesian coordinates it cannot be written in the form of a dot product between del and the vector function", then that would make sense to me, but actually I have seen this dot-product written in a textbook, where $div(\mathbf{F})$ and $\nabla\cdot\mathbf{F}$ are used interchangeably, regardless of the coordinate system. It is also written in this way on wikipedia.

Pre-empting the obvious objection: I am not using the above points to "derive del from the gradient". I am just trying to show that this form of del doesn't seem to give the correct result for the gradient and the divergence simultaneously.

As a final note I am a physics student not a mathematician, so it would be really appreciated if the answer was posed at the difficulty level of the question (or below). Thanks.

Answer 1

Th problem is in your third point.What actually happens is that in cartesian coordinates,we dot $i,j,k$ by $i,j,k$ respectively to get coefficient 1 and are 'constant'(they do not change according to the point)so we 'multiply' term wise and add. This is not the case in general.

The method you mentioned in your third point only works for some coordinate system not all;they are the exception not the general rule.

Answer 2

Consider the expansion of the vector in cylindrical coordinates $${\mathbf F} = F_r\,{\mathbf e}_r +F_\theta\,{\mathbf e}_\theta +F_z\,{\mathbf e}_z$$ When you apply a differential operator (grad, div, curl) you must differentiate the entire expression, the basis vectors as well as the coordinate functions.

For example $$\frac{\partial {\mathbf e}_r}{\partial \theta}={\mathbf e}_\theta$$ This is different from the situation for the Cartesian base vectors, all of whose derivatives are zero.

After accounting for the changing basis vectors, you will obtain valid results.