What is the gradient in polar coordinates?

Click For Summary
The discussion focuses on the representation of the gradient in polar coordinates, specifically seeking clarification on the first-order Laplacian operator. It is emphasized that the Laplacian is a second-order differential operator, and there is no first-order Laplacian. The correct first-order differential operators include the gradient, divergence, and curl, which are denoted by symbols like ∇, ∇·, and ∇×. Caution is advised when applying these operators in curvilinear coordinates, as the expressions differ from those in Cartesian coordinates. For accurate representations, users are encouraged to consult introductory vector analysis textbooks or resources on curvilinear coordinates.
SeM
Hi, on this page: https://en.wikipedia.org/wiki/Laplace_operator#Two_dimensions

the Laplacian is given for polar coordinates, however this is only for the second order derivative, also described as \delta f . Can someone point me to how to represent the first-order Laplacian operator in polar coordinates? I found this at https://math.stackexchange.com/questions/586848/how-to-obtain-the-gradient-in-polar-coordinates, however, I am not sure its correct:

\nabla = \boldsymbol{e_r} \frac{\partial}{\partial r}+ \boldsymbol{e_{\theta}} \frac 1r \frac{\partial}{\partial \theta}.

Thanks!
 
Science news on Phys.org
SeM said:
the first-order Laplacian operator
The Laplacian by definition is a second order differential operator. It is defined as the divergence of the gradient of a scalar field. There is no such thing as a first order Laplacian.

There are three first order differential operators you will encounter in basic vector analysis, the gradient, the divergence and the curl. You will often see them written as ##\nabla##, ##\nabla \cdot##, and ##\nabla \times##. However, you should be very careful here. The reason for expressing them like this is that it becomes correct to schematically write ##\nabla = \vec e_1 \partial_1 + \vec e_2 \partial_2 + \vec e_3 \partial_3## and apply the ##\cdot## and ##\times## as scalar and cross products with the derivatives acting on what is to the right. This only works in Cartesian coordinates. In curvilinear coordinates you will have to find the appropriate expressions and they will not involve writing ##\nabla## as a vector and blindly applying the scalar and cross products.

See https://en.wikipedia.org/wiki/Curvilinear_coordinates#Differentiation for the correct expressions in general coordinates. This is also something you should be able to find in any introductory textbook on vector analysis.

Please also be aware that the "A" tag you put on this thread would imply that you have a graduate level understanding of the subject and expect an answer at that level. Based on the content of the question, the thread level is at most an "I" and I have changed it accordingly.
 
SeM said:
\nabla = \boldsymbol{e_r} \frac{\partial}{\partial r}+ \boldsymbol{e_{\theta}} \frac 1r \frac{\partial}{\partial \theta}.
By the way, this forum supports ##\LaTeX## but you need to put double-$ signs before and after the code:$$
\nabla = \boldsymbol{e_r} \frac{\partial}{\partial r}+ \boldsymbol{e_{\theta}} \frac 1r \frac{\partial}{\partial \theta}.
$$
 
Thanks!
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 6 ·
Replies
6
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 41 ·
2
Replies
41
Views
4K