Differential operators in 2D curvilinear coordinates

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Homework Help Overview

The discussion revolves around the calculation of differential operators in the context of orthogonal curvilinear coordinates, specifically in a 2D setting. Participants express confusion regarding the application of these operators, particularly the curl and Laplace, given the limitations of a two-dimensional coordinate system.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of Lamé coefficients and express uncertainty about how to handle the curl in a 2D context. Questions arise about the necessity of introducing a third dimension and the implications for the scale factor.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the use of existing formulas, but there remains a lack of consensus on how to proceed with the calculations and the meaning of the operators involved.

Contextual Notes

Participants note the challenge of calculating the curl in a two-dimensional space, questioning whether a third coordinate must be introduced and how this affects the scale factor. There is also a mention of the specific nature of the curl's components in relation to the vector field.

Robin04
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Homework Statement


I’m studying orthogonal curvilinear coordinates and practice calculating differential operators.
However, I’ve run across an exercise where the coordinate system is only in 2D and I’m confused about how to proceed with the calculations.

Homework Equations


A point in the plane is given by ##u## and ##v## coordinates where ##u=\frac{\sqrt{x^2+y^2}+y}{2}## and ##v=\frac{\sqrt{x^2+y^2}-y}{2}##
I’m using these formulas to calculate the operators: https://en.m.wikipedia.org/wiki/Orthogonal_coordinates#Differential_operators_in_three_dimensions

The Attempt at a Solution


I’ve calculated the Lamé coefficients (scale factors) and got the following: ##h_u=\sqrt{\frac{v}{u}+1}##, ##h_v=\sqrt{\frac{u}{v}+1}##

In case of the gradient and the divergence I simply just didn’t calculate the third term but I don’t see how to do this with the curl and Laplace.
 
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Robin04 said:

Homework Statement


I’m studying orthogonal curvilinear coordinates and practice calculating differential operators.
However, I’ve run across an exercise where the coordinate system is only in 2D and I’m confused about how to proceed with the calculations.

Homework Equations


A point in the plane is given by ##u## and ##v## coordinates where ##u=\frac{\sqrt{x^2+y^2}+y}{2}## and ##v=\frac{\sqrt{x^2+y^2}-y}{2}##
I’m using these formulas to calculate the operators: https://en.m.wikipedia.org/wiki/Orthogonal_coordinates#Differential_operators_in_three_dimensions

The Attempt at a Solution


I’ve calculated the Lamé coefficients (scale factors) and got the following: ##h_u=\sqrt{\frac{v}{u}+1}##, ##h_v=\sqrt{\frac{u}{v}+1}##

In case of the gradient and the divergence I simply just didn’t calculate the third term but I don’t see how to do this with the curl and Laplace.

For the curl, you need a third dimension, so you could use coordinates ##(x,y,z)## or ##(u,v,z)## and a vector field of the form ##\mathbf{A} = A(x,y) \mathbf{i} + B(x,y) \mathbf{j} + 0 \mathbf{k}##.
 
Ray Vickson said:
For the curl, you need a third dimension, so you could use coordinates ##(x,y,z)## or ##(u,v,z)## and a vector field of the form ##\mathbf{A} = A(x,y) \mathbf{i} + B(x,y) \mathbf{j} + 0 \mathbf{k}##.

And what should I do with the scale factor? Is it 1 in this case?
 
Robin04 said:
And what should I do with the scale factor? Is it 1 in this case?

The link you cite provides all the information you need; just use the formulas given there.
 
Ray Vickson said:
The link you cite provides all the information you need; just use the formulas given there.
I think I'm missing something here. So the curl only exists in three dimensions therefore I must introduce arbitrarily a third coordinate?
 
Robin04 said:
I think I'm missing something here. So the curl only exists in three dimensions therefore I must introduce arbitrarily a third coordinate?

I thought that is what I said in #2.
 
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Ray Vickson said:
I thought that is what I said in #2.

Aah, I see. And does it make sense that only the Z component of the curl is non zero? I’m still struggling to understand what these operators really mean.
 
Robin04 said:
Aah, I see. And does it make sense that only the Z component of the curl is non zero? I’m still struggling to understand what these operators really mean.

For any vector field ##\mathbf{A}(x,y,z)## the curl of ##\mathbf{A}## is pependicular to ##\mathbf{A}##, so if the field lies in the ##(x,y)-##plane the curl points along the ##z## direction.
 
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Ray Vickson said:
For any vector field ##\mathbf{A}(x,y,z)## the curl of ##\mathbf{A}## is pependicular to ##\mathbf{A}##, so if the field lies in the ##(x,y)-##plane the curl points along the ##z## direction.

Thank you very much for your help! :)
 

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