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General curvilinear coordinates

  • Thread starter Benny
  • Start date
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1. Homework Statement

When I work in general curvilinear coordinates and in particular for the computation of line and surface integrals, do I need to do anything apart from working through the 'usual steps?'

2. Homework Equations

If I am correct, computation of line and surface integrals is typically introduced in cartesian coordinates. In such a coordinate system, some basic parameterisations include those of for a circle of radius r eg. (x,y) = (rcos(t),rsin(t)), 0 <= t <= 2pi.

3. The Attempt at a Solution

I know that if I work in say cylindrical or spherical coordinates the expressions for the gradient, divergence and Laplacian are different (and more complicated) than in cartesian coordinates. So do the parameterisations of surfaces and paths become more complicated if I want to compute a line integral in these non-cartesian coordinate systems? If so why would anyone want to use non-cartesian coordinate systems to compute various integrals? Is it because in certain situations, this is the only possible way to compute the integrals?

I'm interested in this because I only really got as far as going through the derivations and computations for alternative forms of the gradient, laplacian etc. I don't feel as if I really learned how to apply useful techniques such as computing integrals in geometries with cylindrical or spherical symmetry apart from trivial cases where the vector field and outward surface normal were both constant.

Any help would be great thanks.
 

AlephZero

Science Advisor
Homework Helper
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I don't fully understand what you are asking, but working in curvilinear coordinates is simpler when (and only when) the functions involved are simpler in those coordinate systems.

The equations modelling physical systems often have cylindrical or spherical symmetry, so many of the terms in the "complicated" expressions for grad etc in curvilinear coordinates may be zero for a particular function.
 
584
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I probably shouldn't have used that template. Anyway, if f(x,y) = x^2 + y^2 then in polar coordinates f(r,theta) = r^2 right? Then would div(grad(f)) = (r^2)'' = 2? What if I have something more complicated like g(x,y) = x^2 + y^2 + 2x + 2y then g(r,theta) = r^2 + 2(cos(theta) + sin(theta)). Would the calculation of div(grad(f)) in polar coordinates be done the same way as for the function f or do I need to make use of the relevant scale factors?
 
459
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Of course you have to use scale factors.
 
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