Laplacian in toroidal coordinates

In summary, the Laplacian operator in toroidal coordinates can be found by using the gradient operator and taking the dot product with itself. The basis vectors in the gradient operator may not be constant in space, so it is important to differentiate them when computing the Laplacian. In some areas of study, it may be more beneficial to use non-orthogonal coordinates based on the natural occurrence of the problem, rather than seeking orthogonality.
  • #1
Einj
470
59
Hi everyone,
I would like to write the Laplacian operator in toroidal coordinate given by:
$$
\begin{cases}
x=(R+r\cos\phi)\cos\theta \\
y=(R+r\cos\phi)\sin\theta \\
z=r\sin\phi
\end{cases}
$$
where [itex]r[/itex] and [itex]R[/itex] are fixed.

How do I do?

More generally how do I find the Laplacian under a generic change of variables?

Thanks
 
Physics news on Phys.org
  • #2
1. For more general coordinates, one of the less obvious features we want to ensure is that the new basis vectors are orthogonal, not just linearly independent of each other. It can be quite difficult to find such a base!
See:
https://en.wikipedia.org/wiki/Toroidal_coordinates
2. In general, the gradient operator in an arbitrary coordinate system will have a scaling factor dependent on the variables, connected to each partial derivative.
For example, we have the spherical gradient operator written as:
[tex]\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta} \frac{\partial}{r\sin\phi\partial{\theta}} +\vec{i}_{\phi}\frac{\partial}{r\partial\phi}[/tex]
Note that the "denominators" here represents the typical arc-lengths associated with a change in the relevant variable. That is a general feature with the gradient operator in any coordinate system.

The Laplacian operator can be seen as the dot product between the gradient operator with itself, but it is critical then to also differentiate the basis vectors, prior to the "dotting process"
 
  • #3
Ok, I get that. I basically write down the metric tensor and so I get my denominators. Now I am a bit confused on how to go from the gradient to the laplacian. In particular, since the basis vectors are no more constant in space I am confused on how to derivate them when we compute [itex]\nabla\cdot\nabla[/itex].
 
  • #4
Einj said:
Ok, I get that. I basically write down the metric tensor and so I get my denominators. Now I am a bit confused on how to go from the gradient to the laplacian. In particular, since the basis vectors are no more constant in space I am confused on how to derivate them when we compute [itex]\nabla\cdot\nabla[/itex].

Here's an example with cylindrical coordinates:
We may split up this in three terms to be summed:
[tex]\vec{i}_{r}\frac{\partial}{\partial{r}}\cdot{(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta} \frac{\partial}{r\partial{theta}}+\vec{i}_{z}\frac{\partial}{\partial{z}})} [/tex]
[tex]\vec{i}_{z}\frac{\partial}{\partial{z}}\cdot{(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta} \frac{\partial}{r\partial{\theta}}+\vec{i}_{z}\frac{\partial}{\partial{z}})} [/tex]
[tex]\vec{i}_{\theta} \frac{\partial}{r\partial{\theta}}\cdot{(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{theta} \frac{\partial}{r\partial{theta}}+\vec{i}_{z} \frac{\partial}{\partial{z}})} [/tex]
Now, the two first lines don't present any trouble, because the basis vectors only depend on the angle, not on either r or z! Furthermore, orthogonality ensures that most of the the vector dot products are zero, and the two expressions yield the contribution to the Laplacian:
[tex]\frac{\partial^{2}}{\partial{r^{2}}}+\frac{\partial^{2}}{\partial{z^{2}}}[/tex]

The last line in the three major terms are much nastier, because the polar basis vectors DO depend on on the angle!
In particular, we see we get the contribution (PRIOR to dotting!) from [tex]\frac{\partial\vec{i}_{r}}{r\partial\theta}=\frac{1}{r}\vec{i}_{\theta}[/tex]

Collecting the non-zero terms when dotted with the angular basis vector yields us the final two contributions to the Laplacian:
[tex]\frac{\partial^{2}}{r^2\partial{\theta^{2}}}+\frac{1}{r}\frac{\partial}{\partial{r}}[/tex]

The same process is valid for any computation of the laplacian
 
  • #5
For more general coordinates, one of the less obvious features we want to ensure is that the new basis vectors are orthogonal

This isn't true. Yes orthogonal coordinates are nice and simpler to work with, but they aren't always the best choice. In Magnet fusion research we often use coordinate systems based off of the equilibrium magnetic field. This "flux coordinate" systems are a generalization a the simple toroidal coordinate system mentioned, are are often non-orthogonal. The use of flux coordinate systems makes many problems analytically tractable.
 
  • #6
the_wolfman said:
This isn't true. Yes orthogonal coordinates are nice and simpler to work with, but they aren't always the best choice. In Magnet fusion research we often use coordinate systems based off of the equilibrium magnetic field. This "flux coordinate" systems are a generalization a the simple toroidal coordinate system mentioned, are are often non-orthogonal. The use of flux coordinate systems makes many problems analytically tractable.

Cool!
:smile:
Giving it one more thought:
Defining coordinates "naturally occurring" in a particular area of study, rather than the abstract nicety condition of orthogonality would often be the simplest for analysis.
 
Last edited:

Related to Laplacian in toroidal coordinates

1. What is the Laplacian in toroidal coordinates?

The Laplacian in toroidal coordinates is a mathematical operator that describes the second-order spatial variation of a scalar field in a toroidal coordinate system. It is commonly used in physics and engineering to solve problems involving toroidal geometries, such as those found in plasma physics and electromagnetics.

2. How is the Laplacian defined in toroidal coordinates?

The Laplacian in toroidal coordinates is defined as the sum of the second partial derivatives of a function with respect to the three coordinate variables: r, θ, and φ. It can be written using the del operator (∇) as ∇² = 1/r² ∂/∂r (r² ∂/∂r) + 1/r² ∂/∂θ (sin θ ∂/∂θ) + 1/r² ∂²/∂φ².

3. What are the applications of the Laplacian in toroidal coordinates?

The Laplacian in toroidal coordinates has many applications in physics and engineering, including the study of toroidal magnetic fields in plasma confinement devices, toroidal resonant cavity modes in electromagnetic waveguides, and the behavior of toroidal particles in accelerators and storage rings.

4. How is the Laplacian solved in toroidal coordinates?

The Laplacian in toroidal coordinates can be solved using separation of variables, where the solution is expressed as a product of three functions, each depending on one of the coordinate variables. These functions can then be solved using standard techniques, such as Fourier series or Bessel functions, and combined to form the overall solution.

5. Are there any special properties of the Laplacian in toroidal coordinates?

Yes, the Laplacian in toroidal coordinates has several special properties that make it useful for solving certain problems. For example, it is a self-adjoint operator, meaning that it is equal to its own adjoint, and it has a complete set of orthogonal eigenfunctions. These properties make it a powerful tool for solving differential equations in toroidal geometries.

Similar threads

Replies
4
Views
560
  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Calculus
Replies
29
Views
924
Replies
6
Views
704
  • Calculus
Replies
8
Views
374
Replies
1
Views
493
Replies
3
Views
799
Replies
2
Views
432
Replies
3
Views
1K
  • Calculus
Replies
7
Views
1K
Back
Top