Laplacian in toroidal coordinates

  • Thread starter Einj
  • Start date
  • #1
470
58
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
 

Answers and Replies

  • #2
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
133
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
470
58
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
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
133
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
346
48
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
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
133
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 Threads on Laplacian in toroidal coordinates

Replies
2
Views
16K
Replies
3
Views
714
  • Last Post
Replies
1
Views
683
  • Last Post
Replies
9
Views
10K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
8
Views
938
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
13
Views
896
Top