I have a non-orthogonal coordinate system defined by $\mathbf x=\mathbf x(r,\beta,z)$, and so I can find the basis vectors as $$ \mathbf g_r=\frac{\partial \mathbf x}{\partial r},\quad\mathbf g_\beta=\frac{\partial \mathbf x}{\partial \beta},\quad \mathbf g_z=\frac{\partial \mathbf x}{\partial z},$$ where $\mathbf g_r$, $\mathbf g_\beta$, and $\mathbf g_z$ are functions of $r$ and $\beta$.

In this coordinate system I have written a system of PDEs, for example $$\frac{\partial v}{\partial r}+\rho(r)\frac{\partial v}{\partial z}=0 $$ for some arbitrary known function $\rho(r)$.

Now I want to replace the $\mathbf g_z$ coordinate direction with a new one, defined as $\mathbf g_n=\mathbf g_r\times\mathbf g_\beta$. I can write $\mathbf g_n=f(r)\mathbf g_r+g(r)\mathbf g_\beta+h(r)\mathbf g_z$, or $\mathbf g_z=\tilde f(r)\mathbf g_r+\tilde g(r)\mathbf g_\beta+\tilde h(r)\mathbf g_n$.

This is where I'm unsure what to do next. Considering the equation for $\mathbf g_z$ in terms of the other directions, I've considered doing the following: $$\begin{align} \frac{\partial}{\partial z}&=\frac{\partial}{\partial r}\frac{\partial r}{\partial z}+\frac{\partial}{\partial \beta}\frac{\partial\beta}{\partial z}+\frac{\partial}{\partial n}\frac{\partial n}{\partial z} \\ &=\tilde f(r)\frac{\partial}{\partial r}+\tilde g(r)\frac{\partial}{\partial \beta}+\tilde h(r)\frac{\partial}{\partial n} \end{align}$$ however this feels wrong. I think I need to use equations for $r$, $\beta$, and $n$ in terms of $r$, $\beta$, and $z$ then do the chain rule for the derivatives.

Given coordinate directions, $\mathbf g_r$, $\mathbf g_\beta$, and $\mathbf g_n$, how can I find these equations for the transformations of the variables in the PDE?

  • $\begingroup$ How do you define $n$? Just because you've written $\mathbf g_n=\mathbf g_r\times \mathbf g_{\beta}$ does not mean that you have defined an $n$. I suppose you want to somehow define $n(r,\beta,z)$ such that $\partial \mathbf x/\partial n = \mathbf g_n$. $\endgroup$ Dec 12, 2014 at 23:34
  • $\begingroup$ I think this is my question. $n$ is a variable (like $r$ and $\theta$ in polar coordinates), and $\mathbf g_n$ is the corresponding coordinate direction (like $\mathbf g_r$ and $\mathbf g_\theta$ in polar coordinates), so if I could work out $n=n(r,\beta,z)$ that would be very helpful, I'd be able to transform my equations into this new coordinate system. $\endgroup$
    – David
    Dec 13, 2014 at 10:19
  • $\begingroup$ hi David, i dont know how to contact you. You are the only one who answered the post math.stackexchange.com/questions/3492367/… ... So i wonder if you could help me solving my question there. Thank you very much! $\endgroup$
    – Senna
    Jan 3, 2020 at 9:57

1 Answer 1


I'll introduce a bit of notation to make the explanation smoother. Let's start with a coordinate system $u^i=(u^1,u^2,u^3)$. The vector field with respect to this notation can be written $\mathbf x = \mathbf x (u^1,u^2,u^3) = \mathbf x (u^i)$. I'll assume that we work in Euclidean space, it's just the coordinate system that is curvilinear. So we can express our vector field with respect to a Euclidean basis as $\mathbf x = x^k \mathbf e_k$. Thus, the coordinates $x^k$ depend on the curvilinear coordinates $x^k=x^k(u^i)$. Note I use the Einstein summation convention, meaning that whenever matching upper and lower indices appear, a sum over all possible index values is implied. In other words

$$\mathbf x = x^k \mathbf e_k = x^1 \mathbf e_1 + x^2 \mathbf e_2 + x^3 \mathbf e_3$$

Introducing the basis that interests you we can now write it in terms of the Euclidean basis :

$$\mathbf g_i = \frac{\partial \mathbf x}{\partial u^i} = \frac{\partial x^k}{\partial u^i} \mathbf e_k$$

We can also define the inverse transformation as

$$\mathbf e_k = \frac{\partial u^i}{\partial x^k} \mathbf g_i$$

in which $\frac{\partial u^i}{\partial x^k}$ is thus a coefficient of the inverse matrix of which $\frac{\partial x^k}{\partial u^i}$ is a coefficient. So

$$\frac{\partial x^k}{\partial u^i}\frac{\partial u^i}{\partial x^l} = \delta^k_l \text{ (*)}$$

with $\delta^k_l$ Kronecker's delta.

If we now want to define a new vector

$$\mathbf g_n = \mathbf g_1 \times \mathbf g_2$$

we can compute its coefficients with respect to the Euclidean basis

$$\mathbf g_n = \frac{\partial \mathbf x}{\partial u^1} \times \frac{\partial \mathbf x}{\partial u^2} = \frac{\partial x^k}{\partial u^1} \mathbf e_k \times \frac{\partial x^l}{\partial u^2} \mathbf e_l = \frac{\partial x^k}{\partial u^1} \frac{\partial x^l}{\partial u^2} \mathbf e_k \times \mathbf e_l \; .$$

Introducing the new notation

$$\mathbf e_k \times \mathbf e_l = \epsilon_{kl}^{\phantom{kl}m}\mathbf e_m$$

in which $\epsilon_{kl}^{\phantom{kl}m}$ is the Levi-Civita symbol, we can write

$$\mathbf g_n = \epsilon_{kl}^{\phantom{kl}m} \frac{\partial x^k}{\partial u^1} \frac{\partial x^l}{\partial u^2} \mathbf e_m = \epsilon_{kl}^{\phantom{kl}m} \frac{\partial x^k}{\partial u^1} \frac{\partial x^l}{\partial u^2} \frac{\partial u^i}{\partial x^m} \mathbf g_i $$

where in the last step we used the inverse transformation to express the new $\mathbf g_n$ in terms of the $\mathbf g_i$.

First remark is that by construction it is clear that $\mathbf g_n$ is not a real vector but a pseudo-vector, as it is the result of a cross product of vectors. So this might lead to some trouble further down the line.

Now, if we suppose that the $$(\mathbf g_1,\mathbf g_2, \mathbf g_n)$$ form a natural basis in their own right, it should be possible to find out how $(u^1,u^2,n)$ depend on the $x^i$ because we sould have a relation of the form

$$(\mathbf g_1, \mathbf g_2, \mathbf g_n)^{\text T}(\nabla u^1,\nabla u^2,\nabla n)=\mathbb{I}_3$$

or applying the inverse of the left matrix to the left of each side

$$(\nabla u^1,\nabla u^2,\nabla n)=\frac{1}{\mathbf g_1\cdot(\mathbf g_2\times\mathbf g_n)}(\mathbf g_2\times\mathbf g_n, \mathbf g_n\times\mathbf g_1, \mathbf g_1\times\mathbf g_2) = \frac{1}{\mathbf g_1\cdot(\mathbf g_2\times\mathbf g_n)}(\mathbf g_1(\mathbf g_2\cdot\mathbf g_2) - \mathbf g_2(\mathbf g_1\cdot\mathbf g_2), \mathbf g_2(\mathbf g_1\cdot\mathbf g_1) - \mathbf g_1(\mathbf g_1\cdot\mathbf g_2), \mathbf g_n)\; .$$

Now, the same formula exists for the $(\mathbf g_1, \mathbf g_2, \mathbf g_n)$ it is in fact formula (*) above but which I'll rewrite in vector form

$$(\nabla u^1,\nabla u^2,\nabla u^3)=\frac{1}{\mathbf g_1\cdot(\mathbf g_2\times\mathbf g_3)}(\mathbf g_2\times\mathbf g_3, \mathbf g_3\times\mathbf g_1, \mathbf g_1\times\mathbf g_2) \; .$$

Note the first two vector components on the left hand side are corresponding, so we'd expect the same on the right hand side, but this leads to a contradiction. Indeed as you can see, the crossproduct of the first components of the right hand sides is

$$(\mathbf g_2\times\mathbf g_n)\times(\mathbf g_2\times\mathbf g_3)=\mathbf g_2\cdot(\mathbf g_3\times\mathbf g_n)\mathbf g_2$$

which is non-zero in general because the three vectors should be linearly independent in general. Indeed $\mathbf g_n$ is orthogonal on $\mathbf g_2$ and because of $\mathbf g_3$ being different from $\mathbf g_n$ there's no reason in general we should expect it to be in the plane spanned by $\mathbf g_2$ and $\mathbf g_3$.

So, on the right hand side, the first components lead to two linearly independent vectors in general, whereas the left hand side told us they should be equal. A clear contradiction. Is there a way to solve this problem?

I suspect that if you make a change of basis in which you consider exactly the reciprocal basis and not just change the last component of the reciprocal basis, that you will have exactly what you need.

EDIT: Just to put out my thought process: starting again from relation (*) but in vector form

$$(\mathbf g_r, \mathbf g_{\beta}, \mathbf g_z)^{\text T}(\nabla r,\nabla \beta,\nabla z)=\mathbb{I}_3$$

or alternatively

$$(\nabla r,\nabla \beta,\nabla z)=\frac{1}{\mathbf g_r\cdot(\mathbf g_ {\beta}\times\mathbf g_z)}(\mathbf g_{\beta}\times\mathbf g_z, \mathbf g_z\times\mathbf g_r, \mathbf g_r\times\mathbf g_{\beta}) \; .$$

If we now suppose that the right hand side of this equation forms a natural basis for some choice of coordinates $(l,m,n)$, then it should be true that

$$\left(\left((\mathbf g_r, \mathbf g_{\beta}, \mathbf g_z)^{\text T}\right)^{-1}\right)^{\text T}(\nabla l,\nabla m,\nabla n)=\mathbb{I}_3$$


$$(\nabla l,\nabla m,\nabla n) = (\mathbf g_r, \mathbf g_{\beta}, \mathbf g_z)$$

On the other hand lets write $(\tilde{\nabla} l,\tilde{\nabla} m,\tilde{\nabla} n)$ for the matrix of derivatives of $(l,m,n)$ w.r.t. $(r,\beta,z)$, then

$$(\nabla r,\nabla \beta,\nabla z)(\tilde{\nabla} l,\tilde{\nabla} m,\tilde{\nabla} n)=(\nabla l,\nabla m,\nabla n)$$

Multiplying both sides by $(\mathbf g_r, \mathbf g_{\beta}, \mathbf g_z)^{\text T}$, we get

$$(\tilde{\nabla} l,\tilde{\nabla} m,\tilde{\nabla} n) = (\mathbf g_r, \mathbf g_{\beta}, \mathbf g_z)^{\text T}(\mathbf g_r, \mathbf g_{\beta}, \mathbf g_z)$$

The right hand side can be worked out to give

$$\left(\begin{array}{ccc} 1+H'(r)^2 & PH'(r) & H'(r) \\ PH'(r) & r^2+P^2 & P \\ H'(r) & P & 1 \end{array}\right)$$

There are obviously no solutions $(l,m,n)$ to those equations.

  • $\begingroup$ This looks good so far. I have a decent working knowledge of this stuff, but I have never learnt it in a proper, structured course, so seeing it presented neatly like this is helpful. Looking forward to the rest. I had to award the bounty because it nearly expired, but please continue with the answer. Thanks! $\endgroup$
    – David
    Dec 14, 2014 at 21:32
  • $\begingroup$ Don't worry, I'll continue. I just need some more time to work it out. $\endgroup$ Dec 14, 2014 at 23:10
  • $\begingroup$ Now, it would be neat if you gave me the example you want to do in detail so that I can see if it works out. $\endgroup$ Dec 16, 2014 at 22:21
  • $\begingroup$ I've edited my question with my specific example. Could you explain the $\nabla u_i$ notation, is that a gradient ($\mathbf g_i u_{,i}$)? Many thanks again for this. $\endgroup$
    – David
    Dec 17, 2014 at 22:00
  • $\begingroup$ That's a gradient, it's $\frac{\partial u^i}{\partial x^k}e^k$. $\endgroup$ Dec 18, 2014 at 9:07

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