# Does a 3-Dimensional coordinate transformation exist such that its scale factors are equal?

Let $\vec r=(x,y,z)$ be the position vector expressed in Cartesian coordinates. Let us define the coordinate transformation as

$\vec r(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w))$

The scale factors are defined by

$h_u=\vert \partial \vec r/\partial u \vert, h_v=\vert \partial \vec r/\partial v \vert, h_w=\vert \partial \vec r/\partial w \vert$

I wonder if a transformation can be defined such that

$h_u=h_v=h_w$

Now a pair of examples in the two dimentional case.

The transformation between elliptic and cartesian coordinates:

$\vec r(u,v)=(cosh(u)cos(v)/2,sinh(u)sin(v)/2)$

$h_u=h_v=\sqrt{cosh^2(u)-cos^2(v)}/2$

The transformation between parabolic and cartesian coordinates.

$\vec r(u,v)=((u^2-v^2)/2,u v)$

$h_u=h_v=\sqrt{u^2+v^2}$

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What does "scale factor" mean in the context of a coordinate system? – Gerry Myerson Feb 28 '13 at 21:34
Hola, Hugo. Bienvenidos! – amWhy Feb 28 '13 at 22:06
OK, how about $r(u,v,w)=(u,v,w)$? – Gerry Myerson Mar 2 '13 at 11:47
By the way, after I edited the spelling of "dimensional", why did you edit it back to "dimentional"? I am not going to get into an edit war on the spelling of a common English word. – Gerry Myerson Mar 2 '13 at 11:51
Thanks a lot @GerryMyerson for your observations. I'm looking for a non-trivial transformation. Somebody have suggested the transformation $\vec r=(u,v,w)/(u^2+v^2+w^2)$ this satisfies the condition of equal scale factors. I wonder if another non-trivial examples exist. Greetings from México. – Hugo Tlahuext Aca Mar 2 '13 at 19:55

Suppose we add another condition: not only $$\left\lVert\frac{\partial\mathbf r}{\partial u}\right\rVert=\left\lVert\frac{\partial\mathbf r}{\partial v}\right\rVert=\left\lVert\frac{\partial\mathbf r}{\partial w}\right\rVert,$$ but also $$\frac{\partial\mathbf r}{\partial u}\cdot\frac{\partial\mathbf r}{\partial v}=\frac{\partial\mathbf r}{\partial v}\cdot\frac{\partial\mathbf r}{\partial w}=\frac{\partial\mathbf r}{\partial w}\cdot\frac{\partial\mathbf r}{\partial u}=0.$$ That is, the coordinate system is orthogonal, which is usually desirable. (In particular, both of your two-dimensional examples have this property.) This means that the Jacobian matrix of the transformation is a multiple of the identity, and the transformation $(u,v,w)\mapsto(x,y,z)$ is a conformal map.
Liouville's theorem states that in three or more dimensions, all such maps are compositions of translations, similarities, orthogonal transformations and inversions. So the space of such coordinate systems is much more restricted than in two dimensions. Nevertheless, we do have a non-Cartesian example, namely inversion: $$(x,y,z)=\frac{(u,v,w)}{u^2+v^2+w^2}.$$ This has $\lVert\partial\mathbf r/\partial u\rVert=\lVert\partial\mathbf r/\partial v\rVert=\lVert\partial\mathbf r/\partial w\rVert=1/(u^2+v^2+w^2)$.
You're right Rahul Narain, i must said that the new coordinate system must be orthogonal. I desire that the scale factor satisfies $h_u=h_v=h_w=(x^2+y^2+z^2)^{1/2}$, where $x=x(u,v,w), y=y(u,v,w), z=z(u,v,w)$ Greetings from México. – Hugo Tlahuext Aca Mar 2 '13 at 19:58
If you do want both orthogonality and equal "scale factors", then Liouville's theorem (linked in my answer) implies that there are no other examples that are not similarity transformations of this one. It has $\lVert\partial\mathbf r/\partial u\rVert=\lVert\partial\mathbf r/\partial v\rVert=\lVert\partial\mathbf r/\partial w\rVert=1/(u^2+v^2+w^2)=x^2+y^2+z^2$, which is pretty close to what you want, anyway. – Rahul Mar 2 '13 at 21:31