# Scale Factors via Ellipsoidal Coordinate System Scale Factors

In Morse & Feshbach (P512 - 514) they show how 10 different orthogonal coordinate systems (mentioned on this page) are derivable from the confocal ellipsoidal coordinate system $(\eta,\mu,\nu)$ by trivial little substitutions, derivable in the sense that we can get explicit expressions for our Cartesian $x$, $y$ & $z$ in terms of the coordinates of some coordinate system by simply modifying the expressions for $x$, $y$ & $z$ which are expressed in terms of ellipsoidal coordinates. Thus given that

$$x = \sqrt{\frac{(\eta^2 - a^2)(\mu^2 - a^2)(\nu^2 - a^2)}{a^2(a^2-b^2)}}, y = ..., z = ...$$

in ellipsoidal coordinates, we can derive, for instance, the Cartesian coordinate system by setting $\eta ^2 = \eta' ^2 + a^2$, $\mu^2 = \mu'^2 + b^2$, $\nu = \nu'$, $b = a\sin(\theta)$ & letting $a \rightarrow \infty$ to get that $x = \eta'$, $y = \mu'$ & $z = \nu'$. Substitutions like these are given to derive a ton of other useful coordinate systems.

I don't see why one shouldn't be able to use the exact same substitutions on the scale factors. Thus given

$$h_1 = \sqrt{\frac{(\eta^2 - \mu^2)(\eta^2 - \nu^2)}{(\eta^2 - a^2)(\eta^2 - b^2)}}$$

I don't see any reason why the exact same substitutions should not give the Cartesian scale factors, namely that $h_1 = 1$ in this case, yet I can't do it with the algebra - I can't get it to work. You get these $a^4$ factors which you just cannot get rid of, thus it seems like one can't derive the scale factors also by mere substitution... Now it might just be late & that I've worked too much, hence my question is:

Is it possible to get the scale factors for orthogonal curvilinear coordinate systems by simple substitutions into the scale factor formulae for the ellipsoidal coordinate system, analogous to the way one can derive the formulae for Cartesian components in terms any 'standard' orthogonal coordinate system by substitutions into the formulae for them expressed in terms of ellipsoidal coordinates? If not, why not?

If there is then the gradient, divergence, Laplacian & curl become extremely easy to calculate in the standard orthogonal coordinate systems, & working separation of variables for all the standard pde's becomes immensely easier, if not I'd have to derive the scale factors by differentiation of completely crazy formulas (at least there's an easy way to remember any formula in $\vec{r}(u^1,u^2,u^3) = ...$ from which we can derive the scale factors thanks to Morse & Feshbach!!!)

Thanks for any help possible.

• Main problem in your attempt is that scale factors are not scalars, but are diagonal components of metric tensor. You are doing a coordinate transformation (curvilinear coordinates to cartesians): for scalars, such as coordinates themselves, substitution works fine. For tensor components, such as scale factors, you need to make the substitution, but also add the necessary components of transformation matrix, then it will work. Oct 23, 2013 at 1:37
• You have no idea how happy that has made me, could you provide any more detail if possible? Maybe even a reference if such a thing exists. Oct 23, 2013 at 15:27

Okay! I apply a transformation, which converts my comment into a slightly more self-contained answer.

The transformations mentioned here are most naturally described by means of tensor calculus or, more generally, differential geometry. It is needed, roughly speaking, when one wants/needs to introduce a coordinate system at each point in space and study the relations between different points and their coordinate systems.

In this case there is a coordinate transformation defined from cartesian system $x^\mu=(x,y,z)$ to ellipsoidal coordinates $x^{\mu'}=(\eta,\mu,\nu)$, and an inverse transformation. $\mu$ and $\mu'$ are indices, which can take values from $1$ to $3$. Trasnformation matrix $t^{\mu}_{~\mu'}$ is defined as $t^{\mu}_{~\mu'}=\dfrac{\partial x^{\mu}}{\partial x^{\mu'}}$. For the given transformation $t^{\mu}_{~\mu'}$ is diagonal, because $x$ depends only on $\eta$, and so on.

Metric tensor with matrix $g_{\mu\nu}$ is a quantity, which defines the scalar product of basis vectors $g_{\mu\nu}\equiv \vec{e}_\mu\cdot\vec{e}_{\nu}$. In cartesian coordinates $x^{\mu}$, therefore, $g_{\mu\nu}=\textrm{diag}\{1,1,1\}$. In primed coordinates, for a given point in space (with its own set of basis vectors) the same definition holds: $g_{\mu'\nu'}=\vec{e}_{\mu'}\cdot\vec{e}_{\nu'}$. From tensor calculus, $g_{\mu'\nu'}$ and $g_{\mu\nu}$ are connected by $g_{\mu'\nu'}=g_{\mu\nu} t^{\mu}_{~\mu'} t^{\nu}_{~\nu'}$, where summation over the same indices is implied. Because $t^{\mu}_{~\mu'}$ is diagonal, $g_{\mu'\nu'}$ is also diagonal, as it should be. Hence $\vec{e}_{\mu'}$ are orthogonal.

Now, in orthogonal bases $\vec{e}_{\mu}$ scale factors are defined as $h_{\mu}=\sqrt{\vec{e}_{\mu}\cdot \vec{e}_{\mu}}$ (which has some physical meaning if one thinks about decomposing a vector in such a basis), here summation is not implied. In cartesians, therefore, $h_1=1$, whereas in ellispoidal coordinates $h_{1'}=\sqrt{\vec{e}_{1'}\cdot \vec{e}_{1'}}=\sqrt{g_{1'1'}}=\sqrt{g_{\mu\nu} t^{\mu}_{~1'} t^{\nu}_{~1'}}=\sqrt{g_{11} t^{1}_{~1'} t^{1}_{~1'}}=t^{1}_{~1'}=t^{1}_{~1'} h_1$. Or, alternatively, one can derive (without using $h_1 = 1$) that $h_1=h_{1'}t^{1'}_{~1} = h_{1'}(t^{1}_{~1'})^{-1}$.

Using the above described, $t^{1}_{~1'}=\dfrac{\eta}{x}=\dfrac{\eta}{\sqrt{\eta^2-a^2}}$. Substituting $t^{1}_{~1'}$ and $h_{1'}$ into $h_1= h_{1'}(t^{1}_{~1'})^{-1}$ and using $\eta\sim a \rightarrow \infty$, one gets $h_1 = 1$.

The answer might be hard to read if you have never studied differential geometry, but the key point is that scale factors are not scalars and a simple variable change doesn't work for them. However, as they are defined for orthogonal systems, the transformation rules that they follow are relatively simple.