Appendix/Mathematics/ScaleFactors

Scale Factors for Orthogonal Curvilinear Coordinate Systems

Here we lean heavily on the class notes and associated references that have been provided by P. A. Kelly in a collection titled, Mechanics Lecture Notes: An Introduction to Solid Mechanics, as they appeared online in early 2021. See especially the subsection of Part III in which the properties of Vectors and Tensors are discussed.

Getting Started

Following Kelly, we will use ${\hat{e}}_i$ and $x_i$ when referencing, respectively, the three (i = 1,3) basis vectors and coordinate "curves" of the Cartesian coordinate system; and we will use ${\hat{g}}_i$ and ${\mathrm{\Theta }}_i$ when referencing, respectively, the three (i = 1,3) basis vectors and coordinate curves of some other, curvilinear coordinate system.

2D Oblique Coordinate System Example

Consider a vector, $\overrightarrow{v}$, which in Cartesian coordinates is described by the expression,

$\overrightarrow{v}$

$=$

${\hat{e}}_1{v}_x+{\hat{e}}_2{v}_y.$

Referencing Figure 1.16.4 of Kelly's Part III, we appreciate that in a two-dimensional (2D) oblique coordinate system where $\alpha$ is the (less than 90°) angle between the two basis vectors, the same vector will be represented by the expression,

$\overrightarrow{v}$

$=$

${\hat{g}}_1{v}^1+{\hat{g}}_2{v}^2.$

The angle between ${\hat{g}}_2$ and ${\hat{e}}_2$ is, (π/2 - α), so we appreciate that,

$v_y$	$=$	$v^2\cos (\frac{\pi }{2}-\alpha )=v^2\sin \alpha$
$\Rightarrow v^2$	$=$	$\frac{v_y}{\sin \alpha }.$

Next, from a visual inspection of the figure, we appreciate that $v_x$ is longer than $v^1$ by the amount, $v^2\cos \alpha$; that is,

$v_x$	$=$	$v^1+v^2\cos \alpha =v_1+\frac{v_y}{\tan \alpha }$
$\Rightarrow v^1$	$=$	$v_x-\frac{v_y}{\tan \alpha }.$

(These are the same pair of transformation relations that appear as Eq. (1.16.3) of Kelly's Part III.)

Covarient:   The set of basis vectors, ${\hat{g}}_1$ and ${\hat{g}}_2$ (note the subscript indices), that are aligned with the coordinate directions, ${\mathrm{\Theta }}_1$ and ${\mathrm{\Theta }}_2$, are generically referred to as covariant base vectors. Contravarient:   A second set of vectors, which will be termed contravariant base vectors, ${\hat{g}}^1$ and ${\hat{g}}^2$ (denoted by superscript indices), will be aligned with a new set of coordinate directions, ${\mathrm{\Theta }}^1$ and ${\mathrm{\Theta }}^2$. This new set of base vectors is defined as follows (see Fig. 1.15.5 of Kelly's Part III): the base vector ${\hat{g}}^1$ is perpendicular to ${\hat{g}}_1$ — that is, ${\hat{g}}^1\cdot {\hat{g}}_2=0$ — and the base vector ${\hat{g}}^2$ is perpendicular to ${\hat{g}}_2$ — that is, ${\hat{g}}_1\cdot {\hat{g}}^2=0$. Further, we ensure that, ${\hat{g}}_1\cdot {\hat{g}}^1=1,$       and,       ${\hat{g}}_2\cdot {\hat{g}}^2=1.$ (Verbatim from p. 138 of Kelly's Part III) A good trick for remembering which are the covariant and which are the contravariant is that the third letter of the word tells us whether the word is associated with subscripts or with superscripts. In "covariant," the "v" is pointing down, so we use subscripts; for "contravariant," the "n" is (with a bit of imagination) pointing up, so we use superscripts.

Continuing with our 2D oblique coordinate system example and appreciating that Kelly has chosen to align the ${\hat{g}}_1$ basis vector with the ${\hat{e}}_1$ (Cartesian) basis vector, we see that the transformation between the two sets of covariant basis vectors is given by the relations,

${\hat{g}}_1={\hat{e}}_1,$

and,

${\hat{g}}_2={\hat{e}}_1\cos \alpha +{\hat{e}}_2\sin \alpha .$

These conditions lead to the following complementary set of contravariant basis vectors:

${\hat{g}}^1={\hat{e}}_1-{\hat{e}}_2\left(\frac{1}{\tan \alpha }\right),$

and,

${\hat{g}}^2={\hat{e}}_2\left(\frac{1}{\sin \alpha }\right).$

Note that, as defined herein, the magnitude (i.e., scalar lengths) of these contravariant basis vectors is not unity; they are, instead,

$|{\hat{g}}^1|\equiv [{\hat{g}}^1\cdot {\hat{g}}^1{]}^{1/2}$	$=$	$\left\{\right[{\hat{e}}_1-{\hat{e}}_2\left(\frac{1}{\tan \alpha }\right)]\cdot [{\hat{e}}_1-{\hat{e}}_2\left(\frac{1}{\tan \alpha }\right)]{\}}^{1/2}$
	$=$	$\{1+\frac{1}{{\tan }^2\alpha }{\}}^{1/2}=\frac{1}{\sin \alpha };$
$|{\hat{g}}^2|\equiv [{\hat{g}}^2\cdot {\hat{g}}^2{]}^{1/2}$	$=$	$\frac{1}{\sin \alpha }.$

Examples

Once expressions for the nine separate direction cosines are known for a system of orthogonal coordinates, then the following hold:

${\hat{\mathbf{g}}}_n$

$=$

$\hat{ı}{\gamma }_{n1}+\hat{ȷ}{\gamma }_{n2}+\hat{k}{\gamma }_{n3};$

and,

$\hat{ı}$	$=$	${\hat{\mathbf{g}}}_1{\gamma }_{11}+{\hat{\mathbf{g}}}_2{\gamma }_{21}+{\hat{\mathbf{g}}}_3{\gamma }_{31},$
$\hat{ȷ}$	$=$	${\hat{\mathbf{g}}}_1{\gamma }_{12}+{\hat{\mathbf{g}}}_2{\gamma }_{22}+{\hat{\mathbf{g}}}_3{\gamma }_{32},$
$\hat{k}$	$=$	${\hat{\mathbf{g}}}_1{\gamma }_{13}+{\hat{\mathbf{g}}}_2{\gamma }_{23}+{\hat{\mathbf{g}}}_3{\gamma }_{33}.$

Hence, the position vector is given by the expression,

$\overrightarrow{\mathbf{x}}=\hat{ı}x+\hat{ȷ}y+\hat{k}z$

$=$

${\hat{\mathbf{g}}}_1({\gamma }_{11}x+{\gamma }_{12}y+{\gamma }_{13}z)+{\hat{\mathbf{g}}}_2({\gamma }_{21}x+{\gamma }_{22}y+{\gamma }_{23}z)+{\hat{\mathbf{g}}}_3({\gamma }_{31}x+{\gamma }_{32}y+{\gamma }_{33}z)$

Cylindrical Coordinates

This is drawn principally from Example #1 (starting on p. 148) of Kelly.

${\lambda }_1$	$=$	$\varpi \equiv [(x^1{)}^2+(x^2{)}^2{]}^{1/2},$
${\lambda }_2$	$\equiv$	${\tan }^{-1}\left[\frac{x^2}{x^1}\right],$
${\lambda }_3$	$\equiv$	$x^3.$

Direction Cosine Components for Cylindrical Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$\varpi \equiv (x^2+y^2{)}^{1/2}$	$1$	$\frac{x}{{\lambda }_1}$	$\frac{y}{{\lambda }_1}$	$0$	$\frac{x}{{\lambda }_1}$	$\frac{y}{{\lambda }_1}$	$0$
$2$	$\phi \equiv {\tan }^{-1}\left[\frac{y}{x}\right]$	${\lambda }_1$	$-\frac{y}{{\varpi }^2}$	$\frac{x}{{\varpi }^2}$	$0$	$-\frac{y}{\varpi }$	$\frac{x}{\varpi }$	$0$
$3$	$z$	$1$	$0$	$0$	$1$	$0$	$0$	$1$

${\hat{\mathbf{g}}}_1$	$=$	$\hat{ı}{\gamma }_{11}+\hat{ȷ}{\gamma }_{12}+\hat{k}{\gamma }_{13}=\hat{ı}\left(\frac{x}{\varpi }\right)+\hat{ȷ}\left(\frac{y}{\varpi }\right)=\hat{ı}\cos \phi +\hat{ȷ}\sin \phi ,$
${\hat{\mathbf{g}}}_2$	$=$	$\hat{ı}{\gamma }_{21}+\hat{ȷ}{\gamma }_{22}+\hat{k}{\gamma }_{23}=-\hat{ı}\left(\frac{y}{\varpi }\right)+\hat{ȷ}\left(\frac{x}{\varpi }\right)=\hat{ı}\sin \phi +\hat{ȷ}\cos \phi ,$
${\hat{\mathbf{g}}}_3$	$=$	$\hat{ı}{\gamma }_{31}+\hat{ȷ}{\gamma }_{32}+\hat{k}{\gamma }_{33}=\hat{k}.$

And the position vector is,

$\overrightarrow{\mathbf{x}}$	$=$	${\hat{\mathbf{g}}}_1({\gamma }_{11}x+{\gamma }_{12}y+{\gamma }_{13}z)+{\hat{\mathbf{g}}}_2({\gamma }_{21}x+{\gamma }_{22}y+{\gamma }_{23}z)+{\hat{\mathbf{g}}}_3({\gamma }_{31}x+{\gamma }_{32}y+{\gamma }_{33}z)$
	$=$	${\hat{\mathbf{g}}}_1[\frac{x^2}{\varpi }+\frac{y^2}{\varpi }]+{\hat{\mathbf{g}}}_2[-\frac{xy}{\varpi }+\frac{xy}{\varpi }]+{\hat{\mathbf{g}}}_{3}z$
	$=$	${\hat{\mathbf{g}}}_1\varpi +{\hat{\mathbf{g}}}_{3}z.$

The line element is,

$d{s}^2$

$=$

$\sum _{i=1}^3{h}_i^{2}d{\lambda }_i^2=d{\lambda }_1^2+{\varpi }^{2}d{\lambda }_2^2+d{\lambda }_3^2.$

In terms of Cartesian basis vector, this is,

$d{s}^2$	$=$	$h_1^2\left[\right(\frac{\partial {\lambda }_1}{\partial x})dx+(\frac{\partial {\lambda }_1}{\partial y})dy+(\frac{\partial {\lambda }_1}{\partial z})dz{]}^2+h_2^2[\left(\frac{\partial {\lambda }_2}{\partial x}\right)dx+\left(\frac{\partial {\lambda }_2}{\partial y}\right)dy+\left(\frac{\partial {\lambda }_2}{\partial z}\right)dz{]}^2+h_3^2\left[\right(\frac{\partial {\lambda }_3}{\partial x})dx+(\frac{\partial {\lambda }_3}{\partial y})dy+(\frac{\partial {\lambda }_3}{\partial z})dz{]}^2$
	$=$	$[{\gamma }_{11}dx+{\gamma }_{12}dy+{\gamma }_{13}dz{]}^2+[{\gamma }_{21}dx+{\gamma }_{22}dy+{\gamma }_{23}dz{]}^2+[{\gamma }_{31}dx+{\gamma }_{32}dy+{\gamma }_{33}dz{]}^2$
	$=$	$\left[\right(\frac{x}{\varpi })dx+(\frac{y}{\varpi })dy{]}^2+[-\left(\frac{y}{\varpi }\right)dx+\left(\frac{x}{\varpi }\right)dy{]}^2+d{z}^2$
	$=$	$(\frac{x}{\varpi }{)}^{2}d{x}^2+(\frac{y}{\varpi }{)}^{2}d{y}^2+2\left(\frac{x}{\varpi }\right)\left(\frac{y}{\varpi }\right)dxdy-2\left(\frac{y}{\varpi }\right)\left(\frac{x}{\varpi }\right)dxdy+(\frac{y}{\varpi }{)}^{2}d{x}^2+(\frac{x}{\varpi }{)}^{2}d{y}^2+d{z}^2$
	$=$	$\left[\right(\frac{x}{\varpi }{)}^2+\left(\frac{y}{\varpi }{)}^2\right]d{x}^2+\left[\right(\frac{y}{\varpi }{)}^2+\left(\frac{x}{\varpi }{)}^2\right]d{y}^2+d{z}^2$
	$=$	$d{x}^2+d{y}^2+d{z}^2.$         Yes!

And, when written in terms of Cartesian coordinates, the "cylindrical" differential volume element is,

$dV=h_1{h}_2{h}_{3}d{\lambda }_{1}d{\lambda }_{2}d{\lambda }_3$	$=$	$[{\gamma }_{11}dx+{\gamma }_{12}dy+{\gamma }_{13}dz][{\gamma }_{21}dx+{\gamma }_{22}dy+{\gamma }_{23}dz][{\gamma }_{31}dx+{\gamma }_{32}dy+{\gamma }_{33}dz]$
	$=$	$\left[\right(\frac{x}{\varpi })dx+(\frac{y}{\varpi }\left)dy\right]\left[\right(-\frac{y}{\varpi })dx+(\frac{x}{\varpi }\left)dy\right]\left[dz\right]$
	$=$	$dz\left\{\right(\frac{x}{\varpi }\left)dx\right[(-\frac{y}{\varpi })dx+\left(\frac{x}{\varpi }\right)dy]+(\frac{y}{\varpi }\left)dy\right[(-\frac{y}{\varpi })dx+\left(\frac{x}{\varpi }\right)dy\left]\right\}$
	$=$	$dz\left\{\right(\frac{-xy}{{\varpi }^2})d{x}^2+(\frac{x^2}{{\varpi }^2})dxdy-(\frac{y^2}{{\varpi }^2})dxdy+(\frac{xy}{{\varpi }^2}\left)d{y}^2\right\}$
	$=$	$\frac{dz}{{\varpi }^2}\{xy(d{y}^2-d{x}^2)+(x^2-y^2)dxdy\}.$

T10 Coordinates

Position Vector

Pulling from our accompanying Table of Direction Cosine Components for T10 Coordinates, the position vector is given by the expression,

$\overrightarrow{\mathbf{x}}$	$=$	${\hat{\mathbf{g}}}_1({\gamma }_{11}x+{\gamma }_{12}y+{\gamma }_{13}z)+{\hat{\mathbf{g}}}_2({\gamma }_{41}x+{\gamma }_{42}y+{\gamma }_{43}z)+{\hat{\mathbf{g}}}_3({\gamma }_{51}x+{\gamma }_{52}y+{\gamma }_{53}z)$
	$=$	${\hat{\mathbf{g}}}_1\left[{\lambda }_1^2{\ell }_{3D}\right]+{\hat{\mathbf{g}}}_2[1+\frac{1}{q^2}-\frac{2}{p^2}]\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}+{\hat{\mathbf{g}}}_3[-x^2(2q^4{y}^2+p^4{z}^2)+q^2{y}^2(p^4{z}^2+2x^2)+p^2{z}^2(x^2-q^4{y}^2)]\frac{{\ell }_{3D}}{{𝒟}}$

Line Element

T10 Example

In terms of Cartesian basis vector, the line element is,

$d{s}^2$	$=$	$h_1^2\left[\right(\frac{\partial {\lambda }_1}{\partial x})dx+(\frac{\partial {\lambda }_1}{\partial y})dy+(\frac{\partial {\lambda }_1}{\partial z})dz{]}^2+h_2^2[\left(\frac{\partial {\lambda }_4}{\partial x}\right)dx+\left(\frac{\partial {\lambda }_4}{\partial y}\right)dy+\left(\frac{\partial {\lambda }_4}{\partial z}\right)dz{]}^2+h_3^2\left[\right(\frac{\partial {\lambda }_5}{\partial x})dx+(\frac{\partial {\lambda }_5}{\partial y})dy+(\frac{\partial {\lambda }_5}{\partial z})dz{]}^2$
	$=$	$[{\gamma }_{11}dx+{\gamma }_{12}dy+{\gamma }_{13}dz{]}^2+[{\gamma }_{41}dx+{\gamma }_{42}dy+{\gamma }_{43}dz{]}^2+[{\gamma }_{51}dx+{\gamma }_{52}dy+{\gamma }_{53}dz{]}^2$
	$=$	$[(x)dx+(q^{2}y)dy+(p^{2}z)dz{]}^2{\ell }_{3D}^2+[(q^{2}y{p}^{2}z)dx+(x{p}^{2}z)dy-(2x{q}^{2}y)dz{]}^2\frac{1}{{{𝒟}}^2}+[-x(2q^4{y}^2+p^4{z}^2)dx+q^{2}y(p^4{z}^2+2x^2)dy+p^{2}z(x^2-q^4{y}^2)dz{]}^2\frac{{\ell }_{3D}^2}{{{𝒟}}^2}$
$\Rightarrow (\frac{{𝒟}}{{\ell }_{3D}}{)}^{2}d{s}^2$	$=$	$[(x)dx+(q^{2}y)dy+(p^{2}z)dz{]}^2{{𝒟}}^2+[(q^{2}y{p}^{2}z)dx+(x{p}^{2}z)dy-(2x{q}^{2}y)dz{]}^2({\ell }_{3D}{)}^{-2}+[-x(2q^4{y}^2+p^4{z}^2)dx+q^{2}y(p^4{z}^2+2x^2)dy+p^{2}z(x^2-q^4{y}^2)dz{]}^2.$

Notice that the coefficient that corresponds to each term is given by the following expressions:

$[d{y}^2]:$		$q^4{y}^2{{𝒟}}^2+x^2{p}^4{z}^2{\ell }_{3D}^{-2}+q^4{y}^2(p^4{z}^2+2x^2{)}^2$
		$q^4{y}^2[q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2]+x^2{p}^4{z}^2[x^2+q^4{y}^2+p^4{z}^2]+q^4{y}^2(p^8{z}^4+4x^2{p}^4{z}^2+4x^4)$
		$q^8{y}^4(4x^2+p^4{z}^2)+6x^2{q}^4{y}^2{p}^4{z}^2+p^8{z}^4(x^2+q^4{y}^2)+x^4(p^4{z}^2+4q^4{y}^2)$
	$=$	$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2.$

$[d{x}^2]:$		$x^2{{𝒟}}^2+q^4{y}^2{p}^4{z}^2{\ell }_{3D}^{-2}+x^2(2q^4{y}^2+p^4{z}^2{)}^2$
	$=$	$x^2(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)+q^4{y}^2{p}^4{z}^2(x^2+q^4{y}^2+p^4{z}^2)+x^2(4q^8{y}^4+4q^4{y}^2{p}^4{z}^2+p^8{z}^4)$
	$=$	$6x^2{q}^4{y}^2{p}^4{z}^2+x^4(4q^4{y}^2+p^4{z}^2)+q^8{y}^4(4x^2+p^4{z}^2)+p^8{z}^2(x^2+q^4{y}^2)$
	$=$	$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2.$

$[dx\cdot dy]:$		$2x{q}^{2}y{{𝒟}}^2+2q^{2}y{p}^{2}z\cdot x{p}^{2}z({\ell }_{3D}{)}^{-2}-2x(2q^4{y}^2+p^4{z}^2)q^{2}y(p^4{z}^2+2x^2)$
	$=$	$2x{q}^{2}y[{{𝒟}}^2+p^4{z}^2({\ell }_{3D}{)}^{-2}-(2q^4{y}^2+p^4{z}^2)(p^4{z}^2+2x^2)]$
	$=$	$2x{q}^{2}y[q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2+p^4{z}^2(x^2+q^4{y}^2+p^4{z}^2)-2q^4{y}^2(p^4{z}^2+2x^2)-p^4{z}^2(p^4{z}^2+2x^2)]$
	$=$	$2x{q}^{2}y[0]=0.$

More Generally

$d{s}^2$	$=$	$h_1^2\left[\right(\frac{\partial {\lambda }_1}{\partial x})dx+(\frac{\partial {\lambda }_1}{\partial y})dy+(\frac{\partial {\lambda }_1}{\partial z})dz{]}^2+h_2^2[\left(\frac{\partial {\lambda }_2}{\partial x}\right)dx+\left(\frac{\partial {\lambda }_2}{\partial y}\right)dy+\left(\frac{\partial {\lambda }_2}{\partial z}\right)dz{]}^2+h_3^2\left[\right(\frac{\partial {\lambda }_3}{\partial x})dx+(\frac{\partial {\lambda }_3}{\partial y})dy+(\frac{\partial {\lambda }_3}{\partial z})dz{]}^2$
	$=$	$[{\gamma }_{11}dx+{\gamma }_{12}dy+{\gamma }_{13}dz{]}^2+[{\gamma }_{21}dx+{\gamma }_{22}dy+{\gamma }_{23}dz{]}^2+[{\gamma }_{31}dx+{\gamma }_{32}dy+{\gamma }_{33}dz{]}^2$
	$=$	$d{x}^2[{\gamma }_{11}^2+{\gamma }_{21}^2+{\gamma }_{31}^2]+d{y}^2[{\gamma }_{12}^2+{\gamma }_{22}^2+{\gamma }_{32}^2]+d{z}^2[{\gamma }_{13}^2+{\gamma }_{23}^{2}d{z}^2+{\gamma }_{33}^2]$
		$+2dxdy[{\gamma }_{11}{\gamma }_{12}+{\gamma }_{21}{\gamma }_{22}+{\gamma }_{31}{\gamma }_{32}]+2dxdz[{\gamma }_{11}{\gamma }_{13}+{\gamma }_{21}{\gamma }_{23}+{\gamma }_{31}{\gamma }_{33}]+2dydz[{\gamma }_{12}{\gamma }_{13}+{\gamma }_{22}{\gamma }_{23}+{\gamma }_{32}{\gamma }_{33}]$
	$=$	$d{x}^2\left[\sum _{s=1}^3{\gamma }_{s1}{\gamma }_{s1}\right]+d{y}^2\left[\sum _{s=1}^3{\gamma }_{s2}{\gamma }_{s2}\right]+d{z}^2\left[\sum _{s=1}^3{\gamma }_{s3}{\gamma }_{s3}\right]+2dxdy\left[\sum _{s=1}^3{\gamma }_{s1}{\gamma }_{s2}\right]+2dxdz\left[\sum _{s=1}^3{\gamma }_{s1}{\gamma }_{s3}\right]+2dydz\left[\sum _{s=1}^3{\gamma }_{s2}{\gamma }_{s3}\right]$
	$=$	$d{x}^2+d{y}^2+d{z}^2.$

The last step of this derivation results from the following series of equations that interrelate the values of various direction cosines:

$\sum _{s=1}^3{\gamma }_{sm}{\gamma }_{sn}$	$=$	${\delta }_{mn},$
[MF53], §1.3, p. 23, Eq. (1.3.1b)

where ${\delta }_{mn}$ is the Kronecker delta function, which is zero when $m$ is not equal to $n$, unity when $m=n$.

   

A similar series of summation expressions — with the order of indices flipped — provides additional interrelationships between the values of the various direction cosines, namely, $\sum _{s=1}^3{\gamma }_{ms}{\gamma }_{ns}$ $=$ ${\delta }_{mn},$ [MF53], §1.3, p. 23, Eq. (1.3.1a) It is particularly easy to validate each member of this set of summation expressions, given that (see above), ${\hat{\mathbf{g}}}_n$ $=$ $\hat{ı}{\gamma }_{n1}+\hat{ȷ}{\gamma }_{n2}+\hat{k}{\gamma }_{n3}.$ Each summation expression derives from the dot-product, ${\hat{\mathbf{g}}}_m\cdot {\hat{\mathbf{g}}}_n$, and the appreciation that (1) the dot product of any unit vector with itself (i.e., $m=n$) gives unity, while (2) the dot product of any unit vector with either of its orthogonal partners (i.e., $m\ne n$) is zero. For any right-handed orthogonal coordinate system, it can also be shown that the following set of nine tabulated expressions details how each one of the $\gamma$'s is related to various algebraic combinations of the others. ${\gamma }_{11}$ $=$ ${\gamma }_{22}{\gamma }_{33}-{\gamma }_{23}{\gamma }_{32},$ ${\gamma }_{12}$ $=$ ${\gamma }_{23}{\gamma }_{31}-{\gamma }_{21}{\gamma }_{33},$ ${\gamma }_{13}$ $=$ ${\gamma }_{21}{\gamma }_{32}-{\gamma }_{22}{\gamma }_{31},$     ${\gamma }_{21}$ $=$ ${\gamma }_{32}{\gamma }_{13}-{\gamma }_{33}{\gamma }_{12},$ ${\gamma }_{22}$ $=$ ${\gamma }_{33}{\gamma }_{11}-{\gamma }_{31}{\gamma }_{13},$ ${\gamma }_{23}$ $=$ ${\gamma }_{31}{\gamma }_{12}-{\gamma }_{32}{\gamma }_{11},$     ${\gamma }_{31}$ $=$ ${\gamma }_{12}{\gamma }_{23}-{\gamma }_{13}{\gamma }_{22},$ ${\gamma }_{32}$ $=$ ${\gamma }_{13}{\gamma }_{21}-{\gamma }_{11}{\gamma }_{23},$ ${\gamma }_{33}$ $=$ ${\gamma }_{11}{\gamma }_{22}-{\gamma }_{12}{\gamma }_{21}.$ [MF53], §1.3, p. 23, Eq. (1.3.2)

Volume Element

${\gamma }_{11}$ $=$ ${\gamma }_{44}{\gamma }_{55}-{\gamma }_{45}{\gamma }_{54},$ ${\gamma }_{14}$ $=$ ${\gamma }_{45}{\gamma }_{51}-{\gamma }_{41}{\gamma }_{55},$ ${\gamma }_{15}$ $=$ ${\gamma }_{41}{\gamma }_{54}-{\gamma }_{44}{\gamma }_{51},$

${\gamma }_{41}$ $=$ ${\gamma }_{54}{\gamma }_{15}-{\gamma }_{55}{\gamma }_{14},$ ${\gamma }_{44}$ $=$ ${\gamma }_{55}{\gamma }_{11}-{\gamma }_{51}{\gamma }_{15},$ ${\gamma }_{45}$ $=$ ${\gamma }_{51}{\gamma }_{14}-{\gamma }_{54}{\gamma }_{11},$

${\gamma }_{51}$ $=$ ${\gamma }_{14}{\gamma }_{45}-{\gamma }_{15}{\gamma }_{44},$ ${\gamma }_{54}$ $=$ ${\gamma }_{15}{\gamma }_{41}-{\gamma }_{11}{\gamma }_{45},$ ${\gamma }_{55}$ $=$ ${\gamma }_{11}{\gamma }_{44}-{\gamma }_{14}{\gamma }_{41}.$

$dV=h_1{h}_4{h}_{5}d{\lambda }_{1}d{\lambda }_{4}d{\lambda }_5$	$=$	$[{\gamma }_{11}dx+{\gamma }_{14}dy+{\gamma }_{15}dz][{\gamma }_{41}dx+{\gamma }_{44}dy+{\gamma }_{45}dz][{\gamma }_{51}dx+{\gamma }_{54}dy+{\gamma }_{55}dz]$
	$=$	$[{\gamma }_{11}dx+{\gamma }_{14}dy+{\gamma }_{15}dz][d{x}^2{\gamma }_{41}{\gamma }_{51}+d{y}^2{\gamma }_{44}{\gamma }_{54}+d{z}^2{\gamma }_{45}{\gamma }_{55}+dxdy({\gamma }_{41}{\gamma }_{54}+{\gamma }_{44}{\gamma }_{51})+dxdz({\gamma }_{41}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{51})+dydz({\gamma }_{44}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{54})]$
	$=$	$d{x}^3{\gamma }_{11}{\gamma }_{41}{\gamma }_{51}+dxd{y}^2{\gamma }_{11}{\gamma }_{44}{\gamma }_{54}+dxd{z}^2{\gamma }_{11}{\gamma }_{45}{\gamma }_{55}+d{x}^{2}dy{\gamma }_{11}({\gamma }_{41}{\gamma }_{54}+{\gamma }_{44}{\gamma }_{51})+d{x}^{2}dz{\gamma }_{11}({\gamma }_{41}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{51})+dxdydz{\gamma }_{11}({\gamma }_{44}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{54})]$
		$+d{x}^{2}dy{\gamma }_{14}{\gamma }_{41}{\gamma }_{51}+d{y}^3{\gamma }_{14}{\gamma }_{44}{\gamma }_{54}+dyd{z}^2{\gamma }_{14}{\gamma }_{45}{\gamma }_{55}+dxd{y}^2{\gamma }_{14}({\gamma }_{41}{\gamma }_{54}+{\gamma }_{44}{\gamma }_{51})+dxdydz{\gamma }_{14}({\gamma }_{41}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{51})+d{y}^{2}dz{\gamma }_{14}({\gamma }_{44}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{54})]$
		$+d{x}^{2}dz{\gamma }_{15}{\gamma }_{41}{\gamma }_{51}+d{y}^{2}dz{\gamma }_{15}{\gamma }_{44}{\gamma }_{54}+d{z}^3{\gamma }_{15}{\gamma }_{45}{\gamma }_{55}+dxdydz{\gamma }_{15}({\gamma }_{41}{\gamma }_{54}+{\gamma }_{44}{\gamma }_{51})+dxd{z}^2{\gamma }_{15}({\gamma }_{41}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{51})+dyd{z}^2{\gamma }_{15}({\gamma }_{44}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{54})]$
	$=$	$dxdydz[{\gamma }_{11}({\gamma }_{44}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{54})+{\gamma }_{14}({\gamma }_{41}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{51})+{\gamma }_{15}({\gamma }_{41}{\gamma }_{54}+{\gamma }_{44}{\gamma }_{51})]+d{x}^3{\gamma }_{11}{\gamma }_{41}{\gamma }_{51}+d{y}^3{\gamma }_{14}{\gamma }_{44}{\gamma }_{54}+d{z}^3{\gamma }_{15}{\gamma }_{45}{\gamma }_{55}$
		$+d{x}^{2}dy[{\gamma }_{14}{\gamma }_{41}{\gamma }_{51}+{\gamma }_{11}({\gamma }_{41}{\gamma }_{54}+{\gamma }_{44}{\gamma }_{51})]+dyd{z}^2[{\gamma }_{14}{\gamma }_{45}{\gamma }_{55}+{\gamma }_{15}({\gamma }_{44}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{54})]+dxd{y}^2[{\gamma }_{14}({\gamma }_{41}{\gamma }_{54}+{\gamma }_{44}{\gamma }_{51})+{\gamma }_{11}{\gamma }_{44}{\gamma }_{54}]$
		$+d{x}^{2}dz[{\gamma }_{15}{\gamma }_{41}{\gamma }_{51}+{\gamma }_{11}({\gamma }_{41}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{51})]+d{y}^{2}dz[{\gamma }_{15}{\gamma }_{44}{\gamma }_{54}+{\gamma }_{14}({\gamma }_{44}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{54})]+dxd{z}^2[{\gamma }_{15}({\gamma }_{41}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{51})+{\gamma }_{11}{\gamma }_{45}{\gamma }_{55}]$

For example, the coefficient of $dxd{z}^2$

${\gamma }_{15}({\gamma }_{41}{\gamma }_{55}+{\gamma }_{45}{\gamma }_{51})+{\gamma }_{11}{\gamma }_{45}{\gamma }_{55}$	$=$	$p^{2}z{\ell }_{3D}\left\{\frac{q^{2}y{p}^{2}z{\ell }_{3D}}{{{𝒟}}^2}\right[p^{2}z(x^2-q^4{y}^2)]+\frac{2x{q}^{2}y{\ell }_{3D}}{{{𝒟}}^2}[x(2q^4{y}^2+p^4{z}^2)\left]\right\}-\{x{\ell }_{3D}\cdot \frac{2x{q}^{2}y{\ell }_{3D}}{{{𝒟}}^2}[p^{2}z(x^2-q^4{y}^2)\left]\right\}$
	$=$	$p^{2}z\left(\frac{{\ell }_{3D}}{{𝒟}}{)}^2\right\{q^{2}y{p}^{2}z[p^{2}z(x^2-q^4{y}^2)]+2x{q}^{2}y[x(2q^4{y}^2+p^4{z}^2)]\}-\{2x^2{q}^{2}y[p^{2}z(x^2-q^4{y}^2)]\left\}\right(\frac{{\ell }_{3D}}{{𝒟}}{)}^2$

And the coefficient of $d{x}^3$ is:

${\gamma }_{11}{\gamma }_{41}{\gamma }_{51}$	$=$	$x{\ell }_{3D}\left(\frac{q^{2}y{p}^{2}z}{{𝒟}}\right)-\frac{{\ell }_{3D}}{{𝒟}}[x(2q^4{y}^2+p^4{z}^2)]$
	$=$	$[q^{2}y{p}^{2}z-(2q^4{y}^2+p^4{z}^2)]\frac{x{\ell }_{3D}}{{𝒟}}$
	$=$	$-[(q^4{y}^2-2q^{2}y{p}^{2}z+p^4{z}^2)+q^4{y}^2+q^{2}y{p}^{2}z]\frac{x{\ell }_{3D}}{{𝒟}}$
	$=$	$-[(q^{2}y-p^{2}z{)}^2+q^{2}y(q^{2}y+p^{2}z)]\frac{x{\ell }_{3D}}{{𝒟}}.$

See Also