In a general coordinate system u^i, i = 1, 2, 3, in three-dimensional Euclidean space, a volume element is given by dV = |e1 du¹ · (e2 du² × e3 du³)|. Show that an alternative form for this expression, written in terms of the determinant g of the metric tensor, is given by dV = √g du¹ du² du³.

Question

In a general coordinate system [latex]{u}^{i}[/latex], i = 1, 2, 3, in three-dimensional Euclidean space, a volume element is given by

[latex]dV = |{\mathbf{e}}_{1} d{u}^{1} · ({\mathbf{e}}_{2} d{u}^{2} × {\mathbf{e}}_{3} d{u}^{3})|[/latex].

Show that an alternative form for this expression, written in terms of the determinant g of the metric tensor, is given by

[latex]dV =\sqrt { g }\ d{u}^{1} \ d{u}^{2} \ d{u}^{3}[/latex].

Show that under a general coordinate transformation to a new coordinate system [latex]{u}^{\prime i}[/latex], the volume element dV remains unchanged, i.e. show that it is a scalar quantity.

Accepted Answer

Working in terms of the Cartesian bases vectors i, j and k, let

[latex]\mathbf{e} _m=\lambda_{m x} \mathbf{i} +\lambda_{m y} \mathbf{j} +\lambda_{m z} \mathbf{k} , ext { for } m=1,2,3 .[/latex]

Then,

[latex]\begin{aligned}& \mathbf{e} _2 d u^2 imes \mathbf{e} _3 d u^3=d u^2 d u^3\left(\lambda_{2 x} \lambda_{3 y} \mathbf{k} -\lambda_{2 x} \lambda_{3 z} \mathbf{j} -\lambda_{2 y} \lambda_{3 x} \mathbf{k} ight. \&\left.+\lambda_{2 y} \lambda_{3 z} \mathbf{i} +\lambda_{2 z} \lambda_{3 x} \mathbf{j} -\lambda_{2 z} \lambda_{3 y} \mathbf{i} ight),\end{aligned}[/latex]

and it follows that

[latex]\begin{aligned}d V= & \mathbf{e} _1 \ d u^1 \cdot\left( \mathbf{e} _2 \ d u^2 imes \mathbf{e}_3 d u^3 ight) \& =d u^1 d u^2 d u^3\left[\lambda_{1 x}\left(\lambda_{2 y} \lambda_{3 z}-\lambda_{2 z} \lambda_{3 y} ight)+\lambda_{1 y}\left(\lambda_{2 z} \lambda_{3 x}-\lambda_{2 x} \lambda_{3 z} ight) ight. \& \left.\quad+\lambda_{1 z}\left(\lambda_{2 x} \lambda_{3 y}-\lambda_{2 y} \lambda_{3 x} ight) ight] \& =d u^1 d u^2 d u^3\left|\begin{array}{ccc}\lambda_{1 x} & \lambda_{1 y} & \lambda_{1 z} \\lambda_{2 x} & \lambda_{2 y} & \lambda_{2 z} \\lambda_{3 x} & \lambda_{3 y} & \lambda_{3 z}\end{array} ight| \& \equiv d u^1 d u^2 d u^3| A |, ext { thus defining } A .\end{aligned}[/latex]

Now consider an element of the matrix [latex]AA ^{ T }[/latex]:

[latex]\left( AA ^{ T } ight)_{m n}=\sum_r A _{m r} A _{n r}=\lambda_{m x} \lambda_{n x}+\lambda_{m y} \lambda_{n y}+\lambda_{m z} \lambda_{n z} .[/latex]

But the elements of the metric tensor are given by

[latex]g _{m n}= \mathbf{e} _m \cdot \mathbf{e} _n=\lambda_{m x} \lambda_{n x}+\lambda_{m y} \lambda_{n y}+\lambda_{m z} \lambda_{n z} .[/latex]

Hence [latex]AA ^{ T }[/latex] = g and, in particular, |A| |[latex]A ^{ T }[/latex]| = | g |. Since |A| = |[latex]A ^{ T }[/latex]|, it follows that |A| = [latex]| g |^{1 / 2}=\sqrt{g}[/latex] and

[latex]d V=d u^1 \ d u^2 \ d u^3| A |=\sqrt{g} \ d u^1 \ d u^2 \ d u^3 .[/latex]

For a transformation [latex]u^{\prime i}=u^{\prime i} \left(u^1, u^2, u^3 ight)[/latex],

[latex]d u^{\prime 1} d u^{\prime 2} d u^{\prime 3}=\left|\frac{\partial u^{\prime}}{\partial u} ight| d u^1 d u^2 d u^3[/latex]

and the covariant components of the second-order tensor [latex]g_{i j}[/latex] transform as

[latex]\begin{aligned}g_{i j}^{\prime} & =\frac{\partial u^k}{\partial u^{\prime i}} \frac{\partial u^l}{\partial u^{\prime j}} g_{k l}, \\Rightarrow \quad g^{\prime} & =\left|\frac{\partial u}{\partial u^{\prime}} ight|\left|\frac{\partial u}{\partial u^{\prime}} ight| g ext { (on taking determinants), } \\Rightarrow \quad \sqrt{g^{\prime}} & =\left|\frac{\partial u}{\partial u^{\prime}} ight| \sqrt{g} .\end{aligned}[/latex]

Thus, the new volume element is

[latex]\begin{aligned}d V^{\prime} & =\sqrt{g^{\prime}} \ d u^{\prime 1} \ d u^{\prime 2} \ d u^{\prime 3} \& =\left|\frac{\partial u}{\partial u^{\prime}} ight| \sqrt{g}\left|\frac{\partial u^{\prime}}{\partial u} ight| \ d u^1 \ d u^2 \ d u^3 \& =\sqrt{g} \ d u^1 \ d u^2 \ d u^3=d V .\end{aligned}[/latex]

This shows that dV is a scalar quantity.

Question 26.25: In a general coordinate system u^i, i = 1, 2, 3, in three-di...

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