Suppose we have a curvilinear coordinate system in $\mathbb{R}^3$, $(q_1, q_2, q_3)$, with $x_i = f_i(q_1, q_2, q_3)$. We define the local basis for this coordinate system by $\hat{e_i} = \frac{\partial\vec r}{\partial q_i}$, and the scale factors in this coordinate system as $h_i = \|\hat{e_i}\|$. Then it is known that the volume element in our coordinate system is $$ dV = h_1h_2h_3dq_1dq_2dq_3 $$ but we also know that the volume element, under a coordinate transformation, is transformed into $$ dxdydz = \left|\frac{\partial(x,y,z)}{\partial(q_1,q_2,q_3)}\right|dq_1dq_2dq_3 $$ where $\frac{\partial(x,y,z)}{\partial(q_1,q_2,q_3)}$ is the Jacobian determinant.

Is my understanding correct in that $h_1h_2h_3$ should equal the absolute value of the Jacobian determinant, or am I misinterpreting something? If my understanding is correct, how can I prove this? I know that if $F = (f_1, f_2, f_3)$, then $$ \begin{pmatrix} \hat{e}_1 \\ \hat{e}_2 \\ \hat{e}_3 \end{pmatrix} = (JF)^\text{T}\begin{pmatrix}\hat{i} \\ \hat{j} \\ \hat{k}\end{pmatrix} $$ (where $JF$ is also the Jacobian matrix, just different notation), but I have no clue how I might use this to show the above equality (assuming it's true).

  • $\begingroup$ Your formula is only valid when the $\hat e_i$ are mutually orthogonal (perpendicular). $\endgroup$ – Ted Shifrin Feb 23 '18 at 22:23
  • $\begingroup$ I had that thought. Upon further thinking, the if the $\hat{e}_i$ are orthogonal, then quantity $h_1h_2h_3$ is equal to the $\|\hat{e}_1\cdot(\hat{e}_2\times\hat{e}_3)\|$, which is also the formula for the volume of a parallelepiped, which is analogous to the Jacobian factor in multivariable substitution (which is also a "volume"). Could you elaborate on this in an answer? $\endgroup$ – user3002473 Feb 23 '18 at 22:32

The column vectors of your Jacobian matrix will be the vectors $\hat e_i$ (by the way, I thought it was customary to use $\hat v$ only for unit vectors $v$!). The determinant of that matrix will be (up to sign) the product of the lengths of those column vectors if and only if the vectors $\hat e_i$ form an orthogonal set. (In general, the absolute value of the determinant is the volume of the parallelogram whose edges are the column vectors.)

Of course, this happens in many important situations (e.g., cylindrical and spherical coordinates) that show up "in real life."

  • $\begingroup$ Thanks for clarifying! And my mistake, I used the hat where I shouldn't have! $\endgroup$ – user3002473 Feb 23 '18 at 23:13

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