Jacobian of $d$-dimensional volume element

Question

This question is in regard to all coordinate transformations where we consider the integration measure or volume element gets scaled by a factor (Jacobian). I can in general write any such transformation matrix, but how come the determinant of the matrix, which is the Jacobian, is coming? I can show that for two dimensional and three dimensional element but in general for any $d$-dimensional case I'm not getting it. Answers with clarifications are welcome.

Answer 1

Realizing that volume-elements are not just products of individual differential elements, but totally anti-symmetrical wedge products of differential 1-forms solve the problem:

$$d^d x : = dx^1 \wedge dx^2 \ldots dx^{n-1} \wedge dx^n$$

where the superscripts are indices (not exponents). With the alternating formalism of differential forms everything works out automatically, in particular Jacobian determinants. Let's take an example:

Spherical coordinates on $S^3$:

$$x^1 = r \sin\chi \sin\theta \cos\phi $$

$$x^2 = r \sin\chi \sin \theta \sin\phi $$

$$x^3 = r \sin\chi \cos \theta $$

$$x^4 = r \cos\chi $$

So in order to get the volume element including the Jacobian we take the differential of $x^i$:

$$dx^1 = r\cos \chi \sin \theta \cos \phi\, d\chi + r\sin\chi\cos\theta\cos\phi\, d\theta - r \sin\chi\sin\theta\sin\phi\, d\phi$$

$$dx^2 = r\cos\chi \sin\theta \sin\phi\, d\chi + r\sin\chi\cos\theta\sin\phi\, d\theta +r \sin\chi\sin\theta\cos\phi\,d\phi$$

$$dx^3 = r \cos\chi \cos\theta\,d\chi - r\sin\chi\sin\theta\, d\theta$$

$$dx^4 = -r \sin\chi \,d\chi$$

In the next step we set up the volume element:

$$ dV =dx^1 \wedge dx^2 \wedge dx^3 \wedge dx^4$$

and plug in the computed differentials. In order to get the result the following rules have to be applied. As the $\wedge$-product is antisymmetric all terms like ($du,dv =d\chi, d\theta$ or $d\phi$)

$$du\wedge du =0 \quad\quad \text{and}\quad\quad du\wedge dv = -dv\wedge du$$

At the end we get:

$$dV = r^3\sin^2\chi \sin\theta d\chi\wedge d\theta\wedge d\phi$$

The Jacobian $J$ is :

$$J = r^3\sin^2\chi \sin\theta $$

So this way it is not necessary to compute the determinant of the transformation matrix like $det(A)=a_{11}a_{22}a_{33}-\ldots $

It is useful in order to learn about the algebra of differential forms in order to fully understand the demonstrated calculation.