# Thread: Jacobian determinant of a direction vector as a function of a twodimensional vector?

1. ## Jacobian determinant of a direction vector as a function of a twodimensional vector?

I have an integral on the following form:

$\displaystyle \int_{\mathbb{R}^2}f(\vec{r})\,\operatorname{d} \vec{r},$

where $\displaystyle \vec{r}$ is a two-dimensional vector. However, I rather want the integral on the form

$\displaystyle \int_{\Omega}g(\widehat{\omega})\,\operatorname{d} \widehat{\omega},$

where $\displaystyle \widehat{\omega} = \vec{r}\,'/|\vec{r}\,'|$ and $\displaystyle \vec{r}\,' = (\vec{r},\, 1),$ i.e. $\displaystyle \vec{r}\,'$ is an extension of $\displaystyle \vec{r}$ from two dimensions to three dimensions where 1 has been added as a third element, and $\displaystyle \Omega$ is the unit hemisphere in which $\displaystyle \widehat{\omega}$ can exist according to its definition. We can see that while the differentials in the two integrals, $\displaystyle \operatorname{d}\vec{r}$ and $\displaystyle \operatorname{d}\widehat{\omega}$, have a different number of dimensions, the domains, $\displaystyle \mathbb{R}^2$ ans $\displaystyle \Omega$ , are both two-dimensional, which makes changing from one of the integrals to the other possible.

Normally when you change differential you use the Jacobian $\displaystyle J$ in the following way

$\displaystyle \int_{\mathbb{A}}f(\vec {F})\,\operatorname{d}\vec{F} \,=\, \int_{\mathbb{B}}f(\vec{x})\,\left|\frac{\text{d} \vec{F}}{\operatorname{d} \vec{x}}\right|\,\operatorname{d} \vec{x} \,=\, \int_{\mathbb{B}}f(\vec{x})\,\left|J_{\vec{F}}( \vec{x})\right|\operatorname{d}\vec{x}.$

where

$\displaystyle J_{\vec{F}}(\vec{x}) \,=\, \frac{\operatorname{d} \vec{F}}{\operatorname{d} \vec{x}} \,=\, \det \begin{bmatrix} \dfrac{\partial F_1}{\partial x_1} & \cdots & \dfrac{\partial F_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial F_m}{\partial x_1} & \cdots & \dfrac{\partial F_m}{\partial x_n} \end{bmatrix}.$

But in my case, when $\displaystyle \vec{r}$ is two-dimensional but $\displaystyle \widehat{\omega}$ is three-dimensional, how do I carry out the corresponding transformation? Since the Jacobian is the size of the hypervolume that is generated by all partial derivatives of $\displaystyle \vec{F}$, the closest I can get to a Jacobian, since the Jacobian itself cannot be formed in this case (the two vectors need to have the same number of dimensions), is

$\displaystyle \frac{\operatorname{d} \widehat{\omega}}{\operatorname{d} \vec{r}} \,=\, \left(\frac{\operatorname{d}\widehat{\omega}}{ \operatorname{d} r_1} \times \frac{\operatorname{d} \widehat{\omega}}{\operatorname{d} r_2}\right) \cdot \widehat{\omega}$

where $\displaystyle \operatorname{d} \widehat{\omega}$ is an infinitesimal surface area on $\displaystyle \Omega$ and $\displaystyle \operatorname{d}\vec{r}$ is the infinitesimal surface area on $\displaystyle \mathbb{R}^2$ that gives rise to $\displaystyle \operatorname{d} \widehat{\omega}$. So my final integral should look like

$\displaystyle \int_{\Omega}f(\widehat{\omega})\,\frac{\text{d} \vec{r}}{\operatorname{d} \widehat{\omega}}\,\operatorname{d}\widehat{\omega } \,=\, \int_{\Omega}f(\widehat{\omega})\,\left(\frac{ \operatorname{d} \widehat{\omega}}{\operatorname{d} \vec{r}}\right)^{-1}\,\operatorname{d}\widehat{\omega} \,=\, \int_{\Omega}f(\widehat{\omega})\,\left(\left( \frac{\operatorname{d} \widehat{\omega}}{\operatorname{d} r_1} \times \frac{\operatorname{d} \widehat{\omega}}{\operatorname{d} r_2}\right) \cdot \widehat{\omega}\right)^{-1}\,\operatorname{d}\widehat{\omega}.$

Now, I know intuitively that this is the way I should do the transformation, but how can I show it mathematically in the simplest way? How can I show that this is my "Jacobian" (although it isn't really a Jacobian) and that I really should use it?