What exactly is the Jacobian in the context of a metric tensor?

Question

Im going over christoffel symbols and tensors and $g$ is defined as the "determinant of a metric tensor" and the square root of $g$ is defined as the Jacobian. My questions are what exactly is a Jacobian and what does it tell us as well as what exactly is meant by "determinant of a metric tensor"?

Answer 1

The Jacobian matrix is a tool used to transform between coordinate systems by taking the rate of change of each component of an old basis with respect to each component of a new basis and expressing them as coefficients that make up an old basis. For example, the Jacobian matrix that transforms polar coordinates to cartesian coordinates in 2 dimensions is given by: $$J_j^i=\frac{\partial{x^i}}{\partial\tilde{x}^j}=\begin{bmatrix}\frac{\partial{x^1}}{\partial\tilde{x}^1} & \frac{\partial{x^1}}{\partial\tilde{x}^2}\\\frac{\partial{x^2}}{\partial\tilde{x}^1}&\frac{\partial{x^2}}{\partial\tilde{x}^2}\end{bmatrix}\Rightarrow \begin{bmatrix}\frac{\partial{x}}{\partial{r}} & \frac{\partial{x}}{\partial\theta}\\\frac{\partial{y}}{\partial{r}}&\frac{\partial{y}}{\partial\theta}\end{bmatrix}=\begin{bmatrix}\frac{\partial}{\partial{r}}(rcos\theta) & \frac{\partial}{\partial\theta}(rcos\theta)\\\frac{\partial}{\partial{r}}(rsin\theta)&\frac{\partial}{\partial\theta}(rsin\theta)\end{bmatrix}=\begin{bmatrix}cos\theta&-rsin\theta\\sin\theta&rcos\theta\end{bmatrix}$$ The components of cartesian basis vectors can now be written as a linear combination of these coefficients and their corresponding polar bases. $$\overrightarrow{e_x}=\frac{\partial{x}}{\partial{r}}\overrightarrow{e_r}+\frac{\partial{x}}{\partial\theta}+\overrightarrow{e_\theta}=cos\theta\overrightarrow{e_r}-rsin\theta\overrightarrow{e_\theta}$$ $$\overrightarrow{e_y}=\frac{\partial{y}}{\partial{r}}\overrightarrow{e_r}+\frac{\partial{y}}{\partial\theta}\overrightarrow{e_\theta}=sin\theta\overrightarrow{e_r}+rcos\theta\overrightarrow{e_\theta}$$ In the case of relating the Jacobian to the metric tensor, the Jacobian can be used to transform the components of one metric to another via the following method: $$\tilde{g}_{ij}=\frac{\partial}{\partial\tilde{x}^i}\cdot\frac{\partial}{\partial\tilde{x}^j}=\left(\frac{\partial{x^a}}{\partial\tilde{x}^i}\frac{\partial}{\partial{x^a}}\right)\cdot\left(\frac{\partial{x^b}}{\partial\tilde{x}^j}\frac{\partial}{\partial{x^b}}\right)$$ $$=\frac{\partial{x^a}}{\partial\tilde{x}^i}\frac{\partial{x^b}}{\partial\tilde{x}^j}\left(\frac{\partial}{\partial{x^a}}\cdot\frac{\partial}{\partial{x^b}}\right)=J_i^aJ_j^bg_{ab}$$ Knowing this, taking the determinant of the metric $g_{ij}$ requires taking the determinant of the Jacobian matrices and $g_{ab}$ as well. $$det\tilde{g}_{ij}=(detJ_i^a)(detJ_j^b)(detg_{ab})$$ Since both Jacobian terms are part of the same matrix and are just written using different indices to differentiate between the components of the old basis, $$(detJ_i^a)(detJ_j^b)=(detJ)^2$$ In the case of our old basis being written in cartesian coordinates: $$detg_{ab}=det\begin{bmatrix}1&0\\0&1\end{bmatrix}=1$$ Therefore the equation of our new transformed metric $g_{ij}$, simplifies to: $$det\tilde{g}_{ij}=(detJ)^2\implies detJ=\sqrt{det\tilde{g}_{ij}}$$ One physical meaning of the determinant of a metric tensor is that its square root is equal to the volume form ω of a group of vectors. This makes intuitive sense upon recognizing how the determinant of an ordinary matrix also gives you the volume form of a group of vectors within a vector space. The metric tensor merely represents the squared value of those vectors involved in the volume form (if you treat the vectors as basis vectors).