# Determinant of the metric tensor

After a change of coordinate system on flat space from $$x\rightarrow y$$, we have the metric tensor:
$$g_{\mu \nu} = \frac{\partial y^{\alpha}}{\partial x^{\mu}} \frac{\partial y^{\beta}}{\partial x^{\nu}}\eta_{\alpha \beta}.$$

Now, after expanding $$y^{\alpha}= x^{\alpha}+\epsilon \xi^{\alpha},$$ I need the determinant $$g$$ in terms of the new variable $$\xi$$. Is there a standard method to do this?

• The determinant is an invariant under coordinate change – Slereah Nov 8 '18 at 17:54
• @Slereah What?! How about changing Cartesian to spherical coordinates, for example? I think you're thinking of orthogonal transformations. – Mike Nov 8 '18 at 18:56
• @Joe Do you know how to calculate the Jacobian? And do you know how to calculate the determinant of the product of two matrices? Have you tried anything? – Mike Nov 8 '18 at 19:00
• @Mike...Yes I do. The determinant of the product is just the product of the det of the two matrices. – Joe Nov 8 '18 at 21:17
• @Mike I think Slereah meant to say $\sqrt{-g} d^4 x$ is invariant under coordinate transformations. $g$ does transform. – Avantgarde Nov 8 '18 at 23:01

Taking the determinant on both sides, you get:

$$g = -\left|\frac{\partial y(x)^\alpha}{\partial x^\beta}\right|^2$$

where $$g = \text{det} (g_{\mu \nu})$$ and $$\text{det} (\eta_{\mu \nu}) = -1$$. On the RHS is the Jacobian (squared) of the coordinate transformation. Can you take it from here?

• @Avantgarde.The Jacobian has to come with partial derivative right? Thanks.. I previously reached the same point too. – Joe Nov 8 '18 at 21:28
• @Joe Yes, thanks! I knew something was amiss when I first wrote it down haha. – Avantgarde Nov 8 '18 at 21:37

Let $$\chi$$ be the coordinate transformation matrix consisting of elements of the form $$\chi = \Big\{\frac{\partial y^\alpha}{\partial x^\beta}\Big\}.$$ The inverse of this matrix $$\chi^{-1}$$ consists of elements of the form: $$\chi^{-1} = \Big\{\frac{\partial x^\beta}{\partial y^\alpha}\Big\}.$$

Therefore we find that the metric $$g$$ (so for example $$\eta=diag(-1,1,1,1)$$) can be transformed as a tensor of rank-$$(0,2):$$

$$g^\prime = (\chi^{-1})^T g \ \chi^{-1}.$$

Taking the determinant we find:

$$\det(g^\prime) = \det((\chi^{-1})^T g \ \chi^{-1}) = \det(g) \det((\chi^{-1})^T)\det(\chi^{-1}) \neq \det(g).$$

The determinant is invariant iff $$\det((\chi^{-1})^T)\ = 1/\det(\chi^{-1})$$.

In general just work out this matrix multiplication and determine $$\det(g^\prime).$$

So we get the following:

$$\det(g^\prime) = - \det(\chi^{-1})^2$$

where $$\det(\chi^{-1})$$ is indeed the Jacobian since $$\det((\chi^{-1})^T)=\det(\chi^{-1})$$ and $$\det(g)=\det(\eta)=-1$$ since you asked for the conversion from flat to non-flat coordinates.