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Mathphys meister
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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.

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 $\det(g^\prime).$

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.

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Mathphys meister
  • 1.2k
  • 1
  • 9
  • 22

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 $\det(g^\prime).$