2
$\begingroup$

The definitions of covariant and contravariant tensor quantities are that they transform as $A' ^i=\frac{\partial x_j}{\partial x'_i} A^j$ and $A'_i=\frac{\partial x'_i}{\partial x_j}A_j$ respectively, and they are related by the metric tensor $g^{\mu\nu}$ so that $A^\mu=g^{\mu\nu}A_\nu$. Is there any way to calculate the coefficients of $g^{\mu\nu}$ from the coordinate transforms listed above (e.g. $g^{\mu\nu}=\left(\frac{\partial x_j}{\partial x'_i}\right)^2$) or is the metric the most intrinsic (read: incalculable) part of the spacetime?

$\endgroup$
  • 1
    $\begingroup$ You can certainly not compute the metric tensor from the coordinate transforms for tensor quantities, which don't have any dependence on the metric tensor and will always look the same. On the other hand, if you are given $A^\mu$ and $A_\mu$ in coordinates, that does gives you 4 linear equations the coefficients of $g_{\mu\nu}$ have to satisfy, so if you have a enough of such pairs, you can reconstruct $g_{\mu\nu}$. Finally, the metrix is the most intrinsic part of the spacetime, but that doesn't necessarily mean that it cannot be reconstructed from derived quanitities. $\endgroup$ – doetoe Jul 28 at 0:12
  • $\begingroup$ Your transformation is wrong. Coordinates should have upper indices and the primes should be on the top of the partial derivative. $\endgroup$ – G. Smith Jul 28 at 0:45
0
$\begingroup$

What we tipically call the metric is $g_{\mu \nu}$, whereas the inverse metric is called $g^{\mu \nu}$ and it is defined through $ g^{\mu \alpha} g_{\alpha \nu} = \delta^{\mu}_{\nu}$. The metric transforms as a 2-rank covariant tensor, i.e. $ g'_{\mu \nu} = \frac{\partial x'_{\mu}}{\partial x_{\alpha}} \frac{\partial x'_{\nu}}{\partial x_{\beta}}g_{\alpha \beta}$. All the information about the geometry of spacetime is encapsulated in $g_{\mu \nu}$. In the context of general relativity, it is obtained from the energy and matter content through Einstein equations: $G_{\mu \nu} = 8 \pi G T_{\mu \nu}$, where $G_{\mu \nu}$ is a very complicated tensor that depends on the first and second derivatives of the metric, and $T_{\mu \nu}$ is the stress-energy tensor. Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.