Tagged Questions
1
vote
1answer
38 views
What is the Lorentz tensor with a superscript and subscript index?
I have been reading about symmetries of systems' actions, e.g. the Polyakov action, and I have encountered Lorentz transformations of the form: $\Lambda^{\mu}_{\nu} X^{\nu}$. I am moderately familiar ...
3
votes
1answer
66 views
Derivation of the volume element (which uses the metric tensor)?
I have often seen $\sqrt{-g}$ in integrals, especially actions, where $g=\mathrm{det}(g_{\mu \nu})$. Does anyone know of a derivation that shows that this is indeed the volume element which must be ...
3
votes
1answer
91 views
When a variation of a tensor is not a tensor?
In a comment about variation of metric tensor it was shown that
$$\delta g_{\mu\nu}=-g_{\mu\rho}g_{\nu\,\sigma}\delta g^{\rho\,\sigma}$$
which is contrary to the usual rule of lowering indeces of a ...
0
votes
1answer
41 views
Contraction of the metric tensor
This is perhaps a simple tensor calculus problem -- but I just can't see why...
I have notes (in GR) that contains a proof of the statement
In space of constant sectional curvature, $K$ is ...
3
votes
2answers
110 views
Difference between slanted indices on a tensor
In my class, there is no distinction made between,
$$
C_{ab}{}^{b}
$$
and
$$
C^{b}{}_{ab}.
$$
All I know, and read about so far, is the distinction of covariant and contravariant, form/vector, etc. ...
2
votes
1answer
41 views
Non-diagonal elements when switching metric signature?
Considering a metric tensor with the signature $(-,+,+,+)$:
$g_{\mu\nu}=
\begin{pmatrix}
-c^2 & g_{01} & g_{02} & g_{03}\\
g_{10} & a^2 & g_{12} & g_{13}\\
g_{20} & g_{21} ...
1
vote
1answer
108 views
Covariant derivative with upper index
I just need clarification, that is, to see that I'm doing the right thing.
When calculating central charge for certain metric, I need to solve an integral that contains Lie brackets etc. And I have ...
7
votes
3answers
617 views
Why is the covariant derivative of the metric tensor zero?
I've consulted several books for the explanation of why
$$\nabla _{\mu}g_{\alpha \beta} = 0,$$
and hence derive the relation between metric tensor and affine connection $\Gamma ^{\sigma}_{\mu ...
2
votes
1answer
92 views
Question from Schutz's
In q. 22 in page 141, I am asked to show that if $U^{\alpha}\nabla_{\alpha} V^{\beta} = W^{\beta}$, then $U^{\alpha}\nabla_{\alpha}V_{\beta}=W_{\beta}$.
Here's what I have done:
$V_{\beta}=g_{\beta ...
