- #1
kdv
- 348
- 6
In that formalism, one writes
[tex] g_{\alpha \beta} dx^\alpha dx^\beta = (-N^2 + N_i N^i) dt^2 + 2 N_j dt dx^j + h_{ij} dx^i dx^j [/tex]
First, a simple question: the lapse function N here is unrelated to the shift vector [itex] N_^i [/tex], right? I mean, clearly [itex] N^2 \neq N_i N^i [/tex] but is there any other relation? I am assuming no.
How are the Latin indices raised and lowered? Is it true that [itex] N_i = h_{ij} N^j [/itex] ?
Now, I have read that a conformal transformation
[tex] g_{\alpha \beta} \rightarrow \Omega^2 g_{\alpha \beta} [/tex]
corresponds to the following transformations of the laspe function and shift vectors:
[tex] N \rightarrow \Omega N, ~~~~N^i \rightarrow N^i ~~~~ h_{ij} \rightarrow \Omega^2 h_{ij} [/tex]
This would make sense if [itex] N_i = h_{ij} N^j [/itex] so that we would have [tex] N_i \rightarrow \Omega^2 N_i [/tex] but I am not sure if I am correct about this way of lowering the indices on N.
Thanks
[tex] g_{\alpha \beta} dx^\alpha dx^\beta = (-N^2 + N_i N^i) dt^2 + 2 N_j dt dx^j + h_{ij} dx^i dx^j [/tex]
First, a simple question: the lapse function N here is unrelated to the shift vector [itex] N_^i [/tex], right? I mean, clearly [itex] N^2 \neq N_i N^i [/tex] but is there any other relation? I am assuming no.
How are the Latin indices raised and lowered? Is it true that [itex] N_i = h_{ij} N^j [/itex] ?
Now, I have read that a conformal transformation
[tex] g_{\alpha \beta} \rightarrow \Omega^2 g_{\alpha \beta} [/tex]
corresponds to the following transformations of the laspe function and shift vectors:
[tex] N \rightarrow \Omega N, ~~~~N^i \rightarrow N^i ~~~~ h_{ij} \rightarrow \Omega^2 h_{ij} [/tex]
This would make sense if [itex] N_i = h_{ij} N^j [/itex] so that we would have [tex] N_i \rightarrow \Omega^2 N_i [/tex] but I am not sure if I am correct about this way of lowering the indices on N.
Thanks