Recent content by archipatelin

I mean globally transformation. I know that the general two-dimensional metric can be transformed into the orthogonal traceless form \[ \left( \begin{array}{cc} g_{00}(t,x) & g_{01}(t,x) \\ g_{01}(t,x) & g_{11}(t,x) \end{array} \right)\rightarrow \left( \begin{array}{cc}...

General four-dimensional (symmetric) metric tensor has 10 algebraic independent components. But transformation of coordinates allows choose four components of metric tensor almost arbitrarily. My question is how much freedom is in choose this components? Do exist for most general metric...

I want to make sure that all solutions of the equation R_{\mu\nu\varkappa\lambda}=0 for any dimension D are isomorphic with tensor in form g_{\mu\nu}=\mbox{diag}(\pm{}1,\pm{}1,\dots,\pm{}1) Or are there other solutions?

Thanks, I found this transformation in form: \rho\equiv{}\rho(t,r) \tau\equiv{}\tau(t,r) Solutions with respect to minkowski metric are: \rho=\frac{2t}{r}\left(1\pm\sqrt{1-\left(\frac{r}{2t}\right)^2}\right) \tau=\frac{1}{2}r\frac{1-\rho^2}{\rho} But it's regular only for...

Yes, it is special case of FLWR metric. But, still don't know this transformation.

A metric consistent with interval: \mathrm{d}s^2=-\mathrm{d}\tau^2+\frac{4\tau^2}{(1-\rho^2)^2}\left(\mathrm{d}\rho^2+\rho^2\mathrm{d}\theta^2+\rho^2\sin(\theta)^2\mathrm{d}\varphi^2\right) get zero for riemann's tensor, therefor must be isomorphic with minkowski tensor. But I don't find thus...

Do know anybody explicit form of variation action quadratic in Riemann tensors (for general dimension)? Link to internet sources? Or computer program for symbolic and tensors algebra, which the variation tell me (preferably open-source)? Thx

This is correct, because I made another big mistake. Weyl tensor isn't governed of analog II. Bianchi identity. Therefore cotton tensor is non-zero generaly for n>2.

A ambiguous variation of Einstein--Hilbert action Variation of EH action is: \delta S_{EH}=\int_{\Omega}{\delta(R\sqrt{-g})dx^4}= \int_{\Omega}{G_{\mu\nu}\delta{g^{\mu\nu}}\sqrt{-g}dx^4}=0, where G_{\mu\nu}:=R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R is symmetric einstein's...

If do you think case when n=3. Yes, I did a big mistake. But I suppose n\geq4. Or Cotton tensor exist only for n=3 space (for n=2 it is also zero)?

Cotton tensor C_{\mu\varkappa\lambda} is define as: \nabla_{\sigma}W^{\sigma}_{\phantom{M}\mu\varkappa\lambda}=-\frac{n-3}{n-2}C_{\mu\varkappa\lambda} where W^{\sigma}_{\phantom{M}\mu\varkappa\lambda} is Weyl tensor and n is dimension of space. Weyl tensor obey II. Bianchi identity (and...

I don't understand it. How this help us? This corrcet for special relativity only. General relativity term is: g_{00}dt^2+2g_{0k}dx^kdt+\gamma_{ik}dx^kdx^i=0 equally, for energy of light is: g_{00}E^2+2g_{0k}p^kE+\gamma_{ik}p^kp^i=0 and your action in general is...

Thank you for your clarification. But, if I understand it well, then your def. of action has this same problems as previous. 1) you cannot use d\tau (the proper time) for integration, because is for light-like vector null, but another affine parametr d\lambda (as wrote Jonathan Scott), which...

Princip stacionary action for propagation of light is apply on thus definition of action: S=\int\!\mbox{d}\tau=\frac{1}{c}\int\!\sqrt{\mbox{d}x_{\mu}\mbox{d}x^{\mu}}=\frac{1}{c}\int\!\sqrt{g_{\mu\nu}\frac{\mbox{d}x^{\nu}}{\mbox{d}\tau}\frac{\mbox{d}x^{\mu}}{\mbox{d}\tau}}\mbox{d}\tau The...

Thank you everybody for help! I see. I thought intuitively (corse wrongly), that they frames are mutually in parallel relation. For special lorentz transformation (movement is on shared axis X), where is contracting just axis X and Y,Z are no changes, relation of parallel is true. But for...