Recent content by archipatelin

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    Coordinate transformation and metric tensor

    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}...
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    Coordinate transformation and metric tensor

    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...
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    General metric with zero riemann tensor

    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?
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    General metric with zero riemann tensor

    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...
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    General metric with zero riemann tensor

    Yes, it is special case of FLWR metric. But, still don't know this transformation.
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    General metric with zero riemann tensor

    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...
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    How to Calculate the Variation of Quadratic Action for Riemann Tensors?

    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
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    Why isn't the Cotton tensor identical zero?

    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.
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    A ambiguous variation of Einstein-Hilbert action

    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...
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    Why isn't the Cotton tensor identical zero?

    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)?
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    Why isn't the Cotton tensor identical 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...
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    Is variation pricip for light-geodetic correct?

    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...
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    Is variation pricip for light-geodetic correct?

    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...
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    Is variation pricip for light-geodetic correct?

    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...
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    General Lorenz transformation is not group

    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...
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