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The stress-energy tensor, is up to multiplicative factors, can be defined by $\frac{\delta S}{\delta g^{\mu\nu}}$, where $S$ is the action and $g_{\mu\nu}$ is the metric. When people talk about the graviton, they talk about quantizing the metric around it's classical solution, so we consider field values $g_{\mu\nu} = g^{(c)}_{\mu\nu} + h_{\mu\nu}$, where ...

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It is not known yet. Gravitons are from quantum mechanics model, while stress-energy tensor is from General relativity (GR) model. Two models are not connected until quantum gravity created. Also, gravitons were never observed, so they are pretty hypothetical. Simultaneously, it is known, that metric tensor is "generated" by stress-energy tensor. ...

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A general vector is written as $v = v^\mu \partial_\mu$. Its norm is defined as $$g(v,v) = g_{\mu\nu}(\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu)(v^\mu\partial_\mu,v^\nu\partial_\nu) = g_{\mu\nu}v^\mu v^\nu (\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu)(\partial_\mu,\partial_\nu) = g_{\mu\nu}v^\mu v^\nu$$ where we have used linearity of the duals and ...

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$\epsilon_{ab}$ are the components of a tensor, so you can raise and lower indices with the metric: \begin{align} \epsilon^a{}_b & = g^{ac} \epsilon_{cb} \\ \epsilon_b{}^a & = g^{ac} \epsilon_{bc}. \end{align} Note that order matters: $\epsilon^a{}_b = -\epsilon_b{}^a$. Since it seems you are just relying on heuristics for when signs get ...

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There appears to be a typo in the coordinate transformation for $r$: it should be $$r = \rho\,\exp(\tau/\alpha).$$ With this in mind, your reasoning is correct; if you work out the partial derivatives, you'll find that $g_{\tau\tau} = 1$, and you can work out $g_{\rho\rho}$ in a similar way. P.S. also note that $t = \tau - ... 0 Aren't you basically asking to the formula of higher dimensional vectors to compute yourself being in the sphere at the center with appropriate lines to all vector points? I think I get what your asking for I asked the same questions when I was younger Anyways here's a hyperlink to 6th dimensional operation. The operator is used in quantum physics, but ... 1 Since the metric$g_{\mu\nu}=g_{\nu\mu}$is symmetric, we must demand that $$\tag{1} \delta g_{\mu\nu}~=~\delta g_{\nu\mu}~=~\frac{1}{2}\left(\delta g_{\mu\nu}+\delta g_{\nu\mu}\right)~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\delta g_{\alpha\beta},$$ and therefore $$\tag{2} ... 2 The singularity comes from the scale factor a(t):$$ds^2 = -dt^2 + [a(t)]^2 ( dr^2 + r^2 d \Omega^2)$$By solving the Friedmann equations for the scale factor we know that:$$a(t) = a_0 t^{\lambda}$$where$\lambda$is some positive number that depends on the matter-radiation ratio of the universe. At$t=0$the scale factor becomes$a(0)=0$. So at ... 1 Wald is a first rate relativist, and as such he is phrasing the concept of general covariance in terms of purely geometrical quantities, rather than resorting to the somewhat imprecise notion of coordinate transformations. In the discussion on pg. 57, he goes on to give an example of what it means to violate the principle of general covariance. In his ... 1 All curves can be parameterised by an affine parameter (commonly written as$\lambda$in the GR books I have). However only for timelike curves does the parameter$\lambda$have a physical meaning i.e. it's the elapsed time$\tau\$ shown on a clock by the observer following the curve. So there's no mathematical difference between parameterising a timelike ...

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Yes, the proper time along a timelike curve in relativity is very much analogous to the length of a curve. Just as the length of a curve is invariant under rotations, the proper time along a curve is invariant under Lorentz transformations. One difference with conventional length is that although a straight line is the shortest length between two points, a ...

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General covariance basically means you can change your coordinate system arbitrarily and express the laws of physics in the new coordinates. Because of this freedom, the relationship between coordinate distances, angles, etc. and physical distances, angles, etc. is variable and is expressed by the metric. So the quoted statement is basically saying that ...

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