How Are Gravitational Waves Derived from Einstein's Field Equations?

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How do you get gravitational waves or gravitons out of the EFE? It certainly doesn't look like a wave equation. Are there some second derivatives hidden in the Einstein tensor?
 
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Yes, there are second derivatives hidden in the Einstein tensor. If you think of what it's made up of: Riemann tensor, which is made up of derivatives of the Christoffel symbols, which are made up of derivatives of the metric tensor.
 
You don't get gravitons out of Einstein's field equations; those only show up when you attempt to generate an effective quantum field theory that includes gravity.
For gravitational waves, the easiest method would be to use the weak-field equations in the transverse gauge, and set the energy-momentum tensor to zero (which corresponds to solutions of the equation infinitely far away from the originating source term). After a few lines of basic tensor analysis, you're left with the curved-spacetime version of the homogeneous wave equation, in terms of the d'Alembertian operator.
 

Thread 'Euclidean geometry and gravity'

I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
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Thread 'Is the variation of the metric ##\delta g_{\mu\nu}## a tensor?'

From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
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