I Cosmic Test Reveals Odd Findings for Einstein's Relativity

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What the popular press writes and a scientific paper says may be somewhat different.
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The paper is more than a year old and has 15 citations, not zero, and not a zillion. That tells you that it is more or less typical among such papers.
 
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Vanadium 50 said:
The paper is more than a year old and has 15 citations

I first read "the paper is more than 15 years old".
Need more coffee...
 
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malawi_glenn said:
I first read "the paper is more than 15 years old".
In some frame, it is.
 
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I think the gist of it is, if you assume a certain class of theories (which include "vanilla" GR) contains the correct theory of gravity, and assume some prior on those theories, and then do a Bayesian estimate of the best fit parameters to predict our current measures of some key cosmological numbers, then GR is not the favoured model. But it's not a resoundingly significant (in the statistical sense) result, and I've no idea how plausible cosmologists in general find their assumptions.
 
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)...
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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