I Rindler Transformation & 't Hooft's Introduction to General Relativity

wnvl2
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I am reading 't Hooft introduction to general relativity.

https://webspace.science.uu.nl/~hooft10 ... l_2010.pdf

In this text 't Hoof derives the Rindler transformation.

Image1.png


A little bit further he writes

Image2.png


My question is, how does he come to that formula $$\rho^{-2}g(\zeta)$$
 
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Gentle shudder at the use of ##it## convention. Did he really do that in 2010? Ugh.

The general way of getting the acceleration is to take the modulus of the four-acceleration of the observer who is stationary in Rindler coordinates. There are possible shortcuts in this case, though. Has he established a relationship between time dilation and gravitational potential? If so you can take its gradient to get the acceleration due to "gravity".
 

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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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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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