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I read in my class notes that the professor said using equivalence principle, it is always possible to choose around any point $p$, a 2-dimensional space-like neighborhood such that the expansion and shear of the null congruence vanishes on this 2-dimensional neighborhood. It does not look trivial to me. How can I prove this?

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  • $\begingroup$ Is Null Geodesic Congruences, Asymptotically-Flat Spacetimes and Their Physical Interpretation (EG: sections 3&4, page 40) or Black holes and Holography (EG: pages 39&40) or Shear free solutions in General Relativity Theory (EG: page 8) the more helpful? $\endgroup$ – Rob Jun 2 '18 at 14:23
  • $\begingroup$ @Rob I looked at them but still it does not look trivial for me. Is it because I can always around a point arrange normal coordinates such that the metric is Lorentzian and its derivative zero: so I can get $k=\partial_{t-r}$ and $N=\partial_{t+r}$ and so the first derivative vanishes? $\endgroup$ – mathvc_ Jun 2 '18 at 14:53
  • $\begingroup$ I found they are not trivial for me to explain, So I thought I'd ask if where I left off was helpful in specifying what level of complexity you wanted for your answer. Much like saying if you stand at the North Pole north directions disappear - easy to say, now for the proof ... Wait until you get to that part of the lesson, ask, answer your own question ... I would need some more study to make a simple answer that I was comfortable with, if you specify the complexity better some genius will be by with the trivial explanation shortly. I'll delete these comments next hour. $\endgroup$ – Rob Jun 2 '18 at 15:15

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