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It is said that we can introduce local inertial coordinates/Fermi normal coordinates for any timelike geodesic. But why only for timelike geodesics? What about null geodesics? Perhaps it has to do with invertibility or something?

Qmechanic
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Bess
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2 Answers2

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We assume OP's question (v2) is the following:

Given a null geodesic on a Lorentzian manifold, do there locally exist Fermi normal coordinates along the null geodesic? (Here the word 'locally' means in some tubular neighborhood.)

The answer is Yes, see. e.g. Ref. 1. (As OP correctly notes, most textbooks deal only with Fermi normal coordinates for timelike geodesics, cf. e.g. Ref. 2 and Ref. 3.)

References:

  1. M. Blau, D. Frank, and S. Weiss, Fermi Coordinates and Penrose Limits, Class. Quant. Grav 23 (2006) 3993, http://arxiv.org/abs/hep-th/0603109

  2. MTW.

  3. E. Poisson, The Motion of Point Particles in Curved Spacetime, (2004), http://www.livingreviews.org/lrr-2004-6

Qmechanic
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But why only for timelike geodesics? What about null geodesics? Perhaps it has to do with invertibility or something?

Physically, Fermi normal coordinates represent the frame of reference of an inertial observer. Relativity doesn't allow observers with lightlike motion, and, yes, one way to understand why they're not allowed is that if you try to extend the Lorentz transformation to $v=c$, it's not one-to-one.

Another way of saying this is that in Fermi normal coordinates, we're trying to get the metric to look like a diagonal matrix with elements $(1,-1,-1,-1)$. This requires that three coordinates be spacelike and one timelike. It's not going to work if one coordinate is lightlike.