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You seem to still suffer not understanding even the fact that there is no such thing asI already did. Go back and re-read a little more closely. Don't hesitate to ask questions about the parts you either don't understand or feel are incomplete. I glossed over some parts, like the derivation of the equation of the null geodesics.

[tex]\frac{\zeta^{\nu}}{d\lambda}[/tex]

in any part of mathematics; let alone physics.

Where? You're making no sense; sorry! On one hand you said you believe that Fermi coordinates have a specific observer like Rindler coordinates. On the other hand you're saying you did prove that a non-Rindler observer sees the same result as Rindler observers do! (while you didn't as can be seen from you setting a consatnt [tex]x[/tex] for your emitter and receiver!) How can you explain this contradiction? More interestingly unreasonable is your leaky understanding of what a Rindler observer is:I already did. Go back and re-read a little more closely. Don't hesitate to ask questions about the parts you either don't understand or feel are incomplete. I glossed over some parts, like the derivation of the equation of the null geodesics.

Three points must be made here:Obviously. I thought that was clear from the fact that the emitter is on the ceiling and the detector is on the floor of an accelerating elevator. Most elevator designers try to avoid situations where the floor accelerates without the ceiling accelerating also.

1- Looking at the acceleration vector of a Rindler observer,

[tex]\nabla_{\arrow{e}_0} \arrow{e}_0=\frac{1}{x}\frac{\partial}{\partial x}[/tex]

it is obvious that these are accelerating with

**constant magnitude**in the direction of [tex]\frac{\partial}{\partial x},[/tex] suggesting [tex]x=const.[/tex]

2- These observers have in Minkowski coordinates hyperbolic world lines with the same asymptotes [tex]T=+X[/tex] and [tex]T=-X[/tex]; the first coincides with the world line of a photon moving along the [tex]X[/tex]-axis in the same direction as the accelerated observer (particle). Such observers are indeed uniformly accelerated that altogether form a Rindler reference frame within this frame physics is exactly the same as when the spacetime is analized using Minkowski coordinates. If the observer is no longer uniformly accelerating, then it is not called Rindler to whom according to the equivalence principle, physics only looks the same in a very small region. This is because the physical laws in a

**local**reference frame in a gravitational field are equivalent to the physical laws in a uniformly accelerated frame.

3- The "time translation symmetry" property of Rindler metric only holds true for Rindler observers (look at the coframe field of Rindler metric and its first component i.e. starting with[tex]\sigma^0=-xt[/tex] one would lead to [tex]d\sigma^0=-xdt[/tex] only iff [tex]x=const.[/tex]) which itself suggests that "physics as a whole has not changed" holds true in this case if this property resembles a corresponding property in Minkowski spacetime (basically "boost symmetry") that is just satisfied for Rindler observers.

Now if you got my point that a Rindler observer appears to be only uniformly accelerating, then stop making jerry-built claims that your example applies to non-Rindler observers as well. If you do not have prior studies on a physical problem, you're not forced to launch yourself into the related discussions.

I never said such nonsense. Either you're 1) escaping from the first fallacy you made here concerning OP's question which Weinberg and Papaterou corrected it or 2) overshadowing the second fallacy that straight line in GR is given by the transfomed formula of the straight line in the Euclidean space or 3) digressing from the main problem by confusing the Kruskal example with Rindler and making some fallacious claim concerning the latter.Altabeh, I notice that you are still unable to demonstrate any experimental measurement which depends on the choice of Rindler or Minkowski coordinates.

I precisely stated everything above and it all falls upon your mind to be able to get what is going on in physics in this case.

AB

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