# Layman's doubts about Gen Relativity

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.
You seem to still suffer not understanding even the fact that there is no such thing as
$$\frac{\zeta^{\nu}}{d\lambda}$$

in any part of mathematics; let alone physics.

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.
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 $$x$$ for your emitter and receiver!) How can you explain this contradiction? More interestingly unreasonable is your leaky understanding of what a Rindler observer is:

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.
Three points must be made here:

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

$$\nabla_{\arrow{e}_0} \arrow{e}_0=\frac{1}{x}\frac{\partial}{\partial x}$$

it is obvious that these are accelerating with constant magnitude in the direction of $$\frac{\partial}{\partial x},$$ suggesting $$x=const.$$

2- These observers have in Minkowski coordinates hyperbolic world lines with the same asymptotes $$T=+X$$ and $$T=-X$$; the first coincides with the world line of a photon moving along the $$X$$-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$$\sigma^0=-xt$$ one would lead to $$d\sigma^0=-xdt$$ only iff $$x=const.$$) 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.

Altabeh, I notice that you are still unable to demonstrate any experimental measurement which depends on the choice of Rindler or Minkowski coordinates.
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.

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

Last edited:
Dale
Mentor
there is no such thing as
$$\frac{\zeta^{\nu}}{d\lambda}$$
Oops, you are correct, I did make a typo. It should of course read:
$$\frac{d\zeta^{\nu}}{d\lambda}$$
Unfortunately I can no longer go back and correct the original.

On one hand you said you believe that Fermi coordinates have a specific observer like Rindler coordinates.
Yes.

On the other hand you're saying you did prove that a non-Rindler observer sees the same result as Rindler observers do!
Yes. I worked the same problem in Rindler coordinates for a Rindler observer and in Minkowski coordinates for a Minkowski observer and they agreed on the result.

DaleSpam said:
you are still unable to demonstrate any experimental measurement which depends on the choice of Rindler or Minkowski coordinates
I never said such nonsense.
Are you now agreeing with me that the measured result of any physical experiment does not depend on the choice of coordinate system used to analyze the experiment?

Assuming that is correct then let me state the following and see if you agree:

1) The result of any physical measurement does not depend on the coordinate system used to analyze the experiment

2) Experimentally, an inertial object is one where an attached accelerometer measures 0

3) By combining 1 and 2 all coordinate systems must agree if an accelerometer reads 0 and therefore all coordinate systems agree if an object is inertial

Last edited:
atyy
I think once again there is just a slight difference in terminology.

Altabeh's point is that a Lorentz inertial frame and a Rindler frame can each be associated with an observer. Even if these two observers are colocated and comoving, some of their measurements will differ (eg. frequency, due to Doppler effects).

Of course the difference between the measurements of the Lorentz inertial observer and Rindler observer will be the same in any coordinate system, since both observers exist in all coordinate systems, which is DaleSpam's point.

And I think we have all agreed that a freely falling observer has zero proper acceleration (following eg. Rindler and Wald's definition) along its entire worldline. With this observer, we can associate Fermi normal coordinates in which the Christoffel symbols are zero along his entire worldline, and so can define the freely falling observer to be "inertial" in a sense.

We also all agree that Fermi normal coordinates are only locally inertial, and that the Christoffel symbols are not zero once they are off the free falling wordline, and in fact even the worldline the second derivatives of the Christoffel symbols are not zero, so there is no global inertial frame associated with a freely falling observer, which is the answer to the third question of the OP, which was Altabeh's point in post #6.

?

Last edited:
Dale
Mentor
there is no global inertial frame associated with a freely falling observer, which is the answer to the third question of the OP, which was Altabeh's point in post #6.
The OP later clarified that he was interested in the "inertialness" of the object itself, not the local coordinate system.

atyy
The OP later clarified that he was interested in the "inertialness" of the object itself, not the local coordinate system.
3)When looked upon from Earth does a falling stone makes non-inertial frame of reference?
Is it safe now to tell that a very small particle in free fall, undergoing translational acceleration (in the kinematic sense) over a very long distance is actually an inertial frame for people on a perfectly still Earth?
This means that an accelerometer at rest gives the value of g at that point and a falling accelerometer records zero acceleration. I think it is a better and intuitive answer to my question.
The accelerometer mechanism is actually functioning in the frame of reference of the object itself. So, its zero reading doesn't shows us that it was in inertial motion (for us).

I've concluded that Altabeh's argument was right. The local flatness theorem holds only for infitesimal regions in space-time curvature and due to the long trajectory of the particle, it is non-inertial from outside. The continuous photon blueshifting/redshifting and apparent modification in c are some phenomenon that exhibit this thing.
The OP's question was a quantum superposition, which as we all know is not relativistically invariant, and hence observer dependent :tongue2:

George Jones
Staff Emeritus
Gold Member
We also all agree that Fermi normal coordinates are only locally inertial
Did you mean to write "We also all agree that for a freely falling observer, Fermi normal coordinates are only locally inertial"?

Rindler coordinates are Fermi normal coordinates.

atyy
Did you mean to write "We also all agree that for a freely falling observer, Fermi normal coordinates are only locally inertial"?

Rindler coordinates are Fermi normal coordinates.
Yes - thanks!

Yes. I worked the same problem in Rindler coordinates for a Rindler observer and in Minkowski coordinates for a Minkowski observer and they agreed on the result.
Thank goodness, you ultimately corrected yourself because I asked you

Could you now prove that a non-Rindler observer would also lead to the same result as in your example?

This shows you're now agreeing that Physics only in the eyes of the specific observer of Rindler coordinates doesn't change.

Are you now agreeing with me that the measured result of any physical experiment does not depend on the choice of coordinate system used to analyze the experiment?
I'm not "now" agreeing with you! This is all because you came to a "right" conclusion above as to what I was trying to say all along since the beginning of this thread. I gave away an example about Schwarzschild coordinates and said that is not "always" believable to say something like this and this had nothing to do with Rindler coordinates or any other experimentally proven result of the coordinate-independence property of any physical phenomena. As I already said, this could be probably true.

1) The result of any physical measurement does not depend on the coordinate system used to analyze the experiment
Do you remember I said there is a global transformation to bring Minkowski into Rindler form? From all observers using this transformation, only to those who are called Rindler the result of any physical measurement does not seem to depend on the coordinate system (Minkowski or Rindler.) That is to say, all observers uniformly accelerating with respect to each other agree on the result that Minkowski observers have measured in their own frame, as your example correctly shows this. The same thing is applicable in case of using Fermi coordinates.

2) Experimentally, an inertial object is one where an attached accelerometer measures 0
Yes but this is not globally possible in GR.

3) By combining 1 and 2 all coordinate systems must agree if an accelerometer reads 0 and therefore all coordinate systems agree if an object is inertial
I think atty responded to this in an awesome way:

Altabeh's point is that a Lorentz inertial frame and a Rindler frame can each be associated with an observer. Even if these two observers are colocated and comoving, some of their measurements will differ (eg. frequency, due to Doppler effects).

And I think we have all agreed that a freely falling observer has zero proper acceleration (following eg. Rindler and Wald's definition) along its entire worldline. With this observer, we can associate Fermi normal coordinates in which the Christoffel symbols are zero along his entire worldline, and so can define the freely falling observer to be "inertial" in a sense.
The OP later clarified that he was interested in the "inertialness" of the object itself, not the local coordinate system.
The OP clearly said in an early post he felt like my answer to his question was right. There was no talk of inertialness, I think.

I think we are now in a position to reach a middle ground on the issue. If no, please let me know.

AB