Layman's doubts about Gen Relativity

[djy]

Not so much correct. Only spacetime around the entire geodesic can be made flat using a special kind of observer and in general such situation cannot be defined for every observer. But if you still back the argument that along any geodesic local flatness can just be guaranteed from the perspective of all observers then there is no more question on your post.

AB

Whether spacetime is flat is frame-independent. It can't be "made flat" by different observers and coordinate systems.

 

[Dale]

I thought your background metric belongs to the spacetime around some gravitating body

Sorry about that. I was not clear in my earlier post.

Yet again this doesn't change anything ever because though the dynamics of spacetime does now seem to have changed

Yes, exactly. Nothing physical is changed, but the coordinate transform leads to non-zero ordinary derivatives along a clearly inertial worldline. Therefore (even in flat spacetime) zero ordinary derivatives do not generally define an inertial worldline, whereas zero covariant derivatives do.

In other words, as you yourself stated, nothing physical has changed, but the result of your formula has. So in general your formula reflects the peculiarities of the coordinate system rather than just the details of the underlying physical situation.

 

[Austin0]

Yes, it is for a specific observer. For the entire trajectory, the observer is a local co-moving co-accelerating observer. For a single point on the trajectory, the observer is a local, instantaneously co-moving (but not necessarily co-accelerating) observer. For local flatness I am using atyy's concept of "ultralocal pseudoflatness" which I rather like the sound of

A local co-moving co-accelerating observer; is that the guy next to him in the elevator? ;-)

Isnt the idea of a comoving but non-accelerating observer somewhat oxymoronic??



How about non inertial observers hovering with thrust and spread out over the total path for local measurements??

 

[Dale]

Altabeh, expanding on my previous comments. In flat spacetime an inertial object has a straight worldline. This is a coordinate-independent geometric statement. You are certainly free to use a standard Minkowski coordinate system. If you do so then the geometric statement that the worldline is straight corresponds to the ordinary derivative expression that you posted.

However, you are certainly not required to use Minkowski coordinates. You could use any other coordinate system you might want and it would not affect the physics nor the underlying geometry at all, but as I showed above it would affect your coordinate-dependent expression. Therefore, your expression is not general (even in flat spacetime) and is in fact a special case of a more general geometric expression.

The covariant derivative expression I posted is a geometric statement and correctly identifies the underlying straightness of the line regardless of the coordinate system used. In Minkowski coordinates it reduces to your coordinate-dependent ordinary derivative expression, thus the covariant derivative is the more general geometric expression that we are looking for.

In curved spacetimes you can no longer produce global Minkowski coordinates, the best you can do is make a first-order approximation to Minkowski coordinates near some specific event or geodesic. If you choose to do so then your special-case formula can be used to determine if an object is inertial. Of course, you need not use those coordinates, you may use any other coordinates without changing the physics. Again, the covariant derivative is a geometric expression which gives the correct result in any coordinate system and reduces to your expression in the first-order approximation coordinates.

The fact that the Minkowski approximation is first-order and fails whenever second-order effects become important only places a limitation on the use of the special case expression which depends on having Minkowski coordinates, not the general expression which does not depend on coordinates.

 

[Altabeh]

Yes, exactly. Nothing physical is changed, but the coordinate transform leads to non-zero ordinary derivatives along a clearly inertial worldline. Therefore (even in flat spacetime) zero ordinary derivatives do not generally define an inertial worldline, whereas zero covariant derivatives do.

First of all, the geodesic equation has more to do with the "physics" of spacetime than the mathematics behind it. You can get the idea behind this from a simple example. In the Eddington-Finkelstein form of the Schwarzschild metric, the singularity of metric disappears which suggests that you're no longer seeing any singularity by doing a coordinate transformation. Now a degenerate metric has been found out to be non-degenerate but what about its physics!? I'll answer this later. Here the intention is to get a "suitable" form for our metric in which the singularity would be able to disappear. When I say "the dynamics of spacetime seem to have changed" this doesn't mean that it has not changed: it has changed only with respect to the eyes of an observer now using a new coordinate system and this is General Relativity. So when we introduce Kruskal coordinates, one can behold a completely strange physical feature (still an open problem) that didn't exist in the original coordinates neither did it in the Eddington-Finkelstein coordinates. Hence you can't judge whether the physics will not undergo a change or not by a coordinate transformation.

Second off, I have a very grand problem with your "inertial" here. The second term in the geodesic equations indeed represents the sum of the inertial and the gravitational acceleration and iff it vanishes you can only decide to say certainly that a "straight line" is followed by an inertially moving particle (observer) or not. Your fallacy would probably originate from the broad use of "straight line" as being inertial in SR; but bringing a transcription of such definition into GR is nothing but nonsense because now the yardstick to measure the dynamics is just the geodesic equation.

In other words, as you yourself stated, nothing physical has changed, but the result of your formula has. So in general your formula reflects the peculiarities of the coordinate system rather than just the details of the underlying physical situation.

I thought I was clear when posting this earlier. I didn't "state" the physical content has not changed but rather I said it has changed in such a way that it seems it has not. When using Rindler coordinates (which is in the given region or the so-called "wedge" the same as Minkowski spacetime), you see that points at the intersections of a coordinate lattice are accelerating away by a constant acceleration with respect to all stationary observers and yet the spacetime is Minkowski-flat but the physics (the dynamics of the particles) has changed. What's up? The observer only uses now a brand-new ruler and clock to measure the Minkowski spacetime so that with such instruments the dynamics "seems" to have changed with respect to a "handcuffed" me that still am looking only at events happening in the Minkowski spacetime of the inside of a cell with a wall clock and the length of the wall as my meter. This seems to be so because I got news from another prisoner that he has more degrees of freedom than me so he can use frames with the built-in meters and clocks from which the inside of his cell looks getting stretched.

Sorry if I was not clear.

AB

 

[Altabeh]

Whether spacetime is flat is frame-independent. It can't be "made flat" by different observers and coordinate systems.

Unfortunately this is nothing but nonsense. You seem to have lost a whole lot of posts in this thread.

AB

 

[Altabeh]

Altabeh, expanding on my previous comments. In flat spacetime an inertial object has a straight worldline. This is a coordinate-independent geometric statement. You are certainly free to use a standard Minkowski coordinate system. If you do so then the geometric statement that the worldline is straight corresponds to the ordinary derivative expression that you posted.

I said that you are not allowed to use the normals of a theory within another as long as there is a very powerful tool to do so. In Rindler metric, straight lines are not defined to be the same as those of Minkowski metric, i.e. d^2x^a/ds^2=0.

However, you are certainly not required to use Minkowski coordinates. You could use any other coordinate system you might want and it would not affect the physics nor the underlying geometry at all,

This is not correct. A coordinate change does not guarantee always that the physics would not change. This is obvious from the large variety of examples I provided you with.

but as I showed above it would affect your coordinate-dependent expression. Therefore, your expression is not general (even in flat spacetime) and is in fact a special case of a more general geometric expression.

I don't think that defining a straight line in GR the way that SR agrees to is more general than my use of the most general tool (geodesic equation) to explain what I mean by inertial or non-inertial.

In curved spacetimes you can no longer produce global Minkowski coordinates, the best you can do is make a first-order approximation to Minkowski coordinates near some specific event or geodesic. If you choose to do so then your special-case formula can be used to determine if an object is inertial. Of course, you need not use those coordinates, you may use any other coordinates without changing the physics. Again, the covariant derivative is a geometric expression which gives the correct result in any coordinate system and reduces to your expression in the first-order approximation coordinates.

What correct result!? Would you please specify it to me?

The fact that the Minkowski approximation is first-order and fails whenever second-order effects become important only places a limitation on the use of the special case expression which depends on having Minkowski coordinates, not the general expression which does not depend on coordinates.

All the EP (which is the starting point for GR) wants to say is that in a sufficiently small region of spacetime all observers agree that a free-falling object has a zero proper acceleration and this can't be extendend to any larger region or to the whole of spacetime or to the entire of the geodesic.

AB

 

[Dale]

A coordinate change does not guarantee always that the physics would not change.

If you honestly believe that simply changing the mathematical numbers used label events can ever actually change what physically happens then the rest of this conversation is pointless. If I am misunderstanding of your view then perhaps you could explain better.

I will state categorically that there is no way for a mathematical operation like a coordinate change to change physics. Rindler coordinates are a good example, no physics is changed by using Rindler coordinates. I would be glad to discuss that further if I am correctly understanding your position.

 

[George Jones]

Whether spacetime is flat is frame-independent. It can't be "made flat" by different observers and coordinate systems.

Unfortunately this is nothing but nonsense. You seem to have lost a whole lot of posts in this thread.

djy is correct.

 

[yuiop]

A local co-moving co-accelerating observer; is that the guy next to him in the elevator? ;-)

Yep

Isnt the idea of a comoving but non-accelerating observer somewhat oxymoronic??

I can see why might think that. A better expression is a local momentarily co-moving non-accelerating observer. Let us say you are standing on a balcony and somebody throws a ball up to you and its apogee coincides with your eye level. For a "moment", the ball appears stationary. You both have dx/dt=0 for that instant and are co-moving. However the ball is accelerating relative to you so the ball has d^2x/dt^2 \ne \ 0 while you have d^2x/dt^2=0 so you are not co-accelerating with the ball. Your proper accelerations are also different. You have proper acceleration while the ball does not (if you do not catch it).

How about non inertial observers hovering with thrust and spread out over the total path for local measurements??

If they stationary, then they will not be co-moving except at the apogee. They have to be local and match the instantaneous velocity of the accelerating particle as it passes to qualify as local momentarily co-moving observers.

 

[Austin0]

A local co-moving co-accelerating observer; is that the guy next to him in the elevator? ;-)

Well then isn't he a more or less useless and redundant observer?

Isnt the idea of a comoving but non-accelerating observer somewhat oxymoronic??

Yep
I can see why might think that. A better expression is a local momentarily co-moving non-accelerating observer.

EARLIER POST RE: FALLING OBJUCT
But as a coordinate frame it appears to me that it would be a series of instantaneously comoving inertial frames, analogous to an accelerating frame in flat spacetime.
Relative to the Earth frame on the surface it [object frame]would have a dynamic metric.
The length contracting over time and the time dilating proportionately.

Would you then perhaps more or less agree with this earlier post"?

How about non inertial observers hovering with thrust and spread out over the total path for local measurements??

If they stationary, then they will not be co-moving except at the apogee. They have to be local and match the instantaneous velocity of the accelerating particle as it passes to qualify as local momentarily co-moving observers.

Understood. I was suggesting as Swarzschild or Earth frame observers for local measurements of length,time and velocity etc. The idea was the were at rest wrt Earth but local

Let us say you are standing on a balcony and somebody throws a ball up to you and its apogee coincides with your eye level. For a "moment", the ball appears stationary. You both have dx/dt=0 for that instant and are co-moving. However the ball is accelerating relative to you so the ball has d^2x/dt^2 \ne \ 0 while you have d^2x/dt^2=0 so you are not co-accelerating with the ball. Your proper accelerations are also different. You have proper acceleration while the ball does not (if you do not catch it).

I would very much like to get into this with you and have thought of taking your suggestion about posting another thread as our discussion was getting interesting in the other thread.

 

[Altabeh]

If you honestly believe that simply changing the mathematical numbers used label events can ever actually change what physically happens then the rest of this conversation is pointless. If I am misunderstanding of your view then perhaps you could explain better.

I will state categorically that there is no way for a mathematical operation like a coordinate change to change physics. Rindler coordinates are a good example, no physics is changed by using Rindler coordinates. I would be glad to discuss that further if I am correctly understanding your position.

I don't understand why "ever" in your post is bold-faced, but let's check what I mean when I say "physics" (the dynamics of the particles) may change. Consider the Kruskal extension of Schwarzschild metric which can be obtained by the following coordinate transformations:

w=\frac{1}{2}e^{r/4m}(\frac{r-2m}{2m}e^{t'/4m}+e^{-t'/4m}),
v=\frac{1}{2}e^{r/4m}(\frac{r-2m}{2m}e^{t'/4m}-e^{-t'/4m}),

with \theta and \phi being unchanged and t' and r are the time and radial coordinates, respectively, used in the Eddington-Finkelstein form of the Schwarzschild metric. Such coordinates "produce" the following feature that is really bizarre: The (t',r) plane is now mapped on half of the (w,v) plane which means using the whole of the (w,v) plane, as a necessity to introduce a complete space, will actually generate a secondary space (t',r). This is an unintuitive feature that can have a physical explanation but definitely there is no analogy of this with the original or the Eddington-Finkelstein coordinates; raising a probable change of physics by doing a simple change in "numbers" and "labels". You find me an "analogy" I quit studying physics. Of course you have to bear in mind that such thing is tied up to the dynamics of the particles (indeed photons) following radial null geodesics of the Schwarzschild metric in the Kruskal coordinates which you can get some information around them in Wald's book (page 152).

If you are fine by my example, then we will go on to the next example, Rindler coordinates and see what makes me say that the "dynamics" in this case looks a little bit different. But for the moment, I have to make one clarification (correction, I believe):

I said that

you see that points at the intersections of a coordinate lattice are accelerating away by a constant acceleration with respect to all stationary observers and yet the spacetime is Minkowski-flat but the physics (the dynamics of the particles) has changed.

As I later clarified this with the cell example, it it obvious that the "physics" remains the same for all Rindler observers. But for everyone else located between the lines of lattice, no!

There is another thing I have to know before jumping into this example: I believe you agree that "Fermi Normal Coordinates" have a specific observer, don't you?

AB

 

[Altabeh]

djy is correct.

If you honestly think he is then I'm okay with that! Unfortunately, I don't think he is!

AB

 

[atyy]

If you honestly think he is then I'm okay with that! Unfortunately, I don't think he is!

AB

I think djy and George Jones both thought you were talking about flatness in the sense that the Riemann curvature tensor is zero. Of course that's true for Minkowski spacetime, regardless if one is in inertial or Rindler coordinates.

However, you are only talking aobut local flatness, which means using Riemann or Fermi normal coordinates, and is of course coordinate dependent.

Regarding "physics" depending on coordinates, I think it's just that DaleSpam and you have different definitions of physics.

eg. In Minkowski spacetime, is the speed of light c in all coordinate systems?

No - not if you use non-inertial coordinates.

Yes - light always follows a null geodesic, and nullness if a coordinate independent property.

 

[Altabeh]

Regarding "physics" depending on coordinates, I think it's just that DaleSpam and you have different definitions of physics.

And you think whose definition is more correct than the other?

AB

 

[atyy]

And you think whose definition is more correct than the other?

AB

I haven't an opinion.

 

[Dale]

Consider the Kruskal extension of Schwarzschild metric ... a probable change of physics by doing a simple change in "numbers" and "labels".

This is a pretty unconvincing example. It is certainly true that there are coordinate charts that do not cover the entire spacetime, but within the region that they do cover they must agree with all other coordinate systems on the result of any physical experiment.

Rindler coordinates and see what makes me say that the "dynamics" in this case looks a little bit different. ... it it obvious that the "physics" remains the same for all Rindler observers. But for everyone else located between the lines of lattice, no!

Do you have even a single example of a specific physical experiment whose predicted result is different depending on the choice of Rindler or Minkowski coordinates? (Within the Rindler wedge, of course)

In Rindler metric, straight lines are not defined to be the same as those of Minkowski metric, i.e. d^2x^a/ds^2=0.

Then how are straight lines defined in the Rindler spacetime?

my use of the most general tool (geodesic equation) to explain what I mean by inertial or non-inertial.

I don't understand this comment. It sounds like you are now agreeing with me that "inertial" means that the covariant derivative is 0 (the geodesic equation).

 

[yuiop]

Well then isn't he a more or less useless and redundant observer?

If we are considering a particle without eyes, arms or legs, then a co-moving co-accelerating observer is very useful if we wish to consider the point of view from the particle's position.

EARLIER POST RE: FALLING OBJUCT
But as a coordinate frame it appears to me that it would be a series of instantaneously comoving inertial frames, analogous to an accelerating frame in flat spacetime.
Relative to the Earth frame on the surface it [object frame]would have a dynamic metric.
The length contracting over time and the time dilating proportionately.

Would you then perhaps more or less agree with this earlier post"?

Yes... more or less

Understood. I was suggesting as Swarzschild or Earth frame observers for local measurements of length,time and velocity etc. The idea was the were at rest wrt Earth but local

It is fairly easy to transform to the frame of local observers that are stationary with respect to the Schwarzschild metric. Even these stationary, accelerating non-inertial observers perceive the (very) local neighbourhood to be aproximately flat. How "local" is debatable and how many angels you can squeeze into this local space is also debatable.

I would very much like to get into this with you and have thought of taking your suggestion about posting another thread as our discussion was getting interesting in the other thread.

If you can formulate a specific clear question on what aspect you are interested in, then I will do my best .

 

[Altabeh]

This is a pretty unconvincing example. It is certainly true that there are coordinate charts that do not cover the entire spacetime, but within the region that they do cover they must agree with all other coordinate systems on the result of any physical experiment.

You got it sort of wrong with my example. This new coordinate system actually talks about a secret property of the original spacetime that can't be discussed by the use of the initial coordinate system and physics as a whole has changed whether you find it unconvincing or not. You seem to only look at my notes without giving a shot at learning from the books and this gives rise to many confusions here.

Do you have even a single example of a specific physical experiment whose predicted result is different depending on the choice of Rindler or Minkowski coordinates? (Within the Rindler wedge, of course)

Why don't you understand me? There is a difference between a Rindler observer and a non-Rindler one. Physics only changes with respect to a specific observer that is not Rindler. Related to this, tell me if you agree that Fermi Normal Coordinates have a specific observer.

Then how are straight lines defined in the Rindler spacetime?

Straight lines are defined in Rindler coordinates by the geodesic equations. This is what I am trying to get you to understand that an straight line in GR isn't defined in the same way as in Special Relativity in general (which you wrongly suggested this could happen by the immediate use of the formula of a line in the secondary coordinates.) From the dynamics of particles following curves in GR, it can be understood whether a curve is an straight line or not. But the obvious fact is that if the background metric is non-flat, then all straight lines are only locally straight. This was all you could get from the following statement:

my use of the most general tool (geodesic equation) to explain what I mean by inertial or non-inertial.

So your

I don't understand this comment. It sounds like you are now agreeing with me that "inertial" means that the covariant derivative is 0 (the geodesic equation).

isn't unfortunately correct.

Anyways, I'm starting to believe this conversation seems to be pointless and if you and I keep going two different ways here, there is not going to be any agreement until one convinces the other that his argument fails to be true.

AB

 

[Dale]

physics as a whole has changed whether you find it unconvincing or not.

Then it shouldn't be so difficult for you to come up with one single example of an experiment where the two different coordinate systems predict different measured results.

Related to this, tell me if you agree that Fermi Normal Coordinates have a specific observer.

Usually I say it the other way around. I.e. that an observer has a coordinate system (where they are at rest). But I am comfortable with your phrasing.

Straight lines are defined in Rindler coordinates by the geodesic equations. This is what I am trying to get you to understand that an straight line in GR isn't defined in the same way as in Special Relativity in general (which you wrongly suggested this could happen by the immediate use of the formula of a line in the secondary coordinates.) From the dynamics of particles following curves in GR, it can be understood whether a curve is an straight line or not. But the obvious fact is that if the background metric is non-flat, then all straight lines are only locally straight.

OK, it sounds to me like we may largely agree. To be clear, do you agree
1) that an inertial object has a geodesic worldline
2) that whether or not a worldline is a geodesic is independent of the coordinates, and
3) that inertial/geodesic worldlines are straight lines in flat spacetime, but
4) you disagree about my use of the word "straight" to describe geodesic worldlines in curved spacetime.

 

[Altabeh]

Then it shouldn't be so difficult for you to come up with one single example of an experiment where the two different coordinate systems predict different measured results.

If a theory is determined to exist mathematically which sounds incompatible with an-already-believed-to-be-true theory whereas any experiment has not yet been carried out to support the idea behind that theory does actually raise a probablity in the whole of the known theory here.

Usually I say it the other way around. I.e. that an observer has a coordinate system (where they are at rest). But I am comfortable with your phrasing.

Okay, now it sounds like we agree that there is also a specific observer for either of the Rindler coordinates (flat spacetime) or the Fermi normal coordinates (the extension of Rindler coordinates into curved spacetimes.)

OK, it sounds to me like we may largely agree. To be clear, do you agree
1) that an inertial object has a geodesic worldline
2) that whether or not a worldline is a geodesic is independent of the coordinates, and
3) that inertial/geodesic worldlines are straight lines in flat spacetime, but
4) you disagree about my use of the word "straight" to describe geodesic worldlines in curved spacetime.

1- Speaking locally, an inertial object has a geodesic worldline from the perspective of all observers;
2- Yes;
3- Yes;
4- In your own way of using "straight", I do disagree (The local problem).

AB

 

[Dale]

If a theory is determined to exist mathematically which sounds incompatible with an-already-believed-to-be-true theory whereas any experiment has not yet been carried out to support the idea behind that theory does actually raise a probablity in the whole of the known theory here.

Sorry, I guess it was not clear what I am asking. I am not asking for a reference to the results of an experiment that has already been carried out. I am asking for an example of any physical experiment that could possibly be set up where the predicted measurement depends on the choice of coordinates.

For example, the frequency of a laser fired from the ceiling and detected on the floor in an accelerating elevator in flat spacetime can be analyzed in both the Rindler coordinates and in the Minkowski coordinates. Both will agree on the detected frequency. Can you think of a counterexample?

1- Speaking locally, an inertial object has a geodesic worldline from the perspective of all observers;
2- Yes;
3- Yes;
4- In your own way of using "straight", I do disagree (The local problem).

Out of those I think that 2 is the most important and since we agree on that I am not inclined to pursue the remainder, particularly 4 which is, in all fairness, fairly loose and sloppy terminology on my part that I nevertheless like to use.

 

[Altabeh]

For example, the frequency of a laser fired from the ceiling and detected on the floor in an accelerating elevator in flat spacetime can be analyzed in both the Rindler coordinates and in the Minkowski coordinates. Both will agree on the detected frequency. Can you think of a counterexample?

From the beginning of this thread, I've been insisting that there are two views on the coordinate-independence thing: One is that there is a probable prediction that stands for the non-invariance of physics when looked from another frame (Kruskal example) which you said you found it unconvincing and the other is that physics changes when looked from another frame (coordinate system) depending on the observer. This means that for example in the two dimensional Rindler coordinates those observers at a constant x undergo a uniform acceleration (they are located on the coordinate lattice stripes) and for these observers nothing physical changes but for an observer located outside the lattice stripes physics looks different and these are the observers that appear to be the non-specific observers of the Fermi normal frame to whom physics seems to change when not talking locally. Of course the Rindler observers are similar to the specific observers of the Fermi frame to whom physics doesn't undergo any change. I hope you got it now.

Out of those I think that 2 is the most important and since we agree on that I am not inclined to pursue the remainder, particularly 4 which is, in all fairness, fairly loose and sloppy terminology on my part that I nevertheless like to use.

Though this thread has gone far away from where it was supposed to go, I don't like to continue arguing about the main question and its answer because earlier in this thread I told you that the physics community has no consensus over the use of "proper acceleration" in GR and thus I'm happy everything is now clear between you and me.

AB

 

[DrGreg]

...I told you that the physics community has no consensus over the use of "proper acceleration" in GR...

As far as I know, "proper acceleration" is a well-established terminology, and those authors who choose to use the term do so consistently with each other. Do you have any evidence of contradictory definitions in reliable sources?

I think all you said in previous posts was that you couldn't find any definitions on the internet or the books you had consulted, but that doesn't mean there is "no consensus" -- if you couldn't find any evidence it proves nothing one way or the other.

 

[Altabeh]

As far as I know, "proper acceleration" is a well-established terminology, and those authors who choose to use the term do so consistently with each other. Do you have any evidence of contradictory definitions in reliable sources?

I think all you said in previous posts was that you couldn't find any definitions on the internet or the books you had consulted, but that doesn't mean there is "no consensus" -- if you couldn't find any evidence it proves nothing one way or the other.

Then you are required to show me where they use such thing in physics if Wald, Papapetrou, Hobson, D'inverno and the other famous authors (physicists) have not talked about it the way you described. Here "consensus" means you and everyone else agree on the definition of a physical concept. Which "authors" do choose to use the term the way you follow and what is the "well-established" definition of proper acceleration here (in GR)?

AB

 

[Dale]

This means that for example in the two dimensional Rindler coordinates those observers at a constant x undergo a uniform acceleration (they are located on the coordinate lattice stripes) and for these observers nothing physical changes but for an observer located outside the lattice stripes physics looks different

This is exactly what you have not shown. You have not shown an example demonstrating that there is any physical experiment that could be performed where the two different coordinate systems would predict different measurements. In fact I gave a counter-example of an experiment where they would predict the same measurement, and I am glad to work it out quantitatively if you doubt it.

 

[Altabeh]

This is exactly what you have not shown. You have not shown an example demonstrating that there is any physical experiment that could be performed where the two different coordinate systems would predict different measurements. In fact I gave a counter-example of an experiment where they would predict the same measurement, and I am glad to work it out quantitatively if you doubt it.

You mean that you doubt that in the two dimensional Rindler coordinates those observers at a constant x undergo a uniform acceleration (they are located on the coordinate lattice stripes) and for these observers nothing physical changes? Or for an observer located outside the lattice stripes physics looks different? (Maybe you have a hard time understanding what the outside of the coordinate lattice is. Well, imagine x\rightarrow 0, then you're tending to a degenerate metric and the outside is where a "degenerate" observer is located.)

I'm going to be glad if you can show me these are incorrect "quantitatively". If you do so, you're then contradicting the fact that Fermi normal coordinates have a specific observer.

AB

 

[Dale]

You mean that you doubt that in the two dimensional Rindler coordinates those observers at a constant x undergo a uniform acceleration (they are located on the coordinate lattice stripes) and for these observers nothing physical changes?

Correct. There is, AFAIK, no physical measurement that can be performed (within the Rindler wedge, of course) whose predicted result depends on the choice of Rindler or Minkowski coordinates.

Or for an observer located outside the lattice stripes physics looks different? (Maybe you have a hard time understanding what the outside of the coordinate lattice is. Well, imagine x\rightarrow 0, then you're tending to a degenerate metric and the outside is where a "degenerate" observer is located.)

Do you mean an observer outside the 1/4 section of spacetime known as the Rindler wedge? If so, I already addressed this previously:

It is certainly true that there are coordinate charts that do not cover the entire spacetime, but within the region that they do cover they must agree with all other coordinate systems on the result of any physical experiment.

Do you have even a single example of a specific physical experiment whose predicted result is different depending on the choice of Rindler or Minkowski coordinates? (Within the Rindler wedge, of course)

IMO it doesn't make any sense to speak of the predictions of a coordinate chart in a region not covered by the chart. In other words, outside of the Rindler wedge the Rindler coordinates do not make any predictions about the results of physical measurements, so it can neither agree nor disagree with the predictions of the Minkowski chart.

Within the region covered by the chart (i.e. the Rindler wedge) they must agree. Do you think that is an incorrect statement?

 

[Altabeh]

Correct. There is, AFAIK, no physical measurement that can be performed (within the Rindler wedge, of course) whose predicted result depends on the choice of Rindler or Minkowski coordinates.

Again this is completely correct iff the observer is Rindler. I wonder if you even know what I mean by pointing at "an specific observer" in either of Rindler or Fermi coordinates. This turns out to be a waste of time if you keep not giving clear answer to my question:

Do you agree that both Rindler and Femi frames have their own specific observers?

Do you mean an observer outside the 1/4 section of spacetime known as the Rindler wedge? If so, I already addressed this previously:

No I don't. I have thoroughly addressed something else. See above.

IMO it doesn't make any sense to speak of the predictions of a coordinate chart in a region not covered by the chart. In other words, outside of the Rindler wedge the Rindler coordinates do not make any predictions about the results of physical measurements, so it can neither agree nor disagree with the predictions of the Minkowski chart.

I'm flabbergasted that you keep talking about the outside of the Rindler wedge. I'm trying to get you to know that the Rindler coordinates within the wedge has a specific observer. Just to this observer physics doesn't seem to change. Got it?

Within the region covered by the chart (i.e. the Rindler wedge) they must agree. Do you think that is an incorrect statement?

Oh God. Who said this is incorrect? You seem to have completely confused the Kruskal example with the Rindler coordinates and have made something nonsense out of the mixture. I'd be glad if you answer my question.

AB

 

[DrGreg]

Then you are required to show me where they use such thing in physics if Wald, Papapetrou, Hobson, D'inverno and the other famous authors (physicists) have not talked about it the way you described. Here "consensus" means you and everyone else agree on the definition of a physical concept. Which "authors" do choose to use the term the way you follow and what is the "well-established" definition of proper acceleration here (in GR)?

AB

I don't understand what your point is. If you are saying that there are some books that do not use term "proper acceleration", yes, of course that is true but it's irrelevant. I'm making no claim that the expression is widely used (though I suspect it might be). I'm claiming that when it is used, it is used consistently. I can only speak of my own experience but, whenever I've seen anyone talk about proper acceleration, it's always been compatible with the definition I gave in post #22 of this thread, with a reference (and repeated below at 3).

It's possible there are alternative definitions, but I've never seen them. Unless I've misunderstood you, you seem to be implying that other (incompatible) definitions exist in reliable sources, so the onus is on you to provide such a source. I am not omniscient, so if such a source can be provided I'll be quite happy to concede this point.

Just to clarify, proper acceleration can be defined as

1. what an accelerometer measures
2. acceleration relative to a free-falling momentarily co-moving observer, measured using "ultra-locally Minkowski" coordinates
3. the scalar magnitude of the 4-acceleration tensor dU^{\mu}/d\tau+\Gamma^{\mu}_{\gamma\nu}U^{\gamma}U^{\nu}, where U is the 4-velocity tensor

All of these formulations are compatible with each other, and all reliable sources I've seen use at least one of them, if they refer to "proper acceleration" at all. (Of course, if the source only ever discusses Minkowski coordinates in flat spacetime, then definitions 2 and 3 can be simplified.)

By the way, the above definition of 4-acceleration is certainly standard and more widespread than "proper acceleration".

By "ultra-locally Minkowski", I mean at the single event where the measurement is taken the metric components equal the Minkowski metric components and all the first-order coordinate-derivatives of the metric components are zero. Such coordinates exist by the equivalence principle.

 

[Dale]

This turns out to be a waste of time if you keep not giving clear answer to my question:

Do you agree that both Rindler and Femi frames have their own specific observers?

I did give a clear answer:

Usually I say it the other way around. I.e. that an observer has a coordinate system (where they are at rest). But I am comfortable with your phrasing.

I'm flabbergasted that you keep talking about the outside of the Rindler wedge. I'm trying to get you to know that the Rindler coordinates within the wedge has a specific observer. Just to this observer physics doesn't seem to change. Got it?

I keep talking about outside of the Rindler wedge because that is the only possible way that what you are saying could be remotely correct. Since you are talking only about within the Rindler wedge then you are simply wrong. What is physically measured does not depend on how you label the events with coordinates. Every observer in any applicable coordinate system will agree on the results of all measurements.

I will work out my example quantitatively, and I encourage you to come up with your own example and do the same.

 

[Altabeh]

I did give a clear answer:

Did I miss it? Pardon, what was it?

I will work out my example quantitatively, and I encourage you to come up with your own example and do the same.

I'm waiting for such "quantitatve" example.

AB

 

[Altabeh]

I don't understand what your point is. If you are saying that there are some books that do not use term "proper acceleration", yes, of course that is true but it's irrelevant.

I'm being curious that what makes you think "it is irrelevant" while earlier in this thread we were clearly addressing the "proper acceleration" in two different ways. Maybe now that we have put aside using the term, it sounds to you that it is irrelevant!

I'm making no claim that the expression is widely used (though I suspect it might be). I'm claiming that when it is used, it is used consistently.

The magnitude of 4-acceleration you mean? I'm trying to dig something pithy from this assertation but I end up asking myself what is your point here? Please explain.

I can only speak of my own experience but, whenever I've seen anyone talk about proper acceleration, it's always been compatible with the definition I gave in post #22 of this thread, with a reference (and repeated below at 3).

Yes I agree that Rindler makes use of "proper acceleration" the way you stated. But this is only him bringing it up and I have not seen any other book even quoting his argument or using such term. But as a mathematical result, clearly you and I know that

\sqrt{g_{ab}|A^aA^b|},

is a scalar invariant. But such thing can't be supported if you don't mention any other reliable source that uses it! Let's say this is only Rindler's idea.

It's possible there are alternative definitions, but I've never seen them. Unless I've misunderstood you, you seem to be implying that other (incompatible) definitions exist in reliable sources, so the onus is on you to provide such a source. I am not omniscient, so if such a source can be provided I'll be quite happy to concede this point.

I simply said that in response to the third question brought up by the OP, I have one answer and it is NO. I tried to show this by quoting from Papapetrou's book that only to a specific observer the motion along the entire of a geodesic sounds to be inertial which means it is not to "all observers" unless locally. Such observer uses Fermi coordinates that can be found in Poinsson's "Relativistic Toolkit". I also tried to show that I have a vivid definition for an inertial object in GR by dragging the discussion of local flatness into consideration and said that only locally all observers agree on the "inertial" property of a free-falling particle. Schutz proves this in a beautiful way but in Weyl's "Space-Time-Matter" this is proved at one single point using an orthogonal transformation. I then used Weinberg's book to clearly address the nonsense claim made by DaleSpam about the OP's question in this thread. You can yourself go and reaad the pages 70-73 to see whether all of my assertations are false or not!

Just to clarify, proper acceleration can be defined as

1. what an accelerometer measures
2. acceleration relative to a free-falling momentarily co-moving observer, measured using "ultra-locally Minkowski" coordinates
3. the scalar magnitude of the 4-acceleration tensor dU^{\mu}/d\tau+\Gamma^{\mu}_{\gamma\nu}U^{\gamma}U^{\nu}, where U is the 4-velocity tensor

Yes these could be used to show what "proper acceleration" is but my definition could also be used, couldn't it? Do you find something "incompatible" in it?

All of these formulations are compatible with each other,

The first and second definitions have a specific observer but the third one is ture for all observers. This is where DaleSpam and I start going in two different directions.

AB

 

[atyy]

Yes I agree that Rindler makes use of "proper acceleration" the way you stated. But this is only him bringing it up and I have not seen any other book even quoting his argument or using such term. But as a mathematical result, clearly you and I know that

\sqrt{g_{ab}|A^aA^b|},

is a scalar invariant. But such thing can't be supported if you don't mention any other reliable source that uses it! Let's say this is only Rindler's idea.
AB

Wald p152 gives the proper acceleration of an observer who is hovering in the Schwarzschild spacetime as a=(1-2M/r)-1/2M/r2. Does this match Rindler's definition?

BTW, I think everyone else (other than you) is no longer discussing the OP's question 3, to which you provided the first right answer in this thread.

 

[Dale]

Did I miss it? Pardon, what was it?

For the 3rd time:

Usually I say it the other way around. I.e. that an observer has a coordinate system (where they are at rest). But I am comfortable with your phrasing.

 

[Altabeh]

For the 3rd time:

The answer would be something like Yes or No. But if this is a Yes, then your answer to the OP's third question is simply false. DrGreg's second definition tells us that

acceleration relative to a free-falling momentarily co-moving observer, measured using "ultra-locally Minkowski" coordinates

where he clarifies "ultra-locally Minkowski" as

By "ultra-locally Minkowski", I mean at the single event where the measurement is taken the metric components equal the Minkowski metric components and all the first-order coordinate-derivatives of the metric components are zero. Such coordinates exist by the equivalence principle.

Using such definition, only to the Fermi observer the spacetime along a (timelike) geodesic is flat thus - if you consider Weinberg or Papapetrou good physicists - the geodesic is a straight line in the sense of special relativity (d^2x^a/ds^2=0) and therefore any motion along such geodesic must be inertial entirely.

AB

 

[Altabeh]

Wald p152 gives the proper acceleration of an observer who is hovering in the Schwarzschild spacetime as a=(1-2M/r)-1/2M/r2. Does this match Rindler's definition?

Not of course. This observer experiences only a gravitational force -thus comparable to an observer following a geodesic- and if it has a generalized proper 4-acceleration defined in DrGreg's latest post, it must be an invariant which isn't. Some people may think since the geodesic is now having a non-zero right-hand side due to this force and so the proper 4-acceleration is non-zero. But one cannot provide any affine transformation to make this force disappear from the geodesic equation. If DaleSpam claims he can, I'm ready to correct myself.

BTW, I think everyone else (other than you) is no longer discussing the OP's question 3, to which you provided the first right answer in this thread.

Thank you for reminding everyone else to discuss it from now on!

AB

 

[Dale]

The answer would be something like Yes or No. But if this is a Yes, then your answer to the OP's third question is simply false. DrGreg's second definition tells us that



where he clarifies "ultra-locally Minkowski" as



Using such definition, only to the Fermi observer the spacetime along a (timelike) geodesic is flat thus - if you consider Weinberg or Papapetrou good physicists - the geodesic is a straight line in the sense of special relativity (d^2x^a/ds^2=0) and therefore any motion along such geodesic must be inertial entirely.

As Dr. Greg mentioned, the three definitions are equivalent. You cannot say that something is true with one definition and false with a second definition if the definitions are equivalent. That is illogical.

 

[atyy]

Not of course. This observer experiences only a gravitational force -thus comparable to an observer following a geodesic- and if it has a generalized proper 4-acceleration defined in DrGreg's latest post, it must be an invariant which isn't. Some people may think since the geodesic is now having a non-zero right-hand side due to this force and so the proper 4-acceleration is non-zero. But one cannot provide any affine transformation to make this force disappear from the geodesic equation. If DaleSpam claims he can, I'm ready to correct myself.

I haven't checked by explicit calculation myself, but Raine and Thomas (http://books.google.com/books?id=O3...warzschild+acceleration&source=gbs_navlinks_s, p42-43) start from proper acceleration defined as a^ua_u (same as Rindler's definition given by DrGreg) and get a=(1-2M/r)^-1/2M/r² (same as the formula on p152 of Wald), so shouldn't that be invariant?

 

[Dale]

This observer experiences only a gravitational force -thus comparable to an observer following a geodesic-

That is not correct, a hovering observer in the Schwarzschild spacetime must be experiencing a non-gravitational force, and is most emphatically not following a geodesic.

 

[DrGreg]

Wald p152 gives the proper acceleration of an observer who is hovering in the Schwarzschild spacetime as a=(1-2M/r)-1/2M/r2. Does this match Rindler's definition?

Yes.

Look at http://people.maths.ox.ac.uk/~nwoodh/gr/index.html , Section 12.1, page 54. Here he shows that the 4-acceleration of the hovering observer is (0,M/r^2,0,0) in Schwarzschild coordinates, and that the magnitude of that 4-vector is

\frac{M}{r^2\sqrt{1 - 2M/r}}​

(All in units where G = c = 1, of course.) He doesn't use the phrase "proper acceleration" but he does describe it as "the acceleration felt by the observer" and "the 'force of gravity' ". And he does use the term "4-acceleration" which is defined as a covariant derivative just like Rindler. He also says the worldline of the observer is not geodesic.

Woodhouse's lecture notes were the basis of his book "General Relativity" (2007) which he later published and the same argument appears in Section 7.3 page 99 of the published book.

 

[DrGreg]

I haven't checked by explicit calculation myself, but Raine and Thomas (http://books.google.com/books?id=O3...warzschild+acceleration&source=gbs_navlinks_s, p42-43) start from proper acceleration defined as a^ua_u (same as Rindler's definition given by DrGreg) and get a=(1-2M/r)^-1/2M/r² (same as the formula on p152 of Wald), so shouldn't that be invariant?

Thanks for that. They explicitly use the term "proper acceleration" (="rest frame acceleration") as the magnitude of the 4-acceleration, defined as in Rindler, so that's another published example. (You missed out a square root, by the way.) And of course it is invariant, it's the magnitude of a tensor.

This observer experiences only a gravitational force -thus comparable to an observer following a geodesic- and if it has a generalized proper 4-acceleration defined in DrGreg's latest post, it must be an invariant which isn't. Some people may think since the geodesic is now having a non-zero right-hand side due to this force and so the proper 4-acceleration is non-zero.

I've read these sentences several times now, but I can't make any sense of them. Sorry.

 

[Altabeh]

That is not correct, a hovering observer in the Schwarzschild spacetime must be experiencing a non-gravitational force.

Nonsense. The formula provided by atty shows the gravitational force felt by a hovering observer over the hole.

I haven't checked by explicit calculation myself, but Raine and Thomas (http://books.google.com/books?id=O3p...gbs_navlinks_s , p42-43) start from proper acceleration defined as auau (same as Rindler's definition given by DrGreg) and get a=(1-2M/r)-1/2M/r2 (same as the formula on p152 of Wald), so shouldn't that be invariant?

I checked it. There is no need to give an affine transformation to make the right-hand side zero because there is none (though comparable to moving along a geodesic, this is not free-fall).

As Dr. Greg mentioned, the three definitions are equivalent. You cannot say that something is true with one definition and false with a second definition if the definitions are equivalent. That is illogical.

I identified the problem and said that there is a fallacy that you have been following up until now and it is that along the entire of a geodesic, the motion seems to be inertial from the perspective of all observers. On the other hand you believe that the Fermi coordinates have a specific observer to which the motion along the geodesic sounds inertial entirely according to Papapetrou (p. 56) and Weinberg (p. 70). Since DrGreg's third definition makes use of Rindler's idea that proper acceleration is zero along any geodesic as measured by ANY observer, how can you explain such chasm?

P.s. Thanks to atty, I am now sure DrGreg's third definition can also be used for proper acceleration in GR.

AB

 

[Altabeh]

I've read these sentences several times now, but I can't make any sense of them. Sorry.

Because the worldline is not a geodesic, the right-hand side is clearly non-zero. I thought first the worldline would be a geodesic so there could have been a possible affine transformation to make the right-hand side zero. But later I came to the idea that the worldline is not a geodesic. You got it now?

AB

 

[Altabeh]

(All in units where G = c = 1, of course.) He doesn't use the phrase "proper acceleration" but he does describe it as "the acceleration felt by the observer" and "the 'force of gravity' ". And he does use the term "4-acceleration" which is defined as a covariant derivative just like Rindler. He also says the worldline of the observer is not geodesic.

So are you ending up with the fact that there is no consensus over ther use of the term "proper acceleration" in GR? Nonetheless, since Wald has also made use of this term the way you described, I made my mind to agree with it as another alternative definition for the proper acceleration in GR.

AB

 

[DrGreg]

I'm being curious that what makes you think "it is irrelevant" while earlier in this thread we were clearly addressing the "proper acceleration" in two different ways.

It's possible I have misunderstood. When I wrote the above it was on the assumption that there are three sets of people

1. those who define proper acceleration the same as Rindler, or a provably compatible way
2. those who define proper acceleration some other way that is incompatible
3. those who don't use the term "proper acceleration" at all.

I thought we were arguing over whether group 2 was empty or not. (Maybe I have misunderstood.) In which case group 3 is irrelevant, it doesn't matter how large or small that group is.

All I was saying was that group 1 is non-empty, I have not discovered anything in group 2, and I don't care about group 3 in this context.

Yes these could be used to show what "proper acceleration" is but my definition could also be used, couldn't it? Do you find something "incompatible" in it?

Sorry, I've lost track of this. Could you give me a clear and precise statement of what your definition is?

 

[Altabeh]

It's possible I have misunderstood. When I wrote the above it was on the assumption that there are three sets of people

1. those who define proper acceleration the same as Rindler, or a provably compatible way
2. those who define proper acceleration some other way that is incompatible
3. those who don't use the term "proper acceleration" at all.

I thought we were arguing over whether group 2 was empty or not. (Maybe I have misunderstood.) In which case group 3 is irrelevant, it doesn't matter how large or small that group is. All I was saying was that group 1 is non-empty, I have not discovered anything in group 2, and I don't care about group 3 in this context.

Oh, this requires you to jump into the early discussions of this thread but whatever the proper acceleration is, our main problem is that how "inertial", related to the OP's question 3, is defined in GR and if some motion keeps looking inertial along the entire of a path, who in GR will be the observer of such phenomenon? Everyone or a specific observer? Do you have any thought or idea on these?

Sorry, I've lost track of this. Could you give me a clear and precise statement of what your definition is?

My definition is that of yours but when the background metric is reduced to the Minkowski metric which means I agree with the SR definition of "proper acceleration" in GR and I think this has already been discussed in detail.

AB

 

[Dale]

Altabeh, here is the example I promised earlier which demonstrates that the predicted measurement does not depend on the choice of Rindler v Minkowski coordinates:

For example, the frequency of a laser fired from the ceiling and detected on the floor in an accelerating elevator in flat spacetime can be analyzed in both the Rindler coordinates and in the Minkowski coordinates. Both will agree on the detected frequency.

Throughout this I will use units where c=1 and capital letters for Minkowski coordinates R=(T,X,Y,Z) and lower case letters for Rindler coordinates r=(t,x,y,z). For completness, the transformation equations are:
T=x\,sinh(t)
X=x\,cosh(t)
Y=y
Z=z

and the metric is:
g(r)=\left(<br /> \begin{array}{cccc}<br /> -x^2 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right) or g(R)=\left(<br /> \begin{array}{cccc}<br /> -1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)

So, let \xi be the (non-geodesic) worldline of the emitter and \zeta be the (non-geodesic) worldline of the receiver and let the emitter emit a wave with one peak that follows the null geodesic \alpha and the following peak that follows the null geodesic \beta, then let a be the intersection of \xi and \alpha, b be the intersection of \xi and \beta, c be the intersection of \zeta and \alpha, and d be the intersection of \zeta and \beta.

Then the time measured by the emitter is:
\frac{1}{f1}=\int_a^b \sqrt{-g_{\mu\nu} \frac{\xi^{\mu}}{d\lambda} \frac{\xi^{\nu}}{d\lambda}} \; d\lambda

The time measured by the receiver is:
\frac{1}{f2}=\int_c^d \sqrt{-g_{\mu\nu} \frac{\zeta^{\mu}}{d\lambda} \frac{\zeta^{\nu}}{d\lambda}} \; d\lambda

So, for a quantitative example in the Rindler frame let's say:
\xi=(t,2,0,0)
\zeta=(t,1,0,0)
\alpha=\left(.693 - .5 ln(4.-\lambda),\sqrt{4.-\lambda},0,0\right)
\beta=\left(.742 - .5 ln(4.41-\lambda),\sqrt{4.41-\lambda},0,0\right)

then in the Rindler frame
\frac{1}{f1}=.0976
\frac{1}{f2}=.0488

So, for the same example in the Minkowski frame:
\xi=(2 sinh(t),2 cosh(t),0,0)
\zeta=(sinh(t),cosh(t),0,0)
\alpha=(\lambda,2.-\lambda,0,0)
\beta=(\lambda,2.1-\lambda,0,0)

then in the Minkowski frame
\frac{1}{f1}=.0976
\frac{1}{f2}=.0488

So the measured frequencies are the same regardless of if we use Minkowski or Rindler coordinates for the analysis. This directly contradicts your assertion that "physics as a whole" is different in the two coordinate systems. I reassert my challenge to you to come up with even a single example (within the Rindler wedge) where the measured result of any experiment depends on the choice of Rindler or Minkowski coordinates.

 

[Altabeh]

Altabeh, here is the example I promised earlier which demonstrates that the predicted measurement does not depend on the choice of Rindler v Minkowski coordinates:Throughout this I will use units where c=1 and capital letters for Minkowski coordinates R=(T,X,Y,Z) and lower case letters for Rindler coordinates r=(t,x,y,z). For completness, the transformation equations are:
T=x\,sinh(t)
X=x\,cosh(t)
Y=y
Z=z

and the metric is:
g(r)=\left(<br /> \begin{array}{cccc}<br /> -x^2 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right) or g(R)=\left(<br /> \begin{array}{cccc}<br /> -1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)

So, let \xi be the (non-geodesic) worldline of the emitter and \zeta be the (non-geodesic) worldline of the receiver and let the emitter emit a wave with one peak that follows the null geodesic \alpha and the following peak that follows the null geodesic \beta, then let a be the intersection of \xi and \alpha, b be the intersection of \xi and \beta, c be the intersection of \zeta and \alpha, and d be the intersection of \zeta and \beta.

Then the time measured by the emitter is:
\frac{1}{f1}=\int_a^b \sqrt{-g_{\mu\nu} \frac{\xi^{\mu}}{d\lambda} \frac{\xi^{\nu}}{d\lambda}} \; d\lambda

The time measured by the receiver is:
\frac{1}{f2}=\int_c^d \sqrt{-g_{\mu\nu} \frac{\zeta^{\mu}}{d\lambda} \frac{\zeta^{\nu}}{d\lambda}} \; d\lambda

So, for a quantitative example in the Rindler frame let's say:
\xi=(t,2,0,0)
\zeta=(t,1,0,0)
\alpha=\left(.693 - .5 ln(4.-\lambda),\sqrt{4.-\lambda},0,0\right)
\beta=\left(.742 - .5 ln(4.41-\lambda),\sqrt{4.41-\lambda},0,0\right)

then in the Rindler frame
\frac{1}{f1}=.0976
\frac{1}{f2}=.0488

So, for the same example in the Minkowski frame:
\xi=(2 sinh(t),2 cosh(t),0,0)
\zeta=(sinh(t),cosh(t),0,0)
\alpha=(\lambda,2.-\lambda,0,0)
\beta=(\lambda,2.1-\lambda,0,0)

then in the Minkowski frame
\frac{1}{f1}=.0976
\frac{1}{f2}=.0488

So the measured frequencies are the same regardless of if we use Minkowski or Rindler coordinates for the analysis. This directly contradicts your assertion that "physics as a whole" is different in the two coordinate systems. I reassert my challenge to you to come up with even a single example (within the Rindler wedge) where the measured result of any experiment depends on the choice of Rindler or Minkowski coordinates.

Fisrt off, I think you pushed into the shadows the whole fallacy you made in the earliest post of yours about "what is inertial in GR" by asserting something that again I do not have any objection to. Second off, why there is no objection to this? (Your example is not mathematically well-written as there are errors in some parts of it. Review again.) Because simply both the emitter and receiver are located at a constant x thus they undergo a uniform acceleration a=1/x; meaning that they are both Rindler observers! Third off, what is a non-Rindler observer to which physics looks different? All observers with the world lines x=x_0,y=y_0, z=z_0 in the Rindler coordinates are in fact the Rindler observers for which the expansion and vorticity vanish*; giving rise to them moving along with each other with a constant distance from one to the other being in the neighbourhood. Such observers would never experience being at a point twice. If you were careful enough to understand the lattice example, the lines that the intersection points lie in them cannot ever cross each other somewhere else and such points are similar to points covering the surface of an inflating bubble.

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

*For a complete discussion of vorticity and expansion, see Poisson, Eric (2004). A Relativist's Toolkit.

 

[Dale]

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

Fisrt off, I think you pushed into the shadows the whole fallacy you made in the earliest post of yours about "what is inertial in GR" by asserting something that again I do not have any objection to. Second off, why there is no objection to this?

What are you talking about?

(Your example is not mathematically well-written as there are errors in some parts of it. Review again.) Because simply both the emitter and receiver are located at a constant x thus they undergo a uniform acceleration a=1/x

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.

Third off, what is a non-Rindler observer to which physics looks different? ...

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

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.