Layman's doubts about Gen Relativity

[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.