Layman's doubts about Gen Relativity

[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:
$$g(r)=\left( \begin{array}{cccc} -x^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$ or $$g(R)=\left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

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:
$$g(r)=\left( \begin{array}{cccc} -x^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$ or $$g(R)=\left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

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.

 

[Altabeh]

I already did. Go back and re-read a little more closely. Don't hesitate to ask questions about the parts you either don't understand or feel are incomplete. I glossed over some parts, like the derivation of the equation of the null geodesics.

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

in any part of mathematics; let alone physics.

I already did. Go back and re-read a little more closely. Don't hesitate to ask questions about the parts you either don't understand or feel are incomplete. I glossed over some parts, like the derivation of the equation of the null geodesics.

Where? You're making no sense; sorry! On one hand you said you believe that Fermi coordinates have a specific observer like Rindler coordinates. On the other hand you're saying you did prove that a non-Rindler observer sees the same result as Rindler observers do! (while you didn't as can be seen from you setting a consatnt $$x$$ for your emitter and receiver!) How can you explain this contradiction? More interestingly unreasonable is your leaky understanding of what a Rindler observer is:

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

Three points must be made here:

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

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

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

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

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

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

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

I never said such nonsense. Either you're 1) escaping from the first fallacy you made here concerning OP's question which Weinberg and Papaterou corrected it or 2) overshadowing the second fallacy that straight line in GR is given by the transfomed formula of the straight line in the Euclidean space or 3) digressing from the main problem by confusing the Kruskal example with Rindler and making some fallacious claim concerning the latter.

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

AB

 

[Dale]

there is no such thing as
$$\frac{\zeta^{\nu}}{d\lambda}$$

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

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

Yes.

On the other hand you're saying you did prove that a non-Rindler observer sees the same result as Rindler observers do!

Yes. I worked the same problem in Rindler coordinates for a Rindler observer and in Minkowski coordinates for a Minkowski observer and they agreed on the result.

you are still unable to demonstrate any experimental measurement which depends on the choice of Rindler or Minkowski coordinates

I never said such nonsense.

Are you now agreeing with me that the measured result of any physical experiment does not depend on the choice of coordinate system used to analyze the experiment?

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

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

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

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

 

[atyy]

I think once again there is just a slight difference in terminology.

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

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

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

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

?

 

[Dale]

there is no global inertial frame associated with a freely falling observer, which is the answer to the third question of the OP, which was Altabeh's point in post #6.

The OP later clarified that he was interested in the "inertialness" of the object itself, not the local coordinate system.

 

[atyy]

The OP later clarified that he was interested in the "inertialness" of the object itself, not the local coordinate system.

3)When looked upon from Earth does a falling stone makes non-inertial frame of reference?

Is it safe now to tell that a very small particle in free fall, undergoing translational acceleration (in the kinematic sense) over a very long distance is actually an inertial frame for people on a perfectly still Earth?

This means that an accelerometer at rest gives the value of g at that point and a falling accelerometer records zero acceleration. I think it is a better and intuitive answer to my question.

The accelerometer mechanism is actually functioning in the frame of reference of the object itself. So, its zero reading doesn't shows us that it was in inertial motion (for us).

I've concluded that Altabeh's argument was right. The local flatness theorem holds only for infitesimal regions in space-time curvature and due to the long trajectory of the particle, it is non-inertial from outside. The continuous photon blueshifting/redshifting and apparent modification in c are some phenomenon that exhibit this thing.

The OP's question was a quantum superposition, which as we all know is not relativistically invariant, and hence observer dependent

 

[George Jones]

We also all agree that Fermi normal coordinates are only locally inertial

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

Rindler coordinates are Fermi normal coordinates.

 

[atyy]

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

Rindler coordinates are Fermi normal coordinates.

Yes - thanks!

 

[Altabeh]

Yes. I worked the same problem in Rindler coordinates for a Rindler observer and in Minkowski coordinates for a Minkowski observer and they agreed on the result.

Thank goodness, you ultimately corrected yourself because I asked you

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

And you answered:

I already did. Go back and re-read a little more closely.

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

Are you now agreeing with me that the measured result of any physical experiment does not depend on the choice of coordinate system used to analyze the experiment?

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

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

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

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

Yes but this is not globally possible in GR.

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

I think atty responded to this in an awesome way:

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

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

The OP later clarified that he was interested in the "inertialness" of the object itself, not the local coordinate system.

The OP clearly said in an early post he felt like my answer to his question was right. There was no talk of inertialness, I think.

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

AB