Understanding Tidal Forces & Rindler Metric

[JesseM]

You seem to not even pay a little attention towards me. These two spacetimes are NOT the same everywhere. The Rindler line-element can be used as a line-element for a flat spacetime but we have circumferences under which Rindler does not reduce to Minkowski and this arises from

1- the existence of a horizon in Rindler metric at $$x=0$$,
2- the so-called "uniform" motion of different observers relative to each other.

Obviously the existence of a horizon means that the Rindler coordinate system does not cover the entire set of events covered in the Minkowski coordinate system. Is that all you mean by "not the same everywhere"? If so I didn't follow that this was what you were saying, though it seems to me you also didn't state it very clearly. Anyway, let's consider a restricted area of spacetime, what I think would be called a "patch" of spacetime in GR, consisting only of the region outside the horizon in Rindler coordinates, and then consider a Minkowski coordinate system (with Minkowski metric) that only covers this patch, not anything beyond it. Would you agree that if we are considering a Minkowski metric defined on a Minkowski coordinate system on this patch, and a Rindler metric defined on a Rindler coordinate system on this patch, then the two spacetimes defined this way are the same everywhere on the patch, that geometrically they represent the "same metric" on this patch?

Yet this does not make the Rindler spacetime be the same as Minkowski everywhere! Why don't you want to agree that the Rindler spacetime does not cover the whole of the Minkowski spacetime and it of course has a horizon and different dynamics? If you exclude this last thing, I would sort of agree with you!

OK, I do certainly agree that the Rindler system doesn't cover the entire region covered by the Minkowski coordinate system, I just meant that they were geometrically identical on the region of spacetime covered by both. Again, if this was all you were objecting to in my saying they were equivalent, then we were having a miscommunication.

On the other hand if you used a metric equation that did not fall into the same equivalence class of describing a flat spacetime geometry, like the Schwarzschild metric, then there would be no possible coordinate transformation such that you'd get all the same physical predictions about coordinate-independent local facts if you used the Schwarzschild metric in Schwarzschild coordinates vs. the Minkowski metric in Minkowski coordinates.

Are you speaking locally? Or globally? You should specify this first!

I think "local" in a GR context usually refers to the infinitesimal neighborhood of a point, whereas "global" refers to the entire spacetime. I'm talking about the in-between case of a patch of spacetime with finite or infinite area, and the idea that two metric equations can be precisely equivalent descriptions of the geometry of spacetime in such a patch, just expressed in different coordinate systems on the patch.

Any spacetime, as EP says, can be made flat approximately in a small region and can be made flat exactly at some given point!

But in a Rindler spacetime, since there is no genuine curvature, it's not just a matter of being "approximately flat" in a small finite-sized patch with the flatness only becoming exact in the limit as the size of the patch shrinks to zero around a point, as would be true in a curved spacetime like the one defined by the Schwarzschild metric. Any patch of spacetime covered by the Rindler coordinate system is exactly flat and one can do a coordinate transformation from Rindler coordinates on this patch to an inertial coordinate system on the same patch where the laws of physics are precisely those seen in Minkowski coordinates with the Minkowski metric. So a Minkowski metric in Minkowski coordinates on this patch and a Rindler metric in Rindler coordinates on this patch are describing the exact same physical geometry and would lead to all the same physical predictions about coordinate-independent facts on this patch. It would not be possible to find any finite-sized patch in Schwarzschild coordinates with the Schwarzschild metric where this sort of exact equivalence with the predictions of the Minkowski metric would hold.

Thus, this notion of an equivalence class of metric equations describing the "same geometry" is a very useful one physically, since it distinguishes between cases where using different metric equations makes no difference at all in terms of any of your coordinate-independent physical predictions, and cases where it does make a difference.

It is sort of useful. But it does not tell everything about the metrics and thus it can't hold a candle to the Petrov classification.

I don't understand enough about GR to follow what the Petrov classification is saying. But if your only objection to my comments about the Rindler metric and the Minkowski metric being equivalent had to do with the fact that the Rindler coordinate system doesn't cover every point covered by the Minkowski coordinate system, what about my suggestion of looking only at a patch covered by both systems? Would you then agree that on this patch both metrics are totally equivalent physically and make all the same coordinate-independent predictions about physical events on this patch? Likewise, if we consider only the patch of spacetime which is outside the event horizon of a nonrotating Schwarzschild black hole (region I of a Kruskal-Szekeres coordinate system), then would you agree that on this patch the Schwarzschild metric and the Kruskal-Szekeres metric and the Eddington-Finkelstein metric describe the exact same spacetime geometry as one another and make identical predictions about physical events on this patch? Assuming you do agree with both of these, is there any technical name for this sort of exact equivalence between different metric equations on patches of spacetime that can have finite or infinite extent (as opposed to EP which normally only works exactly on infinitesimal 'patches')?

Depends how you define the EP I suppose, but certainly it would be possible to construct a "very large" system of rulers and clocks which are free-falling in the pseudo-gravitational field in Rindler spacetime, such that if you use this system to define a new coordinate system, then the equations of laws of physics as defined this coordinate system will be exactly the same as those seen in an SR inertial frame (in fact this free-falling frame is just an SR inertial frame as viewed in Rindler coordinates). Again, please tell me if you disagree with this.

I disagree! In the Rindler metric the situation is the same as any other metric in which the EP is defined only within its common ranges, a small region where we can approximately make the spacetime flat and dreaming of a "very large" region as for the EP to hold in is just impossible because the geodesic equations are position-dependent!

What do you mean by "small region"? Do you disagree that for any "patch" of spacetime that lies within the region covered by the Rindler coordinate system, no matter how large this patch is (even if it extends infinitely far in the direction going away from the Rindler horizon), it would be possible to construct a system of free-falling rulers and clocks filling every point within this patch such that if we use the rulers and clocks to define a new coordinate system, the laws of physics in this system would be identical to those seen in an equivalent patch of Minkowski coordinates in SR? This would be directly implied by my previous paragraph about Rindler and Minkowski metrics making exactly identical physical predictions about events on the patch where both coordinate systems are defined, so I don't see how you can disagree with this unless you also disagree about what I said in that paragraph.

 

[snoopies622]

So what is the resolution to this apparent contradiction?

Hello?

 

[Altabeh]

Would you agree that if we are considering a Minkowski metric defined on a Minkowski coordinate system on this patch, and a Rindler metric defined on a Rindler coordinate system on this patch, then the two spacetimes defined this way are the same everywhere on the patch, that geometrically they represent the "same metric" on this patch?

Yes and this was the whole thing I wanted to say in the first place.

Would you then agree that on this patch both metrics are totally equivalent physically and make all the same coordinate-independent predictions about physical events on this patch?

Only on the patch you're referring to, or speaking technically, the Rindler wedge, Minkowski means Rindler and vice versa.

Likewise, if we consider only the patch of spacetime which is outside the event horizon of a nonrotating Schwarzschild black hole (region I of a Kruskal-Szekeres coordinate system), then would you agree that on this patch the Schwarzschild metric and the Kruskal-Szekeres metric and the Eddington-Finkelstein metric describe the exact same spacetime geometry as one another and make identical predictions about physical events on this patch? Assuming you do agree with both of these, is there any technical name for this sort of exact equivalence between different metric equations on patches of spacetime that can have finite or infinite extent (as opposed to EP which normally only works exactly on infinitesimal 'patches')?

Yes I do and I've not seen a technical term meaning the equivalence class of this type. It sounds like they use "equivalent coordinates" to generally mean all of those coordinates can be replaced by each other in some place that we're referring to as "patches". Otherwise, you know, they can't be equivalent, for instnace, at the surface 2m=r.

What do you mean by "small region"? Do you disagree that for any "patch" of spacetime that lies within the region covered by the Rindler coordinate system, no matter how large this patch is (even if it extends infinitely far in the direction going away from the Rindler horizon), it would be possible to construct a system of free-falling rulers and clocks filling every point within this patch such that if we use the rulers and clocks to define a new coordinate system, the laws of physics in this system would be identical to those seen in an equivalent patch of Minkowski coordinates in SR? This would be directly implied by my previous paragraph about Rindler and Minkowski metrics making exactly identical physical predictions about events on the patch where both coordinate systems are defined, so I don't see how you can disagree with this unless you also disagree about what I said in that paragraph.

The equivalence principle is not all about curvature; it behooves oneself to take into account the geodesic equations as well. In the Rindler metric, all particles following geodesics can't have a zero proper acceleration since the equations are position-dependent and thus the situation is different from the Minkowski spacetime where every particle moves along a straight line because the proper accelerations are all zero. The reason I use a 'small patch' is that there the position differences can be safely neglected and consequently we can suppose the transformation that can revert the mertic back into Minkowski metric.

AB

 

[JesseM]

The equivalence principle is not all about curvature; it behooves oneself to take into account the geodesic equations as well. In the Rindler metric, all particles following geodesics can't have a zero proper acceleration

By "proper acceleration" you aren't just talking about some coordinate-based notion of acceleration, right? Are you claiming here that an accelerometer moving along a geodesic in the Rindler metric would actually measure nonzero G-forces?

Also, are you claiming that if we look at geodesics in the Minkowski metric on the "Rindler wedge" (region of spacetime in Minkowski coordinates that's also covered by Rindler coordinates) and then use the coordinate transformation to figure out the corresponding paths in Rindler coordinates, that these paths will not be the geodesic ones in Rindler coordinates?

 

[bcrowell]

By "proper acceleration" you aren't just talking about some coordinate-based notion of acceleration, right? Are you claiming here that an accelerometer moving along a geodesic in the Rindler metric would actually measure nonzero G-forces?

I won't pretend to speak for Altabeh, but I believe the distinction is that coordinate acceleration is $d^2x/dt^2$, whereas proper accleeration is $d^2x/d\tau^2$, where $\tau$ is the proper time. They're both frame-dependent notions, since x is frame-dependent. In Rindler coordinates, the reason they differ is that clocks at rest with respect to the coordinate lattice run at different rates depending on x. Essentially the proper acceleration is what you measure for falling objects if your lab is kept at rest with respect to the coordinate lattice.

 

[DrGreg]

My understanding is that "proper acceleration" is the acceleration relative to a co-moving free-falling observer (using local distance and time). It's what an accelerometer measures. In tensor terms the scalar magnitude is the magnitude of the 4-acceleration (an invariant quantity), viz

$$\sqrt{\left| g_{\alpha\beta} \, \frac{d^2x^\alpha}{d\tau^2} \, \frac{d^2x^\beta}{d\tau^2} \right|}$$​

(possibly with some c's inserted if you don't assume c=1). I think JesseM is working to the same definition as me, but (as bcrowell suggests) Altabeh is thinking of the 3-vector $d^2 x^i / d\tau^2$ (i=1,2,3), or its magnitude, either way, a coordinate-dependent quantity. If I'm right this explains the confusion.

Proper acceleration in my sense is coordinate-independent, and therefore the same in both Minkowski and Rindler coordinates.

 

[JesseM]

Proper acceleration in my sense is coordinate-independent, and therefore the same in both Minkowski and Rindler coordinates.

So in your sense, geodesics according to the Rindler metric would have zero proper acceleration, right? Also, about my other comment if we look at geodesics in the Minkowski metric on the "Rindler wedge" (region of spacetime in Minkowski coordinates that's also covered by Rindler coordinates) and then use the coordinate transformation to figure out the corresponding paths in Rindler coordinates, would you say that these paths would also be geodesics according to the Rindler metric? That on the Rindler wedge, there is no disagreement between the Minkowski metric and the Rindler metric about whether any given path is a geodesic or not (using the coordinate transformation to identify the 'same path' in each coordinate system)?

 

[Altabeh]

By "proper acceleration" you aren't just talking about some coordinate-based notion of acceleration, right?

As bcrowell suggests, I'm talking about a frame-dependent acceleration which is measured by a comoving observer, i.e. $d^2x^i/d\tau^2$ that shows DrGreg's guess about my way of looking at the proper acceleration is correct! And the dependency of proper acceleration on the position comes from the fact that the second term in the geodesic equations is a function of position in the Rindler metric, Right? This means that if you and I start accelerating from two different positions in the Rindler spacetime, we both will have different proper accelerations due to a discrepancy in our comoving observers' clocks!

Are you claiming here that an accelerometer moving along a geodesic in the Rindler metric would actually measure nonzero G-forces?

The accelerometer can't be ideal so as to not be affected by the gravitational field unless in a small region where we take the EP to hold and thus the accelerometer could be ideal there! Since the gravitational field applies a (uniform!) force on the coordinate lattice in such a way that particles following geodesics would have proper accelerations dependent on the position according to geodesic equations, two different comoving observers cannot measure the same proper accelerations for the particles seen in their own frame.

Also, are you claiming that if we look at geodesics in the Minkowski metric on the "Rindler wedge" (region of spacetime in Minkowski coordinates that's also covered by Rindler coordinates) and then use the coordinate transformation to figure out the corresponding paths in Rindler coordinates, that these paths will not be the geodesic ones in Rindler coordinates?

Yes and it is like that just because the in the Minkowski spacetime the inertial and gravitational accelerations vanish whereas in the Rindler spacetime both accelerations are present through the existence of a non-zero second term in the geodesic equations.

AB

 

[JesseM]

As bcrowell suggests, I'm talking about a frame-dependent acceleration which is measured by a comoving observer, i.e. $d^2x^i/d\tau^2$ that shows DrGreg's guess about my way of looking at the proper acceleration is correct!

OK, but DrGreg said that his own definition was what would actually be measured by a physical accelerometer and is coordinate-independent, do you agree that this is true of his definition?

The accelerometer can't be ideal so as to not be affected by the gravitational field

I didn't say anything about it not being affected by the gravitational field, though. My argument is that the Rindler metric and the Minkowski metric will make all the same predictions about coordinate-independent physical facts on the region of spacetime covered by the Rindler coordinate system, so that if you have an observer moving at constant velocity in Minkowski coordinates who measured no G-forces on his accelerometer, and then you translate his worldline into Rindler coordinates and use the Rindler metric to predict what happens to his accelerometer, you'll also get the prediction that he registers no G-forces. It might be that from the perspective of Rindler coordinates, this has something to do with the pseudo-gravitational field counteracting the coordinate acceleration of the accelerometer, I don't know. But do you disagree with this basic prediction that a constant-velocity path in Minkowski coordinates, when translated into Rindler coordinates using the coordinate transformation, will still be a path that registers zero reading on an ordinary physical accelerometer when we use the Rindler metric to predict the accelerometer's behavior?

Also, are you claiming that if we look at geodesics in the Minkowski metric on the "Rindler wedge" (region of spacetime in Minkowski coordinates that's also covered by Rindler coordinates) and then use the coordinate transformation to figure out the corresponding paths in Rindler coordinates, that these paths will not be the geodesic ones in Rindler coordinates?

Yes and it is like that just because the in the Minkowski spacetime the inertial and gravitational accelerations vanish whereas in the Rindler spacetime both accelerations are present through the existence of a non-zero second term in the geodesic equations.

Well, I feel doubtful about this because I had thought that on the Rindler wedge, Minkowski coordinates with the Minkowski metric and Rindler coordinates with the Rindler metric were just different ways of describing the exact same physical spacetime (both in terms of local events like clock/accelerometer readings and in terms of the geometry of the spacetime), so they couldn't disagree about whether a given path would be a geodesic or not. Hopefully DrGreg will address my question about this so I can get a "second opinion".

 

[bcrowell]

Proper acceleration in my sense is coordinate-independent, and therefore the same in both Minkowski and Rindler coordinates.

I looked in some textbooks and didn't find much to establish what the standard definition would be. I looked in Carroll (online version), MTW, Wald, and Rindler's Essential Relativity. The only one that seemed to define it was Rindler, and Rindler defined it only in the context of SR, in which case the distinction between your definition and mine would be unnecessary.

 

[DrGreg]

I looked in some textbooks and didn't find much to establish what the standard definition would be. I looked in Carroll (online version), MTW, Wald, and Rindler's Essential Relativity. The only one that seemed to define it was Rindler, and Rindler defined it only in the context of SR, in which case the distinction between your definition and mine would be unnecessary.

I didn't get it quite right. I just had a look in Rindler's Relativity: Special, General and Cosmological (2nd Ed 2006, page 214) and his definition is almost, but not quite, what I said. I forgot to account for non-inertial coordinates in the right way. His definition of 4-acceleration is

$$A^\mu = \frac{DU^\mu}{d\tau}$$​

where $U^\mu$ is 4-velocity defined in the usual way $dx^\mu/d\tau$ and "$D/d\tau$" denotes "absolute differentiation" along a worldline, a concept related to covariant differentiation. (Essentially the "directional covariant derivative" in the direction of the tangent vector)

$$A^\mu = \frac{DU^\mu}{d\tau} = \frac{dU^\mu}{d\tau} + \Gamma^\mu_{\alpha\beta}U^\alpha U^\beta$$​

Rindler defines proper acceleration to be the magnitude of this correctly-defined 4-acceleration $\sqrt{|A_{\mu} A^{\mu} |}$.

Setting the proper acceleration to zero gives the geodesic equation

$$\frac{dU^\mu}{d\tau} + \Gamma^\mu_{\alpha\beta}U^\alpha U^\beta = 0$$​

And of course in Minkowski coordinates all the $\Gamma$s are zero.

Geodesics are coordinate independent, so the geodesics in Rindler coordinates are exactly the same as the geodesics in Minkowski coordinates (within the wedge where they both apply).

P.S. The Wikipedia articles on proper acceleration and four-acceleration back all this up. (Citing Wikipedia doesn't prove anything, but it's extra evidence.)

 

[bcrowell]

I didn't get it quite right. I just had a look in Rindler's Relativity: Special, General and Cosmological (2nd Ed 2006, page 214) and his definition is almost, but not quite, what I said. I forgot to account for non-inertial coordinates in the right way. His definition of 4-acceleration is

$$A^\mu = \frac{DU^\mu}{d\tau}$$​

where $U^\mu$ is 4-velocity defined in the usual way $dx^\mu/d\tau$ and "$D/d\tau$" denotes "absolute differentiation" along a worldline, a concept related to covariant differentiation. (Essentially the "directional covariant derivative" in the direction of the tangent vector)

OK. So in that case "proper acceleration," as defined by Rindler, is probably not a useful way to talk about uniform gravitational fields. The frame-independent proper acceleration is guaranteed to be zero for any object that isn't subjected to nongravitational forces. I think the useful thing to talk about is $d^2x/d\tau^2$, which is frame-dependent.

 

[DrGreg]

OK. So in that case "proper acceleration," as defined by Rindler, is probably not a useful way to talk about uniform gravitational fields. The frame-independent proper acceleration is guaranteed to be zero for any object that isn't subjected to nongravitational forces. I think the useful thing to talk about is $d^2x/d\tau^2$, which is frame-dependent.

There is a way of assessing the "acceleration due to gravity" (relative to a coordinate system) using 4-vectors. Take the 4-velocity of a particle that is at rest in your chosen coordinate system. Assuming the usual convention of a 0th timelike coordinate and 1st, 2nd and 3rd spacelike coordinates, the 4-velocity will be

$$U^\mu = \left( \frac{1}{\sqrt{g_{00}}}, 0, 0, 0 \right)$$​

(assuming units in which c=1 and with a +−−− metric signature, for the sake of argument). U¹=U²=U³=0 and $g_{\alpha\beta}U^{\alpha}U^{\beta} = 1$.

Now calculate the 4-acceleration $DU^\mu/d\tau$ (using the correct definition with Christoffel symbols as in my previous post) and its magnitude (the proper acceleration) will be the "acceleration due to gravity".

There is an example of this calculation in Woodhouse's General Relativity p.99. You can see the lecture notes on which the book was based at http://people.maths.ox.ac.uk/~nwoodh/gr/index.html , Section 12.1 page 54. Woodhouse does it for Schwarzschild coords but you can use the same method for Rindler coords.

 

[bcrowell]

Now calculate the 4-acceleration $DU^\mu/d\tau$ (using the correct definition with Christoffel symbols as in my previous post) and its magnitude (the proper acceleration) will be the "acceleration due to gravity".

I must be missing something here. Except for the relatively trivial fact that you're differentiating with respect to a parameter, essentially what it sounds like is that you're taking the covariant derivative of a velocity vector. But the covariant derivative of a velocity vector is zero on a geodesic -- that's one way of defining a geodesic.

 

[DrGreg]

I must be missing something here. Except for the relatively trivial fact that you're differentiating with respect to a parameter, essentially what it sounds like is that you're taking the covariant derivative of a velocity vector. But the covariant derivative of a velocity vector is zero on a geodesic -- that's one way of defining a geodesic.

But my 4-velocity is not the 4-velocity of a free-falling particle (following a geodesic), it's the 4-velocity of a particle permanently at rest in the chosen coordinate system, which, for non-inertial coordinates, is not free-falling and is not following a geodesic.

 

[bcrowell]

But my 4-velocity is not the 4-velocity of a free-falling particle (following a geodesic), it's the 4-velocity of a particle permanently at rest in the chosen coordinate system, which, for non-inertial coordinates, is not free-falling and is not following a geodesic.

Ah, I see. Thanks for the clarification.

 

[Altabeh]

OK, but DrGreg said that his own definition was what would actually be measured by a physical accelerometer and is coordinate-independent, do you agree that this is true of his definition?

Look, the Rindler spacetime has a gravitational field which affects the coordinate lattice, say, uniformly! But the point is the 3-acceleration of a particle, $$d^2x^i/d\tau^2$$, is dependent on position in this spacetime through that factor, $$x^2$$, in the line element and thus it is frame dependent! This acceleration is measured by a comoving observer's accelerometer and in case the observer is at rest, so there is no proper time anymore and we're just left with coordinate acceleration, $$d^2x^i/dt^2$$, which is itself frame-dependent!

I didn't say anything about it not being affected by the gravitational field, though. My argument is that the Rindler metric and the Minkowski metric will make all the same predictions about coordinate-independent physical facts on the region of spacetime covered by the Rindler coordinate system,

and of course the 3-acceleration is not frame independent.

so that if you have an observer moving at constant velocity in Minkowski coordinates who measured no G-forces on his accelerometer, and then you translate his worldline into Rindler coordinates and use the Rindler metric to predict what happens to his accelerometer, you'll also get the prediction that he registers no G-forces.

But you're missing the point that in the Rindler metric we have a gravitational field while there is no such thing in the Minkowski spacetime! The existence of such gravitational field is seen clearly from the fact that the Rindler's Christoffel symbols don't vanish.

It might be that from the perspective of Rindler coordinates, this has something to do with the pseudo-gravitational field counteracting the coordinate acceleration of the accelerometer, I don't know.

YES!

But do you disagree with this basic prediction that a constant-velocity path in Minkowski coordinates, when translated into Rindler coordinates using the coordinate transformation, will still be a path that registers zero reading on an ordinary physical accelerometer when we use the Rindler metric to predict the accelerometer's behavior?

I do because the geodesic equations tell me the accelerometer does not read the same acceleration in the Rindler coordinates as in the Minkowski coordinates. This shows how the 3-acceleration is coordinate-dependent.

Well, I feel doubtful about this because I had thought that on the Rindler wedge, Minkowski coordinates with the Minkowski metric and Rindler coordinates with the Rindler metric were just different ways of describing the exact same physical spacetime (both in terms of local events like clock/accelerometer readings and in terms of the geometry of the spacetime), so they couldn't disagree about whether a given path would be a geodesic or not. Hopefully DrGreg will address my question about this so I can get a "second opinion".

I hope he'd be interested to give us his argument as to how, as you seem to believe in it, in the Rindler spacetime a geodesic is the same as a geodesic in the Minkowski spacetime.

AB

 

[Ich]

I do because the geodesic equations tell me the accelerometer does not read the same acceleration in the Rindler coordinates as in the Minkowski coordinates. This shows how the 3-acceleration is coordinate-dependent.

An accelerometer doesn't measure "3-acceleration", it measures proper acceleration. Free falling particles measure zero proper acceleration, they are moving on geodesics, and they are doing so whether you describe their motion in Rindler- or Minkowski coordinates.

 

[Altabeh]

An accelerometer doesn't measure "3-acceleration", it measures proper acceleration. Free falling particles measure zero proper acceleration, they are moving on geodesics, and they are doing so whether you describe their motion in Rindler- or Minkowski coordinates.

I didn't say an accelerometer measures proper 3-acceleration. I just said since the components of proper 3-acceleration are position-dependent, as the geodesic equations say, then its magnitude is what an accelerometer being under the control of a comoving observer shows and it is of course frame-dependent.

Ab

 

[DrGreg]

I hope he'd be interested to give us his argument as to how, as you seem to believe in it, in the Rindler spacetime a geodesic is the same as a geodesic in the Minkowski spacetime.

First of all, I wouldn't talk of Rindler spacetime and Minkowski spacetime; I'd talk about Rindler coordinates and Minkowski coordinates as two different coordinate systems within the same spacetime.

The property of "being a geodesic" is a geometrical property that doesn't depend on the choice of coordinates; it is an intrinsic property of a worldline. The 4-acceleration

$$A^\mu = \frac{DU^\mu}{d\tau} = \frac{dU^\mu}{d\tau} + \Gamma^\mu_{\alpha\beta}U^\alpha U^\beta$$​

(where U is 4-velocity) is a 4-vector, i.e. a tensor that transforms between coordinate systems in the correct way. (And the symbol $dU^\mu/d\tau$ does not represent a 4-vector.)

Its magnitude, "proper acceleration", $\sqrt{|A_{\mu} A^{\mu} |}$ is therefore a scalar invariant, the same value in all coordinate systems. The geodesic equation -- in any valid coordinate system you like -- is just the condition that the proper acceleration is zero, or equivalently, that the (spacelike) 4-acceleration is the zero 4-vector.

 

[Ich]

I just said since the components of proper 3-acceleration are position-dependent, as the geodesic equations say, then its magnitude is what an accelerometer being under the control of a comoving observer shows and it is of course frame-dependent.

JesseM asked:

But do you disagree with this basic prediction that a constant-velocity path in Minkowski coordinates, when translated into Rindler coordinates using the coordinate transformation, will still be a path that registers zero reading on an ordinary physical accelerometer when we use the Rindler metric to predict the accelerometer's behavior?

He isn't talking about comoving, but free-falling observers. And he isn't talking about coordinate acceleration, but proper acceleration (as measured by an accelerometer).
Your statement that "3-acceleration" is frame dependent, while right, has no relevance whatsoever, and does not back up your disagreement with the statement.
I think there are some misunderstandings that should be clarified.

 

[Altabeh]

JesseM asked:

He isn't talking about comoving, but free-falling observers. And he isn't talking about coordinate acceleration, but proper acceleration (as measured by an accelerometer).
Your statement that "3-acceleration" is frame dependent, while right, has no relevance whatsoever, and does not back up your disagreement with the statement.
I think there are some misunderstandings that should be clarified.

Since the start of this thread, I've been using acceleration as the proper acceleration 3-vector, or similarly, proper 3-acceleration and I clearly made this apparent to everyone when I answered in the affirmative to DrGreg's post:

Altabeh is thinking of the 3-vector $d^2 x^i / d\tau^2$
(i=1,2,3), or its magnitude, either way, a coordinate-dependent quantity.

and my answer was given before JesseM asked the question:

...that shows DrGreg's guess about my way of looking at the proper acceleration is correct.

I don't think the misunderstanding is from me! But I agree that I was talking about comoving observer rather than a free-falling observer!

First of all, I wouldn't talk of Rindler spacetime and Minkowski spacetime; I'd talk about Rindler coordinates and Minkowski coordinates as two different coordinate systems within the same spacetime.

Again, the Minkowski spacetime is not the same as the Rindler spacetime. The first one is only an extended version of the latter. See Wald, R. M. General Relativity, page 151. And talking about Rindler spacetime inspires the application of Rindler coordinates immediately so I don't think this was a necessary note to make!

The property of "being a geodesic" is a geometrical property that doesn't depend on the choice of coordinates; it is an intrinsic property of a worldline. The 4-acceleration

$$A^\mu = \frac{DU^\mu}{d\tau} = \frac{dU^\mu}{d\tau} + \Gamma^\mu_{\alpha\beta}U^\alpha U^\beta$$

(where U is 4-velocity) is a 4-vector, i.e. a tensor that transforms between coordinate systems in the correct way. (And the symbol $dU^\mu/d\tau$ does not represent a 4-vector.)

Edited: Along a geodesic the proper acceleration is zero if such thing is defined to be the magnituse of $DU^\mu/d\tau.$ Otherwise the use of proper acceleration 3-vector combined with a null time-component in the equation $$A=\sqrt{g_{\mu\nu}a^{\mu}a^{\nu}}$$ would not lead to a frame-independent acceleration.

AB

 

[Ich]

I don't think the misunderstanding is from me! But I agree that I was talking about comoving observer rather than a free-falling observer!

I didn't say that the misunderstanding was yours. Your "no" to JesseM's question was clearly wrong, that's why I thought that you are talking about something different, and in that case i think it would have helped if you had read the question carefully.

with [itex] a^{\mu}=\frac{d^2x^{\mu}}{d\tau^2}[/tex] being the proper four-acceleration

That's a coordinate acceleration. It becomes proper acceleration after correcting for coordinate effects (with the help of the Christoffel symbols). That's what DrGreg, JesseM, and I are talking about.

 

[Altabeh]

That's a coordinate acceleration. It becomes proper acceleration after correcting for coordinate effects (with the help of the Christoffel symbols). That's what DrGreg, JesseM, and I are talking about.

This is incorrect! The coordinate acceleration is the derivative of the coordinate velocity with respect to coordinate time, i.e. $$d^2x^{\nu}/dt^2.$$ That is a proper acceleration 3-vector combined with a null time-component.

AB

 

[JesseM]

First of all, I wouldn't talk of Rindler spacetime and Minkowski spacetime; I'd talk about Rindler coordinates and Minkowski coordinates as two different coordinate systems within the same spacetime.

Again, the Minkowski spacetime is not the same as the Rindler spacetime. The first one is only an extended version of the latter.

But if the first is an extended version of the latter, you agree they are "the same spacetime" in the region of the Rindler wedge, right? If so, how can you also argue that they disagree about local physical facts on this wedge, like what set of events a particle with no non-gravitational forces acting on it (i.e. a particle following a geodesic path) will cross through?

Also, do you disagree that, as DrGreg said, an object on a geodesic path feels zero 4-acceleration everywhere along it rather than zero 3-acceleration? And that the magnitude of the 4-acceleration for a given point on a worldline determines what would be measured by a physical accelerometer moving along that worldline? Finally, do you disagree that if we have a path in Minkowski coordinates on the Rindler wedge that has zero 4-acceleration everywhere as calculated by the Minkowski metric (i.e. a path with constant velocity in Minkowski coordinates), then if we use the coordinate transformation to transform this path into Rindler coordinates, the path will still have zero 4-acceleration everywhere along it as calculated using the Rindler metric?

 

[DrGreg]

This is incorrect! The coordinate acceleration is the derivative of the coordinate velocity with respect to coordinate time, i.e. $$d^2x^{\nu}/dt^2.$$ That is a proper acceleration 3-vector combined with a null time-component.

AB

You are right that $d^2x^i/dt^2$ is coordinate acceleration.

What has caused confusion in this thread is your describing $d^2x^i/d\tau^2$ as "proper acceleration". That is not the terminology that everyone else uses and so everyone has been disagreeing with you.

I'm guessing that maybe you thought of this because some authors describe the 3-vector $dx^i/d\tau$ as "proper velocity". That is not a terminology I like; I prefer to call that by its alternative name "celerity". In relativity, most "proper" things are invariant e.g. proper time, proper length and the correct definition of proper acceleration (& I've seen some people describe (rest) mass as "proper mass"). The 3-vector $dx^i/d\tau$ is, of course, coordinate-dependent. The 4-vector $dx^{\nu}/d\tau$ is described as "4-velocity" (rather than "proper velocity").

 

[Altabeh]

But if the first is an extended version of the latter, you agree they are "the same spacetime" in the region of the Rindler wedge, right? If so, how can you also argue that they disagree about local physical facts on this wedge, like what set of events a particle with no non-gravitational forces acting on it (i.e. a particle following a geodesic path) will cross through?

I think once I said that these two are the same when it comes to the Rindler wedge. And about the second question, all problems arise from the way I've learned the implication of "proper acceleration"! Since I've been taking the 3-vector $dx^i/d\tau$ as "proper velocity", so automatically I let $d^2x^i/d\tau^2$ be the "proper 3-acceleration" whereas everyone else makes use of some other definition for "proper acceleration", basically the magnitude of the 4-acceleration $D^2x^i/d\tau^2$. This sort of usage of the proper acceleration would mean that such implication is invariant under the change of coordinates and thus is frame-independent. Along a geodesic, as I said earlier, $D^2x^i/d\tau^2$ vanishes whether on the Rindler wedge or in the Minkowski spacetime. But when we use $d^2x^i/d\tau^2$, then we won't end up getting a frame-independent value for its magnitude. This means that if I looked through the angle you see the whole problem, then I would agree with you!

Also, do you disagree that, as DrGreg said, an object on a geodesic path feels zero 4-acceleration everywhere along it rather than zero 3-acceleration?

The proper 4-acceleration must be zero along any geodesic, but this can't be true for 3-acceleration.

And that the magnitude of the 4-acceleration for a given point on a worldline determines what would be measured by a physical accelerometer moving along that worldline?

Now I'd like to say "YES"!

Finally, do you disagree that if we have a path in Minkowski coordinates on the Rindler wedge that has zero 4-acceleration everywhere as calculated by the Minkowski metric (i.e. a path with constant velocity in Minkowski coordinates), then if we use the coordinate transformation to transform this path into Rindler coordinates, the path will still have zero 4-acceleration everywhere along it as calculated using the Rindler metric?

When it comes to motion along a geodesic, then the proper 4-acceleration $D^2x^i/d\tau^2$ automatically vanishes in any spacetime, leading me to not disagree! But if we are talking about a general path, then the proper 4-acceleration $D^2x^i/d\tau^2$ does not neccessarily vanish on the Rindler wedge for a particle that was already supposed to be moving with a constant velocity in the Minkowski spacetime. This is just because the right hand side of geodesic equations is no longer zero for non-geodesic trajectories (i.e. there is no freely moving particle anymore) in the Rindler coordinates.

AB

 

[JesseM]

Also, do you disagree that, as DrGreg said, an object on a geodesic path feels zero 4-acceleration everywhere along it rather than zero 3-acceleration?

The proper 4-acceleration must be zero along any geodesic, but this can't be true for 3-acceleration.

And that the magnitude of the 4-acceleration for a given point on a worldline determines what would be measured by a physical accelerometer moving along that worldline?

Now I'd like to say "YES"!

"YES" you disagree with that last one about the accelerometer? Here you're talking about an accelerometer moving along that worldline at all times and not an accelerometer held by a temporarily co-moving observer, right? If you do disagree that the magnitude of the 4-acceleration determines the reading of such an accelerometer, what do you think determines the reading of the accelerometer?

Finally, do you disagree that if we have a path in Minkowski coordinates on the Rindler wedge that has zero 4-acceleration everywhere as calculated by the Minkowski metric (i.e. a path with constant velocity in Minkowski coordinates), then if we use the coordinate transformation to transform this path into Rindler coordinates, the path will still have zero 4-acceleration everywhere along it as calculated using the Rindler metric?

When it comes to motion along a geodesic, then the proper 4-acceleration $D^2x^i/d\tau^2$ automatically vanishes in any spacetime, leading me to not disagree! But if we are talking about a general path, then the proper 4-acceleration $D^2x^i/d\tau^2$ does not neccessarily vanish on the Rindler wedge for a particle that was already supposed to be moving with a constant velocity in the Minkowski spacetime.

So if we take a constant-velocity path in Minkowski coordinates (presumably you'd agree this is a geodesic according to the Minkowski metric, and that the Minkowski metric says the 4-acceleration vanishes on this path), then use the coordinate transformation to translate the set of events that constitute the path into Rindler coordinates, you would argue that the 4-acceleration does not vanish along this path according to the Rindler metric?

 

[Altabeh]

"YES" you disagree with that last one about the accelerometer? Here you're talking about an accelerometer moving along that worldline at all times and not an accelerometer held by a temporarily co-moving observer, right? If you do disagree that the magnitude of the 4-acceleration determines the reading of such an accelerometer, what do you think determines the reading of the accelerometer?

An accelerometer held by a free-falling observer is supposed to be reading the magnitude of the proper 4-acceleration of a particle moving along an arbitrary path. But if the observer is moving parallel to the particle at all times (a co-moving observer) then I think both the magnitudes of proper 3-acceleration and proper 4-acceleration can be read by accelerometer in case the two different scenarios for proper acceleration are considered together.

So if we take a constant-velocity path in Minkowski coordinates (presumably you'd agree this is a geodesic according to the Minkowski metric, and that the Minkowski metric says the 4-acceleration vanishes on this path), then use the coordinate transformation to translate the set of events that constitute the path into Rindler coordinates, you would argue that the 4-acceleration does not vanish along this path according to the Rindler metric?

If I can get you right, by velocity' you are presumably referring to the coordinate velocity because in case the "proper" is needed, you state that in your sentences! Since the coordinate velocity is frame-dependent, a coordinate transformation must make it change in particular if the secondary metric is position-dependent (e.g. Rindler metric). In Rindler metric with $$c=1$$ the coordinate velocity $$v$$ along geodesics is given by

$$v=(x/2) a$$

where $$a$$ represents the coordinate acceleration. Thus it is clear that if one seeks out a constant $$v$$, it is necessary to let the particle be hovering at a constant $$x.$$ or be moving along paths with constant $$x.$$ Along non-geodesic trajectories, a constant $$v$$ means that the coordinate $$x$$ must remain constant duing the motion and this tells us that the proper 4-acceleration vanishes. But since we are talking about general paths, the constancy of $$v$$ cannot be always guaranteed in the Rindler metric even if the particle has a constant coordinate velocity in Minkowski spacetime.

AB

 

[JesseM]

An accelerometer held by a free-falling observer is supposed to be reading the magnitude of the proper 4-acceleration of a particle moving along an arbitrary path. But if the observer is moving parallel to the particle at all times (a co-moving observer) then I think both the magnitudes of proper 3-acceleration and proper 4-acceleration can be read by accelerometer in case the two different scenarios for proper acceleration are considered together.

How can a single accelerometer read both the 3-acceleration and the 4-acceleration? It only shows one reading at any given moment, so surely it must agree with one or the other?

If I can get you right, by velocity' you are presumably referring to the coordinate velocity because in case the "proper" is needed, you state that in your sentences! Since the coordinate velocity is frame-dependent, a coordinate transformation must make it change in particular if the secondary metric is position-dependent (e.g. Rindler metric). In Rindler metric with $$c=1$$ the coordinate velocity $$v$$ along geodesics is given by

$$v=(x/2) a$$

where $$a$$ represents the coordinate acceleration. Thus it is clear that if one seeks out a constant $$v$$

No, I said I wanted to look at the wordline of a particle with constant velocity (coordinate velocity) in Minkowski coordinates, but then I made clear that I wanted to translate the same worldline consisting of the same events into Rindler coordinates using the coordinate transformation. This worldline obviously wouldn't have constant coordinate velocity in Rindler coordinates! Since on the Rindler wedge Minkowski coordinates and Rindler coordinates are just different ways of labeling events in the same physical spacetime, surely you agree that we can talk about the same physical worldline as described in the two different coordinate systems? I don't think there was any ambiguity in my description:

So if we take a constant-velocity path in Minkowski coordinates (presumably you'd agree this is a geodesic according to the Minkowski metric, and that the Minkowski metric says the 4-acceleration vanishes on this path), then use the coordinate transformation to translate the set of events that constitute the path into Rindler coordinates

Do you understand what I'm saying now? If so, do you agree that if we take a worldline that has constant coordinate velocity in Minkowski coordinates, then use the coordinate transformation to describe the same worldline in Rindler coordinates, and use the Rindler metric to calculate the 4-acceleration on this worldline, then we'll find it has zero 4-acceleration everywhere?

 

[Altabeh]

How can a single accelerometer read both the 3-acceleration and the 4-acceleration? It only shows one reading at any given moment, so surely it must agree with one or the other?

Our problem is we can't understand each other the way we should! I don't mean an accelerometer can read both of them at the same time! I assume you know that some measurement devices are ad hoc but some aren't, e.g. a clock that shows the current local time in New York and Texas simultananeously which isn't my purpose here! An accelerometer in my opinion can do read both the 3-acceleration and the 4-acceleration but not at the same time and I think I didn't say something like this!

No, I said I wanted to look at the wordline of a particle with constant velocity (coordinate velocity) in Minkowski coordinates, but then I made clear that I wanted to translate the same worldline consisting of the same events into Rindler coordinates using the coordinate transformation.This worldline obviously wouldn't have constant coordinate velocity in Rindler coordinates!

I think I showed this for a particle moving along a geodesic in Rindler coordinates.

Since on the Rindler wedge Minkowski coordinates and Rindler coordinates are just different ways of labeling events in the same physical spacetime, surely you agree that we can talk about the same physical worldline as described in the two different coordinate systems? I don't think there was any ambiguity in my description:

Sorry! This time I confused the proper 4-acceleration with coordinate 4-acceleration. I was looking at your 4-acceleration, talking about proper 4-acceleration but trying to make a meaning of coordinate 4-acceleration out of it!

Do you understand what I'm saying now? If so, do you agree that if we take a worldline that has constant coordinate velocity in Minkowski coordinates, then use the coordinate transformation to describe the same worldline in Rindler coordinates, and use the Rindler metric to calculate the 4-acceleration on this worldline, then we'll find it has zero 4-acceleration everywhere?

Yes!

AB

 

[JesseM]

Our problem is we can't understand each other the way we should! I don't mean an accelerometer can read both of them at the same time! I assume you know that some measurement devices are ad hoc but some aren't, e.g. a clock that shows the current local time in New York and Texas simultananeously which isn't my purpose here! An accelerometer in my opinion can do read both the 3-acceleration and the 4-acceleration but not at the same time and I think I didn't say something like this!

I think "accelerometer" has a standard meaning in physics just like "clock". Any of the various accelerometer designs here should measure 4-acceleration, for example (and the 4-acceleration measured in G-forces should also produce the same effects for an observer following this worldline as would be felt if they were at rest in a gravitational field with the same G-forces). Maybe you could produce a different type of device to measure 3-acceleration locally, but I'm not sure how it would work, and I don't think it would qualify as an "accelerometer".

Do you understand what I'm saying now? If so, do you agree that if we take a worldline that has constant coordinate velocity in Minkowski coordinates, then use the coordinate transformation to describe the same worldline in Rindler coordinates, and use the Rindler metric to calculate the 4-acceleration on this worldline, then we'll find it has zero 4-acceleration everywhere?

Yes!

OK, good. And would you also agree that this worldline is a geodesic according to the Rindler metric? (i.e. it's the worldline that maximizes the proper time between any two events that lie on it)

 

[Altabeh]

OK, good. And would you also agree that this worldline is a geodesic according to the Rindler metric? (i.e. it's the worldline that maximizes the proper time between any two events that lie on it)

Why not? But let's speak more generally and say that the worldline is not just timelike but arbitrary so the proper "quantity" between any two events lying on it is an extremum.

AB

 

[JesseM]

Why not?

Well, because for a long time you were denying my claim that if we looked at a geodesic in Minkowski coordinates, and then used the coordinate transformation to define the same path in Rindler coordinates, then the path would still be a geodesic relative to the Rindler metric. But I guess you were just misunderstanding what I meant when I said things like "same path", not realizing I was talking about using the coordinate transformation to find the new coordinates of the same physical events along a given path which was originally defined in Minkowski coordinates.

But let's speak more generally and say that the worldline is not just timelike but arbitrary so the proper "quantity" between any two events lying on it is an extremum.

Yeah, spacelike and null geodesics in Minkowski coordinates should also map to spacelike and null geodesics in Rindler coordinates, assuming again that we are looking at the same physical set of events in each coordinate system.

Assuming we've settled this issue, perhaps we can revisit my earlier claim that the equivalence principle would hold in arbitrarily large regions of Rindler coordinates? Hopefully you'd agree that in any flat SR spacetime described in Minkowski coordinates, it should be possible to construct a physical network of rulers and synchronized clocks moving inertially relative to this coordinate system (a network of the type that Einstein used to define the concept of an 'inertial frame'), such that if we use the readings on the rulers and clocks to define a new coordinate system, the laws of physics as seen in this coordinate system (which is just a different Minkowski frame) will be those seen in any SR inertial frame. Now, if you agree that Minkowski coordinates (on the Rindler wedge) and Rindler coordinates are just different descriptions of the same physical spacetime, this same physical network of rulers and clocks could also be described in Rindler coordinates, agreed?

Now suppose we have such a physical network, and we want to describe where events in an experiment happen relative to that network in the context of our Minkowski or Rindler coordinate systems. Suppose a particular event E (a collision, say) happens next to the x=12 light-seconds mark on the physical ruler representing the x-axis in this network, and that the physical clock at that mark reads t=8 seconds when E happens next to it. Then if we are using a separate Minkowski coordinate system with coordinates x',t', to describe events in this spacetime, the coordinates of the physical clock at the x=12 l.s. mark on this ruler reading t=8 s may happen at some completely different coordinates in this system, say x'=23 l.s. and t'=100 s. Likewise, in the Rindler coordinate system, the coordinates x'' and t'' of this event will be different as well. However, both the Minkowski coordinate system and the Rindler coordinate system will agree that whatever the coordinates x' and t' (or x'' and t'') of the event of the physical clock at the x=12 l.s. mark reading t=8 s, the same coordinates x' and t' (or x'' and t'') would be assigned to the event E of the collision--the Minkowski coordinate system and the Rindler coordinate system cannot disagree about local facts like whether two events (in this case the event of that clock at that marking reading 8 s and the event E of the collision) coincide at the same point in spacetime or not!

If both the Minkowski coordinate system and the Rindler coordinate system (along with the metric and laws of physics expressed in those coordinate systems) agree in their predictions about what ruler-markings and clock-readings on the physical network coincide with which events in any physical experiment that takes place within the region covered by that physical network, then both coordinate systems should agree in their predictions about what the equations of the laws of physics will be when expressed in the coordinate system defined by the network (not when expressed in terms of their own coordinates). And of course in flat SR spacetime it should always be possible to construct a physical network of rulers and clocks which define a coordinate system where the laws of physics work the same way as in any other inertial coordinate system, right?

This idea is essentially no different than the idea of the equivalence principle, that even if you have a curved spacetime described in some coordinate system like Schwarzschild coordinates, in any local region you should always be able to construct a network of freefalling rulers and clocks such that that the laws of physics as expressed in the coordinates defined by that network (not when expressed in Schwarzschild coordinates in that region) reduce to the same laws seen in an inertial frame in SR (at least to the first order or something). The only difference is that in this case we need not confine the grid of rulers and clocks to a small local region, they can cover any arbitrarily large region of the Rindler wedge and it'll still be true that the laws of physics as expressed in the coordinates defined by the network (not when expressed in Rindler coordinates in that region) will be exactly those seen in any SR inertial frame.

 

[Altabeh]

Well, because for a long time you were denying my claim that if we looked at a geodesic in Minkowski coordinates, and then used the coordinate transformation to define the same path in Rindler coordinates, then the path would still be a geodesic relative to the Rindler metric. But I guess you were just misunderstanding what I meant when I said things like "same path", not realizing I was talking about using the coordinate transformation to find the new coordinates of the same physical events along a given path which was originally defined in Minkowski coordinates.

Well, as DrGreg said, the problem lies in the fact that some authors describe the 3-vector $dx^i/d\tau$ as proper velocity and all my focus on the way I'm expected to treat the problems here automatically had been going to this point and don't get me wrong this was the way I learned and tought people until recently that I changed my mind and started using the magnitude of the 4-acceleration instead of $$d^2x^i/d\tau^2$$ as the "proper" acceleration! If you noticed in my early posts here, I doubted that the gravitational field of Rindler spacetime is uniform just because the proper 3-accceleration I mentioned above is dependent on $$x$$, thus it doesn't allow the spacetime to be accompanied by a uniform field. However, the challenge this caused in mind isn't still settled and I hope further studies in the future will help me ponder the problems more than I do now!

I'm in a hurry to go somewhere, so I don't have enough time to read the whole post. I'll be back soon.

AB