Understanding Tidal Forces & Rindler Metric

[snoopies622]

This diagram seems to show the distance in Rindler coordinates between the top and bottom actually shrinks rather than expands as the ball "falls" in the pseudo-gravitational field..

2. Since the Wely tensor and the Ricci tensor of the Rindler metric are both zero, the sphere must maintain its shape and size.

So what is the resolution to this apparent contradiction?

 

[JesseM]

1-You have to discuss the Rindler metric apart from any other metric e.g. Minkowski.

Again, I think it would be common for physicists to say that they are the "same metric", just expressed in different coordinate systems. For example, googling "same metric" + relativity turns up this Google Books result from A First Course in General Relativity by Bernard Schutz, in which he writes:

For instance, suppose we have a stationary gravitational field. Then a coordinate system can be found in which the metric components are time independent ... Notice that coordinates can also be found in which the same metric has time-dependent components: any time-dependent coordinate transformation from the 'nice' system will do this.

In a footnote at the bottom of the page he also says "It is easy to see that there is generally no coordinate system which makes a given metric time independent", so again he is using 'given metric' to refer to a geometric entity which can be expressed using different equations in different coordinate systems.

We can make any spacetime inertial locally

What do you mean by that exactly? As I understand it the equivalence principle really only applies exactly in an infinitesimal region of curved spacetime, and even there the curvature tensor will not actually be zero as I understand it...there was some discussion of the subtleties of defining the equivalence principle on this thread and this one.

and of course it is not logical to apply a transformation that is going to map all points in the flat spacetime into the Minkowski metric because we can do this to every flat spacetime globally.

Why does that make it "not logical"? Do you get any incorrect predictions about physical results? I would say that every set of metric equations describing a globally flat spacetime is the same metric describing the same spacetime, just expressed in different coordinate systems.

But the result is non-sense in case the initial metric has non-constant components!

Why is it nonsense that a single spacetime geometry (representing a single set of predictions about all coordinate-independent physical measurements like proper time readings at particular events) can be expressed in terms of different metric equations in different coordinate systems, one having constant components and the other having non-constant ones? That's exactly the situation that was described above by Bernard Schutz, where he talked about the "same metric" having time independent components in one coordinate system and time-dependent components in a different one.

2- A coordinate transformation is not allowed to be carried out on a given metric if it makes the gravitational and inertial accelerations go away in the geodesic equations unless it is just applied in a small region where EP holds! For example, I can make the Schwartschild metric be flat at some point and to the highest allowed degree i.e. in a small region this is approximately true! But doing this thing globally kills the second terms in the geodesic equations, leaving us with Minkowski spacetime.

I'm certain you can't find a coordinate transformation between Schwarzschild and Minkowski coordinates such that, when you transform the Schwarzschild metric (expressed in Schwarzschild coordinates) using this coordinate transformation, you get back the Minkowski metric (expressed in Minkowski coordinates). Are you claiming this is possible? If you agree it's not possible, and if you also agree that a mere coordinate transformation isn't going to change your physical predictions about any coordinate-independent facts like proper time along worldlines, then I don't understand what you mean when you say certain coordinate transformations are "not allowed"...do you have a reference to a paper or textbook which states whatever rule you're suggesting here?

Even if two metrics are of the same nature, but since the geodesic equations aren't the same you can't apply a global transformation in such a way that it allows you to get rid of gravitational fields present in the initial metric!

Depends if you define "gravitational field" in terms of actual spacetime curvature (which seems to be the common practice among physicists nowadays) or if you include what this section of the twin paradox page calls "pseudo-gravitational fields". Under the first definition, no gravitational field is present anywhere in the Rindler metric.

If the transformation was between two flat spacetimes with only constant metric components, then you would do this!

The Rindler metric does describe a flat spacetime, even if the coordinate system it's expressed in includes non-constant components.

 

[Altabeh]

Again, I think it would be common for physicists to say that they are the "same metric", just expressed in different coordinate systems. For example, googling "same metric" + relativity turns up this Google Books result from A First Course in General Relativity by Bernard Schutz, in which he writes:

In a footnote at the bottom of the page he also says "It is easy to see that there is generally no coordinate system which makes a given metric time independent", so again he is using 'given metric' to refer to a geometric entity which can be expressed using different equations in different coordinate systems.

Did you find that footnote different from my point here? He is trying to simply say, for instance, the time-dependent scale factor in the FRW metric can't disappear via any coordinate transformation. This is right and I agree to it! But let me give you an example of what point I'm trying to make here.

First of all, I define the conditions that describe the qualities of two, say, flat metrics with certain identical signatures and belonging to the same family in my view:

1- Both metrics must have vanishing Riemann tensors,
2- In both metrics the particles moving on geodesics must have the same dynamical quiddity. I.e. if in one of them the proper acceleration is position-dependent, then so is in the other metric!

Consider the metric

ds^2 = dt^2 - dx^2 = dT^2 + 2X^2dTdX- (1-X^4)dX^2.

For this metric all Christoffel symbols vanish and thus it is an inertial frame for which and the Minkowski metric both conditions meet and thus they belong to the same family (so far it is obvious that they are actually the same by the use of the change of coordinates T = t - x^3/3, X = x.)

But for the Rindler metric the Christoffel symbols do not vanish thus this metric gets out of the family of the Minkowski metric. It may be weird that while these two are basically the same, but they act dynamically differently and this gives rise to why I insist giving up on your idea as a solution to getting constant proper acceleration.

What do you mean by that exactly?

See https://www.physicsforums.com/showpost.php?p=2599439&postcount=63" to know my viewpoint on the equivalence principle.

Do you get any incorrect predictions about physical results? I would say that every set of metric equations describing a globally flat spacetime is the same metric describing the same spacetime, just expressed in
different coordinate systems.

Not exactly the same spacetime in some cases. Spacetime by itself is nothing but a set of events; so I assume you also take into account the dynamics of particles in the "spacetime" as a part of it to give a physical meaning to these events which undergo effects of gravitational fields and curvature.

Why does that make it "not logical"?

Here I mean the statement is vacuously true. As long as you are ignoring the dynamics of particles in the entire spacetime, the result from a mathematical standpoint is correct though it suffers from the "change" of dynamics globally. It looks like you're applying the EP throughout the spacetime.

Why is it nonsense that a single spacetime geometry (representing a single set of predictions about all coordinate-independent physical measurements like proper time readings at particular events) can be expressed in terms of different metric equations in different coordinate systems, one having constant components and the other having non-constant ones?

Can you imagine the EP to occur in a very large region even in a flat spacetime e.g. Rindler spacetime? The geodesic equations are strongly coordinate-dependent in Rindler metric and of course in the cases of gravitational field making the geodesic equations coordinate-dependent, time dilation in turn would depend on position.

Still continues...

AB

 

[JesseM]

Did you find that footnote different from my point here?

Yes, I think it's different from your statement "You have to discuss the Rindler metric apart from any other metric e.g. Minkowski". Schutz would presumably say that the Rindler metric and the Minkowski metric are the "same metric" since either one can be obtained by a coordinate transformation on the other.

He is trying to simply say, for instance, the time-dependent scale factor in the FRW metric can't disappear via any coordinate transformation. This is right and I agree to it!

I agree as well, since the FRW metric is not a stationary one. But he is also saying that for any metric whose components are time-independent in one coordinate system, you can always transform into another coordinate system where the components are time-dependent, yet it is still the "same metric" in both cases--do you agree?

First of all, I define the conditions that describe the qualities of two, say, flat metrics with certain identical signatures and belonging to the same family in my view

It seems like you're using "same family" to discuss something different than the ordinary notion of spacetime geometry, though. Consider differential geometry of ordinary 2-dimensional spatial surfaces, like a sphere and a plane. On a given surface, you can place different coordinate systems and then the equations of the metric describing the curvature of the surface will be different. Still, geometrically a sphere is a sphere and a plane is a plane--you can define the difference between the two in intrinsic terms by talking about things like geodesic length-minimizing paths and the sum of the angles of a triangle formed by three geodesics crossing at three points. So, there is an equivalence class of metric equations which all describe a spherical geometry and thus would all count as the "same metric" in Schutz's terms, just expressed in different coordinate systems on a sphere...likewise there is an equivalence class of metric equations which all describe a planar Euclidean geometry.

Do you agree that this notion of an equivalence class of metrics which which describe the same underlying geometry is meaningful in GR as well, and that this is presumably what Schutz meant when he talked about the "same metric" for a stationary gravity field having either time-dependent or time-independent components depending on what coordinate system it was written down in? And if so, do you agree that in this geometrical sense the Rindler metric is the same metric as the Minkowski metric? Finally, if you are using "same family" to describe a different equivalence class than the geometrical one I'm talking about, is this one that you've come up with yourself or one that you've seen discussed in textbooks or papers? What utility would this alternate equivalence class have?

Do you get any incorrect predictions about physical results? I would say that every set of metric equations describing a globally flat spacetime is the same metric describing the same spacetime, just expressed in
different coordinate systems.

Not exactly the same spacetime in some cases. Spacetime by itself is nothing but a set of events; so I assume you also take into account the dynamics of particles in the "spacetime" as a part of it to give a physical meaning to these events which undergo effects of gravitational fields and curvature.

By "same spacetime" I was referring to this notion of the same spacetime geometry expressed in different coordinate systems (along with the same physical events in both coordinate systems of course). Hopefully you agree this is a meaningful notion?

Why does that make it "not logical"?

Here I mean the statement is vacuously true.

What statement is vacuously true? The original "not logical" statement of yours was "We can make any spacetime inertial locally and of course it is not logical to apply a transformation that is going to map all points in the flat spacetime into the Minkowski metric because we can do this to every flat spacetime globally."

As long as you are ignoring the dynamics of particles in the entire spacetime, the result from a mathematical standpoint is correct though it suffers from the "change" of dynamics globally. It looks like you're applying the EP throughout the spacetime.

What result from a mathematical standpoint is correct? And how am I "applying the EP"? I'm just saying that the Rindler metric equation and the Minkowski metric equation describe the same spacetime geometry (and would be the 'same metric' in different coordinate systems according to the way most physicists like Schutz seem to talk about things), and as such if you use either metric to make predictions about local coordinate-independent physical facts like what readings two clocks show when they pass next to one another, you'll get identical predictions. 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. 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.

Can you please tell me if you disagree with any of the above statements, and if so quote the specific ones you disagree with and explain why?

Can you imagine the EP to occur in a very large region even in a flat spacetime e.g. Rindler spacetime?

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.

 

[Altabeh]

Yes, I think it's different from your statement "You have to discuss the Rindler metric apart from any other metric e.g. Minkowski". Schutz would presumably say that the Rindler metric and the Minkowski metric are the "same metric" since either one can be obtained by a coordinate transformation on the other.

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.

The second statement says that the Rindler metric is in essence the Minkowski spacetime in "expanding coordinates" but one cannot account for it being the same as the Minkowski spacetime at any arbitrary point. Besides this, since the dynamical behavior, or the way geoemetry (gravitational field) allows particles to move, in the two metrics are different, you should be careful when applying an inverse transformation to get Minkowski from Rindler because, for instance, the points (t,0,y,z) are not contained in the Rindler metric and for this reason I can say Rindler spacetime is a compactification of the Minkowski spacetime assuming x to be chosen from \textbf{R}-{0}. Thus such coordinate system is not equivalent to Minkowski everywhere. Agreed?

I agree as well, since the FRW metric is not a stationary one. But he is also saying that for any metric whose components are time-independent in one coordinate system, you can always transform into another coordinate system where the components are time-dependent, yet it is still the "same metric" in both cases--do you agree?

...is the same metric if no event horizon or a point appears where the metric acts irregularly.

It seems like you're using "same family" to discuss something different than the ordinary notion of spacetime geometry, though. Consider differential geometry of ordinary 2-dimensional spatial surfaces, like a sphere and a plane. On a given surface, you can place different coordinate systems

Haven't you ever herad they use the words "bad" and "good" to call coordinate systems? A bad coordinate system is one whose determinant vanishes at some point and the good one has a non-vanishing determinant everywhere. As for a bad coordinate system, I can name the spherical coordinates for which r=0 and \theta = \pi generate singularity. For the Schwartschild metric, yet such singularities are present along with another coordinate singularity on the hyperplane 2m=r. Due to this reason, we can't introduce an orthogonal metric basis that make the metric flat everywhere but only at points with properties mentioned above.

and then the equations of the metric describing the curvature of the surface will be different.

If a flat spacetime is described by another coordinate system, yet the curvature tensor must vanish. This is correct and I don't seem to lay my finger on it, but rather on the statement that in all these coordinates the metric remains ALWAYS the same and I think I gave a vivid example above.

Still, geometrically a sphere is a sphere and a plane is a plane--you can define the difference between the two in intrinsic terms by talking about things like geodesic length-minimizing paths and the sum of the angles of a triangle formed by three geodesics crossing at three points. So, there is an equivalence class of metric equations which all describe a spherical geometry and thus would all count as the "same metric" in Schutz's terms, just expressed in different coordinate systems on a sphere...likewise there is an equivalence class of metric equations which all describe a planar Euclidean geometry.

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! But This is physics and we have to take into account everything related to it!

Do you agree that this notion of an equivalence class of metrics which which describe the same underlying geometry is meaningful in GR as well, and that this is presumably what Schutz meant when he talked about the "same metric" for a stationary gravity field having either time-dependent or time-independent components depending on what coordinate system it was written down in?

Ah, are you interested in asking long question?! Yes, I agree that he is just talking about "equivalence class" of metrics via introducing coordinate systems that do transform the metric into other possible forms. But as I talked about earlier, in GR the curved Riemannian spacetimes have crucial bahaviour when it comes to the transformation! Almost every curved spacetime is only able to be transformed into Minkowski spacetime through an appropriate metric transformation (not coordinate transformation) which is only locally applied as in the case of Schwarzschild metric with Jaccobi coefficients

e_0^0=\frac{1}{c\sqrt{1-2m/r}},
(e_1^1)^2=-(1-2m/r),
(e_2^2)^2=-\frac{1}{r},
(e_3^3)^2=-\frac{1}{r\sin(\theta)}.

Such metric transformation occurs just locally at some given point P.

And if so, do you agree that in this geometrical sense the Rindler metric is the same metric as the Minkowski metric?

I agree due to reason I'm giving below!

Finally, if you are using "same family" to describe a different equivalence class than the geometrical one I'm talking about, is this one that you've come up with yourself or one that you've seen discussed in textbooks or papers? What utility would this alternate equivalence class have?

The most known classification of spacetimes is the Petrov classification and of course other much general classifications (for instance, some classify the spacetimes into categories with different signatures -Lorentzian, Pseudo Riemannian and etc.- by which the metric itself is mostly targeted) exist in the advanced textbooks and articles (just google them!). In the Petrov classification, both Rindler and Minkowski spacetimes have vanishing Riemann tensor so they are of type O thus in the same class! But such classification does not zoom into the dynamics of spacetime as they just look at the Weyl tensor and if it was zero (if the Ricci tensor vanishes so does the Weyl tensor), then the spacetime is of type O. And I think I told you that such classification is accepted at least in my own view.

By "same spacetime" I was referring to this notion of the same spacetime geometry expressed in different coordinate systems (along with the same physical events in both coordinate systems of course). Hopefully you agree this is a meaningful notion?

Yes, of course

What statement is vacuously true? The original "not logical" statement of yours was "We can make any spacetime inertial locally and of course it is not logical to apply a transformation that is going to map all points in the flat spacetime into the Minkowski metric because we can do this to every flat spacetime globally."

This actually refers to the fact that every coordinate transformation has its own "range of working" as in the Schwarzschild metric, you can't apply any coordinate transformation at essential singularities. Transforming coordinates blindly throughout the spacetime is not logical because we may be doing this at a point where the initial metric has a singularity there as in the above case!

However, I just found out that the singularity of Rindler metric is not essential and can be removed by defining the coordinates (u,v)

u=t-\ln(x),
v=t+\ln(x).

Now we get the differential of each coordinate:

du=dt-dx/x,
dv=dt+dx/x.

Multiplying these two equations by each other and rearranging terms give

x^2dt^2-dx^2=x^2dudv=e^{v-u}dudv,

where - \infty<u,v< \infty and we used the fact that x=e^{v-t} and x=e^{t-u}.

Finally introducing U=e^{-u} and V=e^v yields

ds^2=dUdV,

for the Rindler metric with 0<U,V< \infty. But as we can see, there is no longer any singularity at U=0 or V=0. This suggests that we may now extend the spacetime by allowing V and U to be unrestricted and by applying a transformation of the form X=(V-U)/2, T=(V+U)/2 in this last line-element we get the Minkowski metric which is basically the extended spacetime not the Rindler metric itself! You got it?

What result from a mathematical standpoint is correct? And how am I "applying the EP"?

If you're applying a transformation over the entire manifold to make it flat, this is may be correct from a mathematical viewpoint, but since the dynamics of spacetime may also change, it is not true from a physical point of view because it resembles the situation where we apply EP throughout the spacetime! There is a long discussion in this https://www.physicsforums.com/showthread.php?t=377254" that is about answering a fundamental question: Do in a Rindler metric particles traveling along timelike geodesics experience a constant acceleration? The answer as given by bcrowell is yes only if we take instantaneously at rest particles with a comoving instantaneous observer with an ideal clock whose hands are not affected by gravitational field. Such assumption is made after battling with hardships that one needs to go through if he seeks out this property in the Rindler metric, not in the Minkowski metric where you arrive at through a simple coordinate transformation!

I'm just saying that the Rindler metric equation and the Minkowski metric equation describe the same spacetime geometry (and would be the 'same metric' in different coordinate systems according to the way most physicists like Schutz seem to talk about things), and as such if you use either metric to make predictions about local coordinate-independent physical facts like what readings two clocks show when they pass next to one another, you'll get identical predictions.

I think it is now clear that the Minkowski spacetime is the extended Rindler metric! Agreed?

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! Any spacetime, as EP says, can be made flat approximately in a small region and can be made flat exactly at some given point! All the flat metrics are of type O, and fall into the same family but unfortunately there are other metrics within this class that are not flat, but conformally flat.

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.

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!

AB

 

[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 \Gammas 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

<br /> A^\mu = \frac{DU^\mu}{d\tau} = \frac{dU^\mu}{d\tau} + \Gamma^\mu_{\alpha\beta}U^\alpha U^\beta<br />​

(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", <br /> \sqrt{|A_{\mu} A^{\mu} |}<br /> 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 a^{\mu}=\frac{d^2x^{\mu}}{d\tau^2}[/tex] being the proper four-acceleration

<br /> That&#039;s a coordinate acceleration. It becomes proper acceleration after correcting for coordinate effects (with the help of the Christoffel symbols). <i>That&#039;s</i> 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?