Layman's doubts about Gen Relativity

[Dale]

Completely nonsense and irrelevant. What is your reason that $$x=(t,r,\theta,\phi)=\left(t,\sqrt{t^2v^2+r_0^2},\frac{\pi}{2},atan(tv,r_0)\right)$$ is inertial and what does your "inertial" mean here?

That is the parametric equation of a straight line in spherical coordinates where r0 is the radius of closest approach, v is the velocity, and the line lies in the equatorial plane. Any arbitrary straight line may be brought into this form by a suitable rotation, so it is the simplest form without loss of generality.

In flat spacetime this equation, being the equation of a straight line, is inertial by definition. Your equation is wrong because it falsely identifies this worldline as being non-inertial, even locally. Furthermore, as I showed above it falsely identifies uniform circular motion in spherical coordinates as being inertial. Your equation is not even generally valid in flat spacetime, let alone in curved spacetimes.

 

[Dale]

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

On the contrary, it does. Whether or not something is inertial is not a relative quantity, it is absolute and frame invariant.

In GR the worldline of an inertial object is a geodesic in spacetime, and all coordinate systems agree on whether or not a given worldline is a geodesic since it is determined by a covariant equation. Therefore, if an accelerometer reads 0 we know that it is a geodesic (inertial) in all coordinate systems, and vice versa. In our coordinate system (for us) it is an inertial geodesic, just as certainly as it is an inertial geodesic in local coordinates (for it).

 

[atyy]

I think DaleSpam and Altabeh are both right.

A free falling test particle traces a geodesic, which we define to be "inertial motion".

However, the OP was asking about an "inertial frame". A frame is a coordinate system which is a non-local in the sense of involving not just a point, and I believe this is what Altabeh is talking about. I think at best you can make inertial coordinates at every point on a geodesic by Fermi normal coordinates, but not for events not on the geodesic.

 

[Altabeh]

That is the parametric equation of a straight line in spherical coordinates where r0 is the radius of closest approach, v is the velocity, and the line lies in the equatorial plane. Any arbitrary straight line may be brought into this form by a suitable rotation, so it is the simplest form without loss of generality.

Your nonsense here doesn't prove anything. The line you're defining this way in a spherical coordinates is only "locally straight" and itself as a special example can be used to prove my assertation is completely correct by a suitable metric transformation around the trajectory that such "inertial" (it is not so ever) observer follows. Your fallacy here is based on the fact that you've fully forgotten that the spacetime is not flat anymore. Basic stuff.

In flat spacetime this equation, being the equation of a straight line, is inertial by definition. Your equation is wrong because it falsely identifies this worldline as being non-inertial even locally.

Nonsense. See the page 56 of Papapetrou to get where this another non-sequiter fallacy comes from. Your problem is that you try to jump into discussions that you have no prior study on.

Furthermore, as I showed above it falsely identifies uniform circular motion in spherical coordinates as being inertial. Your equation is not even generally valid in flat spacetime, let alone in curved spacetimes.

The problem has been identified above. Your nonsense doesn't seem to be in agreement even with SR, let alone GR. You have not even identified what gives you this motivation to use "non-inertial" or "inertial" for those equations in a non-flat spacetime associated with a spherical coordinates. Read the sources I provided you with.

AB

 

[djy]

Your fallacy here is based on the fact that you've fully forgotten that the spacetime is not flat anymore.

DaleSpam's example with spherical coordinates is in flat spacetime. The coordinate system is curvilinear, which is why the Christoffel symbols don't vanish and why a geodesic is defined by the covariant derivative, not the ordinary derivative.

 

[Dale]

I think DaleSpam and Altabeh are both right.

A free falling test particle traces a geodesic, which we define to be "inertial motion".

However, the OP was asking about an "inertial frame". A frame is a coordinate system which is a non-local in the sense of involving not just a point, and I believe this is what Altabeh is talking about. I think at best you can make inertial coordinates at every point on a geodesic by Fermi normal coordinates, but not for events not on the geodesic.

I think this is a correct assessment of the disagreement between Altabeh and myself. However, the OP clarified in post 45 that he is interested in whether or not the object is inertial and not so much interested in the coordinates.

 

[Dale]

DaleSpam's example with spherical coordinates is in flat spacetime. The coordinate system is curvilinear, which is why the Christoffel symbols don't vanish and why a geodesic is defined by the covariant derivative, not the ordinary derivative.

Precisely and concisely stated.

 

[atyy]

I think this is a correct assessment of the disagreement between Altabeh and myself. However, the OP clarified in post 45 that he is interested in whether or not the object is inertial and not the coordinates.

I see. Yes, a free falling object undoubtedly can be defined to be inertial in a sense which is frame invariant by defining it via the magnitude of its 4-acceleration. However, I also think Altabeh's definition is equivalent, since in post 26 in he specifies dx_a/ds, in Fermi normal coordinates on a geodesic - his emphasis being that that definition makes sense as it reduces to the proper acceleration in flat spacetime. (Of course it wouldn't make any sense if x were taken as a non-inertial coordinate, as you pointed out.)

 

[Dale]

The line you're defining this way in a spherical coordinates is only "locally straight"

It is a geodesic in a flat spacetime. It can't get more globally straight than that.

Your fallacy here is based on the fact that you've fully forgotten that the spacetime is not flat anymore. Basic stuff.

No, the spacetime is flat, changing coordinates does not turn a flat spacetime into a curved one. If you disagree then please tell me which component of the curvature tensor is non-zero.

 

[atyy]

Yes he did a great job but I still don't believe that physics community has an agreement on such definition as Rindler is (let's say) the only one using this. I have wormed partially through Schutz, Letctures on GR by Papapetrou, D'inverno, Weinberg, the first part of Wald,

Wald p152 gives the proper acceleration of an observer who is static in the Schwarzschild spacetime as a=(1-2M/r)^-1/2M/r².

 

[Altabeh]

It is a geodesic in a flat spacetime. It can't get more globally straight than that.

I thought your background metric belongs to the spacetime around some gravitating body (related to the OP's question) which now is only described in spherical coordinates instead of Cartesian coordinates. Yet again this doesn't change anything ever because though the dynamics of spacetime does now seem to have changed, but we don't need any more of metric transformation around the trajectory in some small region to bring the metric into a locally flat phase: there is a globally valid coordinate transformation to do so:

$$\bar{t}=t,$$
$$r=\sqrt{x^2+y^2+z^2},$$
$$\phi=\arccos(z/ \sqrt{x^2+y^2+z^2}),$$
$$\theta=\arcsin(y/ \sqrt{x^2+y^2}).$$ if x>=0 (if x<0, use $$\theta=\pi-\arcsin(y/ \sqrt{x^2+y^2})$$)

This transformation is valid everywhere so along the trajectory of an "inertial" observer you can see that generally the claim given in the page 56 of Papapetrou is true. Another example looking alike yours is the Rindler coordinates for which the same approach happens to be applicable.

No, the spacetime is flat, changing coordinates does not turn a flat spacetime into a curved one. If you disagree then please tell me which component of the curvature tensor is non-zero.

I do agree, of course!

AB

 

[George Jones]

we don't need any more of metric transformation around the trajectory in some small region to bring the metric into a locally flat phaseAB

I don't think this is a good way to use the term "locally flat", and it is certainly not the way that mathematicians use the term. Maybe mathematicians and physicists use the same term to mean two different things. Does Schutz use the term "locally flat" (I won't be able to look at my copy until tomorrow)?

 

[atyy]

I don't think this is a good way to use the term "locally flat", and it is certainly not the way that mathematicians use the term. Maybe mathematicians and physicists use the same term to mean two different things. Does Schutz use the term "locally flat" (I won't be able to look at my copy until tomorrow)?

Poisson uses the term "local flatness" in section 1.6 http://books.google.com/books?id=SO...&resnum=1&ved=0CCUQ6AEwAA#v=onepage&q&f=false

 

[Altabeh]

I don't think this is a good way to use the term "locally flat", and it is certainly not the way that mathematicians use the term. Maybe mathematicians and physicists use the same term to mean two different things. Does Schutz use the term "locally flat" (I won't be able to look at my copy until tomorrow)?

Yes he does call it so several times.

AB

 

[Altabeh]

Poisson uses the term "local flatness" in section 1.6 http://books.google.com/books?id=SO...&resnum=1&ved=0CCUQ6AEwAA#v=onepage&q&f=false

I believe that Poisson himself agrees to what Schutz says about local flatness and his treatment on the problem as you can see this from Poisson mentioning the name of Schutz's book in his references.

AB

 

[atyy]

I guess there could be several objections to "local flatness":

1) the metric is not flat anywhere, not even at a point, since flatness is judged by second derivatives of the metric, not first derivatives (objection to "flat")

2) it's ok to use local flatness for just making the first derivatives zero, but only at the origin of the normal coordinates, so we should not use the term if we talk about a small region (objection to "local")

Any more?

 

[George Jones]

Poisson uses the term "local flatness" in section 1.6 http://books.google.com/books?id=SO...&resnum=1&ved=0CCUQ6AEwAA#v=onepage&q&f=false

I have my hard-copy of Poisson home with me, and, now that you point this out, I remember that Poisson uses the term this way.

Yes he does call it so several times.

AB

I think that it is fairly standard in pure mathematics to use "local" in the following way: A space is locally xxxx if every element in the space is contained in a neighbourhood that is xxxx. For example, a space is locally compact if every element in the space is contained in a neighbourhood that is compact.

 

[yuiop]

I was actually talking about something like a thin rod or rigid cable connecting a point of low potential and a higher point and rotating about a line along its core, although in many ways this has similar results, in that the frequency at the top and bottom cannot actually differ as observed by anyone observer.

I like your example of a long vertical cylidrical rod spinning about its long axis better than my example. Your example, is a clear and simple demonstration that clocks really run slower at lower gravitational potentials. Unfortunately this implies that the speed of light must also really be slower at lower gravitational potentials, to explain why local observers lower down measure the local velocity of light to be c.

...
Couldn't it be said that SR came into being as a rational attempt to make sense of the illogical invariance of light as indicated by Maxwell and MM ??

On a historical sidenote, the MM played no part in Einstein's formulation of SR and is not mentioned in his 1905 paper. We have also shown in other threads that the MM is inconclusive as far as SR is concerned, because it can be explained by a ballistic theory of light. Maxwell's equations are far more significant, in that they make it clear that a ballistic theory of light is ruled out.

BTW I wouldn't say I have bought into the first choice but so far it seems more consistent with the other applications of time dilation.

The problem is that local observer's always measure the speed of light to be c, while a coordinate measurement indicates the speed of light varies with potential. Jonathan Scott's thought experiment indicates that the coordinate speed of light has some objective reality. However, the coordinate speed of light tends to zero at the event horizon, while the local speed of light is always c and this brings about a clear contradiction between the two views at the event horizon. Which is the real measurement? Does light "really" stop at the event horizon?

Are you saying that the relative time dilation at different potentials is not measured?
Is not real?

Well, there is a fine distinction between measured and calculated. I think in general, local observations are considered to be "measured", while comparisons of clock rates at different altitudes (a coordinate measurement) is considered to be "calculated". In other words, the calculated rate of a clock according to a distant observer is considered to be less valid or less real than the measured rate of the clock by a local observer. Above the event horizon, the distinctions are largely philosophical, but if we wish to extend our observations to predicting what happens exactly at the event horizon and below, then we probably need to decide whether coordinate or local measurements are a better representation of physical objectivity.

Isn't the measured dilation differential by your definition "real" and the interpretation of slowing photons "behind the scenes" ?

Yes, the time dilation differential can be measured. Slowing photons "is behind the scenes" if we ignore the fact that we measured the time dilation differential to be real. If we acknowledge that clocks really do run slower at lower potentials, we can synchronise clocks to take this into account and then we would directly measure the local speed of light to vary with potential. The slowing of photons would no longer be behind the scenes using such a clock synchronisation method. In effect, what we do when we measure the vertical speed of light is this. We take a vertical rod and a clock and calibrate the apparatus using the assumption that the speed of light is constant everywhere. We then use said apparatus to measure the speed of light and then declare "lo and behold, the speed of light is constant everywhere!" :tongue:

 

[Altabeh]

DaleSpam's example with spherical coordinates is in flat spacetime. The coordinate system is curvilinear, which is why the Christoffel symbols don't vanish and why a geodesic is defined by the covariant derivative, not the ordinary derivative.

I think I have answered this in post #81. But to shed more light on why bringing a curved metric into a flat form cannot be done through intoducing an explicit coordinate transformation, I want to say that I myself have proved that if one defines a coordinate system with $$n$$ free numbers in any $$n$$ dimensional spacetime the only metric that can be transformed into Minkowski locally is a diagonal one because then we have a system of linear equations in the new coordinates with 0 degrees of freedom which requires you to adjust the free numbers around some arbitrary point according to the coordinates of this point; leading to a perturbed metric and that local flatness is again guaranteed. In general, for a symmetric metric there exists $$n(n+1)/2$$ components and thus equations involving new coordinates which means the number of equations exceeds that of unknowns; leading to an overdetermined system (where the method of least squares can be used to find an approximate solution which is the best one.) So in general local flatness through making use of this way can be weakly evaluated to be true.

AB

 

[Altabeh]

I have my hard-copy of Poisson home with me, and, now that you point this out, I remember that Poisson uses the term this way.


I think that it is fairly standard in pure mathematics to use "local" in the following way: A space is locally xxxx if every element in the space is contained in a neighbourhood that is xxxx. For example, a space is locally compact if every element in the space is contained in a neighbourhood that is compact.

More mathematical . Most of Physicists do prefer to use "locally flat", "locally inertial" or "flat in a sufficiently small region" but they barely utilize terms like "in the neighbourhood of xxxx" or even "in an infinitesiammly small region". For instance, when it comes to EP the use the term "flat in a sufficiently small region" to emphasize that flatness is not bound to a neighbourhood but it may exist within a larger distance or region which is translated as "sufficiently small".

AB

 

[Altabeh]

I think DaleSpam and Altabeh are both right.

A free falling test particle traces a geodesic, which we define to be "inertial motion".

However, the OP was asking about an "inertial frame". A frame is a coordinate system which is a non-local in the sense of involving not just a point, and I believe this is what Altabeh is talking about. I think at best you can make inertial coordinates at every point on a geodesic by Fermi normal coordinates, but not for events not on the geodesic.

Yes. But remember that the Fermi coordinates calls for its own observer; not every observer. So the general answer to the OP's question is no.

AB

 

[yuiop]

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

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

If the object has inertial motion according to an accelerometer "actually functioning in the frame of reference of the object itself", then it has inertial motion according to all observers, whether they are inertial or non inertial, at rest or with relative motion and whether accelerating or not. Inertial motion is a universal observation, as Dalespam has pointed out several times.

Consider a space station orbiting the Earth in a highly elliptical orbit. Its angular and radial coordinates are continually changing and its angular and radial acceleration is continually changing, but to an observer within the space station itself (no windows), the space time is essentially flat to a high degree of accuracy. Even using modern technology, I think it would be extremely difficult for the observer inside the space station to prove that the space station is not stationary in flat space, using only measurements made inside the confines of the space station.

The "long trajectory" of the particle is immaterial. The local space-time is essentially flat along the entire geodesic (i.e. at every point of the trajectory).

An Earth bound observer is a non inertial observer. He knows he is accelerating (as he can measure this using an accelerometer) and he can conclude that the apparent changing motion of the inertially falling stone can be explained entirely by his own acceleration.

 

[Altabeh]

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

Exactly.

 

[atyy]

I think that it is fairly standard in pure mathematics to use "local" in the following way: A space is locally xxxx if every element in the space is contained in a neighbourhood that is xxxx. For example, a space is locally compact if every element in the space is contained in a neighbourhood that is compact.

So your objection is that a point is not a neighbourhood?

How about "ultralocal pseudoflatness"?

 

[Altabeh]

The "long trajectory" of the particle is immaterial. The local space-time is essentially flat along the entire geodesic (i.e. at every point of the trajectory).

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

AB

 

[Altabeh]

How about "ultralocal pseudoflatness"?

A nice one if not so much of a pain to read!

AB

 

[yuiop]

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

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

 

[Altabeh]

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

Well, then let's decide to use that term from now on. That would be even more perfect if it was "ultralocal quasi-pseudo-flatness"!

AB

 

[yuiop]

Well, then let's decide to use that term from now on. That would be even more perfect if it was "ultralocal quasi-pseudo-flatness"!

It does make you wonder if this is an inperfection in GR. If the equations are only "exactly" correct at an infinitesimal point, then at extremes such as at the event horizon or at the singularity there might be a breakdown. Measurements of velocity require a minimum of 2 spatially and temporally separated events. Measurements of acceleration require a minimum of 3 spatially and temporally separated events. Can the measurement of 3 such separated events qualify as being measured "exactly" at one infinitesimal point? It gets worse for third and fourth derivatives.

 

[atyy]

It does make you wonder if this is an inperfection in GR. If the equations are only "exactly" correct at an infinitesimal point, then at extremes such as at the event horizon or at the singularity there might be a breakdown. Measurements of velocity require a minimum of 2 spatially and temporally separated events. Measurements of acceleration require a minimum of 3 spatially and temporally separated events. Can the measurement of 3 such separated events qualify as being measured "exactly" at one infinitesimal point? It gets worse for third and fourth derivatives.

Yes, the EP which gave birth to GR is only heuristic. In GR, it holds only at points, and only for first derivatives (all Christoffel symbols, which are first derivatives of the metric, can be made zero at a point in normal coordinates), but not second derivatives (components of the curvature tensor cannot all be made zero). In a strict mathematical sense, all derivatives are local, since they are limits defined at a point. But in the heuristic sense, higher order derivatives are more nonlocal, as you point out. So yes, the EP only applies locally in two senses 1) at a point 2) and only up to first derivatives at the point. But that's more a problem with trying to derive GR from an EP that is not sharply defined until after we already have the theory. The other motivating principle of GR "general covariance" is also problematic for deriving GR. However, GR itself is fine for the moment, having passed many experimental tests.

 

[djy]

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

AB

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

 

[Dale]

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

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

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

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

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

 

[Austin0]

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

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

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



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

 

[Dale]

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

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

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

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

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

 

[Altabeh]

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

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

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

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

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

Sorry if I was not clear.

AB