I What is the Effect of Gravity on Einstein's Train in Special Relativity?

[PeterDonis]

Converting to Minkowskii coordinates (T,X,Y) we get in those coordinates Peter's result

$U = ( dT/d\tau, dX/d\tau, dY/d\tau) = ( \gamma \cosh \gamma g \tau, \gamma \sinh \gamma g \tau, \gamma v)$

As I said before, this is only valid for $\chi = 1/g$, $\psi = 0$--i.e., . It is not valid in general. The general form of the 4-velocity is given in post #68.

Integrating the four velocity we find the position vs time of the block parameterized by proper time $\tau$

$T(\tau) = (1/g) \sinh \gamma g \tau \quad X(\tau) = (1/g) [ \cosh \gamma g \tau -1 ] \quad Y(\tau) = \gamma v \tau$

This is also not valid generally. (You have also put an offset in the $X$ coordinate which I didn't put in; as I said before, I am putting the floor of the train at $\chi = 1/g$ instead of $\chi = 0$.) The correct general transformation is given in post #68. (For your placement of the spatial origin, you would put $\cosh - 1$ instead of $\cosh$ in the definition of $X$. I kept the spatial origin at $\chi = 1 / g$ to keep as close to the convention of standard Rindler coordinates as possible.)

 

[PeterDonis]

Carrying this out yields the following equations which transform from $\tau, \chi, \psi$ coordinates to T,X,Y, coordinates:

$$
T = (1/g) \sinh \gamma g \tau + \chi \, \sinh \gamma g \tau + \psi \, \gamma v \cosh \gamma g \tau
$$
$$
X = (1/g) [ \cosh \gamma g \tau -1 ] + \chi \, \cosh \gamma g \tau + \psi \, \gamma v \sinh \gamma g \tau
$$
$$
Y = \gamma v \tau + \psi \, \gamma
$$

Ok, but now you have to verify that this gives you the correct 4-velocity. Your formulas give

$$
\frac{\partial T}{\partial \tau} = \gamma \left[ \left( 1 + g \chi \right) \cosh \gamma g \tau + g \gamma v \psi \sinh \gamma g \tau \right]
$$
$$
\frac{\partial X}{\partial \tau} = \gamma \left[ \left( 1 + g \chi \right) \sinh \gamma g \tau + g \gamma v \psi \cosh \gamma g \tau \right]
$$

Those aren't the right formulas; they should be $\partial T / \partial \tau = \gamma g \cosh \gamma g \tau$, $\partial X / \partial \tau = \gamma g \sinh \gamma g \tau$. The only way to get the right formulas for $\partial T / \partial \tau$ and $\partial X / \partial \tau$, and to have them remain valid for $\chi \neq 1 / g$ and $\psi \neq 0$, is to have

$$
T = \chi \sinh \left[ \gamma g \left( \tau + v \psi \right) \right]
$$

$$
X = \chi \cosh \left[ \gamma g \left( \tau + v \psi \right) \right]
$$

For $\chi = 1 / g$, $\psi = 0$, this reduces to $T = (1 / g) \sinh \gamma g \tau$, $X = (1 / g) \cosh \gamma g \tau$, so it is consistent with your restricted formula for the 4-velocity. But it continues to give the right formula for the 4-velocity in general (which, as I've noted, is given in post #68).

 

[PeterDonis]

A simple derivation. In Rindelr coordinates (t,x,y), we can write the motion of the sliding block as y=vt.

Agreed.

We can say $t = \gamma \tau$.

No, we can't, at least not in general. For the center of the block (or the train), which is $\psi = 0$, this works; but in general, a boost in the $y$ direction should give for the transform between $(\tau, \psi)$ and $(t, y)$

$$
t = \gamma \left( \tau + v \psi \right)
$$
$$
y = \gamma \left( \psi + v \tau \right)
$$

This is the substitution I used to obtain my formulas for $T$, $X$, and $Y$ in terms of $\tau$, $\chi$, and $\psi$.

 

[pervect]

Agreed.
No, we can't, at least not in general. For the center of the block (or the train), which is $\psi = 0$, this works; but in general, a boost in the $y$ direction should give for the transform between $(\tau, \psi)$ and $(t, y)$

$$
t = \gamma \left( \tau + v \psi \right)
$$
$$
y = \gamma \left( \psi + v \tau \right)
$$

This is the substitution I used to obtain my formulas for $T$, $X$, and $Y$ in terms of $\tau$, $\chi$, and $\psi$.

While I would agree that my formula doesn't give a 4-velocity other than at the origin of the coordinate systems, I don't think it has to. Basically, what I'm doing is creating Fermi-Normal like coordinates, as on MTW pg 329, except that I'm not using Fermi-Walker transport to transport the basis vectors. Instead I'm allowing the spatial basis vectors to rotate, because it makes the problem tractable and also because it results in a stationary metric we can write in closed form. This is a vast improvement over a non-stationary metric that doesn't have a simple closed-form solution.

The observer constructs his proper reference frame (local coordinates) in a manner analogous to the Riemann normal construction. From each event $P_0(\tau)$ on his worldline, he sends out purely spatial geodesics (geodesics orthogonal to U = $\partial P_0 / d\tau)$ with an affine parameter equal to the proper length...

U is only specified at the origin of the reference frame, it's used to pick out the set of geodesics that are orthogonal to U, defining a space-like hypersurface of events "simultaneous with $P_0$. There is no require that the spatial geodesics remain perpendicular to U in Fermi-Normal coordinates, they just start out perpendicular. In fact, the geodesic is specified by its straightness and it's starting direction, so a curve that remained perpendicular to U would not necessarily remain a geodesic. I gather that you are attempting to find curves that remain perpendicular to U - while it leads to _a_ metric, it doesn't lead to the same metric I found. I haven't checked your work, but assuming that you've not made any mistakes, it would be a perfectly valid metric , but it would not meet the goals I outlined in ttps://www.physicsforums.com/threads/gravity-in-einsteins-train.835994/page-4#post-5254985. I also think that my metric is "simpler" - its easier to find geodesics (which are straight lines in the flat space-time) than non-geodesic curves. I would argue that in a rotating frame of reference one expects the ordinary velocity away from the origin to be $r\omega$, and not zero, which imples that one does not expect the orthogonality condition to hold away from the origin.

It would be useful to have some more papers on how "rotating frames of reference" are usually treated in terms of what sort of coordinates are usually used. The published ones I've found seem to use polar coordinates. Our problem involves acceleration and rotation, so the usual radial symmetry is broken, meaning that we really wan rotating cartesian coordinates rather than rotating polar coordinates. For non published online documents,, http://physics.stackexchange.com/qu...to-a-relativistic-rotating-frame-of-reference seems to have an approach similar to my own.

 

[pervect]

Some more comments on interpreting the metric we got earlier. To recap, the block coordinates are $(\tau, \chi, \psi)$, the inertial coordinates are (T,X,Y), and we've neglected Z because it's not interesting.

The result I get for the resulting metric is

$ds^2 = \alpha d\tau^2 + \beta (\psi d\chi- \chi d\psi) d\tau + d\chi^2 + d\psi^2$

where
\alpha =<br /> -{\frac { \left( \chi\,g+1 \right) ^{2}}{-{v}^{2}+1}}+{\frac {{v}^{2}<br /> }{-{v}^{2}+1}}+{\frac {{g}^{2}{\psi}^{2}{v}^{2}}{ \left( -{v}^{2}+1<br /> \right) ^{2}}}
\beta =\frac{ 2 g v}{1-v^2}

The $\alpha$ term represents overall time dilation. The $-\gamma^2 \left( \chi\,g+1 \right) ^2+\gamma^2 v^2$ terms represents the time dilation due to the linear acceleration and results in "downwards gravity". Note that when v=0 the metric coefficient of $d\tau^2$, which we'll call $g_{00}$, is -$(1+g\chi)^2$, as it is in the Rindler metric. If $\chi$=0 as well, then $g_{00} = -1$. The $\gamma^2 g^2 v^2 \psi^2$ term represents the time dilation due to the rotation and gives rise to an "outward centrifugal force". When $\psi$ becomes large enough, $g_{00}$ becomes equal to zero, making the metric singular. This happens in any rotating frame when the rotation would make an object at rest in the frame move at the speed of light.

The track that the block slides upon, as previously mentioned, is curved in these coordinates, so that the force on the block is always perpendicular to the track.

The $\beta$ cross terms are a consequence of rotation, and give rise to Coriolis forces.

I'm not going to go into the details of deriving the forces, which in GR are represented by the Christoffel symbols, except to say that one can get an intuitive idea of how the metric causes forces by considering the principle of maximal aging (sometimes more accurately called the principle of extremal aging. Various papers and books, such as Feynman's lectures and Thorne's "Exploring Black Holes" can provide more detail on this principle, as can a google search. Basically, though, in GR an objects move on a trajectory that locally maximizes proper time.

This metric I give has flat spatial slices, so if we consider the spatial volume of constant $\tau$, surfaces of constant $\chi$ are flat planes, as are surfaces of constant $\psi$. (I don't believe this is true in Peter's metric).

Because we are just reparameterizing the flat space-time of special relativity, we don't actually need to use the methods of GR, but the mathematical tecnhiques used here are usually taught in GR courses, even though we are applying them to a SR problem.

 

[SlowThinker]

the methods of GR, but the mathematical techniques used here are usually taught in GR courses, even though we are applying them to a SR problem.

Indeed this feels like a not-so-gentle introduction to GR, but it's great to witness a real problem being solved, so I'm grateful for that. Hopefully I'll eventually understand all the steps.

But perhaps, as a sanity check, could we go back to I or B level for a bit? I was thinking about this scenario:
The passengers drop another marble out of the train (or just out of the rocket if that's simpler) and point a laser beam at it. In which direction do they need to aim?
It seems that if the train rotates, they'd eventually have to point upwards? Or does the rotation get slower over time?

Peter's transformation seems OK in this experiment, however checking if his metric is consistent with the transformation is well beyond my skill.

 

[PeterDonis]

While I would agree that my formula doesn't give a 4-velocity other than at the origin of the coordinate systems, I don't think it has to.

I'm not sure I agree, but I think this question can be tabled for now, because even if we restrict to the "origin" (by which I assume you actually mean the worldline of the point on the train/sliding block that I am labeling $\chi = 1/g$, $\psi = 0$), your transformation still does not give the correct 4-velocity components, as I show in post #92. If your transformation gets that wrong, it can't be right, regardless of any other considerations.

Basically, what I'm doing is creating Fermi-Normal like coordinates, as on MTW pg 329, except that I'm not using Fermi-Walker transport to transport the basis vectors.

I have no issue with this; what I tried to do is similar. However, I'm not sure about your interpretation of MTW here; see below.

U is only specified at the origin of the reference frame, it's used to pick out the set of geodesics that are orthogonal to U, defining a space-like hypersurface of events "simultaneous with $P_0$. There is no require that the spatial geodesics remain perpendicular to U in Fermi-Normal coordinates, they just start out perpendicular.

As I read MTW here, the requirement is that spatial geodesics must be orthogonal to U all along the worldline, not just at the point we pick as the spacetime origin of the frame. This can always be done, provided that the coordinates only have to cover a small enough "world tube" around the chosen worldline. What can't always be done is to pick spatial geodesics that are orthogonal to other worldlines in some congruence that describes the entirety of the object of interest (such as the train/sliding block), not just the single worldline we are using to construct the coordinates

In the case under discussion, spatial geodesics are in fact orthogonal to the worldline I have labeled as $\chi = 1 / g$, $\psi = 0$, everywhere along that worldline. But they are not orthogonal to other worldlines in the congruence (worldlines with either a different $chi$ or a different $\psi$).

the geodesic is specified by its straightness and it's starting direction, so a curve that remained perpendicular to U would not necessarily remain a geodesic

Yes; in fact, I haven't checked to see whether integral curves of my $\hat{e}_1$ and $\hat{e}_2$ are geodesics everywhere. I think they are, but I'll have to check to see for sure.

I gather that you are attempting to find curves that remain perpendicular to U

Only along the worldline labeled $\chi = 1/g$, $\psi = 0$. Not elsewhere. Indeed, since the congruence has nonzero vorticity, it is not hypersurface orthogonal, so there are no curves that remain orthogonal to U everywhere.

Some more comments on interpreting the metric we got earlier.

As I said above, the coordinate transformation you used to obtain this metric can't be right, since it gives the wrong 4-velocity components for the restricted case you are considering.

 

[PeterDonis]

Because we are just reparameterizing the flat space-time of special relativity

In addition to my previous comments on this metric (the one quoted in post #95), there is another issue: I have run the metric through Maxima, and it computes a nonzero riemann tensor (actually all of the formulas it spit out were very complicated, so the easiest one to check to verify that it was not zero was the scalar curvature). So this metric cannot be describing flat Minkowski spacetime.

 

[PeterDonis]

Thus we can write

$$
T = (1/g) \sinh \gamma g \tau + \chi (e_1)^0 + \psi (e_2)^0
$$
$$
X = (1/g) [ \cosh \gamma g \tau -1 ] + \chi (e_1)^1 + \psi (e_2)^1
$$
$$
Y = \gamma v \tau + \chi (e_2)^1 + \psi (e_2)^2
$$

This seems to me to be the point at which the logic leading to your coordinate transformation goes wrong. You are assuming here that the $0$ and $1$ components of $(e_1)$ and $(e_2)$ are not functions of $\tau$. (You are also assuming that there is no $\chi$ or $\psi$ dependence in the $\sinh$ and $\cosh$ terms.) Only on that assumption do these formulas match the integrals of the 4-velocity that you give, since those integrals only have one term each that depends on $\tau$. But the final transformation equations you derive have the terms corresponding to $(e_1)^0$, $(e_2)^0$, etc. depending on $\tau$ (because they have factors of $\sinh \gamma g \tau$ or $\cosh \gamma g \tau$ in them)--as they must, because the $0$ and $1$ components of $(e_1)$ and $(e_2)$ do in fact depend on $\tau$. So your derivation ends up with formulas that violate an assumption made at the start.

In fact, if you integrate components of U with respect to $\tau$, you must allow the coefficients, and also the arguments of functions like $\cosh$ and $\sinh$, to be functions of $\chi$ and/or $\psi$. So, for example, to get $U^T = \gamma \cosh \gamma g \tau$, the correct general integral would be of the form

$$
T = f(\chi) k(\psi) \frac{1}{g} \sinh \left[ \gamma g \tau + j(\chi) + h(\psi) \right]
$$

with other possible terms that are not functions of $\tau$; but the above is the only possible function of $\tau$ that can occur in the integral. To get the correct 4-velocity, we then must have $f(\chi) = k(\psi) = 1$ and $j(\chi) = h(\psi) = 0$ at the values of $\chi$ and $\psi$ that describe the worldline we are using to define $U$. I achieved this by setting $f(\chi) = g \chi$, $j(\chi) = 0$ (because it turns out we don't need any $\chi$ dependence in the arguments of the hyperbolic funcions), $k(\psi) = 1$ (because we don't need any $\psi$ dependence in the coefficient) and $h(\psi) = \gamma g v \psi$ and labeling the chosen worldline with the values $\chi = 1 / g$ and $\psi = 0$. If you want to label that worldline by $\chi = 0$, then you would make $f(\chi) = 1 + g \chi$ instead.

Similar logic gives the most general integral of $U^X$ as

$$
X = f(\chi) k(\psi) \frac{1}{g} \cosh \left[ \gamma g \tau + j(\chi) + h(\psi) \right]
$$

plus possible terms that don't depend on $\tau$, but again, this is the only possible $\tau$ dependent term.

 

[pervect]

In addition to my previous comments on this metric (the one quoted in post #95), there is another issue: I have run the metric through Maxima, and it computes a nonzero riemann tensor (actually all of the formulas it spit out were very complicated, so the easiest one to check to verify that it was not zero was the scalar curvature). So this metric cannot be describing flat Minkowski spacetime.

I just got through reinstalling (disk crash) Grtensor and the old version of maple that (appears to be) needed to run it. I found a flat Riemann tensor. I latexified the line element I used in Grtensor just to be sure there were not any transcription errors - it should be the same as my post however.

(-(chi*g+1)^2/(1-v^2) + v^2/(1-v^2) + g^2*psi^2*v^2 /(1-v^2)^2)*d[tau]^2 + (2*g*v/(1-v^2))* (psi*d[chi]-chi*d[psi])*d[tau] + d[chi]^2 + d[psi]^2;<br />

Calculating the Rimeann, I got the following

Covariant Riemann
R(dn,dn,dn,dn) = All components are zero

 

[PeterDonis]

I found a flat Riemann tensor.

I'll re-check in Maxima, it's quite possible I made a transcription error. However, this still doesn't address my comments regarding the incorrect 4-velocity that your transformation formulas lead to.

 

[pervect]

Ok, but now you have to verify that this gives you the correct 4-velocity. Your formulas give

$$
\frac{\partial T}{\partial \tau} = \gamma \left[ \left( 1 + g \chi \right) \cosh \gamma g \tau + g \gamma v \psi \sinh \gamma g \tau \right]
$$
$$
\frac{\partial X}{\partial \tau} = \gamma \left[ \left( 1 + g \chi \right) \sinh \gamma g \tau + g \gamma v \psi \cosh \gamma g \tau \right]
$$

Those aren't the right formulas

They should be the right formuls for $\chi = \psi = 0$, which is where I put the spatial origin of the coordinate system and the block coordinates. I understand that you put the spatial origin at $\chi = 1/g$, I put it at $\chi=0$

With $\psi=\chi=0$ we get $( \gamma \cosh \gamma g \tau, \gamma \sinh \gamma g \tau, \gamma v)$ The spatial origin of the coordinate system must have the correct 4-velocity, as you point out. I.e. the spatial origin of the coordinate system must trace out the wordline of the center of the block.

 

[pervect]

I think that it should be minimally sufficient for the spatial origin of the coordinate system to traces out the coorrect worldline of the observer, and for the Riemann to be all zero, in order for the transform to qualify as "a viewpoint" of an observer moving along some specified worldline in Minkowskii space. Unless I've missed something? One can possibly want additional features, as per my list of desired features, of course, but those two points should be the key ones.

.

 

[pervect]

Indeed this feels like a not-so-gentle introduction to GR, but it's great to witness a real problem being solved, so I'm grateful for that. Hopefully I'll eventually understand all the steps.

But perhaps, as a sanity check, could we go back to I or B level for a bit? I was thinking about this scenario:
The passengers drop another marble out of the train (or just out of the rocket if that's simpler) and point a laser beam at it. In which direction do they need to aim?
It seems that if the train rotates, they'd eventually have to point upwards? Or does the rotation get slower over time?

Peter's transformation seems OK in this experiment, however checking if his metric is consistent with the transformation is well beyond my skill.

Eventually, no signal from the marble will be able to catch up to the accelerating rocket, or the block sliding around it's floor. I'm not sure if this is a sufficient answer, but it's the only one I have at the moment. This is a feature of so-caled hyperbolic motion.

 

[PeterDonis]

They should be the right formuls for $\chi = \psi = 0$, which is where I put the spatial origin of the coordinate system and the block coordinates.

In other words, you think it's ok to add extra terms that depend on $\tau$ to the formulas for $T$ and $X$, as long as those terms vanish at $\chi = \psi = 0$. I'll have to think about that, but I will raise a few initial concerns below.

I think that it should be minimally sufficient for the spatial origin of the coordinate system to traces out the coorrect worldline of the observer

To me, if we're going to physically interpret a chart as "rotating", we have to be able to say what is rotating relative to what. If we define a timelike congruence describing a family of observers, or points in an object, then "rotating" has a simple meaning: the congruence has nonzero vorticity, meaning that a given observer in the congruence sees neighboring observers rotating around him. But since you're only specifying one worldline--the one at your coordinates $\chi = \psi = 0$--I'm not clear on what congruence of worldlines you think describes the block (or the train), so I'm not sure what "rotating" is supposed to mean.

Also, if you're only specifying one worldline, I'm not clear on how to relate what you've done to the rocket--i.e., to a Rindler observer who happens to coincide with your block/train observer at time $T = \tau = 0$.

But more than that, if this is supposed to be "the frame of the block", then it would seem like the points of the block ought to be at rest in the frame. Certainly that's the case for Rindler coordinates: they don't describe the "viewpoint" of just one observer, they describe a common "viewpoint" of a whole family of observers, all of whom are at rest in the chart (constant spatial coordinates), and the variation in their physical measurements is entirely due to their different positions. But again, since you've only specified one worldline, I don't know if the other points of the block are supposed to be at rest in this chart or not.

One way to investigate these possible concerns would be to compute the kinematic decomposition for the congruence of worldlines that are at rest in your chart. As I noted in a previous post, when I do that for my chart, I get nonzero shear and vorticity. The nonzero vorticity seems ok, but the nonzero shear does not; that indicates that the congruence is not rigid, and physically we would hope that we could describe the block (or the train) by a rigid congruence. If the congruence of worldlines at rest in your chart has zero shear (and expansion), that would make it more reasonable as a candidate for describing the block. The problem is that the computation for your chart looks like it will be more complicated than the one for my chart, and I don't know how to get Maxima to do the computation for me.

 

[PeterDonis]

I'll re-check in Maxima, it's quite possible I made a transcription error.

I rechecked, and I did make an error (two, actually). When I correct them, I do get that the metric is flat--vanishing Riemann tensor.

 

[SlowThinker]

Eventually, no signal from the marble will be able to catch up to the accelerating rocket, or the block sliding around it's floor. I'm not sure if this is a sufficient answer, but it's the only one I have at the moment. This is a feature of so-called hyperbolic motion.

I'm aware of this; however I has hoping that it would be possible the other way.
So I used that: the passengers are shining at the marble, even if they can't see it. Obviously their math skills are nearly superhuman

 

[pervect]

In other words, you think it's ok to add extra terms that depend on $\tau$ to the formulas for $T$ and $X$, as long as those terms vanish at $\chi = \psi = 0$. I'll have to think about that, but I will raise a few initial concerns below.
To me, if we're going to physically interpret a chart as "rotating", we have to be able to say what is rotating relative to what. If we define a timelike congruence describing a family of observers, or points in an object, then "rotating" has a simple meaning: the congruence has nonzero vorticity, meaning that a given observer in the congruence sees neighboring observers rotating around him. But since you're only specifying one worldline--the one at your coordinates $\chi = \psi = 0$--I'm not clear on what congruence of worldlines you think describes the block (or the train), so I'm not sure what "rotating" is supposed to mean.

Let's consider a rotating disk. The "observer" would be the worldine at the center of the disk. Let's use $(T_d,X_d,Y_d)$ as the Minkowskii coordinates, and call the "rotating coordinates" $(\tau_d, \chi_d, \psi_d))$ Then we might write the congruence of worldlines by the relations

T_d = \tau_d \quad X_d = \chi_d \cos \omega \tau_d - \psi_d \sin \omega \tau_d \quad Y_d = \chi_d \sin \omega \tau_d + \psi_d \cos \omega \tau_d

Each pair of $\chi_d, \psi_d$ specifies a unqiue worldline, the collection of worldlines specified in this way form the desired congruence. But note that that only for the special "observer" worldline at the center of the disk $\chi_d = \psi_d = 0$ does $\tau_d$ equal proper time. For other members of the congruence $\tau_d$ is an affine parameter of the worldline, but it is not proper time. This does cause us various headaches when we want to compute the vorticity, etc But I'm not aware of any better/simpler way to describe a rotating congruence than above.

In the sliding block case,$\tau$ is analogous to $\tau_d$, $\psi$ is analogus to $\psi_d$ and $\chi$ is analogous to $\chi_d$. Specifying a value for $\psi$ and $\chi$ specifies a unique worldline. But $\tau$ is only proper time for the worldline at $\psi=\chi=0$. Along other worldlines, it's an affine parameter, but it's not proper time.

As an aside, it would probably be worthwhile to calculate the metric resulting from the "rotational transformation" described above. There's a webpage that does the former, but I haven't checked their results personally. http://physics.stackexchange.com/qu...to-a-relativistic-rotating-frame-of-reference. You mention the vorticity and shear, it would be worth calculating those, too.

But more than that, if this is supposed to be "the frame of the block", then it would seem like the points of the block ought to be at rest in the frame.

Points on the rotating disk are "at rest" in the rotating frame, in the sense that $\chi_d$ and $\psi_d$ are constant. In the same sense, points in the sliding block frame are constant in the sense that $\chi$ and $\psi$ are constant. None of the metric coefficients is a function of coordinate time, so the metric is stationary and has a timelike Killing vector field for both the rotating disk and the sliding block.

Certainly that's the case for Rindler coordinates: they don't describe the "viewpoint" of just one observer, they describe a common "viewpoint" of a whole family of observers, all of whom are at rest in the chart (constant spatial coordinates), and the variation in their physical measurements is entirely due to their different positions. But again, since you've only specified one worldline, I don't know if the other points of the block are supposed to be at rest in this chart or not.

I would describe the situation as using a reference worldline of an observer to build a congruence. The congruence is built from the reference worldline, but it's not the same as the reference worldline, it's a congruence.

One way to investigate these possible concerns would be to compute the kinematic decomposition for the congruence of worldlines that are at rest

I haven't attempted this - the main thing I'd watch out for is the fact that it's often assumed that the congruence is generated by a unit timelike vector field, but our time coordinate $\tau$ is a killing vector field, not a unit vector field.

 

[PeterDonis]

I'm not aware of any better/simpler way to describe a rotating congruence than above.

I agree this works for a rotating disk (and I agree with all of your statements and caveats about this example), but that's not what we're talking about here. We're talking about a train, or a sliding block, moving "horizontally" in an accelerating rocket. We shouldn't have to appeal to heuristics based on a rotating disk to specify a congruence for a different physical situation. We should be able to do it directly. See further comments below.

As an aside, it would probably be worthwhile to calculate the metric resulting from the "rotational transformation" described above.

I think that was done quite some time ago. See here:

https://en.wikipedia.org/wiki/Born_coordinates

In the sliding block case, $\tau$ is analogous to $\tau_d$, $\psi$ is analogus to $\psi_d$ and $\chi$ is analogous to $\chi_d$.

But whose $\tau$, $\psi$, and $\chi$? Yours, or mine? That's the question. It can't be both, because they're different--different coordinate transformations to/from Minkowski $T$, $X$, $Y$, and different resulting metrics. Both metrics describe Minkowski spacetime, but the congruences of observers at rest in those two charts are different congruences--different families of worldlines. (They must be, because the two 4-velocity fields derived from the two different transformations are different.) So which one is the right one to describe the block?

I'll defer further discussion of that question to a separate post, because I think computing the kinematic decomposition of the congruence of observers at rest in your chart will help to elucidate it. But see a couple of further comments on that below.

the main thing I'd watch out for is the fact that it's often assumed that the congruence is generated by a unit timelike vector field

That's not an assumption, it's the definition of a congruence. If the congruence is not a unit timelike vector field, the kinematic decomposition doesn't work.

our time coordinate $\tau$ is a killing vector field, not a unit vector field.

No, the coordinate $\tau$ generates both a KVF and a timelike unit vector field. The KVF is $\partial_{\tau}$. The unit timelike vector field is $\partial_{\tau} / | \partial_{\tau} |$, i.e., the KVF divided by its norm. (We also have to restrict the latter to the region in which $\partial_{\tau}$ is timelike, of course, which is not the entire spacetime.)

 

[PeterDonis]

I'll defer further discussion of that question to a separate post

Here is at least the start of the further discussion.

I figured out just now why I have had the feeling all through this thread that there was something nagging at me that I hadn't gotten out into the open. Let me approach it by recapping pervect's reasoning in the previous thread where he derived his "sliding block" metric, since this will establish fixed points on which we both agree.

We are considering a block sliding along the floor of a rocket that is accelerating. Here I'll continue my coordinate conventions, which are that, in a fixed inertial frame with Minkowski coordinates $T, X, Y, Z$, the floor of the rocket has proper acceleration $g$ in the $X$ direction, and the block is sliding, relative to the rocket, in the $Y$ direction. We are for now only considering the bottom of the block, i.e., the surface of the block that is in contact with the floor of the rocket. In other words, we are only looking at events on the hyperbolic "worldsheet" defined by $X^2 - T^2 = 1 / g^2$ (but with the $Y$ coordinate unconstrained); note that I am using my convention for this, not pervect's, which shifts the $X$ coordinate by $1/g$ so that the floor of the rocket is at $X = 0$ at $T = 0$, instead of $X = 1/g$.

In order to specify the motion of the floor of the block, we have to determine what the constraint is. Pervect's constraint, which I agree with, is that, in the fixed inertial frame, the momentum of the block in the $Y$ direction should be constant. This means the ordinary velocity of the block in the $Y$ direction, in the fixed inertial frame, will decrease with time, for reasons which pervect explained in the previous thread (and which I agree with). Pervect expressed this constraint using a constant $K = dY / d\tau$ (where I have switched to $Y$ instead of x and capitalized it to agree with the coordinate convention I'm using, where capital letters refer to inertial coordinates).

I expressed this constraint somewhat differently. My approach was to observe that, in the momentarily comoving inertial frame of the floor of the rocket, the ordinary velocity of the block in the $Y'$ direction (we'll use a prime for the MCIF to distinguish it from the fixed inertial frame above) will also be constant; I called this constant velocity $v$. We can easily show that my constraint is equivalent to pervect's constraint. In the MCIF, we have $\partial T' / \partial \tau = \gamma$, where $\tau$ is the proper time of the block and $\gamma = 1 / \sqrt{1 - v^2}$, and by the chain rule we have $\partial Y' / \partial \tau = ( \partial T' / \partial \tau ) \partial Y' / \partial T' = \gamma \partial Y' / \partial T' = \gamma v$. Finally, since $Y = Y'$, because the MCIF and the fixed inertial frame only differ by a boost in the $X'$ direction, we have $K = \partial Y / \partial \tau = \partial Y' / \partial \tau = \gamma \partial Y' / \partial T' = \gamma v$. So a constant $K$ implies a constant $v$ and vice versa.

Combining all this gives the 4-velocity of the block, expressed in the Minkowski coordinates of the fixed inertial frame (note that this is somewhat different from the expressions I frequently used before, since I have replaced the $\cosh$ and $\sinh$ functions, which did not have Minkowski coordinates as arguments, with their equivalent functions of Minkowski coordinates, so $\cosh g \gamma \tau = g X$ and $\sinh g \gamma \tau = g T$):

$$
U = \gamma g X \partial_T + \gamma g T \partial_X + \gamma v \partial_Y
$$

So far we are all in agreement. But now comes the key point, which was nagging at me before. Everything we've done so far only applies to the floor of the rocket, and to the bottom of the block. But pervect and I have both proposed metrics for the "block frame", and those metrics are not restricted to the floor of the rocket and the bottom of the block. They include the $\chi$ direction, which is the direction of proper acceleration, and is orthogonal to the $\tau \psi$ plane that describes the floor of the rocket and the bottom of the block. So we need to make some sort of assumption about what happens in that direction.

To put the point another way, consider the top surface of the block. It is separated, in the $X$, $X'$, or $\chi$ direction, from the bottom surface of the block and the floor of the rocket. What does the motion of this surface of the block look like, to an observer who is at rest relative to the bottom of the block?

For the rocket, we know the answer to this question: the top of the rocket is at a fixed "ruler distance" from the bottom (which is just the difference in their Rindler $x$ coordinates--note that this is not the same as the difference in their Minkowski $X$ coordinates, which decreases with time), and has less proper acceleration; the latter varies inversely with $x$. So we can describe the 4-velocity of the top of the rocket just as easily as we can that of the bottom. In Minkowski coordinates, it is

$$
U_{r} = \frac{gX}{g \sqrt{X^2 - T^2}} \partial_T + \frac{gT}{g \sqrt{X^2 - T^2}} \partial_X
$$

I have included the explicit factors of $g$ to make clear the point: we cannot just take the 4-velocity of the floor of the rocket, which would be $\sqrt{1 + g^2 T^2} \partial_T + g T \partial_X$, and "extend" it to the rest of the rocket and assume that it applies. In order to properly specify the congruence of worldlines that describes the rocket, we must look at the conditions that the congruence must satisfy, and find a 4-velocity field that satisfies those conditions. The 4-velocity $U_r$ that I wrote above does this for the rocket: as is easily verified by computation, the congruence $U_r$ has zero expansion, shear, and vorticity, and its proper acceleration varies with "altitude" (meaning, with $\sqrt{X^2 - T^2}$, since that is the constant that labels each worldline in the congruence--in Rindler coordinates it is just $x$), whereas the proper acceleration of the congruence $\sqrt{1 + g^2 T^2} \partial_T + g T \partial_X$ does not--it is always $g$. (This latter congruence, btw, is easily seen to be the "Bell congruence", which plays a key role in the Bell spaceship paradox, and of course it is known to have nonzero expansion, which is why the string in the Bell spaceship paradox eventually breaks.)

Once we find the 4-velocity field $U_r$ that describes the rocket, we then need only find a coordinate chart in which integral curves of $U_r$ have constant spatial coordinates, i.e., we want a chart $t, x, y, z$ in which $U_r = \partial_t / | \partial_t |$. This chart will be our desired "rocket frame", a non-inertial frame in which the rocket is at rest. (Of course this chart turns out to be the Rindler chart.) Similarly, once we have found a 4-velocity field $U_b$ that satisfies the conditions for a congruence describing the sliding block, we need only find a chart $\tau, \chi, \psi, z$ in which the integral curves of $U_b$ are given by $U_b = \partial_{\tau} / | \partial_{\tau} |$, and that will be our desired "block frame", a non-inertial frame in which the block is at rest.

The claim I have been making in this thread can now be stated very simply: the 4-velocity field that I presented in previous posts is the desired $U_r$, and the chart I derived in which its integral curves have constant spatial coordinates, is the desired chart for the "block frame" as described above. The one reservation I had, about the congruence having nonzero shear, I have now resolved; I rechecked my computation of the shear and found that I had made a mistake. The shear is actually zero. The vorticity is still nonzero, but that's OK; we have been in agreement that nonzero vorticity is to be expected (though I think the question of what, exactly, it means physically still deserves some discussion). In the notation I have been using in this post, that 4-velocity field is

$$
U_{b} = \frac{gX}{\sqrt{g^2 \left( X^2 - T^2 \right) - v^2}} \partial_T + \frac{gT}{\sqrt{g^2 \left( X^2 - T^2 \right) - v^2}} \partial_X + \frac{v}{\sqrt{g^2 \left( X^2 - T^2 \right) - v^2}} \partial_Y
$$

Conversely, the implied 4-velocity field pervect has been using, which corresponds to taking the 4-velocity field $U = \gamma g X \partial_T + \gamma g T \partial_X + \gamma v \partial_Y$, which is valid at the bottom of the block, and extending it over all of the block, does not work. That should be evident by comparison with the Bell congruence vs. the Rindler congruence above; my 4-velocity field $U_b$ corresponds to $U_r$ above, and pervect's 4-velocity field, which works out to $U = \gamma \sqrt{1 + g^2 T^2} \partial_T + \gamma g T \partial_X + \gamma v \partial_Y$, corresponds to the Bell congruence $U = \sqrt{1 + g^2 T^2} \partial_T + g T \partial_X$ above. So pervect's chart, in which integral curves of his 4-velocity field $U$ have constant spatial coordinates, corresponds to a "Bell chart" in which integral curves of the Bell congruence have constant spatial coordinates. I predict, therefore, that if we can compute the kinematic decomposition of pervect's congruence, we will find that it has nonzero expansion. In other words, objects at rest in his chart will not remain at a constant proper distance from each other; they will "move apart" with time.

One final note: what about the condition that $dY / d\tau$ must be constant along a given block worldline? The formula for $U_b$ above certainly does not make that evident. However, we can see that it is still true by noting that the denominator of all the terms in $U_b$, $\sqrt{g^2 \left( X^2 - T^2 \right) - v^2}$, is in fact constant along each integral curve of $U_b$. So we do in fact have $dY / d\tau$ constant on each worldline. But we do not have $d Y / d\tau = \gamma v$ along each worldline; that is only true on the bottom of the block, where the denominator of each term in $U_b$ becomes $\sqrt{1 - v^2}$, so the last term does become $\gamma v$. But at the top of the block, the denominator is larger, so we have $dY / d\tau < \gamma v$.

What does this mean? It is just a consequence of the fact that, at the top of the block, time "flows" faster than at the bottom. What does this mean for the original constraint we imposed? What is now held constant? The answer is that, in the MCIF of the floor of the rocket, the top of the block should have the same ordinary velocity $v$ in the $Y'$ direction as the bottom does. But since the top of the rocket/block has faster "time flow" than the bottom, that means that $dY / d\tau$ at the top must be less than $dY / d\tau$ at the bottom, which was equal to $\gamma v$. And that is what we see in $U_b$. And, furthermore, we do not see this in pervect's 4-velocity field, where we have $U^Y = \gamma v$ at the top of the rocket as well as at the bottom--with this 4-velocity field, the top of the block will actually be moving faster, in the MCIF of the bottom of the rocket, than the bottom is! (That was the other thing that was nagging at me all through this thread.)

Sorry for the long post to add to all the other long posts in this thread. But I wanted to get all this out while it was fresh in my mind.

 

[PeterDonis]

I rechecked my computation of the shear and found that I had made a mistake. The shear is actually zero.

Just for completeness, here are the corrected results for the vorticity (the shear, as I said, is zero):

$$
\omega_{TX} = - \omega_{XT} = \frac{g v^2}{\left[ g^2 \left( X^2 - T^2 \right) - v^2 \right]^{3/2}}
$$

$$
\omega_{TY} = - \omega_{XT} = - \frac{g^2 v T}{\left[ g^2 \left( X^2 - T^2 \right) - v^2 \right]^{3/2}}
$$

$$
\omega_{XY} = - \omega_{YX} = \frac{g^2 v X}{\left[ g^2 \left( X^2 - T^2 \right) - v^2 \right]^{3/2}}
$$

 

[PeterDonis]

we cannot just take the 4-velocity of the floor of the rocket, which would be $\sqrt{1 + g^2 T^2} \partial_T + g T \partial_X$, and "extend" it to the rest of the rocket

It may be worth expanding on why the "extended" 4-velocity of the floor of the rocket is the formula given in the quote above--i.e., why we cannot just say $U = g X \partial_T + g T \partial_X$ everywhere in the rocket, not just on the floor. The answer is that this $U$ is not a unit vector except on the floor of the rocket. So we have to normalize it somehow. But there are at least two ways to do that.

One way is to take $g X \partial_T + g T \partial_X$ and simply divide it by its norm, which is $g \sqrt{X^2 - T^2}$. That way gives the 4-velocity field $U_r$, i.e., the Rindler congruence. But $U_r$ does not have the same ordinary velocity in the $X$ direction, in the fixed inertial frame, for all worldlines; the ordinary velocity in the $X$ direction is $T / X$, which varies from worldline to worldline. So in the fixed inertial frame, different worldlines in this congruence will be moving at different speeds, at a given time $T$.

The other way is to keep the $X$ component of $U$ constant, and adjust the $T$ component so it is a unit vector. That way gives the 4-velocity field $U = \sqrt{1 + g^2 T^2} \partial_T + g T \partial_X$. This congruence has an ordinary velocity in the $X$ direction of $T / \sqrt{1 + g^2 T^2}$, which is a function of $T$ only, so at a given time $T$ in the fixed inertial frame, every worldline in the congruence is moving at the same speed.

The question is which of the above ways is the "right" way to "extend" the 4-velocity of the floor of the rocket to other altitudes in the rocket. And the counterintuitive thing that this question brings to the fore is that in SR, unlike in Newtonian mechanics, we have to choose which of two desirable properties we want the congruence describing the rocket to have: either the proper distance between the worldlines can be constant, or each worldline can have the same speed in a fixed inertial frame. There is no congruence that has both properties; only the Rindler congruence has the first property, and only the Bell congruence has the second.

The same issue arises in the case of the sliding block, and I believe I have shown that in this case, the 4-velocity field $U_b$, and the metric I derived from it, is the one corresponding to the Rindler congruence in the example above--i.e., the one in which the proper distance between the worldlines in the block is constant. Conversely, I believe that pervect's metric is the one in which the worldlines at rest in the metric correspond to the Bell congruence in the example above--i.e., the one in which each worldline has the same speed at a given time in a fixed inertial frame. So which one is the "right" one depends on which property you think is the right one for the "block frame" to have.

 

[pervect]

I'll take the opportunity to recap a short summary of the procedure by which the transforms were created, one that I edited once before (I did the edits the next day, so they might have been overlooked), and I will edit even more now in an effort to be as clear as I can.

In vector notation we are just saying that the position pointed to by the (spatial) block coordinates $(\chi, \psi)$ at time $\tau$ is equal to $\vec{P_0 }+ \chi \vec{e_\chi} + \psi \vec{e_\psi}$. Here $\vec{P_0}$ is the position of the point-like origin of the block at time $\tau$, $\chi$ and $\psi$ are coordinate values, and $\vec{e_\chi}$ and $\vec{e_\psi}$ are basis vectors. $\vec{P_0}$ was obtained by integrating the 4-velocity of the pointlike origin of the block, U.

$\vec{e_\chi}$ and $\vec{e_\psi}$ are unit spatial vectors, and their components were found by Peter. The coordinate basis in which the components of these vectors were specified is the cartesian (T,X,Y) coordinate basis. I have replaced the numeric indices with symbolic ones, so that $\vec{e_\chi}$ is equivalent to $\vec{e_1}$ and $\vec{e_\psi}$ is equivalent to $\vec{e_2}$ in the original. $\vec{e_\chi}$ has 3 components, which we can write as $(e_\chi)^T, (e_\chi)^X, (e_\chi)^Y)$. $\vec{e_\psi}$ is similar. Hopefully using the symbolic rather than the numeric indices for these vectors will be helpful rather than confusing.

$\vec{e_\chi}$ and $\vec{e_\psi}$ lie entirely in a surface of constant $\tau$, so that in this surface, they are purely spatial vectors. However, due to the relativity of simultaneity, $(e_\chi)^T$ and $(e_\psi)^T$ are not zero, because T is not the same thing as $\tau$.

Multiplying these purely spatial basis vector by the spatial coordinate values gives us a spatial offset. We add this spatial offset to the spatial position of the block to find the cartesian coordinates pointed to by block coordinates $(\chi, \psi)$ at block time $\tau$.

This is the standard way in which we use a set of basis vectors (usually a triad, but for simplicity I've ignored z, so it's a pair of basis vectors) to find spatial coordinates in any affine space - it's not terribly exotic.

 

[PeterDonis]

the position pointed to by the (spatial) block coordinates $(\chi, \psi)$ at time $\tau$ is equal to $\vec{P_0 }+ \chi \vec{e_\chi} + \psi \vec{e_\psi}$.

This works as long as $\vec{e_{\chi}}$ and $\vec{e_{\psi}}$ are orthogonal to the worldline of $\vec{P_0}$. This is only true for the "reference" worldline that you are labeling as $\chi = \psi = 0$. It's not true for other values of $\chi$, and I'm not sure whether it's true even for $\chi =0$ for all values of $\psi$. I'm not sure if your procedure still works if this orthogonality condition is violated.

$\vec{e_\chi}$ and $\vec{e_\psi}$ are unit spatial vectors, and their components were found by Peter.

Yes, but $\vec{e_{\psi}}$ is not orthogonal to $\vec{e_{\tau}}$ everywhere--although $\vec{e_{\chi}}$ is orthogonal to the $\tau \psi$ plane everywhere, so that part is ok.

Also, while the method I used to find $\vec{e_\chi}$ is unambiguous--it's just picking a unit vector in the direction of the proper acceleration--the method I used to find $\vec{e_\psi}$ is not. There are multiple ways to "extend" my definition of $\vec{e_\psi}$ away from the coordinate values $\chi = \psi = 0$ (in your convention). The "canonical" way would be to use $\vec{e_\psi}$ at $\psi = 0$ to determine a spacelike geodesic, and then parallel transport $\vec{e_\psi}$ along that geodesic to other values of $\psi$. However, I don't know if using the formula I gave for $\vec{e_\psi}$ everywhere, not just at $\psi = 0$, satisfies that requirement.

Multiplying these purely spatial basis vector by the spatial coordinate values gives us a spatial offset. We add this spatial offset to the spatial position of the block to find the cartesian coordinates pointed to by block coordinates $(\chi, \psi)$ at block time $\tau$.

Does this still work if the basis vectors are functions of $\tau$? Perhaps it might help to express the basis vectors you are using purely in terms of Minkowski coordinates; I don't know if that has been done anywhere in this thread.

Also, it seems to me that there is something missing from the procedure as you describe it. What 4-velocity vector is supposed to correspond to a pair $(\chi, \psi)$, for pairs other than $\chi = \psi = 0$ (where we agree on what that vector is)? (This kind of gets into the worldline issue that I'll discuss further below.)

This is the standard way in which we use a set of basis vectors (usually a triad, but for simplicity I've ignored z, so it's a pair of basis vectors) to find spatial coordinates in any affine space - it's not terribly exotic.

I understand the procedure, but this leaves open a key question. The block is not just a single point. The whole point of a "block frame", as I said in a previous post, is to find a chart in which each point of the block is at rest, not just the center of the block. In other words, for some range of $\chi$ and $\psi$, each pair $(\chi, \psi)$ labels a unique, fixed point of the block. (This assumes that the set of worldlines so labeled forms a rigid congruence, but as I've shown in previous posts, there does exist a rigid congruence that can be held to describe the block.) So the question is, if we take the points labeled by a given pair $(\chi, \psi)$, for all values of $\tau$, to form a worldline, and consider the set of such worldlines for some range of $\chi$ and $\psi$, do they form a rigid congruence? If they do, then the coordinates $\tau, \chi, \psi$ can indeed be held to describe a "block frame". But if not, then they can't, even if the coordinates are derived using your procedure.

Furthermore, if I hand you a chart $\tau, \chi, \psi$ in which the set of worldlines at constant spatial positions in the chart, labeled by pairs $(\chi, \psi)$ for some range of $\chi$ and $\psi$, forms a rigid congruence, and whose 4-velocity matches the 4-velocity we have agreed describes the center of the block at $\chi = \psi = 0$, then it seems to me that this chart meets all the requirements for a "block frame", whether it was derived using your procedure or not. Given such a chart, one ought to be able to reconstruct it using your procedure, but that doesn't mean your procedure is the only way to get to such a chart. And I claim that the chart I have presented in this thread is exactly such a chart (with one change, that I'm labeling the center of the block by $\chi = 1/g)$ instead of $\chi = 0$).

 

[PeterDonis]

$\vec{P_0}$ was obtained by integrating the 4-velocity of the pointlike origin of the block, U.

I also expressed concern about this before: the integration you did, even if it is correct, is not unique. The coordinate functions that I obtained are also valid integrals of the 4-velocity I gave. (Note that our functions agree for the worldline at the center of the block; the non-uniqueness only comes in when we extend things to other worldlines.) So it seems to me that your procedure does not uniquely determine a coordinate chart. Which brings us back to the question I posed before: which chart is the "right" one to represent the "block frame"?

 

[pervect]

I also expressed concern about this before: the integration you did, even if it is correct, is not unique. The coordinate functions that I obtained are also valid integrals of the 4-velocity I gave. (Note that our functions agree for the worldline at the center of the block; the non-uniqueness only comes in when we extend things to other worldlines.) So it seems to me that your procedure does not uniquely determine a coordinate chart. Which brings us back to the question I posed before: which chart is the "right" one to represent the "block frame"?

Because the derivative of a constant function is zero, integration is never unique, two functions that differ by the same amount have the same derivatives, hence they are both integral of the same function.

When we specify an observer by specifying the observer's worldline, though, it's a matter of convention to say that the the spatial origin of the "observers frame" is at (0,0). When we follow this convention, we fix the value of the integration constant.

 

[pervect]

I'm aware of this; however I has hoping that it would be possible the other way.
So I used that: the passengers are shining at the marble, even if they can't see it. Obviously their math skills are nearly superhuman

I think I've got a better answer as to what happens to a laser pointed "ahead" of the rotating block, in the direction of it's 4-acceleration. The direction of the 4-acceleration (which we've dubbed $\vec{e_X}$) does not rotate with respect to the fixed stars in the Minkowskii frame, so the laser beam won't rotate with respect to the fixed stars either, and will always point in the same direction in the Minkowskii frame. To be really specific, the laser beam emitted from the sliding block will aberrate slightly due to the sliding motion of the block, but it won't rotate.

The way I would describe this in more ambiguous English is to say that the sliding block doesn't rotate relative to the fixed stars, even though it DOES rotate relative to a torque-free gyroscope that's carried along with the sliding block, as described in the paper I quoted much earlier in the thread.

Another way of saying this in ambiguous English - in the presence of frame dragging effects such as Thomas precession, gyroscopes may (and will) rotate with respect to the fixed stars, so the notion of "rotating with respect to the fixed stars" and the notion of "rotating relative to a gyroscope" are different.

My English description of the siutatuion is hopefully more understanadable, though I doubt it's unambiguous :(.

 

[PeterDonis]

Because the derivative of a constant function is zero, integration is never unique, two functions that differ by the same amount have the same derivatives, hence they are both integral of the same function.

When we specify an observer by specifying the observer's worldline, though, it's a matter of convention to say that the the spatial origin of the "observers frame" is at (0,0). When we follow this convention, we fix the value of the integration constant.

That isn't the issue. The issue is that you are integrating a function of multiple variables $(\tau, \chi, \psi)$ with respect to only one of them at a time. That means a "constant" might not be a function of nothing at all; it might be a function of the other variables. The same goes for functions that appear in the arguments of other functions. I pointed this out before, but let's look at it again.

The function we want to integrate has the form $\partial T / \partial \tau = \gamma \cosh \left( \gamma g \tau \right)$, $\partial T / \partial \chi = \sinh \left( \gamma g \tau \right)$, and $\partial T / \partial \psi = \gamma v \cosh \left( \gamma g \tau \right)$ when $\chi = \psi = 0$. Your ansatz for the integral is:

$$
T = \left( \frac{1}{g} + \chi \right) \sinh \left( \gamma g \tau \right) + \gamma v \psi \cosh \left( \gamma g \tau \right)
$$

My ansatz for the integral is:

$$
T = \left( \frac{1}{g} + \chi \right) \sinh \left( \gamma g \tau + \gamma g v \psi \right)
$$

Both of our proposed integrals give the same three partial derivatives at $\chi = \psi = 0$. So your ansatz for the integral is not unique. (Note that there are no integration constants in either of our formulas; as you say, fixing $\chi = \psi = 0$ fixes their values. But that doesn't change the fact that both formulas give the same partial derivatives.) A similar process shows that there are two possible integrals for $X$. (We agree on the transformation for $Y$ so that's not an issue.)

 

[PeterDonis]

The way I would describe this in more ambiguous English is to say that the sliding block doesn't rotate relative to the fixed stars, even though it DOES rotate relative to a torque-free gyroscope that's carried along with the sliding block, as described in the paper I quoted much earlier in the thread.

Another way of saying this in ambiguous English - in the presence of frame dragging effects such as Thomas precession, gyroscopes may (and will) rotate with respect to the fixed stars, so the notion of "rotating with respect to the fixed stars" and the notion of "rotating relative to a gyroscope" are different.

This looks like a good description to me. The only quibble I would have is that the term "frame dragging" is not normally applied to Thomas precession (or even de Sitter precession in Schwarzschild spacetime); it's usually reserved for effects like the Lense-Thirring precession that occur only in stationary but non-static spacetimes (i.e., in somewhat ambiguous English, in cases where spacetime itself, not just an object or a family of observers, is "rotating").

The more technical way of saying it is that the congruence describing the sliding block has nonzero vorticity but is non-rotating relative to a fixed global inertial frame.

 

[SlowThinker]

While in this rotating digression, should not every rotation have a center, that can be identified from within the system?
Where is the center of this rotation? Or is it some funny kind of rotation that is the same everywhere?