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

[PeterDonis]

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?

This is a key reason why I insisted in an earlier post on distinguishing between a rotating disk and the "rotation" of the sliding block.

In the case of the rotating disk, yes, there is a unique center of rotation; that is, there is a point on the disk which is not rotating--it has zero angular velocity, and it feels zero proper acceleration, unlike every other point on the disk. Going along with that, the center of the disk stays at the same point in space, and all the other points on the disk follow paths that are spatially closed: each one comes back to the same point in space once per rotation (all as viewed from a fixed inertial frame).

In the case of the sliding block, however, there is no such center; there is no point of the block that is not "rotating" in the sense pervect described. And, of course, no point on the block stays at the same point in space, and no point on the block has a path that is spatially closed.

What, then, does the term "rotation" refer to that is the same in both cases? That's where the more technical language I used comes in. I said the congruence describing the object (disk or block) has nonzero vorticity. What does that mean?

Consider a particular point on the disk or block, and suppose that it has little rods connecting it to neighboring points. And suppose the point also carries a gyroscope along with it. "The congruence describing the object has nonzero vorticity" means that the set of little rods rotates relative to the gyroscope as the object moves. This is true of both the disk and the block.

 

[pervect]

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?

I would say the later "some funnny kind of rotation" is the closest answser. In my approach, at least, the center of rotation is wherever you put your reference observer, which is to some extent an arbitrary choice. The value of the rotation, in radians/unit proper time, should be independent of your choice of origin of the $\psi$ coordinate, though it DOES depend on your choice of origin to the $\chi$ coordinate . To understand the dependence on the $\chi$ coordinate (which I would informally call heigh). Note that in either the elevator frame or the sliding block frame, two clocks at the same $\psi$ coordinate but different $\chi$ coordinates do not tick at the same rate.

 

[pervect]

An additional comment. If we consider a rotating Newtonian bar, the center of the bar does not accelerate, it's in an inertial frame. The ends of the bar do accelerate, they are not in an inertial frame, so there is a clear choice for the "center of rotation". But in the sliding block case, every point on the block is accelerating, and a torque-free gyroscope placed at any point on the block will rotate relative to the block.

 

[PeterDonis]

I would describe the situation as using a reference worldline of an observer to build a congruence.

There is more than one way of doing this; that's what my point about the non-uniqueness of the integrals illustrates.

If we restrict ourselves to small values of $\chi$ and $\psi$, then our two methods agree (because to first order in their arguments, $\cosh$ is equal to $1$ and $\sinh$ is equal to its argument; and we can use the sum identities for the hyperbolic trig functions to separate out $\cosh \gamma g \tau$ and $\sinh \gamma g \tau$ in my formulas so that we can expand the functions of $\psi$ to first order to get expressions equivalent to yours). So in a small enough world tube about the reference worldline, our two transformations result in the same congruence.

But beyond that small world tube, our methods diverge, and I think the reason for this is that there is no single congruence that has all of the properties that intuitively might seem desirable. Your congruence has the property that surfaces of constant $\tau$ are flat, not just in the small world tube around the reference worldline (where both of our methods result in locally flat surfaces of constant $\tau$--basically this is because our methods are both locally equivalent to Fermi-Walker transport, plus a rotation of basis vectors due to nonzero vorticity), but everywhere. But I suspect that this comes at the price of the congruence being non-rigid; which means that worldlines separated by constant values of $\chi$ or $\psi$ do not stay at the same proper distance from each other.

My congruence is, as I have shown, rigid, which I think is a necessary property to have if we are going to identify worldlines in the congruence with points in the block. But this comes at the price of surfaces of constant $\tau$ not being flat globally. So it looks like we have to pick one or the other; we can't have both.

 

[PeterDonis]

I suspect that this comes at the price of the congruence being non-rigid

In order to get more of a handle on this, I will go ahead and post an analysis of pervect's metric similar to what I've already posted for mine, so we have a good basis for comparison and can see exactly what properties are and are not satisfied by each metric and the associated congruence of worldlines at rest in it. This will take multiple posts as I work through the steps.

The coordinate transformation between Minkowski $T, X, Y$ and pervect's $\tau, \chi, \psi$ is:

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

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

$$
Y = \gamma v \tau + \gamma \psi
$$

And the metric is:$$
ds^2 = - \gamma^2 \left[ \left( 1 + g \chi \right)^2 - \gamma^2 v^2 g^2 \psi^2 - v^2 \right] d\tau^2 + 2 \gamma^2 v g d\tau \left( \psi d\chi - \chi d\psi \right) + d\chi^2 + d\psi^2 + dz^2
$$

The coordinate basis vectors derived from the above are:

$$
\partial_{\tau} = \left[ \gamma \left( 1 + g \chi \right) \cosh \gamma g \tau + \gamma^2 v g \psi \sinh \gamma g \tau \right] \partial_T + \left[ \gamma \left( 1 + g \chi \right) \sinh \gamma g \tau + \gamma^2 v g \psi \cosh \gamma g \tau \right] \partial_X + \gamma v \partial_Y
$$

$$
\partial_{\chi} = \sinh \gamma g \tau \partial_T + \cosh \gamma g \tau \partial_X
$$

$$
\partial_{\psi} = \gamma v \cosh \gamma g \tau \partial_T + \gamma v \sinh \gamma g \tau \partial_X + \gamma \partial_Y
$$

Let's first compute the proper acceleration of the 4-velocity field $U$ of worldlines at rest in this chart. That requires us to divide $\partial_{\tau}$ by its norm, which is $| \partial_{\tau} | = \gamma \sqrt{ \left( 1 + g \chi \right)^2 - \gamma^2 v^2 g^2 \psi^2 - v^2 }$. So the 4-velocity is

$$
U = \frac{1}{\sqrt{ \left( 1 + g \chi \right)^2 - \gamma^2 v^2 g^2 \psi^2 - v^2 }} \left( \left[ \left( 1 + g \chi \right) \cosh \gamma g \tau + \gamma v g \psi \sinh \gamma g \tau \right] \partial_T + \left[ \left( 1 + g \chi \right) \sinh \gamma g \tau + \gamma v g \psi \cosh \gamma g \tau \right] \partial_X + v \partial_Y \right)
$$

The proper acceleration is then

$$
A = U \cdot \frac{dU}{d\tau} = \frac{g \gamma}{\left( 1 + g \chi \right)^2 - \gamma^2 v^2 g^2 \psi^2 - v^2} \left( \left[ \left( 1 + g \chi \right) \sinh \gamma g \tau + \gamma v g \psi \cosh \gamma g \tau \right] \partial_T + \left[ \left( 1 + g \chi \right) \cosh \gamma g \tau + \gamma v g \psi \sinh \gamma g \tau \right] \partial_X \right)
$$

(As a sanity check, we can see that for $\chi = \psi = 0$, $U$ and $A$ reduce to the forms we have previously agreed on for the reference worldline.)

If we then express $A$ in terms of the coordinate basis vectors, we find that it is

$$
A = \frac{g \gamma^2}{\left(| \partial_{\tau} |\right)^2} \left[ \left( 1 + g \chi \right) \partial_{\chi} + g \psi \left( \partial_{\psi} - \gamma \partial_Y \right) \right]
$$

So, for $\psi \neq 0$, $A$ no longer points purely in the $\chi$ direction. (Note also that this is a "mixed" formula, since $\partial_Y$ appears; I won't be able to clean it up until I have the inverse of the above coordinate transformation.)

For completeness, we can also check the unit vectors in the coordinate directions; we find that $\partial_{\chi}$ and $\partial_{\psi}$ are unit vectors everywhere, so we have:

$$
\hat{e}_{\tau} = U = \frac{1}{\gamma \sqrt{ \left( 1 + g \chi \right)^2 - \gamma^2 v^2 g^2 \psi^2 - v^2 }} \partial_{\tau}
$$

$$
\hat{e}_{\chi} = \partial_{\chi}
$$

$$
\hat{e}_{\psi} = \partial_{\psi}
$$

The next thing will be to obtain the inverse of the above coordinate transformation, so I can express $U$ and $A$ above purely in terms of Minkowski coordinates and then compute the kinematic decomposition. (Pervect, if you already have the inverse transformation, it would help to post it, since I'm currently having trouble deriving it.)

 

[PeterDonis]

The next thing will be to obtain the inverse of the above coordinate transformation, so I can express $U$ and $A$ above purely in terms of Minkowski coordinates and then compute the kinematic decomposition.

Well, I haven't yet been able to invert pervect's coordinate transformation, but I have done something that makes it moot as far as computing the kinematic decomposition is concerned: I have found that the 4-velocity field $U$ in his chart is the same as the one in my chart! In other words, his $U$ and my $U$ describe the same congruence of worldlines; the only difference is how we parameterize them. I'll give a sketch of the proof below.

Of course this means that many of the potential issues I had raised in previous posts are moot, since they depend only on the congruence, not on how we parameterize it. (It also gives me a much better handle on comparing the two charts, but I'll save that for a future post.)

Here is how we can see that the two charts both have the same congruence of worldlines at rest in them. For this proof I'll switch to my convention for $X$ and $\chi$, where the floor of the rocket is at $\chi = 1/g$ instead of $\chi = 0$. This will make the formulas simpler since we'll just have $g \chi$ everywhere we used to have $1 + g \chi$. It will also make it easier to see how the proof goes.

So with the new convention, we have the following:

$$
X^2 - T^2 = \chi^2 - \gamma^2 v^2 \psi^2
$$

That expression should look familiar; it occurs in the norm of pervect's $U$. Using it, we can re-express that norm as $| U | = \sqrt{g^2 \left( X^2 - T^2 \right) - v^2}$.

That expression should also look familiar: it is the same as the norm of my $U$, expressed in Minkowski coordinates (see post #81). Guided by that hint, we then look at the rest of the expression for pervect's $U$, and compare it with his coordinate transformation, and find that

$$
U = \frac{1}{\sqrt{g^2 \left( X^2 - T^2 \right) - v^2}} \left( g X \partial_T + g T \partial_X + v \partial_Y \right)
$$

This is, of course, the same formula as I derived in post #81 for my 4-velocity field. So at every point in in Minkowski coordinates, which means at every point in spacetime, pervect's $U$ and my $U$ are the same vector. That means both 4-velocity fields must generate the same congruence of worldlines.

 

[PeterDonis]

his UU and my UU describe the same congruence of worldlines; the only difference is how we parameterize them.

Just to summarize where I am currently on this, the obvious change from my coordinates to pervect's is the $\chi$ coordinate. If we use $\bar{\chi}$ for my coordinate, then we have $\bar{\chi}^2 = \chi^2 - \gamma^2 v^2 \psi^2$. So if we pick a worldline in the congruence that pervect labels with $(\chi, \psi)$, I will label it instead with $\bar{\chi} = \sqrt{\chi^2 - \gamma^2 v^2 \psi^2}$.

What I haven't yet figured out is whether that is the only reparameterization. The fact that we both have the same transformation formulas for $Y$ strongly suggest that it is--i.e., that pervect's $\tau, \psi$ and my $\tau, \psi$ are the same. But I haven't yet been able to get everything else to match up under that assumption.

If we assume for a moment that the above is the only reparameterization, what does it mean? It means that our surfaces of constant $\tau$ are the same, and we label each worldline in the congruence with the same $\psi$, so in each surface of constant $\tau$, our $\psi$ "grid lines" are the same. But our $\chi$ "grid lines" are not; pervect's $\chi$ grid lines and my $\bar{\chi}$ grid lines are tilted with respect to each other (they match only on the reference worldline, the one he labels with $\chi = 0$, $\psi = 0$ and I label with $\bar{\chi} = 1 / g$, $\psi = 0$--but the $1/g$ is just a translation and doesn't affect the tilting of the grid lines, and I have left it out of the discussion above). Furthermore, the "tilt" is not linear; in pervect's chart, his $\chi$ grid lines are straight (because his spatial metric is Euclidean), but my $\bar{\chi}$ grid lines are curved (so the increment of $d\psi$ needed to move along my $\bar{\chi}$ grid lines between the same two worldlines is different, hence my $g_{\psi \psi}$ is not $1$).

So why do my $\bar{\chi}$ grid lines curve, as seen in pervect's chart? Because they have to remain perpendicular to the direction of proper acceleration, so that the proper acceleration vector is always a multiple of $\partial_{\bar{\chi}}$. In pervect's metric, the proper acceleration vector is not always perpendicular to his $\chi$ grid lines. (On the reference worldline, it is, and pervect's $\chi$ grid lines and my $\bar{\chi}$ grid lines are parallel--or, to put it another way, on that worldline, my curved grid lines are just tangent to his straight ones). Also, my $\bar{\chi}$ grid lines are the ones that are at a constant "altitude" in the rocket--for example, the grid line I label $\chi = 1/g$ is the one at the altitude of the floor of the rocket. In pervect's chart, the floor of the rocket, and any other line at a constant altitude in the rocket, will look curved and will not be at a constant $\chi$ coordinate.

I think the above helps to illustrate the differences between our two charts, and what features each one has. But, as I said, I haven't confirmed that the $\chi$ change is the only one, so this is all tentative at this point.

 

[pervect]

In order to get more of a handle on this, I will go ahead and post an analysis of pervect's metric similar to what I've already posted for mine, so we have a good basis for comparison and can see exactly what properties are and are not satisfied by each metric and the associated congruence of worldlines at rest in it. This will take multiple posts as I work through the steps.

The coordinate transformation between Minkowski $T, X, Y$ and pervect's $\tau, \chi, \psi$ is:

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

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

$$
Y = \gamma v \tau + \gamma \psi
$$

I don't _think_ it makes any difference to the subsequent calculations, but I wanted to point out that using the above, $\tau=\chi=\psi=0$ transformed to $X=/1g, T=0$ whereas in my original version there was an additional offset in the transform so that $\tau=\chi=\psi=0$ transformed to $X=T=0$ instead.

The next thing will be to obtain the inverse of the above coordinate transformation, so I can express $U$ and $A$ above purely in terms of Minkowski coordinates and then compute the kinematic decomposition. (Pervect, if you already have the inverse transformation, it would help to post it, since I'm currently having trouble deriving it.)

I don't have an inverse.

 

[pervect]

I think the above helps to illustrate the differences between our two charts, and what features each one has. But, as I said, I haven't confirmed that the $\chi$ change is the only one, so this is all tentative at this point.

I rather belatedly got around to noticing that your $g_{00}$ was only a function of $\chi$. For an observer "at rest", i.e. one with constant $\chi$ and $\psi$ coordinates, the only non-zero component of the 4-velocity U will the $U^{\tau}$ component. Because $g_{00}$ is only a function of $\chi$, when we normalize U by making $g_{00} u^{\tau} u^{\tau}$ have a magnitude of 1 (a value of -1 in the sign conventions I favor), this implies that $U^\tau$ is only a function of $\chi$, and thus $dU/d\tau$, the 4-acceleration, only has components in the $\partial_\chi$ direction.

I tend to think of surfaces of constant $g_{00}$ in explicitly stationary or static metrics (i.e. coordinates in which none of the metric coefficients is a function of time) as equipotential surfaces. I'm not sure I've seen this exact terminology explicitly used in a textbook, though it was inspired by the example of the rotating Earth's geoid being both both an equipotential surface and a surface on which $g_{00}$ is constant. Anyway, as the above argument suggests, choosing coordinates such that equipotential surface are a function only of a single coordinate automatically makes the 4-acceleration of an observer "at rest" in those coordinates perpendicular to the equipotential surface. So expressing the result in coordinate independent terms, we can say that the 4-acceleration of an observer "at rest" in a static or stationary space-time is always perpendicular to an equipotential surface.

So if we consider the floor of the rocket on which the block is sliding, it is an equipotential surface, and this feature is true in all coordinate systems. Your coordinate choices keeps the (appearance?) of the floor of the rocketr "flat" in the sense that the rocket floor has a constant $\chi$ coordinate. If we imagine the rocket having multiple floors at different heights, your transformation would keep all the floors of the rocket "flat" in the sense that each floor would have a characteristic and constant value for $\chi$. This simplicity in the description of the floor comes at the expense of introducing a curved space.

My coordinates make the floor of the rocket "curve", because equipotential surfaces do not have constant $\chi$ coordinate, the surface is a function of both $\chi$ and $\psi$. However, the advantage of my coordinate choice is that it was constructed in such a manner as to make the spatial part of the flat.

 

[PeterDonis]

I don't _think_ it makes any difference to the subsequent calculations, but I wanted to point out that using the above, $\tau=\chi=\psi=0$ transformed to $X=/1g, T=0$ whereas in my original version there was an additional offset in the transform so that $\tau=\chi=\psi=0$ transformed to $X=T=0$ instead.

You're right, there should be an extra constant term in the $X$ transform. It doesn't affect any of the other calculations. (I did all the computations with my convention anyway, where there is just $\chi$ instead of $(1/g) + \chi$ in the transforms and no extra offsets, and the floor of the rocket at $\psi = 0$ is at $\chi = 1/g$ at $\tau = 0$.)

 

[PeterDonis]

Because $g_{00}$ is only a function of $\chi$, when we normalize $U$ by making $g_{00} u^{\tau} u^{\tau}$ have a magnitude of 1 (a value of -1 in the sign conventions I favor), this implies that $U^\tau$ is only a function of $\chi$, and thus $U/d\tau$, the 4-acceleration, only has components in the $\partial_\chi$ direction.

Yes.

I tend to think of surfaces of constant $g_{00}$ in explicitly stationary or static metrics (i.e. coordinates in which none of the metric coefficients is a function of time) as equipotential surfaces.

I called them "surfaces of constant altitude", but "equipotential surfaces" is just as good.

I'm not sure I've seen this exact terminology explicitly used in a textbook

IIRC MTW use it in some places, but I don't think they put a lot of emphasis on it.

choosing coordinates such that equipotential surface are a function only of a single coordinate automatically makes the 4-acceleration of an observer "at rest" in those coordinates perpendicular to the equipotential surface.

No, that is always true, because the 4-acceleration of the observer "at rest" must be in the direction of the gradient of the potential, which is orthogonal to the equipotential surface. That is an invariant statement, independent of coordinates. If you compute the inner product of the 4-acceleration vector in your chart with a vector that is tangent to the equipotential surface, expressed in your chart, you will find that that inner product is zero.

The special feature of my chart, where the equipotential surfaces are surfaces of constant value of a single coordinate $\chi$, is that the vectors tangent to the equipotential surface have no $\partial_{\chi}$ component. So they are linear combinations, in my chart, of $\partial_{\psi}$ and $\partial_z$ only. In your chart, vectors tangent to the equipotential surfaces can have components in all three of the spatial directions (but, as above, they will still be orthogonal to the 4-acceleration everywhere).

So if we consider the floor of the rocket on which the block is sliding, it is an equipotential surface, and this feature is true in all coordinate systems. Your coordinate choices keeps the (appearance?) of the floor of the rocket "flat" in the sense that the rocket floor has a constant $\chi coordinate$. If we imagine the rocket having multiple floors at different heights, your transformation would keep all the floors of the rocket "flat" in the sense that each floor would have a characteristic and constant value for $\chi$. This simplicity in the description of the floor comes at the expense of introducing a curved space.

The term "the floor of the rocket" is ambiguous. In Minkowski coordinates, if we leave out the $Z$ coordinate, the floor of the rocket is described by the "worldsheet" $X^2 - T^2 = 1 / g^2$, with $Y$ unconstrained. This is an infinite set of hyperbolas running in the $Y$ direction, and each one of those hyperbolas is the worldline of a "rocket observer", sitting at rest on the floor of the rocket at some fixed value of $Y$.

To one of those rocket observers (say the one with $Y = 0$ for definiteness, since that's the "reference" observer for Rindler coordinates), the floor of the rocket at some instant of his time is a line in the $Y$ direction "cut" out of the worldsheet at some fixed value of $T$ (and $X$, since $T$ determines $X$ on the worldsheet). This is obviously, once we put the $Z$ dimension back, a flat plane.

Now, since the bottom of the sliding block is always in contact with the floor of the rocket, the congruence of worldlines that describes the bottom of the block (the one that is at rest in both of our charts) must lie in that same worldsheet I described above. But the worldlines sit in that worldsheet at an angle, so to speak. For example, the "reference" observer of the block, the one who is at $\psi = 0$, has a worldline that intersects that of the reference rocket observer above at $T = Y = 0, X = 1/g$ (using my coordinate convention, since that's the one I used for the equation of the worldsheet above). At more and more positive values of $T$, this worldline increases in $X$, but it also increases in $Y$--and for more and more negative values of $T$, the worldline also increases in $X$ (since now it is moving in the $-X$ direction, but decelerating), but it decreases in $Y$. So it curves around the worldsheet at an angle.

Therefore, the intersection of the worldsheet with surfaces of constant $\tau$ in either of our charts (and, as far as I can tell, these are the same surfaces for both charts) will also curve around the worldsheet at an angle; these surfaces will not be the same surfaces I described above as the "floor of the rocket at an instant of time" for the rocket observers. That means the intersection of the worldsheet with surfaces of constant $\tau$ generates surfaces that are in fact curved, not flat, as a matter of geometry. So the curvature is not a matter of appearance.

My chart makes these surfaces, the equipotential surfaces from the point of view of the sliding block, "appear flat" only in the sense that they are surfaces of constant $\chi$; but my metric makes clear that they are in fact curved, geometrically, because $g_{\psi \psi}$ is not $1$. Your chart makes the curvature clear in a different way, by making the equipotential surfaces explicitly curved, i.e., curved surfaces in a Euclidean spatial metric. Either way works, and both agree that the surfaces, geometrically, are curved.

My coordinates make the floor of the rocket "curve", because equipotential surfaces do not have constant $\chi$ coordinate, the surface is a function of both $\chi$ and $\psi$. However, the advantage of my coordinate choice is that it was constructed in such a manner as to make the spatial part of the flat.

Yes, but there's another tradeoff besides making the equipotential surfaces functions of two coordinates. The "apparent" direction in your chart of the proper acceleration vector changes with $\psi$. But, as you noted in a previous post, the actual direction of that vector does not change relative to the "fixed stars"--more precisely, it doesn't change relative to Minkowski coordinates. In those coordinates it always points, spatially, in the $X$ direction. My chart reflects that by keeping the proper acceleration always in the $\chi$ direction.

In short, in a situation like this, as I've said before, it is impossible to have a single chart that directly represents all of the properties we are interested in. We have to pick and choose which properties we want the chart to directly represent, and which ones we want to have it represent only indirectly.

 

[PeterDonis]

What I haven't yet figured out is whether that is the only reparameterization.

On thinking this over, I don't think it can be; at the very least, $\psi$ needs to change also from my metric to pervect's.

Consider pervect's $\chi \psi$ plane at some constant value of $\tau$ (and $z$), and think about how my $\bar{\chi}$ and $\bar{\psi}$ "grid lines" would look. We know that my $\bar{\chi}$ grid lines look curved. But my $\bar{\psi}$ grid lines are everywhere orthogonal to my $\bar{\chi}$ grid lines (because there is no cross term in my metric between those two coordinates). That means my $\bar{\psi}$ grid lines must also curve, as seen in pervect's chart. But pervect's $\psi$ grid lines are straight lines in his chart, so his and mine can't be the same. The best we can hope for is that on the floor of the rocket, his $\psi$ and my $\bar{\psi}$ will be the same.

I'm afraid this quite possibly means that the $\tau$ coordinate needs to transform as well, because, as I observed before, both of us have the same transformation equation for $Y$ in terms of $\tau$ and $\psi$. So the comparison between our two charts in the $\chi \psi$ plane, as above and in previous posts, may only be of limited validity, since it relied on our two charts having the same surfaces of constant $\tau$.

 

[pervect]

I concur that PeterD's time coordinate, which I'll refer to as $\tau^\prime$ in this post, and my time coordinate, which I'll refer to as $\tau$ in this post, are not the same. To check this, I performed the following test.

I took my expressions for $T(\tau,\chi,\psi)$, $X(\tau,\chi,\psi)$, $Y(\tau,\chi,\psi)$, as modified to be compatible with Peter's choice of origin as in post https://www.physicsforums.com/threads/gravity-on-einsteins-train.835994/page-6#post-5260046.

Then I applied Peter's inverse transformation from post https://www.physicsforums.com/threads/gravity-on-einsteins-train.835994/page-4#post-5256530, to get $\tau^\prime(T,X,Y)$.

The result did not appear to simplify to $\tau^\prime = \tau$, but the expression was complicated enough that it was difficult to be absolute sure they were not the same. So what I did was consider the much easier case to test if the surface $\tau=0$ mapped to the surface $\tau^\prime=0$. Substituting $\tau=0$ into my expressions for T,X,Y gives:

$$
T = \gamma \chi \quad X = (1/g + \chi) \quad Y= \gamma \psi
$$

I then substituted these values for T,X,Y into Peter's expression for $\tau^\prime$

$$
\tau^\prime = \gamma (\frac {arctanh(T/X)}{g} - v Y)
$$

I won't give the result here, but it was clearly not equal to zero as it should be if our time coordinates were the same. The results were differ by so much that I wonder if there could be some error - but if so, I haven't spotted it. I do have a simple verbal description of my time coordinate - it's the time coordinate of an inertial frame co-moving with the reference observer. I don't have any such simple verbal description for Peter's time cordainte $\tau^\prime$.

 

[PeterDonis]

I do have a simple verbal description of my time coordinate - it's the time coordinate of an inertial frame co-moving with the reference observer. I don't have any such simple verbal description for Peter's time cordainte $\tau^\prime$.

Both of our time coordinates meet that description, because they both match on the worldline of the reference observer. (If you plug in $\psi = 0$ and $\chi = 1/g$ to my formulas, and plug $\psi = 0$ and $\chi = 0$ into yours, they will match.) They only differ off of that worldline. There is no unique way to define "the time coordinate of an inertial frame co-moving with the reference observer" globally, because there is no such frame globally since the reference worldline is accelerating. We are simply doing the extension of the obvious local time coordinate for the reference observer to a non-inertial frame covering the rest of spacetime (or at least the region of spacetime covered by our scenario) in different ways. Your way makes the surfaces of constant coordinate time Euclidean; my way makes them always orthogonal to the proper acceleration vector.

 

[pervect]

Both of our time coordinates meet that description, because they both match on the worldline of the reference observer. (If you plug in $\psi = 0$ and $\chi = 1/g$ to my formulas, and plug $\psi = 0$ and $\chi = 0$ into yours, they will match.) They only differ off of that worldline. There is no unique way to define "the time coordinate of an inertial frame co-moving with the reference observer" globally, because there is no such frame globally since the reference worldline is accelerating.

Sure there is. While the reference observer is, as you note, accelerating, at any given instant of proper time there is a non-accelerating observer who is co-located with the accelerating observer, and who has zero velocity relative to the accelerating observer. That observer is the co-moving inertial observer, and they have a corresponding inertial frame.

 

[PeterDonis]

While the reference observer is, as you note, accelerating, at any given instant of proper time there is a non-accelerating observer who is co-located with the accelerating observer, and who has zero velocity relative to the accelerating observer. That observer is the co-moving inertial observer, and they have a corresponding inertial frame.

Yes, but it's a different inertial frame at each event on the reference observer's worldline. So there is no such thing as "the" time coordinate of the reference observer's comoving inertial frame. There are an infinite number of such coordinates, one for each event on the reference observer's worldline.

What you may be trying to say here is that $\tau$ is intended to be like the time coordinate in Rindler coordinates, where each surface of constant Rindler time $t$ is orthogonal to each worldline in the Rindler congruence. Of course this isn't possible for the entire congruence $U$ that describes the sliding block, because that congruence has nonzero vorticity; but it is possible for each surface of constant $\tau$ to be orthogonal to the reference observer's worldline. But both of our $\tau$ coordinates have that property, because the cross terms in both of our metrics vanish on the reference worldline.

There is another property that surfaces of constant time in Rindler coordinates have, which is that they are geodesic surfaces, i.e., if we pick any point on such a surface and look at the three spatial basis vectors there, the geodesics determined by those basis vectors (and linear combinations of them) generate the entire surface. However, there is no way to "read off" that property just from the metric; you would need to actually compute the geodesic equation and see. I haven't done that for either of our metrics, so I don't know if either of our charts have the property I just described.

 

[pervect]

What you may be trying to say here is that $\tau$ is intended to be like the time coordinate in Rindler coordinates, where each surface of constant Rindler time $t$ is orthogonal to each worldline in the Rindler congruence.

The congruence is not the Rindler congruence. It's a congruence of _inertial observers_. Obsevers who are not accelerating . They're all relatively at rest in flat space_time. It's not that complicated, really. It's just Einstein clock synchronization. Wiki has a bit on it, https://en.wikipedia.org/wiki/Einstein_synchronization, wiki even mentions the "no-redshift" condition which is one of two necessary conditions for clock synchronization to occur.

Inertial observers in inertial reference frames are also mentioned in http://arxiv.org/abs/0708.2490v1, which I mentioned in post #49 of this threead. I'll quote one section which talks about it briefly.

Because a torque-free gyroscope precesses relative to
the train, it follows that the train has a proper rotation,
meaning that the train rotates as observed from its own
momentary inertial rest frame.

Note that this paper talks about an inertial observer in inertial frame S, which is at rest relative to the track, and another inertial observer in frame S', which frame is moving along the track with some velocity v. They do not see the need to complicate the analysis any more than that.

Note that they make points similar to those I've been attempting to make several times - that the proper frame of the train is rotating, and that in that proper frame, the track is curved.

 

[PeterDonis]

The congruence is not the Rindler congruence.

Which congruence are you talking about? If you mean the congruence of observers at rest in Rindler coordinates, that congruence is accelerated, not inertial. So is the congruence $U$ of observers at rest in either of our metrics. Those are the only congruences we've discussed in this thread, as far as I know.

this paper talks about an inertial observer in inertial frame S, which is at rest relative to the track, and another inertial observer in frame S', which frame is moving along the track with some velocity v

These two inertial observers do not form a congruence. Their worldlines cross. A congruence is a family of worldlines that do not cross anywhere, so every event in the region of spacetime covered by the congruence lies on one and only one worldline in the congruence.

they make points similar to those I've been attempting to make several times - that the proper frame of the train is rotating, and that in that proper frame, the track is curved.

But this "proper frame" is not an inertial frame, and the congruence of worldlines of observers at rest in this frame is not an inertial congruence.

More precisely: if by the "proper frame" you mean an inertial frame in which the train is momentarily at rest, then that frame is not rotating; it can't be. Inertial frames are by definition non-rotating. In this frame, at the instant in which the train is at rest in it, different parts of the track are moving at different velocities in the "vertical" direction--the part towards the rear of the train is moving down, and the part towards the front of the train is moving up. (Fig. 11 in the paper and its accompanying discussion show this.) So in this frame, the track can be thought of as "rotating" around an axis perpendicular to the plane spanned by the "vertical" direction (the direction of proper acceleration) and the direction of the train's and track's relative motion. But the frame itself is not rotating; it's inertial.

Also, in the momentarily comoving inertial frame of the train, the track appears curved. As the paper explains, this can be understood as being due to relativity of simultaneity; in this frame, the events at which the different parts of the track are at the same "vertical" coordinate (the $X$ coordinate in my nomenclature) are not all simultaneous. In other words, whether the track appears curved depends on how you "cut" a surface of constant time out of the entire worldsheet describing the track. The MCIF of the train cuts "at an angle" through that worldsheet, and therefore cuts out a curved surface.

If you want to define a non-inertial "proper frame" in which the train is at rest continuously (which is what I thought both of us were trying to do), then whether or not the track is curved in that frame works the same way as the above: it depends on how the surfaces of constant $\tau$ in that non-inertial frame cut the worldsheet describing the track. As far as I can tell, both of our metrics define surfaces of constant $\tau$ that cut at an angle, so the track appears curved in both of our metrics. Are you saying that the surfaces of constant $\tau$ in one of our metrics (presumably yours) are all the same as surfaces of constant time in some inertial frame? If so, which one?

 

[pervect]

Certain aspects of your (PeterDonis's) metric continue to bother me. The piece I'm missing that would help me the most in resolving my questions is an expression for the inertial coordinates (T,X,Y) as functions of the block coordinates ($\tau,\chi,\psi$), which I don't seem to find. I could be missing this piece due to the length of the thread, if this has actually been posted somewhere, a pointer to it would be appreciated.

That said, let me review the textbook support for the approach I outlined, which may be both clearer and more convincing than my own exosition. It's also much more lengthly, but the length appears to be needed to address all the little issues that keep popping up in our attempts to discuss it here on PF. The approach is basically straight out of MTW's textbook, "Gravitation". The case of the accelerating and rotating observer in flat space-time is first discussed in exercise 6.8 on pg 174. Because it's only an exercise, it's not as easy to follow as the similar but more leisurely discussion of the accelerating and rotating observer in curved space-time starting on page 327,. The title of this discussion is "The Proper reference frame of an accelerated observer", in section $12.6. As a side note, the title of this section shows that these coordinates are important enough to merit a name of their own.

The discussion in section $12.6 gives us the line element of the metric for the proper reference frame to the first order. I won't go into details, but the argument is based on considering certain of the Christoffel symbols along the worldline of the accelerating observer. This approach resulting line element given in 13.32

$$
ds^2 = -(1 + 2a^\hat{j}x^\hat{j}) (dx^\hat{0})^2 -2 (\epsilon_{\hat{j}\hat{k}\hat{l}}x^\hat{k}\omega^\hat{l}) (dx^\hat{0} dx^\hat{j})+\delta_{\hat{j}\hat{k}} (dx^\hat{j}dx^\hat{k})
$$
plus terms of order $O(|x^\hat{j}|^2)$Here $j,k,l$ range from 1 to 3, making $a^j$ and $\omega^l$ both three-vectors, which are respectively the proper acceleration and proper rotation of the frame. $\epsilon_{jkl}$ is the 3-d Leva-Civita symbol, and $\delta$ is the Kronecker delta.

The line element given by MTW agrees with my line element to the first order, where a is the proper acceleration $\gamma^2 g$ of a point on the block, and $\omega$ is the proper rotation, which has a magnitude of $\gamma^2 g v$ (I haven't worked out the sign of the rotation) and is oriented about the $\hat{z}$ axis. This proper rotation is due to Thomas preccession as discussed in http://arxiv.org/abs/0708.2490v1, which also gives us the magnitude quoted above.

This agreement with the textbook results to the first order, plus the flatness of the resulting Riemann tensor, convinces me both that the line element is correct, and that the proper reference frame of the sliding block fits neatly into the paradigm of an "accelerating and rotating observer", i.,e an accelerating observer who chooses to use orthonormal basis vectors, but not to Fermi-walker transport them.

 

[PeterDonis]

The piece I'm missing that would help me the most in resolving my questions is an expression for the inertial coordinates (T,X,Y) as functions of the block coordinates $(\tau,\chi,\psi)$, which I don't seem to find.

It's in post #68, which is indeed quite a while ago now:

https://www.physicsforums.com/threads/gravity-on-einsteins-train.835994/page-4#post-5254985

The inverse transformation is in post #80:

https://www.physicsforums.com/threads/gravity-on-einsteins-train.835994/page-4#post-5256530

[Edit: deleted the rest of what I originally posted here until I can do some more computations.]

 

[PeterDonis]

a is the proper acceleration $\gamma^2 g$ of a point on the block

Note that this is only true on the equipotential surface in which the chosen worldline lies. Other equipotential surfaces have different magnitudes for $a$. And, as your chart explicitly illustrates, the direction of $a$ can, in general, change as well once you are off the chosen worldline (though it doesn't in my chart).

$\omega$ is the proper rotation, which has a magnitude of $\gamma^2 g v$ (I haven't worked out the sign of the rotation) and is oriented about the $\hat{z}$ axis.

This magnitude also varies once you are off the chosen worldline, as does, in general, the orientation of the rotation. Your metric keeps the orientation about the $\hat{z}$ axis, but mine does not. However, to first order, close enough to the chosen worldline, they agree.

 

[PeterDonis]

an accelerating observer who chooses to use orthonormal basis vectors, but not to Fermi-walker transport them.

I am still working on some computations, but I would observe that this statement cannot be true as it stands, since there are nonzero $dx^0 dx^i$ "cross terms" in the line element, which means that the coordinate direction $\partial_0$ is not orthogonal to all of the $\partial_i$ directions. That means the tetrad $\partial_0 / | \partial_0 |, \partial_i$ is not orthonormal (it is normalized, but not orthogonal). On the reference worldline, the "cross terms" vanish, so the above tetrad is orthonormal on that worldline, but it is not orthonormal elsewhere.

 

[andrewkirk]

It occurs to me that this sort of evolving, constructive discussion would really benefit from a wiki rather than this blog format, because then there can be an evolving document that has everything in one place and that continually improves as people like Peter and Pervect contribute to it.

I realize that this is probably wishing for the moon, as wikis are (so I understand) completely different technology from blogs.

 

[SlowThinker]

It occurs to me that this sort of evolving, constructive discussion would really benefit from a wiki rather than this blog format

Not sure about that, wiki is generally the definitive answer rather than a discussion.
In any case, as long as we arrive at the solution, it can be sorted later I just wish I wasn't stuck at page 4

 

[PeterDonis]

this sort of evolving, constructive discussion would really benefit from a wiki

As you suspect, we don't really have the technology for a wiki here at PF. However, the key material from this thread would probably be suitable for a PF Insights article once we have everything sorted out.

 

[pervect]

an accelerating observer who chooses to use orthonormal basis vectors, but not to Fermi-walker transport them.

I am still working on some computations, but I would observe that this statement cannot be true as it stands, since there are nonzero $dx^0 dx^i$ "cross terms" in the line element, which means that the coordinate direction $\partial_0$ is not orthogonal to all of the $\partial_i$ directions. That means the tetrad $\partial_0 / | \partial_0 |, \partial_i$ is not orthonormal (it is normalized, but not orthogonal). On the reference worldline, the "cross terms" vanish, so the above tetrad is orthonormal on that worldline, but it is not orthonormal elsewhere.

Let me quote MTW, bg 174, the beginning of exercise 6.8.

An observer moving along an arbitrarily accelerated worldline chooses not to Fermi-Walker transport his orthonormal tetrad. Instead, he allows it to rotate. The anti symmetric rotation (..) splits into a Fermi-Walker part plus a spatial rotation part.

I don't see much difference between my paraphrase of MTW's remark and the original. I strongly suspect that a good part of the difficulties in our communications arise from a fundamental difference between your approach and MTW's approach. My results with explaining MTW's approach in my own words has not been so successful to date, so rather than attempt to explain further I'll refer you to the text. I'd be happy to try to explain again, anyway, if you think it'll help, but judging by my track record to date I'm thinking it would be better to refer you to the textbook and the fuller context in which the textbook's remarks were made.

If you have the time to study this further, you might want to compare the mathematical equations of how MTW uses the tetrad to find the location of a point.

 

[PeterDonis]

I don't see much difference between my paraphrase of MTW's remark and the original.

I'm not disputing that you've quoted MTW correctly. I'm saying that their use of the term "orthonormal", strictly speaking, can only be correct on the reference worldline, where the "cross terms" in the metric vanish. If there are nonzero "cross terms" in the metric, the coordinate basis vectors cannot be orthogonal. So MTW's terminology here is a little sloppy.

(Note, btw, that MTW only says explicitly that the observer on the reference worldline has an orthonormal tetrad. They do not explicitly say, as far as I can tell, that the tetrad remains orthonormal off the reference worldline, i.e., where some of the $x^i$ spatial coordinates are nonzero. So what they say is not, strictly speaking, wrong; it's just, as I said above, a little sloppy, since they don't point out that "orthonormal" only applies on the reference worldline.)

My results with explaining MTW's approach in my own words has not been so successful to date

You don't need to explain MTW's approach; I understand it, and I think you've captured it pretty well. That's not the issue. The issue is, we want to construct a coordinate chart that covers a region of spacetime beyond the reference worldline and a small "world tube" around it; what properties do we want such a chart to have? And is it even possible for a single chart to have all the properties we would like?

Your chart uses MTW's method, which basically amounts to constructing a chart with the following key properties: surfaces of constant coordinate time in the chart are Euclidean, and all the effects of proper acceleration and rotation can be captured in two spatial vectors $a$ and $\omega$, which appear in the line element in the form that you quoted.

My chart uses a different method, which is the same to first order as yours and MTW's, so it matches within a sufficiently small "world tube" around the reference worldline, but outside that "world tube" it is different. It has a different set of key properties: the proper acceleration always points in one coordinate direction only ($\chi$), and the number of "cross terms" in the line element is minimized (there is only one, $d\tau d\psi$, whereas there are two in your chart).

I'm not saying my method is "right" or yours/MTW's is "wrong"; I'm simply exploring what the differences mean.

 

[pervect]

I have an approximate series conversion from my coordinates to peter's coordinates, to 3'd order, which I've tested simply by trying several random points , converting my coordinates to peter's coordinates, then converting both coordinates to Minkowskii coordinates and making sure they were close. They won't be exactly the same, because it's just a series approximation.

It demonstrates that our coordinates are different by second-order terms, and that none of them are exactly the same when $\psi$ is not zero.

$(\tau,\chi,\psi)$ are my coordinates, $(\tau^\prime, \chi^\prime, \psi^\prime)$ are Peters' coordinates.
<br /> \tau^\prime \approx \tau+{\frac {gv \left( -\chi\,g+1 \right) \chi\,\psi}{{v}^{2}-1}}+1/3\,{\frac {{g}^{2}{\psi}^{3}{v}^{3}}{ \left( -{v}^{2}+1 \right) ^{2}}}
<br /> \chi^\prime \approx \frac{1}{g} +\chi+1/2\,{\frac { \left( \chi\,g-1 \right) g{\psi}^{2}{v}^{2}}{-{v}^{2}+1}}<br />
\psi^\prime \approx <br /> \psi+{\frac {{v}^{2}g \left( -\chi\,g+1 \right) \psi\,\chi}{-{v}^{2}+1}}-1/3\,{\frac {{v}^{4}{g}^{2}{\psi}^{3}}{ \left( -{v}^{2}+1 \right) ^{2}}}<br />

As usual, my origin is at $\chi=\psi=0$, peter's origin has $\chi^\prime$ offset by $1/g$

A quick outline of the conversion process - we have a forward transformation from my coordinates to Minkowskii coordinates, and a reverse transformation from Minkowskii coordinates to Peter's coordinates, so when we compose the two, after correctly compensating for the different origins, we get a conversion from my coordinates to Peter's. The result is rather ugly, hence the series expansion.

 

[SlowThinker]

So I have a fairly basic question. I'm stuck at page 11 of Carroll's Lectures on GR, that's why I don't know the answer It's incredibly dense with definitions and there are no explanations.

Once you have the metric, how do you, say, measure the length of the floor and ceiling, in the $y\ /\ \psi$ direction (as if using a ruler, not radar)? I think I could integrate at $\chi=0\text{ or }1/g, \tau=const$ to get the length of the floor, but how do you find the ceiling?

 

[PeterDonis]

Once you have the metric, how do you, say, measure the length of the floor and ceiling, in the $y\ /\ \psi$ direction (as if using a ruler, not radar)? I think I could integrate at $\chi=0\text{ or }1/g, \tau=const$ to get the length of the floor, but how do you find the ceiling?

First of all, lengths along the floor, or ceiling, or any other constant height in the rocket, are only along the $\psi$ direction in my chart, so you can only integrate at a constant $\chi$ (constant $\tau$ is assumed because you're trying to find length in a surface of constant time) in my chart to find lengths along the floor or any other constant height.

In pervect's chart, the floor/ceiling/any other constant height in the rocket curves, so (except at the center of the block/train) it has a component in the $\chi$ direction as well as the $\psi$ direction. So to find the length along, say, the floor of the rocket in pervect's metric, you would have to find the equation describing the floor of the rocket, and use it and the metric to find the correct integral.

Second, you can't "find" where the ceiling of the rocket is from the information presented so far. That's something that has to be put in by hand; you have to decide, as one of the inputs to the problem, where the ceiling of the rocket is.