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

[pervect]

I wanted to write a bit more about the goals of my analysis.

We start out with the premise that a metric is the best and ultimately the only thing needed to describe a coordinate system. This is semi-philosophical, the source of this idea for me is Misner's "Precis of General Relativity", http://arxiv.org/abs/gr-qc/9508043. So we start out with the goal of finding a metric, a metric which represents our desired coordinate system.

[add]
A useful background (which many but not all readers in the thread will already be familiar with) is the simpler problem of the coordinate system of an accelerated observer - frequently called "Rindler coordinates". A textbook discussion of this can be found in Misner, Thorne, Wheeler's "Gravitation", chapter 6 on "Accelerated observers", pg 163, a discucssion I won't attempt to duplicate. I'll only very briefely summarize it, first one works out the motion of an accelerating observer in Minkowskii space-time, before tackling the more difficult issue of how they might assign coordinates. The general approach I use is similar, with some differences noted below.

Now, there are lots of coordinate systems one could use. The starting point is that the space-time is Minkowskii space time, and thus we know that our desired coordinates from the "viewpoint of the block" will just be a re-parameterization of Minkowskii space-time. But we have some remaining goals to make the specific choice of coordinates represent "the viewpoint of the block" or at least "a viewpoint of the block".

The first goal is that the the spatial origin of the coordinate system (X=0, Y=0, Z=0) represent the center of the sliding block.

The second goal is that the T coordinate should represent the proper time of a clock at the center of the sliding block.

The third goal, which is somewhat a matter of preference, is that we want to have the metric coefficients independent of time. The space-time will always be stationary, the goal here is to make this stationary property explicit by making the metric coefficients independent of time.

While we would prefer to have the spatial axes of our coordinate system non-rotating, this conflicts with the third goal. . It also turns out that the third goal is much easier to achieve than the goal of findine non-rotating (i,e Fermi-normal) coordinates. Thus the textbook approach of Fermi-Walker transporting a triad of basis spatial vectors does not meet our goal, it leads to a different "viewpoint of the block". It also turns out that it's just plain easier to accomplish the third goal of time independent metric coefficients than it is to accomplish the non-rotating goal.

The fourth goal is to make a spatial slice of constant time (dT=0) have the usual Euclidean metric, dX^2 + dY^2 + dZ^2.

Given that we know that the T coordinate is the proper time of the center of the block, we need a simultaneity convention to determine the T coordinate of other points. Because we know that the block is rotating, we adopt the usual approach, the same one we use on the rotating Earth, where we imagine Einstein synchronizing all the clocks at rest in a non-rotating coordinate system to determine the T coordinate. This choice is a result of the fact that it's just not possible to Einstein-synchronize the clocks at rest in a rotating coordinate system, it's the standard approach to creating time coordinates in a rotating coordinate system.

I believe these motivations are sufficient to specify a unique metric, though I won't guarantee it. The reaming issue (besides ensuring uniqueness) is to make sure that all of the above goals are actually met, that there aren't any errors in the calculations or typos or other errors.

I'm not going to rehash the calculations in detail - they're pretty complex and have been previously posted, though unfortunately scatterered and not well-orgainzied. Knowing the end goals and approach might make them easier to follow.

[add]I'll thrown in a few links, though. I start by discussing the motion of the block in Minkowskii coordinates, in the following thread. (Some of the previous posts in the thread might be helpful, too).

https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/#post-4445483

Based on these results, I come up with a set of transformations from Minkowskii coordinates to the "block" coordinates (T,X,Y,Z) that should meet all of the above goals, and work out the resulting metric. The goals underlying the set of transformations I came up with are explained here, owever, and not explained in the original post.

https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/page-2#post-4466113

 

[PeterDonis]

Just to round off another item, I want to look at the behavior of a marble released from the train at some instant, in the fixed inertial frame I used above. The marble's 4-velocity at the instant it is released will be the same as that of the train at the same instant, i.e., it will be $(\gamma_0 (\tau_m) \gamma, \gamma_0 (\tau_m) v_0 (\tau_m) \gamma, \gamma v)$, where $\tau_m$ is the proper time (by the train's clock) when the marble is released. Since the marble is inertial, these 4-velocity components will be constant, so we can write $\gamma_0 (\tau_m) = \Gamma$ and $v_0 (\tau_m) = V$, and we have

$$
U_m = (\Gamma \gamma, \Gamma V \gamma, \gamma v)
$$

as the constant 4-velocity of the marble. The ordinary velocity of the marble in the $Y$ direction is then $\gamma v / \Gamma \gamma = v / \Gamma$, and is constant, whereas the ordinary velocity $v / \gamma_0$ of the train decreases over time, as noted already. So the marble will indeed "get ahead" of the train in the $Y$ direction over time.

If we look at the same 4-velocity in the Rindler chart, we have

$$
u_m = \left[ \Gamma \gamma \gamma_0 \left( 1 - v_0 V \right), - \Gamma \gamma \gamma_0 \left( v_0 - V \right), \gamma v \right]
$$

where $x$ is the instantaneous $x$ coordinate of the marble in the Rindler chart, and will decrease with time, since, as the above makes clear, the marble is moving in the $- x$ direction in this chart. In fact, the ordinary velocity of the marble in the $x$ direction is

$$
v_x = \frac{u_x}{u_t} = - \frac{v_0 - V}{1 - v_0 V}
$$

which is just the relativistic velocity subtraction formula, i.e., $v_x$ is exactly what we expect for the relative velocity of the marble and the rocket in the $x$ direction.

The ordinary velocity of the marbie in the $y$ direction in this chart is

$$
v_y = \frac{u_y}{u_t} = \frac{v}{\Gamma \gamma_0 \left( 1 - v_0 V \right)} = \frac{v}{\Gamma \left( 1 - V \right) \gamma_0 \left( 1 - v_0 \right)} = \frac{v}{D D_0}
$$

where $D = \sqrt{(1 - V) / (1 + V)}$ and $D_0 = \sqrt{(1 - v_0) / (1 + v_0)}$ are the Doppler "redshift factors" for the marble and the rocket, respectively, in the fixed inertial frame. Since $D$ is constant and $D_0 \rightarrow 0$ over time, this shows that $v_y$ increases over time in the Rindler chart, indicating, once again, that the marble gets ahead of the train over time.

 

[SlowThinker]

Thanks Peter and Pervect for your efforts. Sadly,
1. We have reached a point where I'm having trouble following the math, which may be why
2. I'm not entirely convinced that the analysis is correct. Where did Pervect's rotation and curved floor go?

I can only say that now I see why it took the brightest mind of 20th century 10 years to figure this all out

For the past few days I'm trying to use a different approach to measure the geometry of the train, similar to the standard light clock, but adapted to work under constant proper acceleration. It really makes my brain boil, since there is nothing I can rely on: straight lines, angles, simultaneity, height of the train, everything can change. In fact it seems I'll have to incorporate free-falling radioactive clock in the scheme.

Which brings me to this question, hopefully with a simple answer.
On the track where the train is moving, we put marks that are spaced regularly.
The passengers have with them a radioactive material that decays to half between every 2 marks.
Will this condition hold, as the rocket accelerates?

 

[SlowThinker]

Can the result with falling marble be transformed back to gravity field?
Lets say that the planet is still flat and the field is homogeneous at the surface, but gets weaker with altitude.
(Probably the easiest answer would be that a field with these properties is impossible...)
If it's possible to some reasonable accuracy, let's put an observer (Mr. Marble) at high altitude, watching the train.
Will he see the train eventually stop, or at least slow down?

 

[PeterDonis]

Can the result with falling marble be transformed back to gravity field?

No, at least not over a long enough period of time. When spacetime is curved, the long-term trajectories of all of the objects in the scenario will be different.

Lets say that the planet is still flat and the field is homogeneous at the surface, but gets weaker with altitude.
(Probably the easiest answer would be that a field with these properties is impossible...)

The "easiest answer" is the right one. There is no solution in GR that describes a flat planet. Planets (and other gravitating bodies) in GR are spherically symmetric, or roughly so (if they are rotating they won't be exactly spherical, but oblate).

If it's possible to some reasonable accuracy, let's put an observer (Mr. Marble) at high altitude, watching the train.
Will he see the train eventually stop, or at least slow down?

If you want to set up a scenario as close to the flat spacetime one as possible, the marble should not be at some high altitude; it should be released from the train into free fall, so its altitude starts off the same as the train's. For short times after that, its trajectory will look like the flat spacetime one: it will appear, to someone at rest in the train, that the marble is falling downward with acceleration $g \gamma^2$, and initially, it will not "gain" on the train.

Over a longer time, however, the marble's trajectory--and the train's, for that matter--will be curved by the planet's gravity. The first question is: is the speed of the marble and the train greater than the escape velocity from the planet at that altitude? Escape velocity from the Earth's surface is only 11 km/s, or about 1/30,000 of the speed of light, so if we want the train's and marble's speeds to be relativistic relative to the planet, either they will quickly escape (if we use the Earth), meaning we aren't really seeing any effects of curved spacetime, or we need a "planet" which is much larger and has a much higher escape velocity at the altitude of the train--which probably means a supermassive black hole and an altitude fairly close to the horizon. However, if we are fairly close to the horizon, there are no free-fall orbits that don't fall into the hole, so once we release the marble, it will fairly quickly disappear below the hole's horizon, and we can't run a long-term experiment anyway.

 

[PeterDonis]

On the track where the train is moving, we put marks that are spaced regularly.
The passengers have with them a radioactive material that decays to half between every 2 marks.
Will this condition hold, as the rocket accelerates?

Yes. The time dilation factor of the train relative to the rocket is constant; all of the weird effects we have been talking about are relative to a fixed inertial frame, in which both the rocket and the train are accelerating.

 

[PeterDonis]

Based on these results, I come up with a set of transformations from Minkowskii coordinates to the "block" coordinates (T,X,Y,Z) that should meet all of the above goals, and work out the resulting metric.

I'm going to try to do this using the notation I used in my previous posts. We start with the 4-velocity of a point on the floor of the train, which in Minkowski coordinates is

$$
U = \left[ \gamma \cosh \left( g \gamma \tau \right), \gamma \sinh \left( g \gamma \tau \right), \gamma v \right]
$$

where $\tau$ is the proper time along the worldline of the point on train with 4-velocity $U$. We could rewrite this as a function of the coordinate time $T$, but it's easier just to leave everything as a function of $\tau$ and then ask how we would need to transform the coordinates to make the 4-velocity be $u = (1, 0, 0)$, the timelike basis vector of the new chart. (Note that $\gamma$ and $v$ are constants, and that we are leaving out the $Z$ coordinate since nothing changes in that direction.)

Actually, what's easier is to ask what the Minkowski coordinates would have to be as functions of the "train frame" coordinates in order to transform $u$ into $U$. In other words, we want

$$
\partial_{\tau} = \gamma \cosh \left( g \gamma \tau \right) \partial_T + \gamma \sinh \left( g \gamma \tau \right) \partial_X + \gamma v \partial_Y
$$

We also want the second coordinate $\chi$ to have a basis vector that points in the direction of proper acceleration, i.e., we want

$$
\partial_{\chi} = \frac{1}{g \gamma^2} a = \sinh \left( g \gamma \tau \right) \partial_T + \cosh \left( g \gamma \tau \right) \partial_X
$$

Finally, we want the third coordinate $\psi$ to have a basis vector that is orthogonal to both of the above. Writing the three components of this basis vector in Minkowski coordinates as $a$, $b$, and $c$, the two orthogonality conditions are:

$$
a \gamma \cosh \left( g \gamma \tau \right) - b \gamma \sinh \left( g \gamma \tau \right) - c \gamma v = 0
$$

$$
a \sinh \left( g \gamma \tau \right) - b \cosh \left( g \gamma \tau \right) = 0
$$

Using the second condition to eliminate $b$ in the first gives

$$
a \gamma \cosh \left( g \gamma \tau \right) - a \tanh \left( g \gamma \tau \right) \gamma \sinh \left( g \gamma \tau \right) - c \gamma v = 0
$$

This gives $c$ in terms of $a$; we then must have that the vector is a spacelike unit vector, so

$$
a^2 - b^2 - c^2 = - 1
$$

Substituting from the above gives

$$
a^2 \left[ 1 - \tanh^2 \left( g \gamma \tau \right) - \frac{1}{v^2 \cosh^2 \left( g \gamma \tau \right)} \right] = -1
$$

Putting the above all together gives us solutions for $a$, $b$, and $c$, which gives us our third basis vector:

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

It is straightforward to check that each of these is a unit vector (one timelike and two spacelike), and that they are all mutually orthogonal.

I'll do the next step of deriving the form of the metric in the new chart in a follow-up post. (I originally had that in this post, but I need to check it so I've deleted it for now.)

 

[PeterDonis]

[Edit: the metric near the end of this post is not entirely correct; see post #79 for the correct line element.]

This is a continuation of my previous post, where we found three basis vectors, which I will re-label here:

$$
\hat{e}_0 = \gamma \cosh \left( g \gamma \tau \right) \partial_T + \gamma \sinh \left( g \gamma \tau \right) \partial_X + \gamma v \partial_Y
$$

$$
\hat{e}_1 = \sinh \left( g \gamma \tau \right) \partial_T + \cosh \left( g \gamma \tau \right) \partial_X
$$

$$
\hat{e}_2 = \gamma v \cosh \left( g \gamma \tau \right) \partial_T + \gamma v \sinh \left( g \gamma \tau \right) \partial_X + \gamma \partial_Y
$$

The re-labeling is because we should not necessarily expect these three basis vectors, which describe the frame field of "train observers" as seen from the viewpoint of a fixed inertial frame, to be "pure" coordinate partial derivatives $\partial_{\tau}$, $\partial_{\chi}$, $\partial_{\psi}$ everywhere. In particular, by analogy with Rindler coordinates, we should not expect $\hat{e}_0 = \partial_{\tau}$ everywhere; at the very least, we would expect $\hat{e}_0 = \left( 1 / g \chi \right) \partial_{\tau}$ (though, as we'll see, that is not a complete formula for $\hat{e}_0$).

However, we have another issue to deal with. The above formulas were derived, strictly speaking, for a "reference" observer who is at $\chi = 1 / g$, $\psi = 0$ in the train coordinates. But we want general formulas that are valid everywhere in the train chart. So we need to figure out how things vary with $\chi$ and $\psi$ and generalize the above formulas accordingly.

Simply applying a boost with velocity $v$ in the $y$ direction of Rindler coordinates gives us the ansatz:

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

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

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

Computing $\partial_{\tau}$, $\partial_{\chi}$, and $\partial_{\psi}$ and normalizing gives us the following:

$$
\hat{e}_0 = \frac{1}{\gamma \sqrt{g^2 \chi^2 - v^2}} \partial_{\tau} = \frac{g \chi}{\sqrt{g^2 \chi^2 - v^2}} \cosh \left[ g \gamma \left( \tau + v \psi \right) \right] \partial_T + \frac{g \chi}{\sqrt{g^2 \chi^2 - v^2}} \sinh \left[ g \gamma \left( \tau + v \psi \right) \right] \partial_X + \frac{v}{\sqrt{g^2 \chi^2 - v^2}} \partial_Y
$$

$$
\hat{e}_1 = \partial_{\chi} = \sinh \left[ g \gamma \left( \tau + v \psi \right) \right] \partial_T + \cosh \left[ g \gamma \left( \tau + v \psi \right) \right] \partial_X
$$

$$
\hat{e}_2 = \frac{1}{\gamma \sqrt{1 - g^2 \chi^2 v^2}} \partial_{\psi} = \frac{g \chi v}{\sqrt{1 - g^2 \chi^2 v^2}} \cosh \left[ g \gamma \left( \tau + v \psi \right) \right] \partial_T + \frac{g \chi v}{\sqrt{1 - g^2 \chi^2 v^2}} \sinh \left[ g \gamma \left( \tau + v \psi \right) \right] \partial_X + \frac{1}{\sqrt{1 - g^2 \chi^2 v^2}} \partial_Y
$$

For $v = 0$, $\gamma = 1$, this reduces to the standard frame field for Rindler observers. Also, for $\chi = 1 / g$, $\psi = 0$, it reduces to the expressions we derived previously. So it looks like we're on the right track.

Now we compute the coordinate differentials to find an expression for the metric in this chart:

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

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

$$
dY = \gamma v d\tau + \gamma d\psi
$$

Plugging these into the Minkowski metric gives, after some algebra (and after putting back the $z$ coordinate for completeness) [Edit: this is not correct, see post #79 for the correct form]:

$$
ds^2 = - \gamma^2 \left( g^2 \chi^2 - v^2 \right) d\tau^2 + 2 v \gamma^2 \left( 1 - g^2 \chi^2 \right) d\tau d\psi + d\psi^2 + d\chi^2 + dz^2
$$

This makes sense, first of all, because the factor in front of $g_{tt}$ exactly matches the factor in front of $\partial_{\tau}$ in the expression for $\hat{e}_0$ above, verifying that $\hat{e}_0$ is indeed the correct 4-velocity of an observer at rest in this chart. Second, for $v = 0$, $\gamma = 1$, this metric reduces to the Rindler metric, as it should. Third, the purely spatial part of the metric is Euclidean [Edit: no, it isn't; see post #79], as we had hoped.

However, this metric does not look like the one pervect derived; it only has one "cross term", not two. I'll save further comment on what that means, and in general on the physical interpretation of this metric and the associated basis vectors, for a follow-up post.

 

[SlowThinker]

Clearly you are putting a lot of time into this, so I'm trying to follow the math.

We also want the second coordinate $\chi$ to have a basis vector that points in the direction of proper acceleration, i.e., we want

$$
\partial_{\chi} = \frac{1}{g \gamma^2}\mathbf{a}= \sinh \left( g \gamma \tau \right) \partial_T + \cosh \left( g \gamma \tau \right) \partial_X
$$

Finally, we want the third coordinate $\psi$ to have a basis vector that is orthogonal to both of the above. Writing the three components of this basis vector in Minkowski coordinates as $a$, $b$, and $c$, the two orthogonality conditions are:

$$
\mathbf{a} \gamma \cosh \left( g \gamma \tau \right) - b \gamma \sinh \left( g \gamma \tau \right) - c \gamma v = 0
$$

That's nasty, switching the meaning of "a" like that

Simply applying a boost with velocity $v$ in the $y$ direction of Rindler coordinates gives us the ansatz:
$$T = \chi \sinh \left[ g \gamma \left( \tau + v \psi \right) \right]$$
$$X = \chi \cosh \left[ g \gamma \left( \tau + v \psi \right) \right]$$
$$Y = \gamma v \tau + \gamma \psi$$

Where does this come from? Is it pulled out of thin air? Anyway, the expression for Y seems to be somewhat asymmetric compared to the other two, should not it read
$$Y=\gamma\tau+\gamma v\psi \text{ }\,\,\,\,\,\text{?}$$

Computing $\partial_{\tau}$, $\partial_{\chi}$, and $\partial_{\psi}$ and normalizing gives us the following:

$$\hat{e}_0 = \frac{1}{\gamma \sqrt{g^2 \chi^2 - v^2}} \partial_{\tau} = ...$$
$$\hat{e}_1 = \partial_{\chi} = ...$$
$$\hat{e}_2 = \frac{1}{\gamma \sqrt{1 - g^2 \chi^2 v^2}} \partial_{\psi} = ...$$

What is being normalized here? This is a bit over my head

 

[SlowThinker]

It seems that $\hat e_0$ and $\hat e_1$ are being normalized to $|\hat e_0|=1$ and $|\hat e_1|=-1$, but $\hat e_2$ does not quite fit.
Plus, is it even legal to neglect the $\partial_T$ etc. during the normalization?

 

[PeterDonis]

Where does this come from?

As I said, it is taking Rindler coordinates and boosting them in the $y$ direction with velocity $v$. In other words, we take the coordinates $t, x, y, z$ as defined here:

https://en.wikipedia.org/wiki/Rindler_coordinates#Relation_to_Cartesian_chart

Then we use $t = \gamma \left( \tau + v \psi \right)$ and $y = \gamma \left( \psi + v \tau \right)$ to define $\tau$ and $\psi$ (note that this is just a Lorentz transformation), and substitute into the equations given on the Wikipedia page above for $T$, $X$, and $Y$ in terms of $t$, $x$, and $y$ ($z$ is unchanged).

What is being normalized here?

The idea is to get unit vectors in the directions of $\partial_{\tau}$, $\partial_{\chi}$, and $\partial_{\psi}$; we do that by dividing each of those vectors by their norms, which is where the term "normalized" comes from.

It seems that $\hat e_0$ and $\hat e_1$ are being normalized to $|\hat e_0|=1$ and $|\hat e_1|=-1$, but $\hat e_2$ does not quite fit.

You're right, I had left out a factor of $g \chi$ in the first two components of $\hat{e}_2$. Good catch. Fixed now.

 

[PeterDonis]

is it even legal to neglect the $\partial_T$ etc. during the normalization?

I'm not sure what you mean by this, but perhaps you aren't familiar with the notation I'm using, which takes advantage of the fact that there is a one-to-one correspondence between vectors and directional derivatives. So when I write $\partial_T$, that is best viewed not as a partial derivative with respect to $T$, but as a vector that points in the positive $T$ direction. (In this particular case, it's a unit vector, because all of the metric coefficients in the Minkowski metric are $\pm 1$. But in general, as the examples of $\partial_{\tau}$ and the other two vectors in the train chart show, simple vectors of the form $\partial_x$, where $x$ is some coordinate, are not necessarily unit vectors.) The expressions for $\hat{e}_0$ and the other two basis vectors are simply forming a new set of basis vectors by taking linear combinations of the basis vectors in the Minkowski chart ($\partial_T$ and friends).

 

[PeterDonis]

the expression for Y seems to be somewhat asymmetric compared to the other two, should not it read

Y=γτ+γvψ ?​

No. Think of how $t$ and $y$ would transform to $t'$ and $y'$ in a Lorentz transformation in the $y$ direction. There is indeed a symmetry involved, but it's not quite the one you appear to be implicitly assuming should be there.

 

[SlowThinker]

I'm not sure what you mean by this, but perhaps you aren't familiar with the notation I'm using, which takes advantage of the fact that there is a one-to-one correspondence between vectors and directional derivatives. So when I write $\partial_T$, that is best viewed not as a partial derivative with respect to $T$, but as a vector that points in the positive $T$ direction. (In this particular case, it's a unit vector, because all of the metric coefficients in the Minkowski metric are $\pm 1$. But in general, as the examples of $\partial_{\tau}$ and the other two vectors in the train chart show, simple vectors of the form $\partial_x$, where $x$ is some coordinate, are not necessarily unit vectors.) The expressions for $\hat{e}_0$ and the other two basis vectors are simply forming a new set of basis vectors by taking linear combinations of the basis vectors in the Minkowski chart ($\partial_T$ and friends).

Indeed this is the first time I see partial derivatives of coordinates. What puzzles me, if the $\partial_T$ is not 1 (and it probably won't be), how can the $\hat e_0$ be normalized? It looks as if the normalization expects all the derivatives to be $\partial_T=\partial_X=\partial_Y=1$. If you say it's correct, I guess I'll just play along.
Edit: Maybe $\hat e_0$ can be seen as the first row in some matrix $\hat E$? You get $(\partial_\tau, \partial_\chi, \partial_\psi)^T=\hat E \cdot (\partial_T, \partial_X, \partial_Y)^T$. It looks fishy to me but at least I can see why you want the rows of $\hat E$ to have unit length.

 

[PeterDonis]

this is the first time I see partial derivatives of coordinates.

Then I would strongly recommend taking some time to learn differential geometry, at least to the extent that it is used in physics. What I did is not intended to be obvious without that background knowledge. Even Einstein had to spend several years being taught differential geometry by his friend Marcel Grossman before he could come up with General Relativity.

I learned differential geometry from Misner, Thorne, and Wheeler, but that may not be the best source. Carroll's online lecture notes on GR have a good discussion of it in Chapter 2:

http://arxiv.org/abs/gr-qc/9712019

He talks about the equivalence between vectors and directional derivatives, which is the key thing I'm using here.

It looks as if the normalization expects all the derivatives to be $\partial_T=\partial_X=\partial_Y=1$.

No, they're not the number $1$. They are unit vectors--vectors of length $1$ (more precisely, of squared length $\pm 1$), but in different directions. So when we form linear combinations of them, we are doing vector addition, not number addition.

For a simpler example, consider a two-dimensional Euclidean plane. In standard Cartesian coordinates $x, y$, the basis vectors are $\partial_x$ and $\partial_y$; these are unit vectors in the $x$ and $y$ directions. The metric in this chart is

$$
ds^2 = dx^2 + dy^2
$$

Now suppose we form the vector

$$
v = \frac{1}{\sqrt{2}} \left( \partial_x + \partial_y \right)
$$

The length of this vector is

$$
| v | = \sqrt{\left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2} = 1
$$

so it is a unit vector, but pointing in a different direction, making an angle of 45 degrees with both $\partial_x$ and $\partial_y$.

 

[pervect]

I'm going to try to do this using the notation I used in my previous posts. We start with the 4-velocity of a point on the floor of the train, which in Minkowski coordinates is

$$
U = \left[ \gamma \cosh \left( g \gamma \tau \right), \gamma \sinh \left( g \gamma \tau \right), \gamma v \right]
$$

where $\tau$ is the proper time along the worldline of the point on train with 4-velocity $U$. We could rewrite this as a function of the coordinate time $T$, but it's easier just to leave everything as a function of $\tau$ and then ask how we would need to transform the coordinates to make the 4-velocity be $u = (1, 0, 0)$, the timelike basis vector of the new chart. (Note that $\gamma$ and $v$ are constants, and that we are leaving out the $Z$ coordinate since nothing changes in that direction.)

Reviewing my previous posts, I don't seem to be getting the same four-velocity U. Could you quickly review your notation for me? I'll try to translate my notation (which wasn't well documented) to yours to compare the results. We have three coordinate systems of some interest.

1) Minkowskii coordinates. I used (t,x,y,z) for these. What symbols did you use?

2) Rindler coordinates. These would be the coordinates in the rocket frame, an intermediate set of coordinates. I didn't really have a notation for them, an unfortunate oversight.

3) Block coordinates (the co-ordinates we are finding the metric in, the ones that are supposed to represent the viewpoint of the block.) We can regard the metric we find as a concise definition of the block coordinates.

Additionally, the acceleration of the rocket and rocket floor is in the "x" direction, the motion of the block relative to the floor is in the "y" direction, and we can pretty much ignore z?

Also is "v" the coordinate velocity of the block relative to the floor (in the Rindler frame) and can we assume c=1 and gamma = 1 / sqrt(1-v^2)?

 

[PeterDonis]

Maybe $\hat e_0$ can be seen as the first row in some matrix $\hat E$?

No. Vectors aren't matrices. They can be treated as one-dimensional arrays of components, if we have picked a particular coordinate chart; but their components will change if we change charts.

In the simple example I gave in my last post, the vector $v$ has components $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$ in the Cartesian chart. But if we switch to polar coordinates, the same vector $v$ now has components $\left( 1, 0 \right)$, since the coordinate basis vectors are now $\partial_r$ and $\partial_{\theta}$, and $v$ points in a purely radial direction.

 

[PeterDonis]

Reviewing my previous posts, I don't seem to be getting the same four-velocity U.

The one you quoted was preliminary only, it's not valid everywhere. The correct 4-velocity is the vector $\hat{e}_0$ in post #68. Note that as I expressed it there, it gives components in both the train (or sliding block) chart and the Minkowski chart, but expressed as linear combinations of the basis vectors, with components as functions of the train chart coordinates.

1) Minkowskii coordinates. I used (t,x,y,z) for these. What symbols did you use?

I used $T, X, Y, Z$, because that's the notation used in the Wikipedia page on Rindler coordinates.

2) Rindler coordinates. These would be the coordinates in the rocket frame, an intermediate set of coordinates. I didn't really have a notation for them, an unfortunate oversight.

I only mention these a couple of times, but where I did, I used $t, x, y, z$, again because that's the notation used in the Wikipedia page on Rindler coordinates.

3) Block coordinates (the co-ordinates we are finding the metric in, the ones that are supposed to represent the viewpoint of the block.) We can regard the metric we find as a concise definition of the block coordinates.

I used $\tau, \chi, \psi, z$, to avoid any possible confusion with Rindler coordinates.

Additionally, the acceleration of the rocket and rocket floor is in the "x" direction, the motion of the block relative to the floor is in the "y" direction, and we can pretty much ignore z?

Yes, for both Rindler coordinates (lower-case x, y, z in my notation) and Minkowski coordinates (capital X, Y, Z in my notation).

Also is "v" the coordinate velocity of the block relative to the floor (in the Rindler frame) and can we assume c=1 and gamma = 1 / sqrt(1-v^2)?

Yes.

Edit: I should also add that I put the floor of the rocket/train at $\chi = 1 / g$, whereas you are putting it at your $X = 0$. So my metric wouldn't be exactly the same as yours; mine is a kind of Rindler-like chart for the train, whereas yours is a kind of not-quite-Fermi normal chart (since you don't Fermi-Walker transport all your basis vectors). So they should be closely related, but not necessarily identical.

 

[PeterDonis]

Plugging these into the Minkowski metric gives, after some algebra

I just realized that I did some of the algebra wrong. The coefficient of $d\psi^2$ can't be 1, because the vector $\partial_{\psi}$ is not a unit vector; as we can read off from the definition of $\hat{e}_2$, the norm of $\partial_{\psi}$ is $\gamma \sqrt{1 - g^2 \chi^2 v^2}$ (and we can check this by direct computation from the coordinate transformation given). So the coefficient of $d\psi^2$ in the metric should be the square of that, and re-checking my algebra I find that it is. So the correct line element is

$$
ds^2 = - \gamma^2 \left( g^2 \chi^2 - v^2 \right) d\tau^2 + 2 v \gamma^2 \left( 1 - g^2 \chi^2 \right) d\tau d\psi + \gamma^2 \left( 1 - g^2 \chi^2 v^2 \right) d\psi^2 + d\chi^2 + dz^2
$$

So the spatial part of the metric is not, in fact, Euclidean; it is curved in the $\psi$ direction. I'll leave further comment to a follow-up post.

 

[PeterDonis]

Ok, I've been promising a follow-up on the physical meaning of the metric I derived, so here it is.

First, let's invert the coordinate transformation to get the "train frame" coordinates in terms of the Minkowski coordinates. This turns out to be:

$$
\tau = \gamma \left( \frac{1}{g} \tanh^{-1} \frac{T}{X} - v Y \right)
$$

$$
\chi = \sqrt{X^2 - T^2}
$$

$$
\psi = \gamma \left( Y - \frac{v}{g} \tanh^{-1} \frac{T}{X} \right)
$$

We note that $\chi$ is the same as the Rindler $x$ coordinate; it corresponds to the Minkowski $X$ coordinate of a train observer's worldline when $T = 0$, i.e., when the rocket is at rest in the Minkowski frame. However, $\tau$ and $\psi$ are more complicated; we can see from the above that they are linear combinations of the Rindler $t$ and $y$ coordinates, of the sort we would expect from a Lorentz boost in the $y$ direction with velocity $v$.

Next, let's look at the three metric coefficients $g_{\tau \tau}$, $g_{\tau \psi}$, and $g_{\psi \psi}$. All three of these go to zero at particular values of $\chi$, i.e., at particular altitudes. For $g_{\tau \tau}$, this value is

$$
\chi = \frac{v}{g}
$$

This is below the floor of the rocket/train (fortunately!), which is at $\chi = 1 / g$, since $v < 1$. However, it is not at $\chi = 0$, which is where the Rindler horizon is in standard Rindler coordinates. In other words, the "Rindler horizon" for the train observers (I put it in quote because they are not standard Rindler observers) is a shorter distance below them than the Rindler horizon is for the rocket observers.

For $g_{\tau \psi}$, the value at which it vanishes is

$$
\chi = \frac{1}{g}
$$

In other words, on the floor of the rocket/train, the metric is orthogonal; non-orthogonality only appears above or below that altitude. What's more, if we plug $\chi = 1 / g$ into $g_{tt}$ and $g_{\psi \psi}$, we find that those metric coefficients become $-1$ and $1$, respectively. So on the floor of the rocket/train, this metric is actually the Minkowski metric. This tells us that our metric is indeed similar to Fermi normal coordinates for a train observer on the floor of the train ("similar" because the spatial origin is not on the worldline--the worldline is at $\chi = 1 / g$ instead of $\chi = 0$).

For $g_{\psi \psi}$, the value at which it vanishes is

$$
\chi = \frac{1}{v g}
$$

This tells us that at some altitude above the floor of the rocket/train, the vector $\partial_{\psi}$ becomes null instead of spacelike. This is actually to be expected at some point because of the non-orthogonality of $\partial_{\tau}$ and $\partial_{\psi}$. But we don't really have enough information at this point to understand what it means, physically.

To better understand what is going on, it is useful to compute the kinematic decomposition of the congruence describing the train observers. I'll do that in yet another follow-up post.

 

[PeterDonis]

As the next follow-up, I'll compute the kinematic decomposition of the "train observer" congruence. It will be useful to re-express the 4-velocity $\hat{e}_0$ purely in terms of Minkowski coordinates. This will simplify computing covariant derivatives since all of the connection coefficients are zero in the Minkowski chart.

We have the inverse coordinate transformation from my previous post, but it will help to have the following additional formulas, which are straightforward to derive:

$$
\cosh \left[ g \gamma \left( \tau + v \psi \right) \right] = \frac{X}{\sqrt{X^2 - T^2}}
$$

$$
\sinh \left[ g \gamma \left( \tau + v \psi \right) \right] = \frac{T}{\sqrt{X^2 - T^2}}
$$

Armed with all that, we can rewrite the 4-velocity as follows:

$$
\hat{e}_0 = U = \frac{1}{\sqrt{g^2 \left( X^2 - T^2 \right) - v^2}} \left( g X \partial_T + g T \partial_X + v \partial_Y \right)
$$

As a first check, let's compute the proper acceleration of this; it is $A^a = U^b \partial_b U^a$, which in components gives, after some algebra:

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

If we refer back to the unit basis vectors, we will see that we have

$$
A = \frac{g^2 \chi}{g^2 \chi^2 - v^2} \hat{e}_1
$$

So the proper acceleration does indeed always point in the $\chi$ direction, as expected. (Also, for $\chi = 1 / g$, the magnitude of $A$ is $g \gamma^2$, which is consistent with our previous results.)

Next, we compute the tensor $A_a U_b + \partial_b U_a$, which will give us the rest of the kinematic decomposition. Note that the indexes are lowered; that means that the $T$ components of vectors will have their signs flipped from what is written above (but the product $A_T U_T$ will have two minus signs compared to $A^T U^T$, so the sign flips cancel). Since all quantities are functions of $T$ and $X$ only (and the $Y$ component of $U$ is constant), we will have the six potentially nonzero components. After computation, the results are:

$$
A_T U_T + \partial_T U_T = 0
$$

$$
A_T U_X + \partial_X U_T = \frac{g v^2}{\left( g^2 \left( X^2 - T^2 \right) - v^2 \right)^{3/2}}
$$

$$
A_T U_Y = - \frac{g^2 v T}{g^2 \left( X^2 - T^2 \right) - v^2}
$$

$$
A_X U_T + \partial_T U_X = - \frac{g v^2}{\left( g^2 \left( X^2 - T^2 \right) - v^2 \right)^{3/2}}
$$

$$
A_X U_X + \partial_X U_X = 0
$$

$$
A_X U_Y = \frac{g^2 v X}{g^2 \left( X^2 - T^2 \right) - v^2}
$$

There are no nonzero diagonal components, so the expansion scalar is zero. The symmetric and antisymmetric parts give the shear $\sigma$ and vorticity $\omega$:

$$
\sigma_{TY} = \sigma_{YT} = - \frac{1}{2} \frac{g^2 v T}{g^2 \left( X^2 - T^2 \right) - v^2}
$$

$$
\sigma_{XY} = \sigma_{YX} = \frac{1}{2} \frac{g^2 v X}{g^2 \left( X^2 - T^2 \right) - v^2}
$$

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

$$
\omega_{TY} = - \omega_{YT} = - \frac{1}{2} \frac{g^2 v T}{g^2 \left( X^2 - T^2 \right) - v^2}
$$

$$
\omega_{XY} = - \omega_{YX} = \frac{1}{2} \frac{g^2 v X}{g^2 \left( X^2 - T^2 \right) - v^2}
$$

So this congruence has both nonzero shear and nonzero vorticity (unlike the Rindler congruence, for which expansion, shear, and vorticity all vanish). I'll comment further on those in one more follow-up post.

 

[andrewkirk]

I have been away from this thread for a while. as I've been trying to work through it myself. So I'm hopelessly out of date with what's been written here. Hence the following question may be nonsensical or out of context, but I'll ask it anyway. It's about this:

it is taking Rindler coordinates and boosting them in the $y$ direction with velocity $v$. In other words, we take the coordinates $t, x, y, z$ as defined here:

https://en.wikipedia.org/wiki/Rindler_coordinates#Relation_to_Cartesian_chart

Then we use $t = \gamma \left( \tau + v \psi \right)$ and $y = \gamma \left( \psi + v \tau \right)$ to define $\tau$ and $\psi$ (note that this is just a Lorentz transformation)

My understanding is that a Lorentz transformation is defined, and has the quoted form, for a transformation between two inertial frames. A frame of Rindler coordinates is not inertial. Doesn't that disqualify it from being able to handle a velocity boost with just a Lorentz transformation?

 

[PeterDonis]

My understanding is that a Lorentz transformation is defined, and has the quoted form, for a transformation between two inertial frames. A frame of Rindler coordinates is not inertial. Doesn't that disqualify it from being able to handle a velocity boost with just a Lorentz transformation?

I was using sloppy terminology; I should have said that the transformation from $t, y$ to $\tau, \psi$ has the same form as a Lorentz transformation, if you just look at the transformation equation. But of course it isn't really a Lorentz transformation, as you say, since the metric in both charts is not the Minkowski metric (because they're not inertial coordinates). The point is that, heuristically, we can view the "train coordinates" as Rindler coordinates that have been "boosted" in the $y$ direction with velocity $v$; that is what motivates the coordinate transformation I gave. But my derivation certainly does not rest on the claim that the transformation is an actual Lorentz transformation between inertial frames; obviously it's not.

 

[andrewkirk]

I should have said that the transformation from $t, y$ to $\tau, \psi$ has the same form as a Lorentz transformation, if you just look at the transformation equation.

Can you please link to where you derived the transformation? Sorry to be a nuisance but this thread has grown so much, and contains so much maths that I've lost track of what's where.

In my own meanderings I had been pursuing the line of trying to get things in 'train coordinates' (by which I mean a coordinate system in which the centre of the train remains at the origin) by first getting the 'rocket coordinates' as Rindler coordinates and then applying a velocity boost. But I hit a brick wall when I realized I couldn't use the Lorentz transformation for that last step. It looked like I'd have to work out the transformation from first principles using reflected beams of light etc, and that looked like so much work that I left it to try another avenue.

Thank you.

 

[SlowThinker]

Can you please link to where you derived the transformation? Sorry to be a nuisance but this thread has grown so much, and contains so much maths that I've lost track of what's where.

From post #67 it follows quite straight forward, with a few references to page 3.
I have to admit I find some things hard to swallow, but Peter seems confident in those steps, so I'm just trying to follow.

I'm plowing through the Carroll's book, but it's not an easy read. After 11 pages of definitions, my morale is a bit low.

 

[PeterDonis]

Can you please link to where you derived the transformation? Sorry to be a nuisance but this thread has grown so much, and contains so much maths that I've lost track of what's where.

It's not a nuisance, I'm having trouble keeping track myself of all the stuff I've posted.

Post #67, which SlowThinker referred to, is the start of a derivation, but the basis vectors given there are only correct for $\chi = 1 / g, \psi = 0$ (which corresponds to a "reference observer" at rest on the floor of the train, who is at Minkowski coordinate $Y = 0$ at $T = 0$). The correct general basis vectors and the coordinate transformation are given in post #68, but the line element given there is not completely correct. Post #79 gives the correct line element. Post #80 gives the inverse coordinate transformation.

As far as "derivation" goes, I haven't by any means given a complete one; I've left out a lot of algebra, and in any case a coordinate transformation isn't really "derived", it's guessed, first, and then the consequences of the guess are worked out to see if they make sense and meet the original requirements. The key steps in my reasoning are:

(1) We want a transformation that makes the 4-velocity of a "train observer" (strictly speaking, one on the floor of the train--see item 4 below) look, heuristically, like the 4-velocity of a Rindler observer boosted in the $y$ direction with velocity $v$. The transformation in post #68, and its inverse in post #80, do that; that's what the computations in post #68 of the 4-velocity confirm (and the further computations in post #81, expressing the 4-velocity purely in terms of Minkowski coordinates, provide further confirmation).

(2) In the new coordinate chart, the 4-velocity of a "train observer" should be a unit timelike vector purely in the $\tau$ direction, i.e., it should be $\partial_{\tau} / \sqrt{- g_{\tau \tau}}$. This is obvious from the computations referred to above.

(3) In the new coordinate chart, the proper acceleration of a "train observer" should be purely in the $\chi$ direction, i.e., it should be $a \partial_{\chi}$, where $a$ is the magnitude of the proper acceleration. I verified this when I did the computations underlying post #68, but I didn't actually post a confirmation until post #81. Note that this requirement implies, since the 4-velocity and proper acceleration must be orthogonal, that the $\tau$ and $\chi$ coordinates are orthogonal (no $d\tau d\chi$ "cross terms" in the metric), which can easily be verified by looking at the line element in post #79.

(4) On the floor of the train, which corresponds to $\chi = 1 / g$, all of the stuff above should reduce to what was given in post #67--i.e,. the metric should be orthogonal (i.e., the $d\tau d\psi$ cross term should vanish), the 4-velocity should be simply $\partial_{\tau}$ (i.e., $\partial_{\tau}$ should be a unit vector, meaning that the $\tau$ coordinate measures proper time for observers on the floor of the train), the spatial part of the metric should be Euclidean (i.e., $\partial_{\psi}$ should also be a unit vector), and the form of $\partial_{\tau}$ and $\partial_{\psi}$ should make evident the "Lorentz boost" in the $y$ direction compared to Rindler coordinates. These can all be verified by looking at the posts referenced above.

So the coordinate transformation I guessed in post #68 turns out to work, in the sense that it meets the requirements above for a "train frame". But, as I noted in post #79, this chart does not have at least one property that pervect was trying to achieve: the spatial part of the metric is not Euclidean everywhere. (It is on the floor of the train, as noted above, but not elsewhere--by contrast, the spatial part of the metric in Rindler coordinates is Euclidean everywhere.) Also, other than on the floor of the train, the $\tau$ and $\psi$ coordinates are not orthogonal, which means that the basis vectors $\hat{e}_0$ and $\hat{e}_2$ do not exactly look like Rindler basis vectors "boosted" in the $y$ direction with velocity $v$.

 

[pervect]

I am still re-working through the problem - however, I don't see how the spatial slices can possibly be curved. To be curved, the spatial displacement vectors in a surface of constant coordinate time $\tau$ would have to not commute, i.e. $\hat{\chi} + \hat{\psi}$ would not be equal to $\hat{\psi} + \hat{\chi}$, similar to the way that going in a great circles on a sphere one mile east, followed by going one mile north, doesn't wind up at the same spot as going one mile north first, then one mile east second. But I believe we we (and MTW in similar derivations for Rindler coordinates) set out by assuming that the space was an affine space, where we could add displacement vectors linearly and without regard to order.

Additionally, I'd expect the off-diagional terms in a rotating metric to look like $d\phi \, dt$ in polar coordinates. Considering that $\phi = \arctan y/x$ in cartesian coordinates, then the cartesian equivalent of the $d\phi$ term is $({x\,dy - y\,dx}) / ({x^2+y^2})$, which has the form that I originally had but you objected to.

That said, your expression for the 4-velocity looks right to me, and I haven't been able to convince myself yet that it's the same as the expressio in my original posts, so I'm not convinced that they're right, either.

 

[PeterDonis]

I don't see how the spatial slices can possibly be curved. To be curved, the spatial displacement vectors in a surface of constant coordinate time $\tau$ would have to not commute, i.e. \hat{\chi} + $\hat{\psi}$ would not be equal to $\hat{\psi} + \hat{\chi}$, similar to the way that going in a great circles on a sphere one mile east, followed by going one mile north, doesn't wind up at the same spot as going one mile north first, then one mile east second

I agree this is a good test for curvature, so I'll see if I can compute it. I was inferring spatial curvature simply on the basis of the fact that $g_{\psi \psi}$ is not $1$, but it's possible that that is a coordinate artifact, so you're right that we should test it by computing invariants.

I believe we we (and MTW in similar derivations for Rindler coordinates) set out by assuming that the space was an affine space

I didn't assume that, at least I don't see that I did--except in the obvious sense that the underlying manifold is still Minkowski spacetime.

I'd expect the off-diagional terms in a rotating metric to look like $d\phi dt$ in polar coordinates.

If the rotation--by which we really mean nonzero vorticity of the congruence of observers at rest in the chart--were generated in the "usual" way, by observers following worldlines whose spatial projections were closed curves--more precisely, whose spatial projections were closed orbits of a spacelike Killing vector field--I would agree. But the rotation of this congruence is not being generated that way; the spatial projections of the worldlines in the congruence are not closed orbits of a spacelike KVF.

I admit this is just an intuitive, heuristic argument; but in any case, I have checked and re-checked the part of the computation that relates to a possible $d\tau d\chi$ cross term in the metric, and I am convinced that that term vanishes. Or, to put it another way, I am convinced that the vectors $\partial_{\tau}$ and $\partial_{\chi}$ are orthogonal everywhere; this is easy to verify from the forms of the basis vectors given in post #68, and as far as I can tell, everything else is consistent with those forms of the basis vectors.

 

[PeterDonis]

I have just run the metric I derived through Maxima and confirmed that its Riemann tensor vanishes (Maxima actually couldn't quite derive this on its own, I had to check some output by hand to confirm that terms cancelled), which is good because it means it actually does describe Minkowski spacetime! Also this gives the connection coefficients in the "train frame", which I will be using to check the computations I did in the Minkowski chart. For reference, here they are (note that I have not included some which are related to the below by symmetry in the lower indexes):

$$
\Gamma^{\tau}{}_{\tau \chi} = \frac{\gamma^2}{\chi}
$$

$$
\Gamma^{\tau}{}_{\chi \psi} = \frac{\gamma^2 v}{\chi}
$$

$$
\Gamma^{\chi}{}_{\tau \tau} = \gamma^2 g^2 \chi
$$

$$
\Gamma^{\chi}{}_{\tau \psi} = \gamma^2 v g^2 \chi
$$

$$
\Gamma^{\chi}{}_{\psi \psi} = - \gamma^2 v^2 g^2 \chi
$$

$$
\Gamma^{\psi}{}_{\tau \chi} = - \frac{\gamma^2 v}{\chi}
$$

$$
\Gamma^{\psi}{}_{\chi \psi} = - \frac{\gamma^2 v^2}{\chi}
$$

The first and third of these reduce to the familiar values for Rindler coordinates if we set $v = 0$ (and therefore $\gamma = 1$); the others all vanish, as they should. Also, the third can be used to check the proper acceleration of a train observer, whose 4-velocity is $\hat{e}_0 = (1 / \sqrt{-g_{\tau \tau}}) \partial_{\tau}$; this gives

$$
A = \frac{\Gamma^{\chi}{}_{\tau \tau}}{- g_{\tau \tau}} \partial_{\chi} = \frac{g^2 \chi}{g^2 \chi^2 - v^2} \partial_{\chi}
$$

which agrees with what was computed in post #81 (since $\partial_{\chi} = \hat{e}_1$).

 

[pervect]

I worked through the problem using Peter's notation.

I agree with Peter's 4-velocity.

A simple derivation. In Rindelr coordinates (t,x,y), we can write the motion of the sliding block as y=vt. We can say $t = \gamma \tau$. This gives us the 4-velocity in Rindler coodinates as

U = ($dt/d\tau, dx/d\tau, dy/d\tau$) = ($\gamma, 0, \gamma \, v$).

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

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

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

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

Peter's values for $(\hat{e}_i)$ in https://www.physicsforums.com/threads/gravity-in-einsteins-train.835994/page-4#post-5254985 check out. Thus we can write

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

The notation here is slightly different than Peter's (it is the same notation as used in MTW). Here $(e_i)^0$ is the $\partial_T$ component of the basis vector $e_i$ in Peter's notation, where i=0,1,2. Similarly $(e_i)^1$ is the $\partial_X$ comonent and $(e_i)^2$ is the $\partial_Y$ component of said basis vector.

[edit for consistenh]
If this seems daunting, it's not. In vector notation we are just saying that the position at time $\tau$ is $\vec{P_0 }+ \chi \vec{e_\chi} + \psi \vec{e_\psi}$, where $\vec{P_0}$ is the position of the block at time $\tau$, $\vec{e_\chi}$ and $\vec{e_\psi}$ have at time $\tau$ the component values that Peter found. We have replaced the numeric indices (0,1,2) that Peter used with symbolic indices ($\tau, \chi, \psi)$. - We we multiply these purely spatial basis vector by the spatial coordinate value ($\chi, \psi$) to get the spatial displacement from the block position. Thus, by adding the initial block position $\vec{P_0}$ to the spatial displacement vector we calculate, we find the spatial position of the point specified by coordinates $(\chi, \psi)$ at time $\tau$. This all happens on a surface of constant coordinate time $\tau$. Of course, because of the relativity of simultaneity, a surface of constant coordinate time $\tau$ isn't a surface of constant cordinate time T, this is the reason our spatial vectors have components in the $\partial_T$ direction. This process specifies a momentarily co-moving spatial frame of reference, though it's a rotating co-moving frame and not an inertial co-moving frame.

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

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

The remaining part requires a lot of computer algebra, plus a fair amount of manual simplification and collection of terms, but is concetually easy. We find the metric in the new coordinates by using the chain rule to find dT, dX, and dY in terms of $d\tau$, $d\chi$, $d\psi$. Then we calculate the line element -dT^2 + dX^2 + dY^2

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 physical interpretation of this is fairly straightforwards, it's just an expression for time dilation (the alpha coefficient) and rotation (the beta coefficient), with the spatial part of the metric being perfectly ordinary Euclidean space.