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

[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
$$\alpha = -{\frac { \left( \chi\,g+1 \right) ^{2}}{-{v}^{2}+1}}+{\frac {{v}^{2} }{-{v}^{2}+1}}+{\frac {{g}^{2}{\psi}^{2}{v}^{2}}{ \left( -{v}^{2}+1 \right) ^{2}}}$$
$$\beta =\frac{ 2 g v}{1-v^2}$$

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.

 

[PeterDonis]

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

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

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

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

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

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

 

[PeterDonis]

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

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

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

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

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

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

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

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

 

[PeterDonis]

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

Agreed.

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

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

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

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

 

[pervect]

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

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

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

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

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

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

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

 

[pervect]

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

The result I get for the resulting metric is

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

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

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

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

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

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

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

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

 

[SlowThinker]

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

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

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

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

 

[PeterDonis]

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

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

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

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

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

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

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

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

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

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

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

Some more comments on interpreting the metric we got earlier.

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

 

[PeterDonis]

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

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

 

[PeterDonis]

Thus we can write

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

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

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

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

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

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

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

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

 

[pervect]

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

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

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

Calculating the Rimeann, I got the following

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

 

[PeterDonis]

I found a flat Riemann tensor.

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

 

[pervect]

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

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

Those aren't the right formulas

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

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

 

[pervect]

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

.

 

[pervect]

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

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

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

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

 

[PeterDonis]

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

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

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

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

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

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

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