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

[PeterDonis]

My questions are based on the understanding that the X,Y,T frame is the momentarily comoving (inertial) reference frame (MCRF) of the rocket at the time the beams are emitted and the X',Y',T' frame is the MCRF of the train at the time that the rearward-pointing beam is emitted from (wlog) the front of the train.

I suppose my first question is: have I understood that correctly?

Not quite. You have the X, Y, T frame correct. The X', Y', T' frame is the MCRF of the train at the instant the beams are emitted. Both beams are emitted from the same source at the center of the train; one beam travels towards the rear of the train, the other travels towards the front of the train. The event of the beams being emitted is not quite at the common spacetime origin of the two frames; it is at an $X$ and $X'$ coordinate of $\delta$, a small distance above the floor of the rocket. But since this distance is not in the $Y$ direction, it doesn't bring in any simultaneity issues; the event of the beams being emitted is at time zero in both frames, $T = T' = 0$.

The light beams are emitted from different places

No, they're not. See above. This is not quite the standard Einstein train and light flashes thought experiment, which is what you may be thinking of.

Rindler coordinates are mentioned but, as far as I can tell, they are not used in any calculations. Is that correct?

Yes. I only mention them to note that, once all of the analysis is done in the MCRF's, transforming to Rindler coordinates is relatively trivial and obviously does not affect any of the conclusions.

 

[andrewkirk]

Both beams are emitted from the same source at the center of the train; one beam travels towards the rear of the train, the other travels towards the front of the train.

That was my initial hypothesis. But I discarded it because the time taken to reach the destination was given as $L \sqrt{(1 - v)/(1 + v)}$ rather than $\frac{L}{2} \sqrt{(1 - v)/(1 + v)}$, and the train's length is $L$. Should the references to time be divided by 2?

 

[SlowThinker]

That was my initial hypothesis. But I discarded it because the time taken to reach the destination was given as $L \sqrt{(1 - v)/(1 + v)}$ rather than $\frac{L}{2} \sqrt{(1 - v)/(1 + v)}$, and the train's length is $L$. Should the references to time be divided by 2?

The length of the train car is 2L in Peter's derivation. Probably he just forgot the 1/2 factor somewhere or decided to drop it and didn't mention it.

 

[PeterDonis]

The length of the train car is 2L in Peter's derivation.

Yes, I assumed that each end of the train was at distance $L$ (in the train frame) from the light source at the center. That was because I was too lazy to fiddle around with extra factors of $1/2$.

 

[andrewkirk]

A couple of questions. Would a marble placed on the floor of the train roll towards the back? Would a string attached to a penduluum-like mass hung from the ceiling of the train make an acute angle facing the rear wall, and an obtuse angle facing the front wall, rather than hang straight down?

No. "Gravity" still points straight down inside the train.

What do we mean by 'down'?

Say we interpret 'down' to mean 'in the negative direction of the X' axis of the train's MCRF'. If gravity for the train passenger points in that 'down' direction then a plumb bob in the train will align with the X' axis and a dropped object in the train will always have an acceleration 3-vector in the train's MCRF that points along the negative X' axis.

What about a ball placed on the train floor? Assuming the above, it will roll if the X' axis is not normal to the floor. We know from Peter's calculations that the floor does not align with the Y' axis. So either the Y' axis is not orthogonal to the X' axis, or the ball will roll backwards in the train. My recollection is that orthogonal spatial axes of inertial frames remain orthogonal when given a velocity boost. Am I misremembering that?

I know I'm being lazy and not doing the maths on the above. I will do it, but I thought I'd first see if anybody has already done it.

 

[PeterDonis]

What do we mean by 'down'?

I should take my own advice and actually do the math, so let's do that now. In the MCRF of the rocket (in which the train is moving at $v$ in the $Y$ direction), the 4-acceleration of the rocket has components $a^a = (a^T, a^X, a^Y) = (0, g, 0)$. So we just need to apply a Lorentz transformation into the MCRF of the train, i.e., a boost in the $Y$ direction with velocity $v$. But both $a^T$ and $a^Y$ are zero, so the Lorentz transformation leaves the 4-acceleration components unchanged. Hence, the proper acceleration in the train's MCRF is still in the $X'$ direction with magnitude $g$. That means "down" does mean "in the negative $X'$ direction in the train's MCRF".

Now let's see what happens to a ball placed on the train's floor. Again, we start in the rocket's MCRF, in which, by hypothesis, the ball at time $T = 0$ has velocity $v$ in the $Y$ direction and proper acceleration $g$ in the $X$ direction. So the ball's 4-velocity at $T = 0$ is $u^a = (\gamma, 0, \gamma v)$, and after some small time $\delta$ it will be $(\Gamma, g \delta, \Gamma v)$. To find what $\Gamma$ is, we use the fact that the 4-velocity must have magnitude $-1$, so we must have $- 1 = - \Gamma^2 + g^2 \delta^2 + \Gamma^2 v^2 = - \Gamma^2 \left( 1 - v^2 \right) + g^2 \delta^2$. This gives

$$
\Gamma = \sqrt{\frac{1 + g ^2 \delta^2}{1 - v^2}} = \gamma \sqrt{1 + g^2 \delta^2}
$$

Since $g \delta$ is small, we can use $\sqrt{1 + g^2 \delta^2} \approx 1 / \sqrt{1 - g^2 \delta^2}$, and multiply the square roots out to obtain

$$
\Gamma \approx \frac{1}{\sqrt{1 - v^2 - g^2 \delta^2}}
$$

This is what we expect for an object with an $X$ component of velocity of $g \delta$ and a $Y$ component of velocity of $v$, i.e., at least to this order of approximation, the marble's $Y$ velocity is unchanged, which means if we boost to the MCRF of the train, the marble will not have moved in the $Y$ direction.

However, as you can see, this is only an approximation; as you will see if you work things out to the next order, $\Gamma$ will actually be slightly smaller than the value given above. What is actually going on? As SlowThinker pointed out, and as I agreed in a previous post, what is happening is that, viewed from a fixed inertial frame, the train's $Y$ velocity decreases as it accelerates in the $X$ direction (this must happen in order to conserve momentum in the $Y$ direction). But this affects the train itself, not just objects inside it. So over a long enough time of observation, yes, you will see the marble's $Y$ velocity decrease, when viewed in a fixed inertial frame. But you will also see the train's $Y$ velocity decrease by the same amount! So, relative to the train, the marble will stay put; it will not "roll backwards".

 

[SlowThinker]

Now we just need to find what a falling marble is doing, after a long time, as viewed from the train.
Is it "legal" so just say that the marble from train looks the same as the train viewed from the marble, and thus the marble will fall down at first and then start to lag?
I'm pretty sure it is right, but it's interesting that even though there is asymetry in the acceleration (only the train feels force), the relative view is symmetric.
That's some heavy stuff going on.

 

[PeterDonis]

Now we just need to find what a falling marble is doing, after a long time, as viewed from the train.

We already figured this out: the falling marble will gradually get ahead of the train, because the falling marble is inertial, so it has a constant velocity relative to a fixed inertial frame; but the train, relative to a fixed inertial frame, will have a gradually decreasing velocity in the $Y$ direction.

Is it "legal" so just say that the marble from train looks the same as the train viewed from the marble

"Looks the same" is not how I would describe it; but the two views (marble from train and train from marble) must be consistent.

the marble will fall down at first and then start to lag?

No, it will fall down at first and then get ahead. See above.

 

[SlowThinker]

We already figured this out: the falling marble will gradually get ahead of the train, because the falling marble is inertial, so it has a constant velocity relative to a fixed inertial frame; but the train, relative to a fixed inertial frame, will have a gradually decreasing velocity in the $Y$ direction.

That's the marble's view.
But from the train, the marble is moving (after some time) fast down ($-X$ direction), and this falling speed approaches $c$. By the same argument as before, its movement in the $Y$ direction must slow down, to avoid exceeding $c$.

 

[PeterDonis]

from the train

The train's frame, over a long period of time, is not an inertial frame, so your argument does not work; it is only valid in an inertial frame.

 

[SlowThinker]

The train's frame, over a long period of time, is not an inertial frame, so your argument does not work; it is only valid in an inertial frame.

You must be correct here. In a non-inertial frame, distant objects can exceed $c$.
For example, when I quickly accelerate to 0.9c, a star in that direction quickly halves its distance from me.

But does that mean that, when I drop an object from a plane (and neglect air resistance), the object will fall ahead of me? Although I won't be able to see it.
That offers a new perspective on black hole time dilation: the object already fell in, I just can't see it yet. I've heard that before but could not understand; now I think I might.

 

[PeterDonis]

does that mean that, when I drop an object from a plane (and neglect air resistance), the object will fall ahead of me?

No. Remember that all my analysis is for the case of a train inside an accelerating rocket in flat spacetime. That case is only locally the same as the case of a train on a flat planet--i.e., the two only have the same properties within the MCRF I was using. The conclusion about the train's $Y$ velocity decreasing as it accelerates in the $X$ direction is not derived within a single MCRF; you need to look at longer-term behavior. The longer-term behavior of the curved spacetime case (a train on a planet) is different from the flat spacetime case.

(Also, in the curved spacetime case, a planet won't be flat over long distances; it will be spherical, at least if we ignore its rotation. So analyzing the behavior of the curved spacetime case outside of a single local inertial frame adds several complications that we have not discussed in this thread.)

 

[PeterDonis]

That offers a new perspective on black hole time dilation: the object already fell in, I just can't see it yet.

This is actually a valid perspective, but not for the reason you are thinking. The analysis that shows light taking a long time to get from a point of emission close to the hole's horizon, out to a distant observer, can be done using only one spatial dimension (the radial dimension). There is no need to bring in a second spatial dimension perpendicular to it, as we did in the train case to show the dropped marble gradually getting ahead of the train.

 

[pervect]

I'm still having difficulty with the fully mathematical approach to GR (via metric and tensors), so I'm making thought experiments to get a feel for some issues.
Let's have a train moving at relativistic speed on a flat planet (so that the train goes straight). Also the gravity is supposed to be constant, say $g$.
The passengers set up an experiment, in which light is sent from the center of a train car horizontally forward and backward and they measure the height where it hits the walls.
There is a view which I call "stationary preferred", in which the light will be seen to fall with acceleration $g$ by a stationary observer. So the light will hit the back wall at nearly the original height, but it will hit the front wall a good bit lower - since it takes much more time to reach the front wall.
Another view is "passenger preferred", in which the light hits the walls nearly at the original height, same on back and front wall. It seems both views cannot be correct, and the "passenger preferred" is incorrect.

Some questions:
1. Is the "stationary preferred" view correct?
2. Does gravity create a preferred reference frame? Did I rediscover the Lens-Thirring effect?
3. If a passenger drops something, will it fall with acceleration $g$ as viewed by a stationary observer? This would mean that a passenger would feel her weight increased $\gamma$-times. But some physicists say that the weight of objects in Einstein's train is not changed (http://arxiv.org/abs/physics/0504110 page 6).
4. I realize that objects dropped in the train will not fall perfectly vertically, since they would eventually exceed the speed of light for a stationary observer. Is there a simple way to compute the trajectory?

(4) seems to be similar to the case of a particle in a uniform electric field...

I've been away a bit, and missed this thread,but I've written about some very similar issues in the past, and posted some relevant references. The specific paper of interest is http://arxiv.org/abs/0708.2490v1

I'll quote the relevant section, as the relevance of the paper may not be apparent at first glance.

There are two important non-intuitive points I want to make.

1) A torque-free gyroscope mounted on the moving train (also called a sliding block in some of my PF posts if one wishes look them up) will rotate due to Thomas precession. This implies that the "rest frame of the train" is rotating.

2) The "accelerating flat platform" , when viewed from the "rest frame of the train" (to be precise and use the language of the paper, a momentarily co-moving inertial frame) is not a flat platform, but a curved platform. The "flatness" of the platform depends on the observer. The Lorentz transform is linear, but the path of the train is curved in space-time, due to the proper acceleration of the train. Some of this space-time curvature due to the proper acceleration of the platform appears as spatial curvature of the platform when we do a Lorentz transform to a momentarily co-moving inertial frame.

Here are the applicable sections from the paper in question.

As an application where both the reference frame and
the gyroscope accelerates we will consider a gyroscope on
a train that moves along an upward accelerating platform
as shown in Fig. 3

A gyroscope with a torque free suspension on the train
will precess clockwise for v >0

Figure 11 is helpful in understanding "why" the train frame is a rotating frame.

[add]One of the sliding block threads is https://www.physicsforums.com/threads/more-on-the-sliding-block.737614/, in which I eventually figure out that because the train-frame is rotating, there is no such thing as a born-rigid train that started out at rest and accelerates along the platform. Thus, some of the confusion here (at least in my own thinking) involves carrying over traditional ideas of "rigid objects". If we consider the problem of what happens to a born-rigid train that is initially at rest, and accelerates on the "flat" platform, the conclusion we come to is that such a motion cannot be born-rigid, as the train was initially not rotating before it accelerated, and is rotating after the acceleration.

 

[PeterDonis]

One of the sliding block threads is https://www.physicsforums.com/threads/more-on-the-sliding-block.737614/, in which I eventually figure out that because the train-frame is rotating, there is no such thing as a born-rigid train that started out at rest and accelerates along the platform.

Thanks, pervect, I knew we had had a previous thread on this topic but I couldn't find it. Now I just have to read through it again to remember what was said.

 

[SlowThinker]

I've been away a bit, and missed this thread,but I've written about some very similar issues in the past, and posted some relevant references. The specific paper of interest is http://arxiv.org/abs/0708.2490v1

Welcome to the thread mister
I need to quit early today, so I went through the thread but didn't have time to read the paper yet.
Just 3 quick questions:
1) Are you saying that a ball on the floor will actually roll forward?
2) What does the parabolic shape mean? If I put a heavy weight in the back of the train, will it accelerate the train?
Or is it that if the passengers use rulers to measure
2a) the distance from the center of the floor to the top corners of the car,
2b) the distance from the center of the ceiling to the bottom corners of the car,
they will be different? Note that we can't use radar distance, since the time dilation at the floor is significantly higher than at the ceiling.
3) Is the asymmetry between forward and backward direction described by the Thomas precession? Does the paper explain why the backward ray hits the wall at almost unchanged height, in the presence of the parabolic deformation of the train?

I'll read through the paper during the weekend, thanks.

 

[pervect]

Welcome to the thread mister
I need to quit early today, so I went through the thread but didn't have time to read the paper yet.
Just 3 quick questions:
1) Are you saying that a ball on the floor will actually roll forward?
2) What does the parabolic shape mean? If I put a heavy weight in the back of the train, will it accelerate the train?

I believe the answer to both questions is no, though it's based on my own calculations rather than the paper I cited. My results indicate that the parabolic surface is an equipotential surface. So a ball or a block anywhere on the surface has the same potential, so there is no tendency for it to slide one way or the other. You can think of the equipotential surface as having some terms due to the gravity, and other terms due to the rotation. For an easy to understand similar problem, imagine using potential methods to work out the curved shape of the surface of the water in a rotating bucket.

Or is it that if the passengers use rulers to measure
2a) the distance from the center of the floor to the top corners of the car,
2b) the distance from the center of the ceiling to the bottom corners of the car,
they will be different? Note that we can't use radar distance, since the time dilation at the floor is significantly higher than at the ceiling.

I haven't worked out what passengers would measure with a ruler. I do believe I have a metric which would reasonably represent "a point of view" of an observer on the train which could answer this question. The metric, written in coordinates (T,X,Y,Z) can be transformed by a coordinate change to flat space-time, it remains only to confirm that (X=Y=Z=constant) represents "sliding block motion". This was worked out in https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/page-2#post-4466113, but it hasn't (as far as I know) been checked by anyone else to confirm it's accuracy. I suspect it's not terribly easy to follow the way I wrote it :(.

Letting

$g_{aa}=−1−(1+K^2)((1+Zg)^2−(gKX)^2−1)$

and

$g_{bb}=2K\sqrt{1+K^2}$

where K is some constant and a function of the velocity v of the sliding block.

The metric I feel that should describing the "point of view" of an observer on the sliding block in coordinates (T,X,Y,Z) would be

$d\tau^2 = g_{aa}dT^2+g_{bb}(XdZ−ZdX)dT+dX^2+dY^2+dZ^2$

Breaking thus up, we have a time dilation term $g_{aa}$, some terms that represent the rotation $g_{bb}$, and a flat spatial metric in X,Y,Z.

3) Is the asymmetry between forward and backward direction described by the Thomas precession? Does the paper explain why the backward ray hits the wall at almost unchanged height, in the presence of the parabolic deformation of the train?

I believe it should, the train rotates a bit while the light travels, raising one end and lowering the other. But I haven't thought it through in any great detail.

 

[pervect]

Thanks, pervect, I knew we had had a previous thread on this topic but I couldn't find it. Now I just have to read through it again to remember what was said.

Some of the posts/ threads were
https://www.physicsforums.com/threads/supplees-submarine-paradox.646874/#post-4133710
https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/#post-4444415
https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/#post-4445483
https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/page-2#post-4466113

But the "more on the sliding block thread" actually references a paper, rather than my own personal work, so it's probably cleaner.

 

[PeterDonis]

I am still digesting the previous threads and papers, but one item deserves some comment now. Let's describe the 4-velocity and 4-acceleration of a point on the train (the bottom of the train, to be precise) in the MCRF of the rocket at some instant of time. We will use $(T, X, Y)$ as the coordinates in this MCRF, and we have, by hypothesis, that the train moves with velocity $v$ in the $Y$ direction, while the rocket is accelerating with proper acceleration $g$ in the $X$ direction. Therefore, a point on the train, in this MCRF, will have 4-velocity $u = (\gamma, 0, \gamma v)$, where $\gamma = 1 / \sqrt{1 - v^2}$, and 4-acceleration $a = (0, g, 0)$.

If we now transform into the MCRF of the train at the same instant, all we are doing is boosting by $v$ in the $Y$ direction. This boost does not affect components of 4-vectors in the $X$ direction at all, so in the train (primed) frame, we will have $u' = (1, 0, 0)$ and $a' = (0, g, 0)$.

The reason I bring this up is that the above seems to indicate that the proper acceleration felt by an object at rest with respect to the train, which is the magnitude of $a$ or $a'$ (they must be the same since magnitudes of 4-vectors are invariants), is $g$, the same as the proper acceleration felt by an object at rest with respect to the rocket. (I've used this conclusion in previous posts in this thread.) This seems to contradict what has been said in previous threads, that the proper acceleration felt by an object at rest with respect to the train should be $\gamma g$, i.e., a factor of $\gamma$ (or perhaps more than one such factor) larger than that felt by an object at rest with respect to the rocket. I'm not sure right now how to resolve this apparent contradiction.

 

[SlowThinker]

a point on the train, in this MCRF, will have 4-velocity $u = (\gamma, 0, \gamma v)$, where $\gamma = 1 / \sqrt{1 - v^2}$, and 4-acceleration $a = (0, g, 0)$.

so in the train (primed) frame, we will have $u' = (1, 0, 0)$ and $a' = (0, g, 0)$.

he magnitude of $a$ or $a'$ (they must be the same since magnitudes of 4-vectors are invariants)

My guess would be that $a'=(\gamma v g, \gamma g, 0)$. But it's really just a guess.

What puzzles me is that people in the Sliding block thread agreed that the train is curved. You can't really boost a flat train from the rocket frame, to a curved train in the train's frame, can you?
So, either the train is curved, and pretty much all math up to this point would be invalid, since the boosting could not be used.
Or, the train has to be flat.

 

[PeterDonis]

I'm not sure right now how to resolve this apparent contradiction.

Still not sure, but here's some more math to pile on: [Edit: corrected some formulas to include extra factor of $\gamma$.]

In the MCRF of the rocket, the 4-velocity of the train is $u = (\gamma, 0,\gamma v)$, per my previous post. If we then boost this in the $X$ direction, to give an $X$ velocity $v_0$ (so that we're now looking at things in an inertial frame in which the rocket is, at some instant, moving at $v_0$ in the $X$ direction), the 4-velocity of the train becomes $U = (\gamma_0 \gamma, \gamma_0 v_0 \gamma, \gamma v)$, where $\gamma_0 = 1 / \sqrt{1 - v_0^2}$.

The proper acceleration $A$ in this fixed inertial frame can be found by taking the derivative of $U$ with respect to $\tau$ along the worldline of the train (more precisely, of some particular point on the train, which is what all these things refer to). We can simplify the process of taking these derivatives by computing them for $\gamma_0$, and $\gamma_0 v_0$ in advance. We use the fact that $\gamma_0 = \cosh \left( g \gamma \tau \right)$ and $\gamma_0 v_0 = \sinh \left( g \gamma \tau \right)$ to further simplify things (note the extra factor of $\gamma$, because $\tau$ for the train is time dilated by $\gamma$ relative to $\tau_0$ for the rocket), and obtain:

$$
\frac{d \gamma_0}{d \tau} = g \gamma \sinh \left( g \gamma \tau \right) = g \gamma \gamma_0 v_0
$$

$$
\frac{d \gamma_0 v_0}{d \tau} = g \gamma \cosh \left( g \gamma \tau \right) = g \gamma \gamma_0
$$

Finally, we note that $\gamma v$, the $Y$ component of $U$, is constant; it does not change. What changes is the ordinary velocity $v_Y$ of the train in the $Y$ direction in the fixed inertial frame we are now working in; this is given by $v_Y = U_Y / U_T = \gamma v / \gamma_0 \gamma = v / \gamma_0$.

Putting all of the above together, we have

$$
A = \frac{d}{d \tau} \left( \gamma_0 \gamma, \gamma_0 v_0 \gamma, \gamma v \right) = \left( g \gamma_0 v_0 \gamma^2, g \gamma_0 \gamma^2, 0 \right) = g \gamma^2 \left( \gamma_0 v_0, \gamma_0, 0 \right)
$$

If we boost this back to the MCRF of the rocket, we end up with $a = g \gamma^2 \left( 0, 1, 0 \right)$, which obviously has magnitude $g \gamma^2$, not $g$. (We could also compute this magnitude, more laboriously, from the equation for $A$ above in the fixed inertial frame.) So now the question is, why is the proper acceleration of the train, in the rocket's MCRF, $(0, g \gamma^2, 0)$ instead of $(0, g, 0)$, as I had thought it was before?

 

[PeterDonis]

My guess would be that $a'=(\gamma v g, \gamma g, 0)$. But it's really just a guess.

Close, but not quite. The proper acceleration has to be orthogonal to the 4-velocity, so since the 4-velocity of the train has a component in the $T'$ direction, the proper acceleration can't have a component in that direction. The question is, as my previous post says, why the $X'$ component of the proper acceleration in the train's MCRF is $\gamma^2 g$ [edit: corrected to add extra factor of $\gamma$] instead of $g$. The standard heuristic answer is "time dilation", but I'm still trying to see where exactly that appears in the math.

 

[andrewkirk]

I'm wrestling with some maths about this at present, but since I have nothing that I dare expose on that front at present, I thought I'd contribute the following non-mathematical perspective that helps me think about why the train seems to be tilted, in its frame.

Imagine the rocket is initially inertial, with the train traveling inertially in the Y direction at speed v. Then the rocket starts to accelerate in the X direction with constant acceleration g.

In the rocket's MCRF, the acceleration commences at the same time at all points on the rocket floor.
But in the train's frame, acceleration commences earlier towards the front than the back.
I imagine this, from the train's viewpoint, as a set of rocket engines attached to the base of the rocket, aligned along the Y axis. From the train's point of view, the frontmost burner lights up first, followed by the next, and so on towards the back. So the front end of the train starts accelerating upwards before the back end does, and this tilts the train, in its own reference frame.

Where the analogy might break down is that this should cause the tilt to constantly increase, rather than stabilise. That's because the staged firing of the rocket engines will impart an angular momentum on the train (and the rocket floor) that will be preserved. Another way to look at that is that, if the front of the train starts accelerating one second before the rear, the difference in their X' coordinates will constantly increase, as it is $\frac{1}{2}gt^2-\frac{1}{2}g(t-1)^2=gt-\frac{1}{2}$. This doesn't seem consistent with the formal calcs above that show the tilt is constant.

This is of course a big hand-wave, and no doubt changes when one applies the appropriate Lorentz transformations. But it helps me visualise it.

 

[PeterDonis]

FWIW, the $\gamma^2$ factor in the proper acceleration of the train also pops out easily in Rindler coordinates. We have the 4-velocity of the rocket as $(1, 0, 0)$ in these coordinates. and its 4-acceleration is $(0, g, 0)$; this is derived using the full formula for the covariant derivative:

$$
a^a = u^b \nabla_b u^a = u^b \partial_b u^a + \Gamma^a{}_{bc} u^b u^c
$$

The partial derivative term is zero, and the only connection coefficient that matters is $\Gamma^X{}_{TT} = 1 / X = g$ (since the $X$ coordinate of the rocket in this chart is $1 / g$), so we have $a^X = g u^T u^T = g$, with all other components of $a$ zero.

For the train, the only change is that the 4-velocity is now $(\gamma, 0, \gamma v)$, but the same formula for $a^X$ still applies and it is still the only nonzero term (the partial derivative term is still zero and no other connection coefficients matter), so we have $a^X = g u^T u^T = g \gamma^2$.

I think the issue I was having before is that, by only looking at the MCRF at one instant, we lose the information about the global behavior of the worldlines that is necessary in order to do a correct computation of the proper acceleration. To do that computation correctly, we need at least enough information about the worldlines to properly compute derivatives, and in the MCRF we don't have that: we have constant 4-velocity components (to the approximation used in the MCRF), and zero connection coefficients (again, to the approximation used in the MCRF), so we've basically thrown away the information we would need to even see any proper acceleration in the math. We can still use information about the proper acceleration to do computations in the MCRF--but we can't use the MCRF to determine what the proper acceleration is; we have to do that some other way, and then plug the result into computations in the MCRF by hand if we need it.

 

[PeterDonis]

In the rocket's MCRF, the acceleration commences at the same time at all points on the rocket floor.
But in the train's frame, acceleration commences earlier towards the front than the back.

Yes.

Where the analogy might break down is that this should cause the tilt to constantly increase, rather than stabilise. That's because the staged firing of the rocket engines will impart an angular momentum on the train (and the rocket floor) that will be preserved.

The staged firing does exert a torque, but the torque is not continuous; at least, it isn't if the train and rocket are of finite length in the $Y$ direction! Assuming that is the case, there will be a finite period of time, relative to the train, during which the staged rocket firings are occurring; at the end of that time, all of the rockets are firing and there is no longer any torque being exerted. The "tilt" at this point, relative to the train, is constant; it doesn't continue to increase. (It's important to note, here, that "relative to the train", as pervect's analysis shows, implies a rotating frame; that is the frame in which the "tilt" is constant once the staged firing is completed. However, this "frame" has some issues lurking, which I plan to do a follow-up post to explore.)

 

[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?