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

[SlowThinker]

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...

 

[PeterDonis]

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.

Ok so far.

Another view is "passenger preferred", in which the light hits the walls nearly at the original height, same on back and front wall.

No, it doesn't, because in this frame, the "gravitational field" is not the same as in the stationary frame, due to the relativity of simultaneity. Try analyzing this experiment in a rocket in flat spacetime that is accelerating upward with acceleration $g$ but has a sideways velocity as well; by the equivalence principle, this should be the same as the train moving horizontally on a flat planet.

 

[jartsa]

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?

Yes.

2. Does gravity create a preferred reference frame? Did I rediscover the Lens-Thirring effect?

I guess it does.

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).

You are right, paper is wrong. Consider a long circular train wrapping around a planet. That train has an increased rest mass when it has some internal motion, like spinning around its center of mass. Load more mass on a train - suspension springs compress - suspension springs = scale.

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?

The projectile follows a geodesic. Let me consult Wikipedia about geodesic ... oh it's complicated.

 

[jartsa]

No, it doesn't, because in this frame, the "gravitational field" is not the same as in the stationary frame, due to the relativity of simultaneity. Try analyzing this experiment in a rocket in flat spacetime that is accelerating upward with acceleration $g$ but has a sideways velocity as well; by the equivalence principle, this should be the same as the train moving horizontally on a flat planet.

No, it works like this: I stand still and watch a transparent accelerating rocket, inside of which the experiment of sending light beams to the walls is conducted. The light beams hit the walls at the same altitude. Now I start to accelerate orthogonally to the rocket acceleration. This causes the rocket wall that is farthest away from me to gain lead compared to other walls, in my frame. Light beams still hit the same spots on the walls.

(The synchronized motion of walls in one frame is a non-synchronized motion of walls in another frame)

Now the case with gravity: A transparent train and I are standing on the surface of a planet and inside the train the experiment of sending light beams to the walls is conducted. The light beams hit the walls at the same altitude. Now I start to accelerate orthogonally to the gravitational acceleration. This does not cause the train wall that is farthest away from me to gain lead compared to other walls, in my frame. Light beams do not hit the same spots on the walls.

 

[Nugatory]

Now the case with gravity: A transparent train and I are standing on the surface of a planet and inside the train the experiment of sending light beams to the walls is conducted. The light beams hit the walls at the same altitude. Now I start to accelerate orthogonally to the gravitational acceleration.

I think you've misunderstood the problem (or how to apply the equivalence principle to the problem). The question asks about a train moving at constant speed through a gravitational field that is perpendicular to the direction of movement. Acceleration enters into the picture only if when you solve the problem by applying the equivalence principle, and in that case the acceleration you consider is parallel to the direction of the field, not perpendicular.

 

[Markus Hanke]

Would the value of g be the same in both reference frames ? I am thinking that, if I move on a geodesic on the surface of a planet ( i.e. approximately a great circle ) at relativistic speeds, the circumference of that planet should appear length-contracted to me. On the other hand, the total rest mass of the planet should remain the same, so the density of the planet as calculated by me should increase the faster I go in my train. What am I missing ?

 

[SlowThinker]

Would the value of g be the same in both reference frames ? I am thinking that, if I move on a geodesic on the surface of a planet ( i.e. approximately a great circle ) at relativistic speeds, the circumference of that planet should appear length-contracted to me. On the other hand, the total rest mass of the planet should remain the same, so the density of the planet as calculated by me should increase the faster I go in my train. What am I missing ?

0) To me it's clear that gravity in the train is something like \gamma g. Although things don't fall "down".
1) Circumference is length-contracted, but the radius of the planet is not. The tracks are going down very fast, and you need a strong centripetal force to stay at them. The centrigufal force beats gravity even at very low speeds (about 0.00001 c on Earth).
2) Perhaps the length contraction can be seen as the source of increased gravity in the train: more mass under the train is squashed into smaller length, making its gravity stronger. But confirming this would require "getting hands dirty" with actual math

 

[Markus Hanke]

Perhaps the length contraction can be seen as the source of increased gravity in the train

That is what I was wondering, but my intuition tells me that this can't be right, because it would mean we can increase g without bound by just going closer and closer to c. Besides, the source of gravity is not mass density alone but the full energy-momentum tensor, which as a geometric object is of course invariant between frames. I think the reasoning of increased density leads to greater g in the train is a Newtonian one that doesn't really apply in this relativistic situation; the other components of the energy-momentum tensor have to be taken into account also.

EDIT : The above is actually silly, since we are dealing with the situation outside the planet in vacuum, so the energy-momentum tensor vanishes there ! Please ignore. Just having one of those days

 

[jartsa]

I think you've misunderstood the problem (or how to apply the equivalence principle to the problem). The question asks about a train moving at constant speed through a gravitational field that is perpendicular to the direction of movement. Acceleration enters into the picture only if when you solve the problem by applying the equivalence principle, and in that case the acceleration you consider is parallel to the direction of the field, not perpendicular.

Well is there not enough accelerating and moving happening? And still the light hits the same spots of the walls. Some acceleration may be there because it is very easy to see what effect motion has when there is first motionlessness.

Or could a train moving inside an accelerating rocket be a different thing than a rocket that is accelerating and moving sideways? Hmm.. Oh dear, yes it is, because the rocket floor keeps the train aligned with the floor.

So, when the train is accelerating along the floor it feels the floor tilting. When the train moves along the floor, does it have to use energy to keep the velocity constant on the tilted floor? Hmm ... No because if the floor is tilted then the rocket nozzles are equally tilted and 'gravity' is perpendicular to the floor. When the train accelerates, the whole rocket turns according to the train, that's how it is, probably.

 

[SlowThinker]

I stand still and watch a transparent accelerating rocket, inside of which the experiment of sending light beams to the walls is conducted. The light beams hit the walls at the same altitude. Now I start to accelerate

That's the problem. The gravitational field creates a preferred frame of reference. You moving on the surface of the planet with a standing train, is not the same as you standing and train moving.

 

[SlowThinker]

Or could a train moving inside an accelerating rocket be a different thing than a rocket that is accelerating and moving sideways? Hmm.. Oh dear, yes it is, because the rocket floor keeps the train aligned with the floor.

So, when the train is accelerating along the floor it feels the floor tilting.

I don't think so, both parts. I'm pretty sure the train would keep its speed without using engines. That's the interesting part: things fall down and back, but passengers would stand vertically.

 

[PeterDonis]

Since nobody has taken up my suggestion in post #2, I'll do it myself. This is a scenario where, as we will see, it's important to actually do the math, instead of making heuristic guesses, because what happens is somewhat counterintuitive.

So we have a "rocket" which is accelerating in some specific direction, which we'll call the $x$ direction, with proper acceleration $g$ (more precisely, this is the proper acceleration of an observer at rest at the bottom of the rocket, but we can consider the rocket's height to be small enough that $g$ is constant everywhere inside it). This rocket is very wide in one of the transverse directions, which we'll call the $y$ direction, and we have a train moving inside the rocket in that direction (the $y$ direction) with speed $v$ as measured by an observer at rest in the rocket, and rest length $L$. At some instant, a pair of light beams is emitted by a source moving with the train, in the $+y$ and $-y$ directions. The question is, what is the motion of these beams?

Let's first look at things in an inertial frame in which the rocket is momentarily at rest at the instant the light beams are emitted; this instant will be $T = 0$ (we will use capital letters for coordinates in this frame to avoid confusion). In this frame, we can assign coordinates as follows: the bottom of the rocket is at $X = 1 / g$ at $T = 0$; the light source is at $X = 1 / g + \delta$ at $T = 0$; and the beams are emitted in the $\pm Y$ direction. The train's speed is $v$ in this frame (since the rocket is momentarily at rest in the frame), and the length of the train in this frame is $L / \gamma = L \sqrt{1 - v^2}$. In this frame, the backward-moving light beam will take time $L \sqrt{(1 - v)/(1 + v)}$ to hit the back end of the train, and the forward-moving light beam will take time $L \sqrt{(1 + v)/(1 - v)}$ to hit the front end of the train. The bottom of the rocket moves a distance $\frac{1}{2} g T^2$ in time $T$, provided $T$ is small enough (we are assuming it is, i.e., that the height $\delta$ of the light source and the rest length $L$ of the train are small enough), so the distances moved will be $\frac{1}{2} g L (1 - v)/(1 + v)$ and $\frac{1}{2} g L (1 + v)/(1 - v)$, and the height above the bottom of the rocket at which the two light beams strike will therefore be $\delta$ minus these two distances. Obviously this final height will be smaller for the forward-moving beam than for the backward-moving beam.

Translating the above into Rindler coordinates, which correspond to the "stationary" frame, is simply a matter of reinterpreting the heights $\delta - \frac{1}{2} g L (1 - v)/(1 + v)$ and $\delta - \frac{1}{2} g L (1 + v)/(1 - v)$ as the $x$ coordinates (lower case $x$ since we are now talking about Rindler coordinates) of the light beams at the instant they strike the back and front ends of the train, respectively. That is, instead of the bottom of the rocket accelerating in the $+ X$ direction at $g$, we now have the light beams accelerating in the $- x$ direction at $g$. Everything else is the same, since we have $t = T$ to a good enough approximation for the duration of the light beams' travel.

The question now is, how do we translate all this into the rest frame of the train? Let's first try to translate into an inertial frame moving at $v$ in the $Y$ direction relative to the inertial frame we used above. The events of interest have the following coordinates in the original inertial frame:

Light beams emitted: $T = 0$, $X = \delta$, $Y = 0$. Bottom of rocket at this instant (directly below light source): $T = 0$, $X = 0$, $Y = 0$. (This event is the common origin of both frames.)

Back light beam hits back of train: $T = L \sqrt{(1 - v)/(1 + v)}$, $X = \delta$, $Y = - L \sqrt{(1 - v)/(1 + v)}$. Bottom of rocket at this instant (directly below back of train): $T = L \sqrt{(1 - v)/(1 + v)}$, $X = \frac{1}{2} g L (1 - v)/(1 + v)$, $Y = - L \sqrt{(1 - v)/(1 + v)}$.

Front light beam hits front of train: $T = L \sqrt{(1 + v)/(1 - v)}$, $X = \delta$, $Y = L \sqrt{(1 + v)/(1 - v)}$. Bottom of rocket at this instant (directly below front of train): $T = L \sqrt{(1 + v)/(1 - v)}$, $X = \frac{1}{2} g L (1 + v)/(1 - v)$, $Y = L \sqrt{(1 + v)/(1 - v)}$.

We now simply do a Lorentz transformation at speed $v$ in the $Y$ direction, to obtain:

Light beams emitted: $T' = 0$, $X' = \delta$, $Y' = 0$. Bottom of rocket at this instant (directly below light source): $T' = 0$, $X' = 0$, $Y' = 0$. (This event is the common origin of both frames.)

Back light beam hits back of train: $T' = L$, $X' = \delta$, $Y' = - L$. Bottom of rocket at this instant (directly below back of train): $T' = L$, $X' = \frac{1}{2} g L (1 - v)/(1 + v)$, $Y' = - L$.

Front light beam hits front of train: $T' = L$, $X' = \delta$, $Y' = L$. Bottom of rocket at this instant (directly below front of train): $T' = L$, $X' = \frac{1}{2} g L (1 + v)/(1 - v)$, $Y' = L$.

Notice that the $X'$ coordinates are the same in this frame as the $X$ coordinates were in the original inertial frame, so the height differences are unchanged; the backward-moving beam still hits higher above the bottom of the rocket than the forward-moving beam. And this is true even though, in this frame, both beams hit at the same time! How is this possible?

To see what we missed, let's look at the events describing the points on the bottom of the rocket that will be directly below the back and front of the train when the light beams hit those ends of the train, at the instant the light is emitted. In the original inertial frame, these events are:

Bottom of rocket below back of train: $T = 0$, $X = 0$, $Y = - L \sqrt{(1 - v)/(1 + v)}$.

Bottom of rocket below front of train: $T = 0$, $X = 0$, $Y = L \sqrt{(1 + v)/(1 - v)}$.

Now we transform these events into the new inertial frame, moving at $v$ in the $Y$ direction:

Bottom of rocket below back of train: $T' = v L / (1 + v)$, $X' = 0$, $Y' = - L / (1 + v)$.

Bottom of rocket below front of train: $T' = - v L / (1 - v)$, $X' = 0$, $Y' = L / (1 - v)$.

Now we see what is happening: in the train frame, the rocket is "tilted" in the $Y'$ direction--the rocket below the back end of the train is at $X' = 0$ at a later time than the rocket below the front end of the train. So when the light beams hit the ends of the train (which happens at both ends at $T' = L$ in this frame), the bottom of the rocket at the back end is at a lower $X'$ coordinate, and hence the beam hits at a higher height above the bottom of the rocket.

Translating this into the actual rest frame of the train should again be simple: we just reinterpret acceleration of the rocket in the $+ X$ direction as acceleration of the light beams in the $- x$ direction. Everything else stays the same, including the above conclusion that the backward-moving beam hits at a higher height above the bottom of the rocket.

 

[phinds]

Very impressive, Peter, but I have to say ... you have WAYYY to much spare time on your hands

 

[andrewkirk]

Yes, I'm really looking forward to working through Peter's calcs (as I find this puzzle intriguing). But I have to put it off until I've got enough time to sit down and really concentrate on it.

 

[SlowThinker]

Back light beam hits back of train: $T' = L$, $X' = \delta$, $Y' = - L$. Bottom of rocket at this instant (directly below back of train): $T' = L$, $X' = \frac{1}{2} g L (1 - v)/(1 + v)$, $Y' = - L$.

Front light beam hits front of train: $T' = L$, $X' = \delta$, $Y' = L$. Bottom of rocket at this instant (directly below front of train): $T' = L$, $X' = \frac{1}{2} g L (1 + v)/(1 - v)$, $Y' = L$.

The math (and effort ) is impressive indeed, and seems to be correct. However, I'm still having trouble interpreting it, and can't quite agree with your conclusion.
First, it's clear that there is some kind of tilt or rotation even before the last part, in the part I quoted. The last paragraph, transforming simultaneous events in one frame into another frame, is meaningless.
Anyway, I can't accept the resolution that the train is running uphill, and that's why light hits the front wall lower. The passengers are sending light parallel to the floor, even if the train was rotated or even tilted.
Further, let's consider a ball on a horizontal table in the train. We'll view it from the only reasonable frame (the one you started with), where the train is moving, and the rocket accelerating. The ball or table is never accelerated in any direction other than +X, so there's no reason it would ever fall from the table. So the gravity points "down", even if things may fall in funny ways.
Same reasoning applies to the question whether the train would slow down. Sometimes I think it should, but the argument with a ball on table looks correct.

 

[PeterDonis]

I'm still having trouble interpreting it, and can't quite agree with your conclusion.

In other words, your intuition is leading you to a different conclusion from the math. That's why I put the warning at the beginning of my post that the conclusion would be counterintuitive. Unfortunately, that doesn't make it wrong; one of the first lessons everyone who studies relativity has to learn is that if your intuitions conflict with the theory, your intuitions are wrong and need to be retrained.

it's clear that there is some kind of tilt or rotation even before the last part, in the part I quoted.

You mean, because the two $X'$ coordinates are different? Yes, that in itself is enough to confirm that, contrary to intuition, the backward-moving light beam does hit the train higher above the bottom of the rocket than the forward-moving light beam does. See below.

The last paragraph, transforming simultaneous events in one frame into another frame, is meaningless.

Why? What's wrong with doing a simple Lorentz transformation? That's all I did.

However, as I said above, the calculation prior to that, showing the two different $X'$ coordinates for the bottom of the rocket underneath the back and front of the train, is already sufficient to show that your intuition, that in the "train frame", the two light beams should hit at the same height, is wrong. And it's already obvious that intuition has to be wrong somewhere, because the "stationary frame" and the "train frame", according to intuition, give different answers for direct observables--how far above the bottom of the rocket each light beam hits. The only question is, which intuition is wrong--the "stationary frame" intuition (that the backward-moving light beam hits higher) or the "train frame" intuition (that both beams hit at the same height). I am simply showing that it's the latter intuition that's wrong.

The passengers are sending light parallel to the floor

Huh? That's impossible; the floor is accelerating upward. More precisely: the floor is accelerating upward in the $X$ direction, and the light is moving in the $Y$ direction, in the original inertial frame (the one in which the rocket is momentarily at rest at $T = 0$). That means that, at an instant of time in that inertial frame, both light beams are the same distance from the floor of the rocket. (This may be what you were trying to say by saying "parallel to the floor", but it's very important to be precise, as you'll see in a moment.)

Now: since the train is moving in the $Y$ direction in the original inertial frame, then when we transform to an inertial frame in which the train is at rest (more precisely, it's momentarily at rest at $T' = 0$--it is accelerating in the $X'$ direction, but we're assuming the acceleration is small enough that we can approximate things using the inertial "train frame"), we will find that it is impossible, in the new frame, for the light beams to be the same distance from the floor of the rocket (except at the instant $T' = 0$ when the beams are emitted). This is an obvious consequence of relativity of simultaneity: events which are spatially separated cannot be simultaneous in two different frames. Since we know by hypothesis that events at which the two light beams are at the same distance from the floor of the rocket are simultaneous in the original inertial frame, such events cannot be simultaneous in the new inertial frame (the "train frame"). That is the missing piece we need to understand why the backward-moving light beam hits higher in the train frame.

the gravity points "down", even if things may fall in funny ways.

Yes, that's correct. But it's irrelevant to my argument. I am not saying that gravity in the train frame points in a different direction. In fact, if you look at my calculations, I am making use of the fact that the acceleration in the train frame is in the $X'$ direction only.

Once again, the key point has to do with relativity of simultaneity; that's where the counterintuitive feature comes in. There is nothing counterintuitive about the direction of gravity in this scenario; if you are focusing on that, you are missing the point.

 

[SlowThinker]

Thanks Peter, perhaps you missed this sentence in my very first post.

It seems both views cannot be correct, and the "passenger preferred" is incorrect.

And yes, I'm trying to improve my intuition.

If we return to the original 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?
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?

the answers to 1-3 would be Yes, and to 4 "no easy way"?

To clear the last few points,

The last paragraph, transforming simultaneous events in one frame into another frame, is meaningless.

Why? What's wrong with doing a simple Lorentz transformation? That's all I did.

What is simultaneous in the stationary frame, has no meaning for the passengers. That's why "meaningless". I am not saying it is wrong or anything like that.

The passengers are sending light parallel to the floor, even if the train was rotated or even tilted.

Huh? That's impossible; the floor is accelerating upward.

Actually the key word was "sending". What happens after, is the point of this discussion.
Sorry if I sound like nit-pick, but I spent 3 hours choosing the right words for those 12 lines

 

[PeterDonis]

perhaps you missed this sentence in my very first post.

I understand that you said that in the OP, but when you say you "can't agree" with my conclusion, that's tantamount to not agreeing with your own statement in the OP.

the answers to 1-3 would be Yes, and to 4 "no easy way"?

The answers to 1 and 3 are yes. The answer to 2 is "sort of" and "no" (since 2 is really two questions). Gravity does not create a "preferred frame", but it does break spatial isotropy--the direction of the gravitational field is different from the directions perpendicular to the field. The Lense-Thirring effect is due to the rotation of the central gravitating body, not transverse motion of the test object; we are ignoring the Earth's rotation here (certainly there is no rotation in my flat spacetime scenario with the accelerating rocket).

As for question 4, what does "perfectly vertically" mean? Does it mean with respect to the train, or with respect to the stationary observer? Objects dropped from rest relative to the train do fall perfectly vertically relative to the train; they don't relative to the stationary observer (since they have the train's horizontal velocity relative to that observer).

Also, I'm not sure what you meant by "eventually exceed the speed of light" in the previous post you quoted. We have not analyzed the long-term motion of any of these objects; we have only done an analysis in a single momentarily comoving inertial frame. We have to restrict the analysis that way if we want it to apply to the case of a train moving horizontally on a flat planet, since the equivalence principle only applies in a single local inertial frame.

Obviously on a real planet, objects dropped from the train will not fall indefinitely; even if the objects are idealized as not interacting with the planet's substance, they will eventually fall through it and come out the other side, and at that point they will be accelerating in the opposite direction. (Also, of course, a real planet is not flat but spherical, so the train's path will curve.) We can't use a flat spacetime model to approximate any of this behavior; we would have to use a full-blown curved spacetime model. At that point, I would hope we would all agree that our unaided intuition is inadequate to the task, and we would have to take the time to grind through the math.

What is simultaneous in the stationary frame, has no meaning for the passengers.

I don't know what you mean by this. The fact that events that are simultaneous in the stationary frame cannot be simultaneous in the train frame is a fact for both stationary observers and train passengers. If the train passengers know that a certain pair of events--such as, say, the events of both light beams (backward and forward moving) being at a certain distance $X$ above the rocket's floor--are simultaneous in the stationary frame, then those train passengers know that those two events cannot be simultaneous in the train frame. That means the light beams cannot be at the same distance above the rocket's floor at the same time in the train frame. So facts about simultaneity of events in the stationary frame can certainly be used to deduce facts about the simultaneity (or lack thereof) of events in the train frame. That is what my argument does.

the key word was "sending".

In other words, you meant the words "parallel to the floor" to only apply at the instant the beams are emitted? Sure, I already assumed that: the beams are emitted in the $Y$ direction, which is also the $Y'$ direction. But that, in itself, says nothing about their respective heights above the rocket floor at any other instant other than the instant they are both emitted.

 

[SlowThinker]

I understand that you said that in the OP, but when you say you "can't agree" with my conclusion, that's tantamount to not agreeing with your own statement in the OP.

I understood your conclusion as if you're suggesting that the floor of the train looks tilted to passengers, and that's why there is difference between front and back wall. And I could not agree with that.

As for question 4, what does "perfectly vertically" mean? Does it mean with respect to the train, or with respect to the stationary observer? Objects dropped from rest relative to the train do fall perfectly vertically relative to the train; they don't relative to the stationary observer (since they have the train's horizontal velocity relative to that observer).

Also, I'm not sure what you meant by "eventually exceed the speed of light" in the previous post you quoted.

If a ball is dropped in a tall train, its horizontal speed (Y) would stay constant, while the vertical speed (X) would increase. Even if X is limited by the speed of light, $\sqrt{Y^2+X^2}$ would exceed $c$ long before $X$ would, since $Y$ is already close to $c$. So, a dropped object will lose some $Y$ speed and lag behind the train. The difference will be small at first, but increase as the vertical speed gets higher. If the train is moving fast enough, it might be noticable even at a low height.

We can't use a flat spacetime model to approximate any of this behavior; we would have to use a full-blown curved spacetime model. At that point, I would hope we would all agree that our unaided intuition is inadequate to the task, and we would have to take the time to grind through the math.

In fact I was hoping to eventually extend the experiment to unlimited height. I'm trying to understand why a black hole has a horizon, while a homogeneous gravitational field can be infinitely deep.
If someone could translate our results into the GR formalism, it would be awesome, but perhaps worth an Insight article rather than a forum post. I can't quite imagine how someone would use a metric to derive these results, much less finding the metric.

In other words, you meant the words "parallel to the floor" to only apply at the instant the beams are emitted? Sure, I already assumed that: the beams are emitted in the $Y$ direction, which is also the $Y'$ direction. But that, in itself, says nothing about their respective heights above the rocket floor at any other instant other than the instant they are both emitted.

"Parallel to the floor" also describes direction, not just the height (that is understood to be the same for both rays). As in, X is constant to first order in Y.

 

[PeterDonis]

I understood your conclusion as if you're suggesting that the floor of the train looks tilted to passengers, and that's why there is difference between front and back wall. And I could not agree with that.

Then how else would you interpret the math? The math gives an unequivocal answer, but how to describe it in English is a different question. I'm not suggesting that the "tilted" description in English is "right"; I'm saying that the math is right, and offering the "tilted" description in English as a way of (heuristically) understanding why the math works out the way it does. But there may be other ways of understanding it.

If a ball is dropped in a tall train, its horizontal speed (Y) would stay constant, while the vertical speed (X) would increase. Even if X is limited by the speed of light, $\sqrt{Y^2+X^2}$ would exceed c long before X would, since Y is already close to c. So, a dropped object will lose some Y speed and lag behind the train.

No, that's not what will happen, because your argument about $\sqrt{Y^2 + X^2}$ applies just as much to the train as to the dropped object; if one "lags", the other will "lag" by the same amount.

Actually, what will happen is that the proper acceleration in the $X$ direction will require a larger force to be exerted by the rocket's engine than it would if the train were not present inside the rocket; the force required will go to infinity at some $X$ velocity less than $c$, instead of at $X$ velocity equal to $c$ as it would if the train were not present. So the "lag" will actually be in the rocket itself.

I'm trying to understand why a black hole has a horizon, while a homogeneous gravitational field can be infinitely deep.

A "homogeneous gravitational field" can't be infinitely deep. It has a horizon, just as a black hole does. The horizon in the flat spacetime case is called the Rindler horizon.

"Parallel to the floor" also describes direction

It describes the direction the light emitter is pointing, which is the $Y$ direction. But that in no way prevents the two light beams from being at different distances from the floor at the same instant in the train frame.

Part of the problem here, to me, is that you are trying to reason with English words instead of math. You can't do that; English words are too imprecise. That's why we use math in physics: to make sure we are stating things precisely, so that we know exactly what observations we are predicting.

As in, X is constant to first order in Y.

Yes, this is true in the sense that $Y$ is equal to $T$ for a light beam, and $X$ is a quadratic function of $T$. But we already know that $T$ changes by enough for the variation in $X$ to be visible (because, by hypothesis, we can observe the "bending" of the light beams), so the fact that that variation is quadratic doesn't affect any of the conclusions.

 

[SlowThinker]

Then how else would you interpret the math?

To me, it looks as if the gravity works in a funny way. Such as when you're in a centrifuge, you feel the "gravity" pointing "down", but when you stand up, you get pushed to the side; and when you move hands, it feels strange. But when you don't move, everything is fine.
We agreed that the floor is perpendicular to the gravity, just like it would be in a centrifuge.

So, a dropped object will lose some $Y$ speed and lag behind the train.

No, that's not what will happen, because your argument about $\sqrt{Y^2 + X^2}$ applies just as much to the train as to the dropped object; if one "lags", the other will "lag" by the same amount.

You surely must be wrong here. Do you agree that if I jump out of the train and out of the rocket as well, then from my perspective, my Y-position will eventually start to get ahead of the train?
The passenger perspective is more difficult to figure out, but I'm pretty sure that the same reasoning applies, and the passengers would feel getting ahead of me.

Actually, what will happen is that the proper acceleration in the $X$ direction will require a larger force to be exerted by the rocket's engine than it would if the train were not present inside the rocket; the force required will go to infinity at some $X$ velocity less than $c$, instead of at $X$ velocity equal to $c$ as it would if the train were not present. So the "lag" will actually be in the rocket itself.

Probably you meant "if the train were not moving"? But this can't be right either. Surely the rocket can accelerate like any other object, and only feel trouble reaching $c$, not $\sqrt{c^2 - X^2}$. There is only one invariant speed.

A "homogeneous gravitational field" can't be infinitely deep. It has a horizon, just as a black hole does. The horizon in the flat spacetime case is called the Rindler horizon.

I should have asked this question in the beginning, instead of trying to figure out time dilation in a deep field using a relativistic train

Part of the problem here, to me, is that you are trying to reason with English words instead of math. You can't do that; English words are too imprecise.

You are right in a sense. I'd prefer to use pictures instead of math (or before math), but drawing pictures on a computer is not easy. That, and it also seems like heavy math is needed to arrive at simple results.
I can read other's math (unless they start pulling new symbols out of thin air), but writing my own often leads to a dead end.

Yes, this is true in the sense that $Y$ is equal to $T$ for a light beam, and $X$ is a quadratic function of $T$. But we already know that $T$ changes by enough for the variation in $X$ to be visible (because, by hypothesis, we can observe the "bending" of the light beams), so the fact that that variation is quadratic doesn't affect any of the conclusions.

If the light was sent in opposite directions, but at an angle, it would still fit all your definitions so far:
1) the starting height is the same
2) close to start, it's close to starting height
3) it's a part of a parabola
but it's also important the the light starts horizontally. But I agree that your computations used a ray that does start horizontally, so the conclusions hold.

 

[PeterDonis]

To me, it looks as if the gravity works in a funny way.

Remember that "gravity" in relativity is not a force; it's just the geometry of spacetime. Yes, the geometry of spacetime can have counterintuitive properties. But, as I've said before, that just means your intuitions need to be retrained.

You surely must be wrong here.

Don't guess. Do the math. Or at least do the analysis. See below.

Do you agree that if I jump out of the train and out of the rocket as well, then from my perspective, my Y-position will eventually start to get ahead of the train?

No. Look at things in the inertial frame in which the rocket is momentarily at rest at the instant you jump. At that instant, you jump out of the train and the rocket, which means you start following an inertial worldline whose $Y$ velocity, in this frame, is the same as that of the train. Since your worldline is inertial, its $Y$ velocity cannot change; since the train's $Y$ velocity is also constant, your $Y$ position will always be the same as the train's, in this frame.

To see how things look in your rest frame, just boost by velocity $v$ in the $Y$ direction; this puts you in the instantaneous "train frame", the inertial frame in which the train is momentarily at rest at the instant you jump out. In this frame, neither you nor the train have any $Y$ velocity at the instant you jump; and since the train is accelerating in the $X$ direction, its $Y$ velocity doesn't change (and of course yours doesn't because you are at rest in this frame). So in this frame also, you and the train will always have the same $Y$ position.

this can't be right either.

Please, stop guessing and do the math, or at least the analysis. You are confusing yourself by guessing based on your intuition, instead of actually doing the analysis and looking at the conclusions it leads you to.

Surely the rocket can accelerate like any other object

It's not the rocket we're talking about, it's the train; since the train is contained in the rocket, accelerating the rocket means accelerating the train. If we consider an idealized rocket that is infinitely wide, and an idealized train that is infinitely long, then the limiting factor is the train's speed, not the rocket's; the train's speed will always be larger than the rocket's (because the train has an extra component of velocity in the $Y$ direction), so if the train's speed is limited to $c$, the rocket's speed must be limited to less than $c$, as seen from a fixed inertial frame. This is your own argument; I have just restated it to make the conclusion clearer.

As for the rocket, remember that neither its speed nor the train's speed will ever reach $c$, so we can always boost to an inertial frame in which the rocket is momentarily at rest and the train is moving at velocity $v$ in the $Y$ direction. In this frame, of course the rocket can accelerate. So it's not that the rocket stops accelerating; it's just that, viewed from a fixed inertial frame, the rocket's speed can't asymptote to $c$, because the train is contained inside the rocket, so the train's speed constrains the rocket's speed in that fixed inertial frame.

If the light was sent in opposite directions, but at an angle

This could be done, yes, but it would be a different scenario from the one I analyzed and that we have been discussing. Change the scenario and of course you can change the observables. But you can't specify one scenario and then reason as though it were a different scenario.

In other words: we could certainly specify, from the start, a scenario in which the two light beams are sent out in such a way that they both hit their respective ends of the train at the same height above the bottom of the rocket. But for that to happen, the light beams would have to follow different worldlines from the ones I used in my analysis. It would be instructive to set up this scenario and analyze it, but unfortunately I don't have time to do that right now, so for the time being I'll leave it as an exercise.

 

[mairzydoats]

Hello

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?

PeterDonis's analysis seems to suggest the answers to these would be yes, but I may be misinterpreting.

Thanks!

 

[mairzydoats]

... and another question:

Wouldn't it actually require a y-directed force just to keep this train moving at v in the y-direction?

I ask because if the accelerating rocket's floor were the only force acting on the train, then in order for the train to conserve its momentum in the y-direction as it is accelerated in the x-direction, its velocity in the y direction would decrease, no? So holding the train's y-velocity constant would mean an increase to its y-momentum, which would in turn require a constant infusion of energy, hence a y-directed force on the train, right?

Guess I'm falling for same trap PeterDonis has criticized throughout this thread: I didn't do the analysis or math. I'm just extrapolating this from another staff mentor's answer to my question in the thread I started, "need some help understanding electromagnetics and relativity". The scenario in my question is a little analogous to this one. The field E plays the part of the rocket and the charged particle plays the part of the train.

 

[jartsa]

Hello

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?

PeterDonis's analysis seems to suggest the answers to these would be yes, but I may be misinterpreting.

Thanks!

Here's my analysis:

The floor and the marbles on the floor recide approximately at the same point in space. If some forces are affecting the marbles, then the same forces are affecting the floor. Floor and marbles move in sync.

On the other hand the floor and the ceiling are forced to move in sync by the walls, except that the walls are Lorentz contracting, so the floor and the ceiling are not co-moving, if observed by an inertial observer. Therefore the ceiling is pulling the floor forwards making the marbles on the floor to roll backwards. The strength of this effect depends on the floor-ceiling mass ratio and floor-ceiling distance.

EDIT: As seen from an inertial frame, the longitudinal motion of the floor slows down as the transverse motion of the floor speeds up. Longitudinal means along the floor's long axis. This is closely related to time dilation.EDIT2: It may be important that as the walls are Lorentz contracting and the floor is moving towards the ceiling, horizontal laser beams shot by one laser gun on the floor and other laser gun on the ceiling will cross each other. This is the effect known as aberration, or relativistic beaming.

 

[PeterDonis]

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.

Wouldn't it actually require a y-directed force just to keep this train moving at v in the y-direction?

No. Once again, look at things in an inertial frame in which the rocket is instantaneously at rest. The train is moving at $v$ in the $Y$ direction in this frame. It gets accelerated in the $X$ direction, but that doesn't affect its motion in the $Y$ direction at all.

I ask because if the accelerating rocket's floor were the only force acting on the train, then in order for the train to conserve its momentum in the y-direction as it is accelerated in the x-direction, its velocity in the y direction would decrease, no?

Why do you think that would be the case? The acceleration of the rocket only affects the train's momentum in the $X$ direction, not the $Y$ direction.

 

[PeterDonis]

the floor and the ceiling are not co-moving, if observed by an inertial observer.

Why do you think this is the case?

 

[mairzydoats]

Why do you think that would be the case? The acceleration of the rocket only affects the train's momentum in the $X$ direction, not the $Y$ direction.

https://www.physicsforums.com/threa...nding-electromagnetics-and-relativity.835867/ post #2:

"Increasing the momentum in the y direction while keeping the momentum in the x direction fixed will actually decrease the velocity in the x direction"

 

[SlowThinker]

Don't guess. Do the math. Or at least do the analysis.

So I've spent quite a few hours trying to derive the speed of the rocket as viewed by me jumping out of the train, but still can't arrive at anything like x=cosh(t). So I'll just take that for granted. The speed may or might not be $u=\frac{t}{\sqrt{t^2/c^2+1/g^2}}$, the important thing is that the speed approaches $c$.

In the rocket, we set up light clock, where the light travels in the same direction as the tracks (perpendicular to acceleration). It has tilted mirrors, so that the light stays in the clock during acceleration.
We know that if the train is running at speed $v$, and the distance of clock mirrors is $d$, then everytime the clock bounce, the train went distance $v\ d/c$, perhaps with a small but constant correction to account for the light following a parabola.
After my jump, as the rocket is gaining speed, the light in the light clock is bouncing more and more parallel with the flight direction, and it takes more and more time to bounce. So it takes more and more time for the train to move $v\ d/c$. So the train is going slower and slower. Simply put, the time on the rocket is Lorentz dilated.

On the other hand, I am flying at constant speed $v$. So the train is lagging behind.

This is not very mathematical but perhaps it qualifies as analysis?

 

[PeterDonis]

https://www.physicsforums.com/threa...nding-electromagnetics-and-relativity.835867/ post #2:

"Increasing the momentum in the y direction while keeping the momentum in the x direction fixed will actually decrease the velocity in the x direction"

Ah, ok. This is going beyond the analysis I did, which was restricted to a single local inertial frame--i.e., it was restricted to a time short enough that the train's motion can be idealized as constant speed in the $Y$ direction plus constant acceleration in the $X$ direction. I have not extended the analysis to cover a long period of time, since I was only concerned with explaining why the two light beams do not hit their respective ends of the train at the same heights.

Your comment does, however, bring up a point I was mistaken about. I had said this in a previous post, in response to SlowThinker:

it's not that the rocket stops accelerating; it's just that, viewed from a fixed inertial frame, the rocket's speed can't asymptote to $c$, because the train is contained inside the rocket, so the train's speed constrains the rocket's speed in that fixed inertial frame.

When I posted this, I had forgotten the effect you refer to: that over a long period of time, if the rocket keeps accelerating in the $X$ direction, the speed of the train in the $Y$ direction, as seen in a fixed inertial frame, will have to decrease, in order to conserve momentum in the $Y$ direction. (Heuristically, this is because the relativistic mass of the rocket/train increases, as seen in a fixed inertial frame, and the train's momentum is relativistic mass times velocity, so if relativistic mass increases, velocity must decrease.) So in the limit, the rocket can indeed have its speed asymptote to $c$ in the $X$ direction, because the speed of the train in the $Y$ direction goes to zero in the limit.

 

[PeterDonis]

I've spent quite a few hours trying to derive the speed of the rocket as viewed by me jumping out of the train, but still can't arrive at anything like x=cosh(t).

That's because that's not the speed of the rocket in your rest frame if you jump out of the train. It would be the speed of the rocket in the rest frame of someone who jumped out of the rocket, but who was stationary in the rocket before jumping, not moving with the train. But if you jump out of the train, your rest frame is a different one, which is boosted by $v$ in the $Y$ direction.

the train is going slower and slower.

On the other hand, I am flying at constant speed $v$. So the train is lagging behind.

Yes, I now agree that this will be the case. I wasn't looking at longer term effects. See my response to mairzydoats just now.

This is not very mathematical but perhaps it qualifies as analysis?

Yes, the light clock provides a good way of seeing, physically, why the train has to slow down, seen from a fixed inertial frame, as the rocket accelerates.

 

[jartsa]

Why do you think this is the case?

The floor and the ceiling are not co-moving in an inertial observers frame because in the inertial observers frame there is this kind of momentary situation: The ceiling moves at speed 10 m/s relative to the observer. The floor moves at speed 10.1 m/s relative to the observer.

(Because walls are Lorentz contracting at rate 0.1 m/s at that moment)

 

[PeterDonis]

(Because walls are Lorentz contracting at rate 0.1 m/s at that moment)

Lorentz contracting in which direction? Do you mean the $X$ direction, because of the acceleration of the rocket? How are you deriving the "rate of Lorentz contraction"? And what does this have to do with motion in the $Y$ direction?

 

[jartsa]

Lorentz contracting in which direction? Do you mean the $X$ direction, because of the acceleration of the rocket? How are you deriving the "rate of Lorentz contraction"? And what does this have to do with motion in the $Y$ direction?

Yes, I mean the $X$ direction, because of the acceleration of the rocket.

Let's say our train inside the rocket moves back and forth. It's a clock now. Clock's ticking must slow down as speed of clock increases. Ticking is back and forth motion in this case. Now let's put many identical train-clocks at different X-coordinates. The trains-clocks must go out of sync just like any other clocks at different altitudes would do.

I don't want to derive the "rate of Lorentz contraction" right now ... But if a rocket has proper acceleration a₁ and a rock towed by that rocket has proper acceleration a₂ then there is a "proper contraction acceleration" of a₂ - a₁ I mean the end of the towing rope accelerates with proper acceleration a₂ - a₁ towards the other end of the rope because of the contraction of the rope.

Previous paragraph may be safely ignored : ) Actually "rate of Lorentz contraction" is the difference of lengths at different moments, divided by time between the moments, and measured by an observer that is not accelerating. Well that's just a definition, not a derivation.

 

[andrewkirk]

I'm working through Peter's analysis and have a couple of questions.

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?

Light beams emitted: $T = 0$, $X = \delta$, $Y = 0$. Bottom of rocket at this instant (directly below light source): $T = 0$, $X = 0$, $Y = 0$. (This event is the common origin of both frames.)
...
Light beams emitted: $T' = 0$, $X' = \delta$, $Y' = 0$. Bottom of rocket at this instant (directly below light source): $T' = 0$, $X' = 0$, $Y' = 0$. (This event is the common origin of both frames.)

I don't think we can have both of those can we? The light beams are emitted from different places, so their instant of emission cannot be simultaneous in both the primed (train) and unprimed (rocket) frames can they?

I suspect there's no need for the emissions to be simultaneous, as the key measurement is heights, not times, but it may affect the notations and coordinate representations used. In what follows it's assumed they are simultaneous in both frames, which confuses me since as I understand it they cannot be.

To see what we missed, let's look at the events describing the points on the bottom of the rocket that will be directly below the back and front of the train when the light beams hit those ends of the train, at the instant the light is emitted. In the original inertial frame, these events are:

Bottom of rocket below back of train: $T = 0$, $X = 0$, $Y = - L \sqrt{(1 - v)/(1 + v)}$.

Bottom of rocket below front of train: $T = 0$, $X = 0$, $Y = L \sqrt{(1 + v)/(1 - v)}$.

Are the $X=0$ figures correct? Won't the X coordinate be positive when the beam hits the other end, since the bottom of the rocket will have accelerated away from the origin of the X,Y,T frame during the beam's flight?

Next, Rindler coordinates are mentioned but, as far as I can tell, they are not used in any calculations. Is that correct? If not, where are they used other than in the para in which they are introduced? Do I have to master Rindler coordinates in order to understand the derivation?

I expect I'll have more questions before I fully understand this, but I'd better make sure I understand these before moving on.

Thank you.

 

[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.