# I Gravity on Einstein's train

1. Oct 4, 2015

### 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...

2. Oct 4, 2015

### Staff: Mentor

Ok so far.

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.

3. Oct 5, 2015

### jartsa

Yes.

I guess it does.
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.

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

4. Oct 5, 2015

### jartsa

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.

5. Oct 5, 2015

### Staff: Mentor

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.

6. Oct 5, 2015

### 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 ?

7. Oct 5, 2015

### SlowThinker

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

8. Oct 5, 2015

### Markus Hanke

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

Last edited: Oct 5, 2015
9. Oct 5, 2015

### jartsa

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.

Last edited: Oct 5, 2015
10. Oct 5, 2015

### SlowThinker

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.

11. Oct 5, 2015

### SlowThinker

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.

12. Oct 5, 2015

### Staff: Mentor

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.

13. Oct 5, 2015

### phinds

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

14. Oct 5, 2015

### 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.

15. Oct 5, 2015

### SlowThinker

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.

16. Oct 5, 2015

### Staff: Mentor

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.

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.

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.

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.

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.

17. Oct 5, 2015

### SlowThinker

Thanks Peter, perhaps you missed this sentence in my very first post.
And yes, I'm trying to improve my intuition.

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

To clear the last few points,
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.

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

18. Oct 5, 2015

### Staff: Mentor

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 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.

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.

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.

19. Oct 5, 2015

### SlowThinker

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.
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.

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.

"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.

20. Oct 6, 2015

### Staff: Mentor

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.

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.

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.

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.

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.