# B A guy in the train

#### jekozo

Hi guys,

I have read about these twins where one of them stayed on the Earth and the other traveled with a rocket. So I thought about the other case where a person stays on the trains station and the train is passing by with a near speed of light - if the guy in the train drops a ball on the table - it will fall for let say 1 sec and it will fall just vertically, while the guy one the station will see the ball falling at the same time but will fall let say a meter away.
So I thought - what if the guy on the train drops the ball outside the window - then what happens?

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#### Dale

Mentor
This question requires general relativity since it involves gravity. It cannot be answered at the B level.

Do you want an A level answer or would you like to change your question to avoid gravity?

#### jekozo

This question requires general relativity since it involves gravity. It cannot be answered at the B level.

Do you want an A level answer or would you like to change your question to avoid gravity?
Hi Dale, I would love an A level answer

#### Nugatory

Mentor
So I thought - what if the guy on the train drops the ball outside the window - then what happens?
No matter how you attack the problem, the trajectory of the ball will be the same whether it's dropped inside the train or outside (if we ignore air resistance, which may be different).

#### Ibix

An A level answer to this is to refer you to the geodesic equations. This is unnecessary for the reason @Nugatory points out - the initial conditions are the same whether you drop the ball inside the train or out (assuming we're doing this in a vacuum so the wind of the train's passage can be ignored) and the laws of physics are the same so the result is the same.

I think the underlying assumption here is that there's something "different" about being inside the train because the train is moving. This isn't correct. The only relevant thing here is the motion of a body - in this case the ball. If it starts the experiment moving horizontally at 99% of the speed of light, its motion following that will be the same whether it is inside the train or outside.

Incidentally, if the ball falls for 1s and lands 1m along the platform, it is not doing 0.99c horizontally. More like a slow walking pace.

#### jekozo

No matter how you attack the problem, the trajectory of the ball will be the same whether it's dropped inside the train or outside (if we ignore air resistance, which may be different).
I was actually thinking more about time- how it will be different now for the two participants

#### Ibix

I was actually thinking more about time- how it will be different now for the two participants
As noted already, there is no difference between "passenger drops a ball" and "passenger sticks an arm out of the window and drops a ball", until the ball hits the floor. The passenger and trackside observer have different notions of "time", it is true, but this is true and unchanged in both experiments, and makes no difference to the ball anyway.

#### sweet springs

The only relevant thing here is the motion of a body - in this case the ball. If it starts the experiment moving horizontally at 99% of the speed of light, its motion following that will be the same whether it is inside the train or outside.
I was actually thinking more about time- how it will be different now for the two participants
In IFR of the train station let two balls start falling at the same time and the same place with initial vertical speed 0 and with horizontal speed each 0 and say 0.99c. Newtonian mechanics says they touch the station ground simultaneously. I would like to know how Relativity says.

#### Ibix

Newtonian mechanics says they touch the station ground simultaneously.
Only in the limit of a very, very large planet - it takes the ball half a second to fall a metre, so this planet has to be flat to very high precision over a distance of 150,000km for your result to apply.
I would like to know how Relativity says.
I think you can do it in Schwarzschild coordinates by declaring the initial r coordinate of the ball to be $r_0$ and insisting that the "acceleration due to gravity" be $g$. That gives you a relationship between $r_0$ and $R_S$ that lets you eliminate the latter. Then you plug your initial conditions into the geodesic equations and consider the limit as $r_0\rightarrow\infty$, which is the case of a "sufficiently large" planet.

The maths appears to be messy.

#### sweet springs

Thanks.　　Gravity seems to break symmetry in translation of horizontal direction movement.

#### Ibix

Thanks.　　Gravity seems to break symmetry in translation of horizontal direction movement.
I'm not sure what you mean. Gravity defines what you mean by "horizontal". It's just that there's no easy way to describe a uniform field in GR, so you need to do something like the above to make a non-uniform field "uniform enough" to do your experiment.

#### sweet springs

Thanks. Well, train and station are often used in explaining SR. The systems are perfectly relative and have no privileges. Absolute rest has no meaning.
But gravity effect on each system in different way in general. GR breaks equity of system of relative motion in SR. Sometimes gravity may appoint a kind of absolute rest system against itself.

#### Ibix

That's not gravity breaking the symmetry, though. That's the fact that the planet is at rest with respect to either the station or the train, and which one it is matters for this scenario where it doesn't for the usual ones. It's possible to set up a Newtonian scenario where it makes no difference because you can pick one where the forces happen to be uniform and invariant. That's much harder to do in GR, not least because the stress-energy tensor of a moving body is not the same as that of a stationary one.

#### pervect

Staff Emeritus
Hi guys,

I have read about these twins where one of them stayed on the Earth and the other traveled with a rocket. So I thought about the other case where a person stays on the trains station and the train is passing by with a near speed of light - if the guy in the train drops a ball on the table - it will fall for let say 1 sec and it will fall just vertically, while the guy one the station will see the ball falling at the same time but will fall let say a meter away.
So I thought - what if the guy on the train drops the ball outside the window - then what happens?
I hope you'd agree that, with the exception of effects due to air resistance, it doesn't matter if you drop a ball inside a building, or outside a building. The ball falls in the same manner, indoors or outdoors.

The walls just don't matter as to how the ball drops, except that they make keep out the wind.

The same basic logic applies to the moving train. The presence or absence of walls doesn't affect the motion of the balls.

Perhaps you're trying to ask some other question, but it's not clear what it might be. So I'll answer the question you asked, and invite you to ask a better or different question if you have other concenrs.

Some adjustments might need to be made to your question if we start exploring other more complex questions, but as far as the effects of the walls go, it's pretty easy to answer that they just don't matter to how the ball falls.

#### FactChecker

Gold Member
2018 Award
@jekozo, Instead of saying that he dropped the ball (which brings gravity into it), say that there is no gravity and he tossed the ball down. That removes the gravity "can of worms". Also, ignore air resistance and say that it is in a vacuum. Then the answer is that it makes no difference whether the ball is in or outside of the train. There are still two different reference frames which map the ball trajectory differently.

#### PeterDonis

Mentor
In IFR of the train station let two balls start falling at the same time and the same place with initial vertical speed 0 and with horizontal speed each 0 and say 0.99c. Newtonian mechanics says they touch the station ground simultaneously. I would like to know how Relativity says.
It's easier to get rid of the planet and have the train station accelerating upward in empty space. Then you can solve the problem using SR (you don't need GR because spacetime is flat), and the solution is that the two balls hit the floor of the train simultaneously (in the frame of the train station).

#### PeterDonis

Mentor
Instead of saying that he dropped the ball (which brings gravity into it), say that there is no gravity and he tossed the ball down.
Unless the train station is accelerating upward, this not only gets rid of gravity, it gets rid of dropping altogether: if the train station floating freely in empty space and you let go of a ball, it will just float beside you.

#### sweet springs

It's easier to get rid of the planet and have the train station accelerating upward in empty space. Then you can solve the problem using SR (you don't need GR because spacetime is flat), and the solution is that the two balls hit the floor of the train simultaneously (in the frame of the train station).
Falling ball phenomena can be interpreted and used as a mechanism of clock. Vertical falling ball consuming time to floor is equal in any of these pseudo IFRs, i.e. reference frames of train, train station, etc.

Instead of saying that he dropped the ball (which brings gravity into it), say that there is no gravity and he tossed the ball down.
Then ball movement phenomena with same toss ( or impulse?, I am not sure of defining real settings.) care of not pseudo but real IFR. For these any pseudo or real IFRs i.e. train, train station, etc., due to time dilation of LT one's own vertical falling ball touch the floor first and the other's balls follow. So may I say that in a IFR vertical falling ball touch the ground level first then balls having horizontal initial velocity follow. The faster, the later.

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#### PeterDonis

Mentor
Vertical falling ball consuming time to floor is equal in any of these pseudo IFRs, i.e. reference frames of train, train station, etc
"Time to floor" is ambiguous. Simultaneity is different in different frames, so you can't say that the two balls will hit the floor "at the same time" in all frames; that will only be true in the frame of the train station, as I said. Nor can you say that the proper time of the two balls from being dropped to hitting the floor is the same: it isn't, because the balls are moving relative to each other. So you need to be very careful how you make statements about "time".

#### pervect

Staff Emeritus
The simplest point of view is to look at the situation in an inertial frame of reference. Then the balls are following geodesics which are straight lines, and the elevator floor is accelerating up to meet them. Later, we will interpret this as the balls "falling" in the non-inertial frame, but we'll analyze this in an the inertial frame of the balls.

Let's say that the floor is accelerating upwards in the z direction, the balls are lined up in a grid on the x and y directions and the balls all have the same height z at any time t.

Then one can say that if the floor is also flat, the floor will strike all the balls at the same time, regardless of their x and y positions, and if the floor is not flat, then it will not.

This is all well and good, but it turns out that the flatness of the floor is frame-depenent. If you pick another frame of reference, also inertial, but moving in , say, the x direction, the Lorentz contraction reads

$$x' = \gamma(x - \beta t) \quad t' = \gamma(t - \beta x/c)$$

where $\beta$ is the normalized velocity of the motion in the x direction which is equal to v/c

By out definition of the problem, z is a constant for the floor at any time t. But this defintion implies that z cannot be constant at any time t' in the moving (moving in the x direction) inertial frame we've been discussing. This happens because t' is not equal to t, they differ because of the relativity of simultaneity, the term that makes t' a function of both t and x.

See for instance figures 3 and 11 in https://arxiv.org/abs/0708.2490v1

The paper is about another effect (one which is also interesting but irrelevant to this thread, and effect called Thomas Precession). However, it illustrates some interesting facts about the floor, we can say that if the floor is flat in an a specific inertial frame that we declare to be "at rest" in the x and y directions, it will not be flat in another inertial frame that's moving in the x direction. (Or in the y direction, for that matter, but the case above analyzed the motion in the x direction)

To snip the relevant images from a similar post I did a while back:

The Lorentz transformation is linear, so one expects it to map straight lines into straight lines. But the accelerating worldline of the elevator floor is not a straight line. On a space-time diagram, it's curved. This curvature on the space-time diagram is enough to make the spatial flatness of the floor depend on one's frame of reference.

#### sweet springs

So are the reference frame of train station or rocket floor are special among other y-moving frames with speed v ? In my previous post I assumed they are all relative and no one play a role of "absolute rest".

#### jartsa

Why don't the vertical velocities of the rail ends result in a tilted rail?

How has the left end of the rail advanced more than the middle of the rail?

Hmmm ... ok I have figured it out, the rail is both curved and tilted. Because that's what the result of the depicted motions is.

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#### pervect

Staff Emeritus
So are the reference frame of train station or rocket floor are special among other y-moving frames with speed v ? In my previous post I assumed they are all relative and no one play a role of "absolute rest".
Perhaps I'll think of a better way to explain it more intuitively, but at the moment all I have time to do is go through what the math says.

The ship with the flat floors is accelerating in such a manner that a gyroscope does not precess, with some acceleration g.

The block sliding across the "flat floor" of this accelerating ship has a different value of proper acceleration, and a gyroscope mounted on the block does precess, and it has a different, higher value for it's proper acceleration. $\gamma^2 g$, as I recall.

For the first point, see https://arxiv.org/abs/0708.2490v1

Maybe more later, I have some distractions pulling me away.

#### pervect

Staff Emeritus
I suppose the best approach is to draw a space-time diagram with some automated software. We will first consider a specific inertial frame.

First, we need to write the equations. To make things simple, we will approximate the acceleration of the elevator with the Newtonian formula, $z = t^2 / 2$. This will be a good aproximation for small t.

If you want to do the full relativistic version, with a change of origin so that z=1 when t=0 rather than z=0 when t=0, we can write:

$$z = \sqrt{t^2+1}$$

For small t this is approximately 1 + 1/2 t^2, for large t it apprroaches t.

Then our non-boosted elevator has a mathematical description of

$$z = \frac{1}{2} t^2$$

which plots like this. It's just a parabola in the {z,t} plane, with the value of z independent of x.

On this diagram, the surface of the elevator is the set of points {z,x} that satisfy the above relationship at some particular time t. We will consider specifically t=0, then the set of points composing the elevator is z=0. We can represent this by plotting z vs x, and finding that z is constant, the floor of the elevator is a straight line. It's a straight line and not a plane because we've surpressed the y dimension in our analysis, so we can draw a 3d space-time diagram.

Next, we apply the lorentz transforms. We will write

$$t' = \gamma (t - \beta x) \quad x' = \gamma(x - \beta t) \quad z'=z$$
$$t = \gamma(t' + \beta x) \quad x = \gamma(x' + \beta t) \quad z=z'$$

Now, on this space-time diagram, the surface of the non-boosted elevator at t=0 is the set of points {z,x} that satisfy our relation for z. It's easy enough to see that z is not a function of x, so if we plot z versus x, we get a straight line.

On the boosted space time diagram, we want the surface of the elvator at t'=0, rather than t=0. So we write z' as a function of t' and x', which gives us the following

$$z'= z = \frac{1}{2} t^2 = \frac{1}{2} \gamma^2 ( t' + \beta x')^2$$

We can see that at t'=0

$$z' = \frac{1}{2} \gamma^2 \beta^2 x'^2$$

which is the floor of the elevator that was constructed to be flat in our first coordinate system. We can see that it's no longer flat in the new coordinate system.

With $\beta=3/5$ and $\gamma=5/4$, the complete space-time diagram for the elevator looks like this.

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#### sweet springs

that will only be true in the frame of the train station, as I said.
Let me say the balls free fall from the same point and same time with various horizontal speed, they hit the ground simultaneously in the train station (pseudo) IFR. In y-moving (pseudo) IFR, i.e. train IFR, the balls hit the ground in different times. By this experiment we can identify which y-moving (pseudo) IFR is "at rest".

I wonder in train IFR initial speed of these balls are horizontal and the vertical falling ball go down straight, not curved.

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