Twins Paradox: Dropping a Ball from a Moving Train

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In summary: I think this is what you're trying to say.If you drop a ball on a table, it will fall at the same time (inertial frame of reference) as someone watching it from the ground. If you drop the ball on a train, the passenger on the train and the person watching it from the ground will have a different idea of when the ball falls, due to the train's motion.
  • #1
jekozo
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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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  • #2
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?
 
  • #3
Dale said:
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
 
  • #4
jekozo said:
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).
 
  • #5
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.
 
  • #6
Nugatory said:
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
 
  • #7
jekozo said:
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.
 
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  • #8
Ibix said:
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.
jekozo said:
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.
 
  • #9
sweet springs said:
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.
sweet springs said:
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 let's 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.
 
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  • #10
Thanks.  Gravity seems to break symmetry in translation of horizontal direction movement.
 
  • #11
sweet springs said:
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.
 
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  • #12
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.
 
  • #13
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.
 
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  • #14
jekozo said:
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.
 
  • #15
@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.
 
  • #16
sweet springs said:
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).
 
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  • #17
FactChecker said:
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.
 
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  • #18
PeterDonis said:
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.

FactChecker said:
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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  • #19
sweet springs said:
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".
 
  • #20
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:

achment-php-attachmentid-66480-d-1392063282-20-png.png


achment-php-attachmentid-66481-d-1392063282-20-png.png


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.
 
  • #21
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".
 
  • #22
pervect said:

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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  • #23
sweet springs said:
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.
 
  • #24
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.

[add]
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.

?hash=f631d5810b4e1d4428970a1b76185079.gif

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.

241669
 

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  • #25
PeterDonis said:
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.
 
  • #26
sweet springs 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".

Let's say all this happens inside an accelerating rocket. (Or inside a rocket hovering in a gravity field)

1: When a free falling (= inertial) observer looks at a free falling ball, he sees that the ball is moving inertially.

2: When a free falling observer looks at the rocket's walls he sees that the walls are moving non-inertially. The walls accelerate downwards. If the walls have any horizontal speed that speed is affected by the gravitational time dilation that is increasing as the walls are moving to lower gravitational potential.

If we want to know how a falling observer sees the distance between the walls and the falling ball changing, those two rules above give us the answer. Right?Oh, now I understand that "pseudo-inertial frame" is probably the frame of an observer standing on the surface of a planet. So I talked about a very different situation.
 
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  • #27
From https://en.wikipedia.org/wiki/Rindler_coordinates Geodesics, if my poor math succeeds to work
[tex](\frac{d\hat{x}}{d\tau})^2=\frac{\hat{H}^2}{\hat{x}^2}-1[/tex]
where H is the initial height with 0 vertical speed and P,Q horizontal speed
[tex]\hat{x}=\frac{x}{\sqrt{1+P^2+Q^2}}[/tex]
[tex]\hat{H}=\frac{H}{\sqrt{1+P^2+Q^2}}[/tex]

I cannot solve it analytically. From this Horizontal speed change scale but does not seem to affect x-movement. But if we set floor or ground at x=H_0<H, since it is not scaled like x or H, I suspect P and Q matters on ball hitting timing. Your correction and teaching are appreciated.
 
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  • #28
sweet springs said:
From https://en.wikipedia.org/wiki/Rindler_coordinates Geodesics, if my poor math succeeds to work
[tex](\frac{d\hat{x}}{d\tau})^2=\frac{\hat{H}^2}{\hat{x}^2}-1[/tex]
where H is the initial height with 0 vertical speed and P,Q horizontal speed
[tex]\hat{x}=\frac{x}{\sqrt{1+P^2+Q^2}}[/tex]
[tex]\hat{H}=\frac{H}{\sqrt{1+P^2+Q^2}}[/tex]

I cannot solve it analytically. From this Horizontal speed change scale but does not seem to affect x-movement. But if we set floor or ground at x=H_0<H, since it is not scaled like x or H, I suspect P and Q matters on ball hitting timing. Your correction and teaching are appreciated.

This doesn't look quite right - perhaps the Wiki's abbreviated notation is throwing you.

Let's start with the setup. The geodesic equations are equations in four variables, t,x,y,z, each of which is a function of proper time, ##\tau##. The path of the geodesic worldine is thus specified by:

$$t(\tau) \quad x(\tau) \quad y(\tau) \quad z(\tau)$$

Not that the "dot" operator in the wiki represents differentiation by proper time ##\tau##, not the time coordinate t.

Then the geodesic equations from the wiki are:

$$\frac{d^2 t}{d\tau^2} + \frac{2}{x} \frac{dt}{d\tau}\frac{dx}{d\tau} = 0 \quad \frac{d^2x}{d\tau^2} + x \frac{dt}{d\tau}\frac{dt}{d\tau} = 0$$

Wiki's notation is that ##\dot{x} = \frac{dx}{d\tau}## and ##\dot{t} = \frac{dt}{d\tau}##, because t and x are functions of proper time, and the dot operator rerpesents differentiation with respect to proper time.

Wiki skips over the next step, which is to note that the first geodesic equation in ##\ddot{t}## is equivalent to

$$\frac{d \left( x^2 \frac{dt}{d\tau} \right)} { d\tau} = 0$$

which can be seen by applying the chain rule

or in Wiki's dot notation

$$\frac{d( x^2 \dot{t} ) }{d\tau} = 0$$

This relation leads to the result that
$$\dot{t} = \frac{dt}{d\tau} = \frac{E}{x^2}$$

here E is an arbitrary constant, the point is that if something doesn't vary with proper time, it's constant, and we call that constant E. It turns out to be related to the energy. The justification for this likes in the Killing vectors of the space-time, but to explain further would be a digression.

Wiki then suggests solving the equation by imposing the line element condition for a 4-velocity, rather than using the remaining geodesic condition. With the -+++ metric convetion being used, the magnitude of a four-velocity is always -1. This means that

$$-x^2 \left( \frac{dt}{d\tau} \right)^2 + \left( \frac{dx}{d\tau} \right)^2 + \left( \frac{dy}{d\tau} \right)^2 + \left( \frac{dz}{d\tau} \right)^2 = -1$$

We can apply our previous definition of E to simplify this even further. Wiki also notes that the geodesic equation ##\ddot{y}=0## leads to ##\dot{y} = P## and ##\ddot{z}=0## leads to ##\dot{z}=Q##. This means that ##\frac{dy}{d\tau}## and ##\frac{dz}{d\tau}## are constant, you may remember this from previous posts. The argument can again be based on Killing vectors - in this case, the Killing vectors are the conserved y and z components of momentum rather than energy.$$ -\frac{E^2}{x^2} + \left( \frac{dx}{d\tau} \right)^2 + P^2 + Q^2 = -1$$

This should be more tractable to solve, but I haven't done that, I've just restated what Wiki did, hopefully with a more understandable notation and a bit less tersely.
 
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  • #29
It's a bit messy, and I might have made an error, even with the symbolic algebra package I'm using,but with suitable initial conditions, I get the for the solution of the geodesic equations.

Following wiki <link>, I've set a=0 and the speed of light c=1, so the metric I'm using is:

$$-x^2 dt^2 + dx^2 + dy^2 + dz^2$$

This leads to the geodesic equations from my previous post, for completeness I'll duplicate them here:

$$\frac{d^2 t}{d\tau^2} + \frac{2}{x} \frac{dt}{d\tau}\frac{dx}{d\tau} = 0 \quad \frac{d^2x}{d\tau^2} + x \frac{dt}{d\tau}\frac{dt}{d\tau} = 0$$

Recall that t,x,y, and z are the Rindler coordinates, while ##\tau## is proper time. t,x,y,z are all functions of ##\tau##.

There are several versions of the Rindler metric. This version is one of the simplest to work with. Note that the plane given by setting the coordinate x=0 is a coordinate singularity, the metric is singular there as the coefficient of dt^2 is zero. This is often called the Rindler horizon. A particle would need infinite proper acceleration to stay on the Rindler horizon.

If one measure time t in years and distance x in light years, the plane x=1, which is one light year from the Rindler horizon, has a proper acceleration of 1 light year/year^2, which is approximately 10 m/s^2, one Earth gravity. So we can consider x=1 the location of the floor of the elevator, and with the convenient choice of units of measuring time in years and distance in light years, the proper acceleration of the floor of the elevator is 1 Earth gravity.

The solutions I'm getting are then:

$$ x(\tau) = \sqrt {{\frac {{E}^{2}}{{P}^{2}+1}}- \left( {P}^{2}+1 \right) {\tau}^{2}}$$
$$t(\tau) = -\frac{1}{2}\,\ln \left( E - \tau - \tau\,{P}^{2} \right) +\frac{1}{2}\,\ln \left( E+\tau+\tau\,{P}^{2} \right)$$
$$y(\tau) = P \, \tau$$
$$z(\tau)=0$$

P and E are constants of motion. Physically, P relates to the linear momentum in the y direction, E to the energy. I've assumed that there is no velocity or momentum in the z direction.

The maximum height (height is the x coordinate here) occurs at ##\tau=0## and is given by ##E/\sqrt{1+P^2}##. So, if one knows the maximum height, one can compute E. Seting the maximum height to occur at ##\tau=0## was a simplifying choice of the initial conditions I made.

t=0 also occurs at ##\tau=0##. ((correction from first draft)).

I'd recommend re-verify that these equations actually satisfy the geodesic equations - baring typos, though, my symbolic algebra program claims that they do.

[add]
To anticipate the next question, we can find an expression for ##\tau## as a function of t
$$\tau = \frac{E}{P^2+1} \tanh t$$

Using this, we can find x as a function of t, rather than of tau

$$x = \sqrt{\frac{E^2}{P^2+1}(1-\tanh^2 t)} = x_{max} \sqrt{1 - \tanh^2 t}$$

where ##x_{max}## is the maximum value of x, which occurs at ##\tau=0##. So we can see that as a function of coordinate time t, only the maximum height matters.
 
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  • #30
Thank you so much for thorough analysis, perverct.

#1
pervect said:

y(τ)=Pτy(τ)=Pτ​

y(\tau) = P \, \tau
#2
pervect said:

τ=EP2+1tanhtτ=EP2+1tanh⁡t​

\tau = \frac{E}{P^2+1} \tanh t
These formula made me think of y-moving FR in the following way.

Let (t,x,y) be Rindler coordinate and (t',x',y') be coordinate of y-moving FR against Rindler FR.
In Rindler FR, x' axis rod moves y-direction with constant speed, say.
[tex]y=Vt[/tex]
This formula means that in Rindler FR the different part of x' axis rod has different local y- speed according to its x' value written on the axis. The larger, the slower. Light speed at x=0 limit, at the event horizon.

Let the #1 ball and the y-moving FR x' axis has same speed at initial time in Rindler FR by choosing appropriate value of P and V. The ball is at rest in the y-moving FR at t=t'=0. The two trajectories in Rindler FR, ##y=P\tau## and ##y=Vt## do not coincide as #2 suggests. It means that in y-moving FR the ball DO NOT fall along x' axis, i.e. keeping y'=0.

So may I say that among y-moving FRs there is one special FR where the ball put on its x-axis with no y-speed falls along x axis, vertically. This is Rindler FR.
 
  • #31
In other words in y-moving FR against Rindler FR , direction of gravity acceleration does not coincide with x' axis and horizontal direction if it means orthogonal to gravity acceleration does not coincide with y' axis.

So as far as falling or x-motion is concerned, Rindler FR and y-moving FR are not relative in the sense of SR.

cf. I recall the discussion I joined in the thread
https://www.physicsforums.com/threads/gravity-on-einsteins-train.835994/
Back to OP address, applying Rindler case to Earth gravity case, train FR and train station FR are NOT RELATIVE in the sense of SR, as far as falling motion by gravity is concerned. Gravity acceleration is vertical or along x-axis in train station FR. It is not vertical nor along x' axis but perpendicular in train FR.
In other thread I call this gravity breaks "symmetry" among FRs. I now think relativity ( in the sense of SR) is a better word. Searching which FR show orthogonal direction of gravity acceleration against given floor or ground of constant X, we can identify train station FR which plays a special role in the case.
 
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  • #32
sweet springs said:
In other words in y-moving FR against Rindler FR , direction of gravity acceleration does not coincide with x' axis and horizontal direction if it means orthogonal to gravity acceleration does not coincide with y' axis.

I just did a quick calculation to confirm my recollection this is not true. The acceleration 4-vector is given by
$$a^b = u^a \nabla_a u^b$$

It turns out that this vector points in the same direction for both a point at rest on the elevator floor, and a point sliding along the floor.

The most notable physical effect in what passes for the frame of reference for the sliding block is not a change in the direction of gravity, but rather Thomas precession. This is not a change in "gravity", it's more subtle. It means that a gyroscope that initially points "up", towards some distant fixed star, will not continue to point "up" if it is mounted on a sliding block. This is in contrast to the gyroscope mounted on the floor of the elevator, which will not precess and will continue to point towards the distant fixed star.

For a paper which describes the effect above, see https://arxiv.org/abs/0708.2490v1, which takes a non-tensor approach, though it's still not terribly easy to follow. I believe I've mentioned that paper previously.

The Thomas precession may not seem like something that one is necessarily interested in, but it's not really possible to fully understand the sliding block frame without understanding this property it has. See more below.

The approach I suggest to understanding what we might mean by "the frame of a sliding block" in the spirit of General Relativity is to write a metric. This spawned a very long thread, which I think you're aware of, where we explored just what metric would be associated with a sliding block. In the end, we came up with several different candidates for what one might mean by "a frame of a sliding block". The particular chocie I espoused was "the proper reference frame" of the sliding block, but there are other candidates that also have their merits. I won't rehash that here without some guidance, the thread was about 11 pages long.

It is only necessary to define a metric if one wants to understand what I call an "extended" frame of reference. In a purely local sense, one can define a frame of reference as a set of four vectors. However, this defintion of a "frame of reference" is only at a point, it does not encompass any experiment that has a non-zero volume. The problem of dropping a ball is an example of a problem that requires a frame of reference that encompasses a non-zero volume and not just a poit. So if we wanted to drop a ball in the sliding block frame, we'd need a metric.

The gyroscope precession is relevant to the path a ball dropped on the sliding block would take, because there will be generalized forces on the sliding block, "fictitious forces", if you will, that are essentially coriolis forces. These fictitious forces cause the gyroscope to precess, and they'd deflect the path of any falling body in the sliding block frame.

Defining a metric allows pretty much anything that GR can calculate to be calculated. Also, any complete description of a set of coordinates must allow one to compute the metric (otherwise it's not a complete description). So defining a metric is both necessary and sufficient in the context of GR to define a set of coordinates. There's a paper by Misner that mentions this, that I can quote. Outside the context of GR, mathematicans may demand more than this, but for instance the IAU defines the set of coordinates used to describe the ICRF, the International Celstial Reference Frame, by writing down it's metric.

From the GR point of view, the space-time itself is flat in the Minkowskii frame, the Rindler frame, and whatever frame one decides to use to represent the sliding block. To be more specific, the Riemann curvature tensor vanishes, and a tensor is a coordinate independent entity, so that if it vanishes in one coordinate system it vanishes in all coordinate systems.

Basically, the different "frames" are just different coordinate choices. It's perfectly possible and possibly much easier to use the Minkowskii frame to do all the calculations, then convert it to the frame of choice.

"The frame of choice" must be sufficiently well defined to allow one to translate the results from the Minkowskii frame to the new frame, however. One way of doing this is to define a map (called a diffeomorphism) that associates points between the Minkowskii frame and whatever frame one chooses to use. But this is not necessary, the "frame" can be defined by it's metric without any reference at all to Minkoskii space.

The machinery of GR is sufficiently powerful for one to do the computations in any frame one desires. It may be easier to do all the calculations in the Minkowskii frame and convert back to the new frame, but one can also do the calculations in an arbitrary frame of reference. The procedure for doing so starts write down the metric associated with the new frame. The metric defines the coordinates that one has chosen, and also allows one to use tensor methods (such as the geodesic equations) to compute quantites of physical interest in a manner that's native to that frame.
 
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  • #33
PeterDonis said:
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).

I am thinking of the message again. In IFR where the train station is accelerating upward x, the balls with various initial y-velocities released from an air point in a moment hit the train floor simultaneously. In another IFR which moves y direction with ##V_y>0## against the original IFR where the moving train keep the same y position, due to different synchronization of clocks from the original IFR, these balls hit the floor in different timings. The balls with negative y initial speed in new IFR hit the floor. The faster, the sooner. Then the balls with positive y initial speed hit the floor. The faster, the later.
 
  • #34
sweet springs said:
In another IFR which moves y direction with Vy>0 against the original IFR where the moving train keep the same y position,

Ambiguity lies in this postulate. Does the train, with no friction and no engine working, keep constant y-position in this new IFR? Suggested by Peter conservation of y momentum of the train results decreasing y- speed in the original IFR, that should result the train does not keep the y-position in the new IFR,i.e. the train once stays still in the new IFR gradually starts moving in negaive y-direction.

So in FR of moving train, that is not a IFR, a ball with zero initial y velocity move positive y-direction and hit positive y displaced point on the floor. In the train FR a free fall ball with no initial y-velocity does not fall vertically but is distracted to positive y direction.
 
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  • #35
sweet springs said:
Suggested by Peter conservation of y momentum of the train results decreasing y- speed in the original IFR, that should result the train does not keep the y-position in the new IFR,i.e. the train once stays still in the new IFR gradually starts moving in negaive y-direction.
[ed]
As for
i.e. the train once stays still in the new IFR,
I would like to add for precision,
i.e. the train once stays still as for y direction in the new IFR,

It is about the case of train station accelerating upward in empty space. Does it also stand in the case of train station and running train on Earth ? With idealized constant gravity acceleration g on Earth, does a free fall ball released from ceiling of the train, which is moving in positive y direction with no friction and no engine working on rail, hit not straight down but a y forward displaced point on the floor ?
Does the animation for familiar lihgt pulse zigzag in train, e.g. https://www.physicsforums.com/threads/light-in-a-moving-train.884811/page-2#post-5565955, stands only with no gravity in a strict sense ?
 
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<h2>1. What is the Twins Paradox?</h2><p>The Twins Paradox is a thought experiment in physics that explores the concept of time dilation in special relativity. It involves two identical twins, one of whom travels in a high-speed vehicle while the other stays on Earth. When they are reunited, the traveling twin will have aged less due to the effects of time dilation.</p><h2>2. How does dropping a ball from a moving train relate to the Twins Paradox?</h2><p>The thought experiment of dropping a ball from a moving train is often used to explain the concept of time dilation in the Twins Paradox. In this scenario, the ball is dropped from the train as it moves at a high speed, and the observer on the train and the observer on the ground will perceive the ball's motion differently due to time dilation.</p><h2>3. How does time dilation occur in the Twins Paradox?</h2><p>Time dilation occurs in the Twins Paradox due to the effects of special relativity. When an object moves at high speeds, time moves slower for that object compared to a stationary observer. This means that the traveling twin in the thought experiment will experience time passing slower than the stationary twin on Earth, resulting in a difference in their ages when they are reunited.</p><h2>4. Can the Twins Paradox be observed in real life?</h2><p>While the Twins Paradox is a thought experiment, the concept of time dilation has been observed and confirmed in experiments involving high-speed particles and atomic clocks. However, the effects of time dilation would only be noticeable at extremely high speeds or in extreme gravitational fields, making it difficult to observe in everyday life.</p><h2>5. How does the Twins Paradox challenge our understanding of time?</h2><p>The Twins Paradox challenges our understanding of time by showing that it is not a constant, but rather can be affected by factors such as speed and gravity. It also raises questions about the relativity of time and how it can be perceived differently by different observers. The thought experiment highlights the complex nature of time and the need for a deeper understanding of the universe.</p>

1. What is the Twins Paradox?

The Twins Paradox is a thought experiment in physics that explores the concept of time dilation in special relativity. It involves two identical twins, one of whom travels in a high-speed vehicle while the other stays on Earth. When they are reunited, the traveling twin will have aged less due to the effects of time dilation.

2. How does dropping a ball from a moving train relate to the Twins Paradox?

The thought experiment of dropping a ball from a moving train is often used to explain the concept of time dilation in the Twins Paradox. In this scenario, the ball is dropped from the train as it moves at a high speed, and the observer on the train and the observer on the ground will perceive the ball's motion differently due to time dilation.

3. How does time dilation occur in the Twins Paradox?

Time dilation occurs in the Twins Paradox due to the effects of special relativity. When an object moves at high speeds, time moves slower for that object compared to a stationary observer. This means that the traveling twin in the thought experiment will experience time passing slower than the stationary twin on Earth, resulting in a difference in their ages when they are reunited.

4. Can the Twins Paradox be observed in real life?

While the Twins Paradox is a thought experiment, the concept of time dilation has been observed and confirmed in experiments involving high-speed particles and atomic clocks. However, the effects of time dilation would only be noticeable at extremely high speeds or in extreme gravitational fields, making it difficult to observe in everyday life.

5. How does the Twins Paradox challenge our understanding of time?

The Twins Paradox challenges our understanding of time by showing that it is not a constant, but rather can be affected by factors such as speed and gravity. It also raises questions about the relativity of time and how it can be perceived differently by different observers. The thought experiment highlights the complex nature of time and the need for a deeper understanding of the universe.

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