# B A guy in the train

#### jartsa

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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#### sweet springs

From https://en.wikipedia.org/wiki/Rindler_coordinates Geodesics, if my poor math succeeds to work
$$(\frac{d\hat{x}}{d\tau})^2=\frac{\hat{H}^2}{\hat{x}^2}-1$$
where H is the initial height with 0 vertical speed and P,Q horizontal speed
$$\hat{x}=\frac{x}{\sqrt{1+P^2+Q^2}}$$
$$\hat{H}=\frac{H}{\sqrt{1+P^2+Q^2}}$$

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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#### pervect

Staff Emeritus
Science Advisor
From https://en.wikipedia.org/wiki/Rindler_coordinates Geodesics, if my poor math succeeds to work
$$(\frac{d\hat{x}}{d\tau})^2=\frac{\hat{H}^2}{\hat{x}^2}-1$$
where H is the initial height with 0 vertical speed and P,Q horizontal speed
$$\hat{x}=\frac{x}{\sqrt{1+P^2+Q^2}}$$
$$\hat{H}=\frac{H}{\sqrt{1+P^2+Q^2}}$$

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.

• sweet springs

#### pervect

Staff Emeritus
Science Advisor
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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• sweet springs

#### sweet springs

Thank you so much for thorough analysis, perverct.

#1

y(τ)=Pτy(τ)=Pτ​

y(\tau) = P \, \tau
#2

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.
$$y=Vt$$
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.

#### sweet springs

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

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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#### pervect

Staff Emeritus
Science Advisor
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.

• sweet springs

#### 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).
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.

#### sweet springs

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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#### sweet springs

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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#### jartsa

Here in this video there's a rail whose leftmost point starts accelerating upwards, the next point does exactly the same thing at the same time in the frame of the rail. In our frame the point next to the leftmost point imitates the leftmost point with a small delay.

I did not consider anything else than that change of simultaneity when I made the video. That should suffice, right?

https://ibb.co/WtgNJD8

The cause of the acceleration might be that the rail starts to emit radiation from it's underside. The direction of the emitted radiation would be changing in our frame, it seems. I mean in the frame of our monitors.

Oops, I did not consider that the change of orientation causes a change in the simultaneity. I mean when a clock bolted on the the left side of the rail moves faster than a clock bolted on the right side of the rail, the clocks become less out of sync in our frame.

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#### sweet springs

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.
[EDIT]
The balls with positive y initial speed in new IFR hit the floor. The faster, the sooner. Then the balls with negative y initial speed hit the floor. The faster, the later.

I took it wrong.

#### sweet springs

Here in this video there's a rail whose leftmost point starts accelerating upwards, the next point does exactly the same thing at the same time in the frame of the rail. In our frame the point next to the leftmost point imitates the leftmost point with a small delay.
Thanks for animation. Your point seems due to different synchronization at a moment floor is not horizontal flat but perpendicular. Figure 11 in Post #20 and figure in Post #24 seem the train is at the bottom. I am confusing. Which is right shape of the floor the train people conceive, perpendicular or bottom ?

#### jartsa

Thanks for animation. Your point seems due to different synchronization at a moment floor is not horizontal flat but perpendicular. Figure 11 in Post #20 and figure in Post #24 seem the train is at the bottom. I am confusing. Which is right shape of the floor the train people conceive, perpendicular or bottom ?
In my animation the direction of the acceleration is exactly vertical. Points experince a vertical velocity change, but no horizontal velocity change.

The correct force that will cause that particular acceleration is not a vertical force, but some other force. See this for details: https://en.wikisource.org/wiki/The_Direction_of_Force_and_Acceleration

So the floor is tilted (not horizontal) and the force that pushes the floor, or a ball standing on the floor, is tilted (not vertical). If the floor and the force are tilted the same amount, then the ball does not start rolling on the floor.

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#### jartsa

In this video there's a nozzle of a rocket that starts firing simultaneously at all point in its rest frame. Let's say it's a photon rocket. The photons come from a big container above the nozzle.

Now let's consider the photon gas inside the container. At the beginning of the video the net momentum of the gas points to the right. At the end it points to the right and upwards.

We can see that at the beginning the gas below the nozzle has gained some downwards momentum, or it has lost some upwards momentum - let's use the latter phrasing. So the gas has lost some upwards momentum, which has gone into the rocket.

At the end the gas below the nozzle has approximately no momentum at all, because of the redshift of the expelled photons. So the gas has lost some upwards momentum and some rightwards momentum, which have gone into the rocket.

So we have: At the beginning the gas pushes the rocket upwards. At the end the gas pushes the rocket upwards and rightwards.

That rightwards push is the supposedly interesting point here. Like, maybe I made an error and there is no rightwards push. Last edited:

#### sweet springs

We can see that at t'=0

z′=12γ2β2x′2z′=12γ2β2x′2​

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.
Thanks. I interpret this as follows.

#1 In this train (I?)FR where train is at still in x' direction and also in z' direction, time advances in positive x' and it retrogresses in negative x' direction by change of synchronicity from that of (I?)FR of rocket.

#2 The rocket have approached from far positive z' decreasing its speed, stop now at its smallest z' position and will return to far positive z' increasing speed.

Combining #1 and #2, the left ward x' floor shows past z' position and the rightward z' floor shows future z' position. Both are higher than position now so the floor is parabola shaped.

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#### sweet springs

(contd.)

#3 In new IFR where the train is momentary at still in X' direction and moving in positive Z' direction with acceleration, time advances in positive X' and it retrogresses in negative X' direction by change of synchronicity from that of original IFR where rocket is accelerating with no x movement.

#4 In original IFR the rocket approached from far positive Z decreasing its speed, stopped at its smallest Z position and is returning to far positive z' increasing speed.

Combining #3 and #4, the left ward X' floor shows past Z' position and the rightward Z' floor shows future Z' position. With right side high and left side low the floor is tilted, together with higher order non linear component in a strict sense.

I would like to make a naive conjecture that the train is on slope right side up. Gravity have component left ward to let train start toward negative X'. This force let train FR to transfer from one instantaneous IFR to another instantaneous IFR where X' position of train is at still in a moment. The slope would become milder as rocket accelerated so that X momentum is conserved in original IFR, i.e.
$$\beta_{X}^2=\frac{u_{X}^2}{1+u_{X}^2}(1-\beta_{Z}^2)$$
where
$$u_{X}=const$$
representing conserved x momentum of train multiplied by mc.
$\beta_{Z}$ approaches to 1 as time goes. Thus train speed $\beta_{X}$ approaches down to zero.

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#### sweet springs

To snip the relevant images from a similar post I did a while back:
Figure 3 of right drop angular momentum would be interpreted by slope in post #42. In new IFR angular momentum is horizontal but floor is tilted or slope right side up. They make an acute angle. I should appreciate to know the estimation of the angle in the paper.

#### sweet springs

I just did a quick calculation to confirm my recollection this is not true. The acceleration 4-vector is given by

ab=ua∇aubab=ua∇aub​

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.
I will be happy if I can confirm that this is calculation between my original IFR and new IFR. FR where train keep same x position is not a IFR. Though not a IFR, this FR is naturally realized system under artificially conditioned Z acceleration. In this FR free fall ball trajectory deviates from vertical line.

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#### sweet springs

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 ?
Now I think I am prepared to say YES to my question. In FR of constant gravity on Earth, we should mind that Earth as source of gravity is at rest in a very rough sense. For another FR of train where Earth is moving right or leftward, earth gravity works not only vertical but has some right or left component. This component of force drag and let train move together with Earth. Or in the FR where Earth is still, the force reduces train speed.

Synchronized rocket engines FR or Still Earth FR plays a role of, say "absolute rest" FR. In these cases relativity does not work the way as enjoyed in SR.

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#### pervect

Staff Emeritus
Science Advisor
The Earth moving past a stationary point is a different and more complex problem than the one I worked out with Einstein's train on Einsteins elevator.

I'd suggest the Olson and Guarnio paper, "Measuring the active gravitational mass of a moving object ", if you can find a copy somewhere that is not paywalled, or request one from a local library. If one imagnes a static dust cloud, and a heavy object (such as the Earth, or a black hole), flying through the cloud, there will not be a constant gravitational field while the object passes through the cloud, but a time-varying one. It is convenient to look at the total velocity change of a particle in the dust cloud induced by the passage of the heavy object.

This is convenient especially because the space-time is flat before and after the passage of the heavy object, though it is not flat while the object is flying through the cloud. The presence of curved space-time confuses the GR calculations for the reader not intimately familiar with the techniques needed to deal with curvature, and some of the logical consequences of it's presence. Unfamiliarity with curvature and the associated formalisms results in mistakes that are not easy to communicate, because the person unfamiliar with the detailed mathematical expression of the theory lacks the necessary specialized knowledge on how to correctly solve the problem. Unfortunately rather than obtaining the necessary background knowledge to really understand the theory, which is quite a long process, it is not uncommon for someone to try and take shortcuts and compute what they think GR ought to say according to their imperfect understanding of the theory, rather than go through or look up the actual calculations and the mathematics behind them. Unfortunately, it is rather unlikely that a person not familiar with the formalisms of GR who does some intuition based non-rigorous calculations will get the correct answer to this problem. I would thus suggest looking up the result of a correct calculation, such as the paper I referenced, to avoid this pitfall.

Olson and Guarino find both the Newtonian and General relativistic formula for the velocity change imparted to particles in the dust cloud due to the passage of the moving body in their paper. At any point in the cloud, we will have some component of the resulting velocity change that is perpendicular to the line of motion of the heavy body, and another component that is parallel to this line. Both components are non-zero. Olson and Guarino work out the magnitude of the total velocity change in both the Newtonian case and the GR case, and compare the two. Because of the flatness of space-time before and after the passage of the massive body, the total velocity change is perfectly well defined. Relative velocity of objects at different locations in the presence of strong curvature is not, in general, necessarily well defined. This is due to the fact that the tangent spaces at two different points in a manifold are different spaces, a concept that is not easy to communicate to someone without the necessary background. To compare the two velocities, one must transport a vector from one tangent space to the other. The definition of curvature basically says that this transport process is path-dependent. Using the "before" and "after" approach, there is a natural way to map the "after passage" space to the "before passage" space, however.

Olson & Guarino's abstract gives the summary of the result of the calculation of the induced velocity, which is that the GR predictions are rather similar to the Newtonian ones with an adjustment to the mass of the body that is passing through the cloud. The "adjusted" or "effective" mass, however, is not equal to the relativistic mass of the moving body. As the abstract states, in the ultra-relativistic case, the "adjusted" mass is nearly twice the relativistic mass. This is rather similar to other cases in which neglecting curvature effects gives 2:1 sorts of errors in the naieve calculations.

• sweet springs

#### sweet springs

So we have: At the beginning the gas pushes the rocket upwards. At the end the gas pushes the rocket upwards and rightwards.
Let me check if I catch the situation of the video. In IFR of this video the rocket has rightward initial velocity and zero upward velocity. At time zero the engines start in a synchronized way of proper FR of the rocket.

Does the rocket floor tilt in a way right side up and left side down ?

How dose engines or exhausted photon gas push the rocket side wards? I think gas push the rocket only upwards and no side force is necessary to reduce rightward speed of the rocket. rightward component of 4-velocity
$$\frac{\beta_x}{\sqrt{1-\beta_x^2-\beta_z^2}}$$
is kept constant by $\beta_x$ and $\beta_z$ changes with time even under no force applied.

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#### jartsa

Let me check if I catch the situation of the video. In IFR of this video the rocket has rightward initial velocity and zero upward velocity. At time zero the engines start in a synchronized way of proper FR of the rocket.
Yes.
Does the rocket floor tilt in a way right side up and left side down ?
No. According to us clocks at the rear are ahead of clocks at the front, if the rocket crew has sychronized the clocks. The rear is the left side, as the thing moves to the right. Left side starts moving upwards first.

Here's yet another video:

The crew of that very powerful spaceship must be able to get to the planet during their lifetime, allthough the planet is very far away. So the crew must become time dilated according to us. When the rockets fire, everything moving inside the ship starts slowing down. That's time dilation.

The spaceship must hit the target without any steering, because the planet is static and straight ahead of the ship in the ship frame. When the rockets fire, the ship does not start to move slower to the right. It would miss the planet if it did.

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#### sweet springs

#1
Well, I am still in confusion right side up or left side up.
Say right end engine and left end engine make a impulse thrust at the same tome of rocket proper FR,
in video IFR which is observed, right first left follows or left first right follows?

...Ah, in rocket proper (I)FR, the video IFR move leftward and observe left first right follows. Thanks.

#2
Here's yet another video:
It's very instructive. Thanks. In your vieo for transverse direction
$$v_{star}=v_{rocket}=const.$$
Thus about transverse momentum
$$p_{star}=const.$$
$$p_{rocket} > p_{0\ roclet}$$
,where 0 means initial value, due to increasing $\gamma$ by acceleration of rocket. This means that rocket engine push the rocket side way to increase transverse momentum. So I still keep puzzled how rocket thrusts cause transverse force.

...Ah, now I think I got it. Exhausted photon gas goes not straight down but tilted direction. It means transverse momentum keep given to rocket. My post #47 was wrong. Thanks.

#3
Do you think video IFR and train FR behave the same way, i.e. constant transverse velocity ?

I think train FR has reducing transverse velocity thus changing video IFRs to smaller velocity one as its instantaneous IFR.
In video IFR, the right ward moving train against the rail on the floor with same velocity with video IFR is at rest at a moment but train is climbing up slope of the tilted floor with relative speed against floor. Next moment train reduce right ward speed by transforming kinetic energy to gravitational potential energy. Does such a conjecture work?

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#### jartsa

This means that rocket engine push the rocket side way to increase transverse momentum. So I still keep puzzled how rocket thrusts cause transverse force.

...Ah, now I think I got it. Exhausted photon gas goes not straight down but tilted direction.

Tilted to the left or tilted to the right?

If below the rocket there is a planet that co-moves with the upper planet, then the expelled gas must keep hitting that planet. Which means that the expelled gas must move to the right the same way as the planets move to the right.

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