Twins Paradox: Dropping a Ball from a Moving Train

[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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[pervect]

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]

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.

 

[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.

 

[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?

 

[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.

 

[jartsa]

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?

A toy train going around on the floor of an accelerating rocket probably looks weird in some frames, because of frame dependent forces. We just need to understand forces, then everything will be quite simple. What you said does not sound simple enough. If we managed to understand a moving lever with two rockets attached to its ends ... that would be great.
https://physics.stackexchange.com/questions/65963/right-angle-lever-paradox-in-special-relativity
My point is that there is something to ponder about forces.

 

[sweet springs]

Tilted to the left or tilted to the right?

In video IFR, exhausted photon goes in direction "from right-up to left-down."
So rocket gets reverse momentum of direction "from left-down to right-up" that is interpreted as "upward plus right ward". Thus rightward momentum is supplied to rocket.

Rains fall from sky to Earth vertical. In train window we see rain drops fall tilted. It does not harm flowers on Earth get vertical rain drops. So dose this case and no problem, I think.

 

[sweet springs]

A toy train going around on the floor of an accelerating rocket probably looks weird in some frames, because of frame dependent forces.

I agree that such a circle motion is much more complicated to analyze than the problem I have been thinking, i.e. a moving train on straight rail on the floor of accelerating rocket with no friction and no engine working as suggested by Peter in post#16 to understand the case of a train moving on surface of Earth.

 

[jartsa]

@sweet springs - See this video:


It's again a spacecraft that travels to a distant twin planet system using two photon rockets. The fuel contains angular momentum. The spacecraft inherits the angular momentum of the photon gas that redshifts out of existence when the redshift is very large. I mean almost redshifts out of existence.

The angular momentum of the spacecraft increases and the rotation does not slow down.

This is very much related and based to post #48.

Note that there is almost nothing in the middle of the spacecraft . If the fuel was stored there, then the angular momentum of the fuel would be smaller, and the change of rotation would be different. If food and water was stored there, then the rotational inertia of the ship would be different, and the change of rotation would be different.

 

[sweet springs]

It's again a spacecraft that travels to a distant twin planet system using two photon rockets. The fuel contains angular momentum. The spacecraft inherits the angular momentum of the photon gas that redshifts out of existence when the redshift is very large. I mean almost redshifts out of existence.

Rotating rocket pair lose energy, momentum and angular momentum carried out by exhausted photon gas. In this exhaust process angular velocity of the pair would be kept constant.

 

[jartsa]

Rotating rocket pair lose energy, momentum and angular momentum carried out by exhausted photon gas. In this exhaust process angular velocity of the pair would be kept constant.

Let me ask a question:

We have a photon rocket moving at enormous speed. The fuel in that rocket has enormous momentum.

When the fuel leaves the rocket through the nozzle, does the fuel

A) Take its enormous momentum with it.
B) Leave its enormous momentum behind it

We are not interested about some small or medium sized momentum changes now. Who gets that enormous momentum?

 

[jbriggs444]

Who gets that enormous momentum?

The "fuel" leaving the rocket is a beam of light, right? So it is moving rearward at c regardless of what frame of reference we consider, right? It most definitely does not carry away any forward momentum.

So it is clear that the rocket and the unexpended fuel retain any enormous momentum they started with.

 

[jartsa]

The "fuel" leaving the rocket is a beam of light, right? So it is moving rearward at c regardless of what frame of reference we consider, right? It most definitely does not carry away any forward momentum.

So it is clear that the rocket and the unexpended fuel retain any enormous momentum they started with.

Yeah, it was a very well aimed beam of light, which when leaving the spacecraft left its forwards momentum to the motor, which sent the forwards momentum all around the spacecraft .

The 'motor' may the 'nozzle' that I have mentioned, or it may be a laser, doesn't really matter.If the fuel was used to heat the spacecraft , then also in that case the forwards momentum of the fuel would have been converted to forwards momentum of the spacecraft . Well that's dumb, but the momentum changes are about the same when heating the ship and when propelling the ship.

 

[jbriggs444]

but the momentum changes are about the same when heating the ship and when propelling the ship.

If you use fuel on board the ship to heat the ship, momentum does not change at all.

 

[jartsa]

If you use fuel on board the ship to heat the ship, momentum does not change at all.

Heating: Fuel's momentum becomes ship's momentum, fuels rest mass becomes ship's rest mass.

Propelling: Fuel's momentum becomes ship's momentum, fuels rest mass seems to disappear. Result: same momentum as before but in a smaller rest mass.

 

[jbriggs444]

Propelling: Fuel's momentum becomes ship's momentum, fuels rest mass seems to disappear. Result: same momentum as before but in a smaller rest mass.

You are spitting momentum out the back. The remaining ship gains +p while the exhaust stream gains -p.

 

[jartsa]

You are spitting momentum out the back. The remaining ship gains +p while the exhaust stream gains -p.

Bob stands behind a powerful photon rocket. The pilot turns the engines on. Bob flies backwards, rocket flies forwards. Bob says: "I gained momentum -p, while the rocket gained momentum p."

Joe is moving at the same velocity as the center of mass of the exhaust of the photon rocket. Joe says the momentum of the exhaust is zero. Joe says that during that event described above the photon rocket gained momentum p, while it spat out momentum zero. I mean the remaining photon rocket gained momentum p.

(Bob gained momentum -p by colliding to the photon gas cloud at high speed, according to Joe)

Joe also says that as the rocket gains even more speed, its exhaust starts having momentum in the direction of the motion. That is caused by the relativistic beaming effect.

The exhaust photon gas has a rest frame because the rocket does not spit out perfectly collimated light.

 

[jbriggs444]

Joe is moving at the same velocity as the center of mass of the exhaust of the photon rocket.

Joe cannot move that fast.

Edit: But it seems that you want to contemplate a rocket exhaust that is not collimated. You want to have it both ways. You want to have a photon exhaust and you want to have a massive exhaust.

Let me see if I can follow your scenario with that understanding.

We have Bob standing behind the rocket. Bob does not define an inertial frame because he just got bowled off of his feet by the photon exhaust. But we ignore that and adopt the frame of reference in which he is initially at rest. Rocket accelerates one way. Bob accelerates the other. Momentum is conserved. Rocket gains +p, Bob ends up with the -p.

That part of the scenario seems straightforward enough.

But now we adopt the frame of Joe who is moving rearward at nearly the speed of light. The rocket emits a burst of light. From Joe's frame of reference, the burst has zero total momentum. Accordingly, from Joe's frame of reference, the rocket cannot have gained any momentum. However, it has lost mass. So it must have gained velocity.

I do not agree with your analysis which claims that the rocket has gained momentum from Joe's frame.

 

[PeterDonis]

When the fuel leaves the rocket through the nozzle, does the fuel

A) Take its enormous momentum with it.
B) Leave its enormous momentum behind it

Neither; the correct answer is

C) The rocket gains some forward momentum, because the exhaust carries away some rearward momentum.

We are not interested about some small or medium sized momentum changes now.

You should be. The exhaust that is emitted in a short period of time does not carry "enormous momentum"; it carries a small amount of rearward momentum, and the rocket gains a small amount of forward momentum.

Do the analysis first in the rocket's instantaneous rest frame just before a small packet of exhaust is emitted. The rocket + fuel starts out with zero momentum in this frame. After the small packet of exhaust is emitted, the rest mass of the rocket + fuel is slightly smaller (because some fuel got burned to make the exhaust), and the rocket + remaining fuel has some forward momentum, which is balanced by the small packet of exhaust having some rearward momentum.

Now transform to a frame in which the rocket, just before the small packet of exhaust is emitted, is moving forward at some relativistic velocity $v$. In this frame, the forward momentum of rocket + fuel just before the small packet of exhaust is emitted is some very large value. After the small packet of exhaust is emitted, the forward momentum of rocket + fuel has increased by a small amount, which is balanced by the small packet of exhaust + fuel having some rearward momentum. But both changes are much smaller than in the instantaneous rest frame. Why? Because the rearward momentum of the exhaust, in this frame, is decreased by the Doppler factor (since the exhaust is photons), so the forward momentum gained by the rocket + remaining fuel must be decreased by the same factor, since they both have to balance.

Another way of viewing this is to note that the same proper acceleration of the rocket equals its coordinate acceleration in the instantaneous rest frame, but this transforms to a much smaller coordinate acceleration in the frame in which the rocket is moving forward at relativistic velocity. And smaller coordinate acceleration means smaller increment of momentum.

 

[PeterDonis]

The exhaust photon gas has a rest frame because the rocket does not spit out perfectly collimated light.

In my previous post I was ignoring this complication. Including it just means, as @jbriggs444 pointed out, that there is now a frame, the rest frame of the exhaust, in which the momentum gain of the rocket is zero, because the exhaust has zero momentum. (If the rocket exhaust is perfectly collimated photons, there is no such frame.)

I personally don't see the point of adding this complication.

 

[jartsa]

Let me see if I can follow your scenario with that understanding.

We have Bob standing behind the rocket. Bob does not define an inertial frame because he just got bowled off of his feet by the photon exhaust. But we ignore that and adopt the frame of reference in which he is initially at rest. Rocket accelerates one way. Bob accelerates the other. Momentum is conserved. Rocket gains +p, Bob ends up with the -p.

That part of the scenario seems straightforward enough.

But now we adopt the frame of Joe who is moving rearward at nearly the speed of light. The rocket emits a burst of light. From Joe's frame of reference, the burst has zero total momentum. Accordingly, from Joe's frame of reference, the rocket cannot have gained any momentum. However, it has lost mass. So it must have gained velocity.

I do not agree with your analysis which claims that the rocket has gained momentum from Joe's frame.

Well it seems we agree now. I meant that "rest of the rocket" gained momentum, when I said rocket gained momentum in Joe's frame.

"Rest of the rocket" means rocket minus the fuel that was burned during the time period that we are interested about.
During the burning, which takes time t, the nozzle or the laser exerts a constant force F on the ship.

The impulse is F *t

In Joe's frame the impulse is $F * \gamma t$ So it's larger.

 

[PeterDonis]

there is now a frame, the rest frame of the exhaust, in which the momentum gain of the rocket is zero, because the exhaust has zero momentum

Actually, I realized that this is not correct, because the frame referred to is the rest frame of the exhaust after it is emitted. But the exhaust is not at rest in this frame before it is emitted; it is moving forward. (More precisely, the fuel that will become the exhaust is moving forward.) So the exhaust loses a small amount of momentum in this frame; the rocket (plus the remaining unburned fuel) must therefore gain the same small amount of momentum in this frame.

 

[jartsa]

Actually, I realized that this is not correct, because the frame referred to is the rest frame of the exhaust after it is emitted. But the exhaust is not at rest in this frame before it is emitted; it is moving forward. (More precisely, the fuel that will become the exhaust is moving forward.) So the exhaust loses a small amount of momentum in this frame; the rocket (plus the remaining unburned fuel) must therefore gain the same small amount of momentum in this frame.

Yes.

Let's use light as 'fuel'. I mean we put photon gas in a tank, then we let it gradually escape through a nozzle. This time the exhausted light is perfectly collimated.

In the launchpad frame the photon gas 'fuel' has momentum $\gamma mv$ where v is speed of rocket and m is rest mass of fuel.

The exhausted photon gas has momentum $−mc \frac {1}{ \gamma (β+1) }$ in the launchpad frame.

(-mc is the momentum of the exhaust in the rocket frame, the term after it is the redshift factor)Let us set c=1. Now at rocket speed 0.87 the 'fuel' accelerates backwards, first it accelerates to zero momentum in the launchpad frame, then it accelerates to the final momentum in the launchpad frame.

The momentum change in the launchpad frame during the first phase of the acceleration is
$\gamma mv$ = 2∗0.87 = 1.74 m

The momentum change during the second phase of the acceleration is ... at the end in the launchpad frame there should be a beam of light with momentum:

$redshiftfactor∗mc$

so during the second phase of acceleration the momentum change is 0.267 m in the launchpad frame.

The latter change of momentum is much smaller than the former one.

 

[jbriggs444]

the rocket (plus the remaining unburned fuel) must therefore gain the same small amount of momentum in this frame.

Which is exactly the same as a conventional rocket burning and expelling fuel when moving at its own exhaust velocity.

 

[sweet springs]

Let's use light as 'fuel'. I mean we put photon gas in a tank, then we let it gradually escape through a nozzle. This time the exhausted light is perfectly collimated.

In the launchpad frame the photon gas 'fuel' has momentum γmvγmv\gamma mv where v is speed of rocket and m is rest mass of fuel.

The exhausted photon gas has momentum −mc1γ(β+1)−mc1γ(β+1)−mc \frac {1}{ \gamma (β+1) } in the launchpad frame.

It might be interesting to think of another setting say particle-antiparticle annihilation, e.g. electron and positron, propulsion photon rockets. Say one of two generated photons goes back. We may set mirror at the engine to reflect the other photon which go forward and make use of it to get momentum $2mc$.
Thus exhausted photons have energy $2mc^2$ and momentum $-2mc$ in IFR of rocket just before exhaustion where m is mass of particle annihilated and rocket goes plus direction. So in this IFR rocket gets momentum $2mc$ with no distinction of body and fuel in tank and loses mass $2m$ of matter and antimatter in fuel tanks.