Transverse inertial motion of rockets in acceleration

[sweet springs]

Hello. I have a question on inertial motion transverse to gravity force, e.g. blocks of dry ice floating on the smooth and flat floor on Earth. For simple mathematics I explain my problem in Rindler case.

Let a group of N rockets gather at origin (0,0,0,0) of a IFR and let them start with constant X-proper acceleration.
We set #N rocket have initial velocity of transverse direction, say Y, as $v_{N}<<c$.
In the IFR the law of inertial motion of Y direction tells that Y position of #N rocket according to time T is
$$Y_N=v_N T$$

IFR people observe clocks on rocket walls tick slower and slower due to increasing X-speed of rockets, e.g. rocket clock ticks for first 1 meter Y-displacement 3 minutes but for next 1 meter it ticks only 2 minutes.

As for X-acceleration in IFR, does TOR not assure constant v_Y for inertial motion but deceleration due to v <c ? In rocket systems does inertial motion in transverse direction appear as acceleration?

 

[Ibix]

As you say, this is just Rindler motion. You've just decided to analyse it from an inertial frame that, compared to the usual inertial frame, is moving orthogonally to the acceleration. The rockets don't care what frame you use. Their experience is just of proper acceleration. Any motion perpendicular to their acceleration direction is an artifact of your choice of coordinates and physically irrelevant.

 

[sweet springs]

So let me understand the point. Push dry ice blocks on flat and smooth floor of accelerating rocket. Dry ice blocks
a. keep constant speed in Rocket system
b. keep constant y-speed in the IFR
c. Both a. and b.
d. Neither a. nor b.
Which one is right to understand?

 

[jartsa]

As for X-acceleration in IFR, does TOR not assure constant v_Y for inertial motion but deceleration due to v <c ? In rocket systems does inertial motion in transverse direction appear as acceleration?

$v<c$
so according to Pythagorean formula:
$v_x^2 + v_y^2 < c^2$
so:
$v_x^2<c^2-v_y^2$
so:
$v_x<\sqrt{c^2-v_y^2}$

That right hand side is a x-speed that the rocket can approach but never reach. As the x-speed approaches that value, various relativistic effects approach infinity.

(In this scenario the $v_y$ stays constant in the IRF. Because there was just a constant proper x-acceleration, no y-accelaration)

 

[jartsa]

So let me understand the point. Push dry ice blocks on flat and smooth floor of accelerating rocket. Dry ice blocks
a. keep constant speed in Rocket system
b. keep constant y-speed in the IFR
c. Both a. and b.
d. Neither a. nor b.
Which one is right to understand?

I guess push means a short push, and then the block slides freely.

Well, the block can be pushed to such speed that it moves across the floor almost as fast as a light pulse moves across the floor.

What happens then can be best seen from an inertial frame:

An inertial observer says it takes more and more time for a light pulse to travel across the rocket floor, as the rocket increases its speed.

And an inertial observer says it takes more and more time for a block to travel across the rocket floor, as the rocket increases its speed.

Both things are clocks that time dilate, like clocks are supposed to do. I mean the light pulse and the block are clocks.

Observers in an accelerating rocket do not notice anything odd about clocks inside the rocket.

 

[PeterDonis]

d. Neither a. nor b.

This is the correct one.

What is conserved here is the momentum of the rockets in the $y$ direction. See my response to @jartsa below.

That right hand side is a x-speed that the rocket can approach but never reach. As the x-speed approaches that value, various relativistic effects approach infinity.

This is wrong. You are incorrectly assuming that the $y$ velocity of the rockets in the IFR stays the same. It doesn't. As the rockets accelerate, their masses, as seen from the IFR, increase. That means their $y$ velocities, as seen from the IFR, decrease, because their $y$ momentum must remain constant. So as their $x$ velocities increase, their $y$ velocities decrease, as seen from the IFR, and there is no problem with the $x$ velocities getting as close to $c$ as you wish.

 

[PeterDonis]

Because there was just a constant proper x-acceleration, no y-accelaration

This is wrong as well. The correct statement is that there is no component of 4-force in the $y$ direction. But there is coordinate acceleration in the $y$ direction, because of the increase in energy of the rocket as seen in the IFR.

One way of seeing it is to note that the velocity in the $y$ direction is $dy / dt$, but what is constant is $dy / d\tau$, where $\tau$ is the rocket's proper time. By the chain rule, we have

$$
\frac{dy}{dt} = \frac{dy}{d\tau} \frac{d\tau}{dt}
$$

and since $dy / d\tau$ is constant (because momentum in the $y$-direction is constant) and $d\tau / dt$ is decreasing, $dy / dt$ is decreasing.

 

[Ibix]

[Option d] is the correct one.

Is this correct? The rocket's Rindler frame acts pretty much like the surface of a planet, and (neglecting friction) the blocks will slide at constant speed, surely. The increasing time dilation of this frame compared to an inertial frame implies that the y speed in the inertial frame decreases, consistent with your point about conservation of y momentum. So isn't option a the correct choice?

 

[pervect]

I've worked this out in the past - at the moment I don't have time to go hunting for the thread(s). It was a precussor to more thourough examination of the "sliding relativisitc block". Anyway, as I recall, if y is the transverse direction to the acceleration, $\tau$ is the proper time of a clock on the sliding block, and T is the "Rindler coordinate" time , which is the proper time of a clock on the floor of the rocket, (note that this is different from the proper time of a clock on the block), then dy/d$\tau$ is constant, and dy/dT is also constant, (though it's a different constant than the previous one). dy/dt is not constant.

"Keep a constant speed in the rocket system" is a bit ambiguous, but if one interprets this as saying"dy/dT is constant",T being the Rindler coordinate time, it would be a correct statement.

This exposition assumes that the "block" is actually a point particle, so it does not have a height.

With this assumption, it turns out that $dy/d\tau = (dy/dT) / \sqrt{1 - (dy/dT)^2/c^2}$, but it's critical that the height of the block in the direction of the acceleration) be negligible otherwise $dy/d\tau$ will vary with height. This formulation assumes that the T coordinate is the proper time of a clock on the floor of the rocket, clocks at different heights in the rocket will run at different rates.

 

[jartsa]

This is wrong. You are incorrectly assuming that the yy velocity of the rockets in the IFR stays the same. It doesn't. As the rockets accelerate, their masses, as seen from the IFR, increase. That means their yy velocities, as seen from the IFR, decrease, because their yy momentum must remain constant. So as their xx velocities increase, their yy velocities decrease, as seen from the IFR, and there is no problem with the xx velocities getting as close to cc as you wish.

When the rocket expels propellant to the -x direction, then the mass of the rocket without the fuel increases.

The propellant is propelled to the opposite direction than the rocket, so propellant's mass decreases.

So, what about the y-momentum of the propellant?

 

[PeterDonis]

The rocket's Rindler frame acts pretty much like the surface of a planet, and (neglecting friction) the blocks will slide at constant speed, surely.

A Rindler frame is only like the surface of a planet locally--and locally means locally in time as well as space. We are talking about long-term effects that are not local in time.

isn't option a the correct choice?

Hm, yes, it could be with an appropriate interpretation of "speed in the rocket system".

 

[PeterDonis]

what about the y-momentum of the propellant?

It has none. Neither has the rocket. The only thing that has y-momentum in the scenario is the block, and the block never expels any exhaust, so its y-momentum remains constant.

 

[PeterDonis]

I've worked this out in the past - at the moment I don't have time to go hunting for the thread(s).

This one seems relevant:

https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/#post-4445483
Although I know there was a much longer "relativistic sliding block thread", and I can't find it.

 

[sweet springs]

What is conserved here is the momentum of the rockets in the y direction.

Thanks. I restate y-momentum in the IFR is
$$m\frac{v_{Y}}{\sqrt{1-v^2}}=m\frac{v_{Y}}{\sqrt{1-v_X^2-v_Y^2-v_Z^2}}=m\ v_{Y0}$$
where c=1 and v_Y0 is constant initial Y-velocity. v_Y decreases according to increase of v_X as time goes.

 

[sweet springs]

"Keep a constant speed in the rocket system" is a bit ambiguous, but if one interprets this as saying"dy/dT is constant",T being the Rindler coordinate time, it would be a correct statement.

Thanks. For this inertial motion I would add frames of reference of y, z, and t, not including x careful to your notice, where a dry ice block is at rest are relative in TOR sense and have no privileges to others.

 

[jartsa]

Not trying to be argumentative or anything ... but if a moving uranium nucleus splits in two parts, then the center of mass of the system continues at the original velocity.

Momentum is conserved because the system's mass does not change.

A rocket is like that nucleus. Right?

 

[PeterDonis]

A rocket is like that nucleus. Right?

No, because the uranium nucleus splits into pieces flying off at angles other than 180 degrees. The rocket goes in the $+x$ direction, and its exhaust goes in the $-x$ direction. So the rocket and its exhaust never have any $y$ momentum. Nor does anything else in the problem except the block. So the block's $y$ momentum must be constant.

 

[jartsa]

Not trying to be argumentative or anything ... but if a moving uranium nucleus splits in two parts, then the center of mass of the system continues at the original velocity.

Momentum is conserved because the system's mass does not change.

If a particle moves into an electric field, and then gets ripped apart by the field, then the mass of the system increases, and the velocity of the center of mass of the system must change - to conserve momentum.

 

[pervect]

Post 90 in "Gravity on Einstein's Train", <<link>>, has the 4-velocities in Rindler and Minkowskii coordinates.

That's the very long thread we mentioned previously. Post 90 is somewhere near the middle.

In Rindelr coordinates (t,x,y), we can write the motion of the sliding block as y=v t. We can say t=$\gamma \, \tau$. This gives us the 4-velocity in Rindler coodinates as

U = (dt/dτ,dx/dτ,dy/dτ) = ($\gamma$,0,$\gamma$v).

I'm not going to duplicate the rest here, but I also write the 4-velocity in Minkowskii coordinates, and integrate the four-velocity to get the position as a function of time.

To clarify, in this post the direction of the acceleration is the x direction, the direction the block slides is the y direction. The z direction is omitted.