B Twins Paradox: Dropping a Ball from a Moving Train

[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.

 

[jartsa]

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 γmv\gamma mv where v is speed of rocket and m is rest mass of fuel.

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

When I say photon gas has mass m, I mean that a small parcel of the photon gas has mass m, not all of it.

I didn't edit the post to add this clarification, because it might have changed the latex to something else. That happened once.

 

[jartsa]

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.

Exactly the same kind of light comes out of the rear of that matter-antimatter rocket as came out of the rear of my photon-gas-fuel rocket. At least if the photon-gas-fuel was manufactured near the launchpad by combining matter and anti-matter.

The mass losses of the rockets are the same too. Because photon gas produced by annihilating 1 kg of matter-antimatter has rest mass 1 kg. And when that photon gas is accelerated to speed v, it has momentum $\gamma mv$.

It takes 1 kg of matter-antimatter to accelerate 1 kg of matter, or 1kg of matter-antimatter, or 1 kg of photon-gas to speed 0.87c, using some 100% efficient device.

 

[sweet springs]

1 kg of photon-gas

When photons are exhausted exactly to same direction, IFR where total momentum of photons is zero cannot exist, so mass of photons is not defined. Mass does not conserve but energy (and momentum) conserves.

 

[jbriggs444]

IFR where total momentum of photons is zero cannot exist, so mass of photons is not defined.

The mass is perfectly well defined. It is the invariant length of the energy-momentum four-vector, $mc^2 = \sqrt{E^2-p^2c^4}$. This is zero for a perfectly collimated light pulse in vacuum.

 

[jartsa]

So, a ball dropped in a an accelerating rocket continues its coordinate-motion at the coordinate-velocity it had at the moment it was dropped. And there is no change of the sideways coordinate-motion of the rocket. So the falling ball stays above a spot on the rocket floor.

Does the absence of change of sideways speed of the rocket need an explanation? Well, people that don't know relativity say the sideways momentum of a guy on the rocket is constant, while people that know relativity say that it changes. The latter group of people might want to know where the momentum comes from when the momentum of the guy on the rocket increases.

Well, as the sideways momentum of the fuel has been increasing since the launch of the rocket, the fuel has a lot of sideways momentum. My suggestion is that the sideways momentum that goes to the guy comes from the fuel, as the fuel turns to the exhaust.

As the fuel turns to the exhaust, there is a mass defect. There should be a momentum defect too, right?
By that I mean that the exhaust has less mass than the fuel, and the exhaust has less sideways momentum than the fuel. (This "momentum defect" makes sense if the rocket fuses hydrogen and expels helium, not so much if it's a photon rocket)

 

[jbriggs444]

Does the absence of change of sideways speed of the rocket need an explanation?

The entire premise is flawed. "Sideways" is not an invariant direction.

 

[jartsa]

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.

It's not like that at all. Luckily even I can calculate something using the formulas here:
http://www.math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.htmlMomentum depends on time at the launchpad like this:
In the launcpad frame $p = \gamma mv$ = $\sqrt{1+(\frac{at}{c})^2} mv$Momentum depends on distance like this:
In the launcpad frame $p = \gamma mv$ = $(\frac{ad}{c^2} +1 ) mv$The increase of gamma does not slow down, so the increase of momentum does not slow down.

There is also a table of gammas at different times on that page, interestingly the gamma and the time in years seem to be almost the same numbers at the end of the table.
Hey, if we use this: $v= \frac {at} { \sqrt{1+(\frac{at}{c})^2}}$

we get:

$p = \gamma mv$ = $\sqrt{1+(\frac{at}{c})^2} mv = mat$

 

[PeterDonis]

It's not like that at all.

Sure it is. Instead of giving irrelevant formulas, you should just calculate the coordinate acceleration as a function of coordinate time. You even manage to find a relevant formula that can be used for that:

$$
v = \frac{at}{\sqrt{1 + (at)^2}}
$$

(I'll use units where $c = 1$ since it avoids cluttering up the formulas and doesn't change any of the conclusions.)

Now we just evaluate $dv / dt$:

$$
\frac{dv}{dt} = \frac{a}{\sqrt{1 + (at)^2}} - \frac{1}{2} \frac{at (2 a^2 t)}{\left( 1 + (at)^2 \right)^{3/2}} = \frac{a}{\left( 1 + (at)^2 \right)^{3/2}}
$$

This is obviously a decreasing function of $t$, so coordinate acceleration decreases as $v$ increases, just as I said. And since the rate of change of momentum, in the launchpad frame, is just the coordinate acceleration times $m$, the rate of change of momentum in the launchpad frame also decreases as $v$ increases, just as I said.

 

[jartsa]

Sure it is. Instead of giving irrelevant formulas

Well I think the formulas are good for at least something momentum related.

When an ideal photon rocket moves at speed very close to c, we know that its momentum increases only very slowly, because the light that it emits contains very small amount of momentum, and also because its velocity changes slowly and its mass stays almost constant.

This is the momentum of mass m that has experienced constant proper acceleration a for coordinate time t:
$p=mat$
The mass m has been accelerated - it has not accelerated by itself like a rocket. The m is constant, the a is constant, t increases constantly, so p increases constantly.

So the cargo of that aforementioned photon rocket has momentum $mat$ at time t. That momentum increases at constant rate. While the momentum of the rocket increases at decreasing rate. Note that I'm kind of agreeing there in that last sentence.

 

[PeterDonis]

Note that I'm kind of agreeing there in that last sentence.

Exactly. You are switching between only counting the momentum of the "cargo" (the part of the rocket that isn't unburned fuel and so will never be ejected as exhaust) and counting the momentum of the entire rocket (including unburned fuel). Those are two different things that will give different answers.

 

[jartsa]

Exactly. You are switching between only counting the momentum of the "cargo" (the part of the rocket that isn't unburned fuel and so will never be ejected as exhaust) and counting the momentum of the entire rocket (including unburned fuel). Those are two different things that will give different answers.

So if a railroad car moves along the rails at speed 0.999 c , and heat energy is transferred to the rail from the car through the wheels, what happens?

No momentum comes comes out of the car in the rest frame of the rails, but some part of the car loses momentum in that frame - that part which cools. So to conserve momentum some other parts of the car must gain momentum in the rail frame.

So when that heat transfer is happening the momentum of the cargo in the car is increasing in the rail frame.

Does that sound odd?

 

[PeterDonis]

No momentum comes comes out of the car in the rest frame of the rails

Sure it does; the heat energy is being transferred by friction, which slows the car down, so its momentum in the rail frame decreases.

 

[jartsa]

Sure it does; the heat energy is being transferred by friction, which slows the car down, so its momentum in the rail frame decreases.

No no, there is no friction. Let's forget heat energy. There is a battery on board, its energy is used to electrify the two rails. Electric current goes through the wheels to the rails. I mean energy goes from the battery to the rails.

Now battery loses forwards momentum. something should gain the same amount of forwards momentum.

 

[PeterDonis]

There is a battery on board, its energy is used to electrify the two rails.

Why? If the battery is on board it can just power the train.

Now battery loses forwards momentum. something should gain the same amount of forwards momentum.

The electric current in the rails has momentum.

 

[jartsa]

Why? If the battery is on board it can just power the train.
The electric current in the rails has momentum.

Oh.

Well how about this: A sowing machine moves without friction at constant speed 0.9c on a field relative to the soil. Every now and then it sticks a potato on the soil.

The sowing machine has a "potato tank" where the potatoes are stored. When we observe from the frame of the soil the momentum of the "potato tank" we can see that that momentum is decreasing.

Now, where can we see increasing momentum required by the law of conservation of momentum?

This sowing machine places the potato in the soil so that the potato does not move relative to the soil. IOW the potato has zero momentum in the frame of the soil.

 

[PeterDonis]

When we observe from the frame of the soil the momentum of the "potato tank" we can see that that momentum is decreasing.

This sowing machine places the potato in the soil so that the potato does not move relative to the soil. IOW the potato has zero momentum in the frame of the soil.

These two statements are inconsistent. Obviously a scenario based on inconsistent premises is going to give nonsensical outcomes.

 

[PeterDonis]

At this point the OP of the thread has long been answered and no further useful content is being added. Thread closed.

 

[Nugatory]

Now, where can we see increasing momentum required by the law of conservation of momentum?

The potato is moving relative to the Earth while it is in the tank. Planting it requires that the Earth exerts a force on it to arrest this forward motion; and therefore the potato exerts an equal and opposite force on the Earth which imparts momentum to the earth.

This is the same problem as throwing a sticky ball at a brick wall so that it sticks instead of rebounding: the total momentum of the ball+wall system is conserved.