Twins Paradox: Dropping a Ball from a Moving Train

[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.