What if particles are moving near light speed and one slows?

[Ignorantsmith12]

TL;DR Summary

As recommended, I read the introductory article on the twin paradox and several other threads on it. However, I am still confused and, therefore, cannot explain which particle would age more if one particle in a group of fast particles slowed down and then sped up again.

I tried to solve this problem independently, but given how confused I am about the twin paradox and the concept of relativity of simultaneity, I have no confidence in my solution.

Suppose a collection of particles has been moving very close to the velocity of light relative to the Cosmic microwave background restframe since the Big Bang (is that possible?). Imagine that one of them somehow gets stuck in Earth's atmosphere and is stopped, but then after, I don't know, some years, it gets taken into a particle accelerator and blasted back into space at its original velocity. Forgetting that this would never happen, which particle would age more according to the calculations of an earth-based observer? Which particle would age more from the perspective of the particles that never slowed?

Adding another wrinkle, imagine the above scenario, except the particles have been moving in a circle with a diameter of several light-years so that the particle that had been slowed can rejoin its fellows when it is blasted back into space as the other particles circle back. Who gets older faster then, and according to who's reference frame?

PS The thread made pick a prefix. I chose graduate as I have my BA but I have no formal education in physics.

 

[anuttarasammyak]

Forgetting that this would never happen, which particle would age more according to the calculations of an earth-based observer? Which particle would age more from the perspective of the particles that never slowed?

In QM particles have no age. They do not career information when they were born or where they have been. Once they are detected, an observer know their expectancy life which depends on what elementary particle it is and its speed in IFR of the observer.

Following your scinario but with clocks gas of light speed instead of particles, the trapped on Earth clock's time is larger than that of almost light speed free motion clocks , if all the clock showed same time at the same origin point in the begining of the universe.

Not SR but GR should be applied to Big bang and other features of the universe, the particles or clocks keep loosing speed wrt FLRW coordinate, for instance, but SR is still applicable in local IFRs which is regarded almost at rest wrt FLRW metric coordinates as the previous paragraph says.

 

[Ibix]

The key point about the twin paradox is that the twins must meet again for there to be an unambiguous answer. The particles (as @anuttarasammyak notes, particles don't really age in any meaningful sense, but there's no theoretical reason why they couldn't be accompanied by a clock) don't meet again in your first scenario, so it is not a twin paradox scenario and the answer wil be different for the two frames you specify. To answer it you simply need to consider the Lorentz gamma factor of the two clocks in the Earth frame and in the particle frame - what results do you get?

In the scenario where the unslowed clock travels in a large circle (perhaps due to a magnetic field affecting the particle it's attached to) then the slowed and unslowed clocks meet again and there is an unambiguous answer. It's easy to calculate using the Earth's rest frame - again, just use the gamma factors. However, the particles moving in a circle are not moving inertially. It is possible to do a calculation from their perspective and get the same answer, but it is much harder and not very enlightening.

 

[anuttarasammyak]

In the scenario where the unslowed clock travels in a large circle (perhaps due to a magnetic field affecting the particle it's attached to) then the slowed and unslowed clocks meet again and there is an unambiguous answer. It's easy to calculate using the Earth's rest frame - again, just use the gamma factors. However, the particles moving in a circle are not moving inertially. It is possible to do a calculation from their perspective and get the same answer, but it is much harder and not very enlightening.

In the scenario of clocks gas like cosmic back ground radiation lights, we can compare catched and Earth trapped clock's time and gas clock of the same place and find the former ticks more. This I thought as compensation of compareing the same clocks.

 

[Ibix]

In the scenario of clocks gas like cosmic back ground radiation lights, we can compare catched and Earth trapped clock's time and gas clock of the same place and find the former ticks more. This I thought as compensation of compareing the same clocks.

I'm not sure what you mean by "gas clock", but I think you mean doing something like using the CMB temperature as a clock. Yes you can do that, but you are just picking one inertial frame's synchronisation convention and using that because it happens to be convenient to implement. The unslowed particle frame answer is equally valid.

(I assume we're not working on cosmological scales here, so spacetime can be treated as flat.)

 

[anuttarasammyak]

I'm not sure what you mean by "gas clock", but I think you mean doing something like using the CMB temperature as a clock.

Gas whose molecules are clocks, I mean. The universe is filled with moving clocks as with CMB photons. Now I perceive clock gas and a molecule of clock gas are more appropriate English.

Say in meeting of three molecules on Earth
Earth trapped molecule ticks t1
A molecule which was aside with it when it was trapped, go through and is reflected back to Earth by a wall which is at rest at FLRW coordinate ticks t2
A molecule passing by ticks t3,
$$t_2=t_3<t_1$$
So we do not need the same elastic-collision-relfected-back mlecule to check the fact that Earth trapped molecule ticks more. This is the compensation I meant.

 

[Ibix]

Gas whose molecules are clocks, I mean. The universe is filled with moving clocks as with CMB photons.

Well, CMB photons aren't clocks because they follow null paths. And in a gas of timelike particles the particles have different speeds and hence different tick rates, and they change speeds in collisions. So I don't know how you would actually measure time with such a setup. And it doesn't get around the fact that (even if you can define a clear notion of time this way) you are simply picking one frame and declaring it more important. Things like the CMB rest frame are similar to the Earth's surface rest frame in our everyday lives - very important because there's a lot of mass at rest in that frame, but not significant in terms of fundamental physics.

There's nothing wrong with picking an inertial frame and preferring it, but it's a choice.

Say in meeting of three molecules on Earth
Earth trapped molecule ticks t1
A molecule which was aside with it when it was trapped, go through and is reflected back to Earth by a wall which is at rest at FLRW

I think you've just built a true twin paradox scenario within the OP's non-meeting-twins scenario. So there's a unique answer to your vetsion, but not the OP's. OP's particles will say the tick rate of the bounced clock changed when its speed changed. The Earth frame says it changed direction and not speed so the tick rate is not affected. So there's a unique answer to how old your particles are at their second meeting, but not to how old the OP's particles that continued straight on are at the same time.

 

[Ignorantsmith12]

...In the scenario where the unslowed clock travels in a large circle (perhaps due to a magnetic field affecting the particle it's attached to) then the slowed and unslowed clocks meet again and there is an unambiguous answer. It's easy to calculate using the Earth's rest frame - again, just use the gamma factors. However, the particles moving in a circle are not moving inertially. It is possible to do a calculation from their perspective and get the same answer, but it is much harder and not very enlightening.

I take "not very enlightening" to mean not much time dilation occurs to any of the particles.

 

[Nugatory]

I take "not very enlightening" to mean not much time dilation occurs to any of the particles.

It means that the math is way messier and more complicated but provides no new insights into the physics. It’s sort of like choosing to use Roman instead of Arabic numerals to balance your checkbook - harder and more error-prone, even if you do it right you’ll just get the same answer when you’re done, and you won’t have learned anything new about your financial situation from the extra effort.

 

[Ibix]

I take "not very enlightening" to mean not much time dilation occurs to any of the particles.

It means two things. First, that how much difference in ages there will be is trivial to compute in the Earth frame (as I said, decide how long Earth clocks say it'll be and then divide by the gamma factor to get how long the particle clocks say it'll be), and the answer is an invariant so it's the same result in any frame. Second, it means that you can do the calculation in the particles' rest frame too but, as Nugatory says, pretty much the only reason to do it is to prove that you can. You might do it as a step towards learning the maths of GR because you need a lot of the same maths to do it but without all the extra complications of curved spacetime.