Time Dilation and thermal motion

[sweet springs]

[Mentors' note: This thread was split off from another thread about the cosmic ray muon measurements]

The particle at rest in the moving ship is moving against the Earth.
The particle at rest on the Earth is moving against the moving ship.
A scientist in the moving ship measures the life time of the particle at rest in the ship.
Another scientist on the Earth measures the life time of the particle at rest on the Earth.
They get the same half life time of 1 second.

If we catch the particles of half lifetime 1 second in space and would like to bring them to the Earth, how should we do?
In case we would like to keep meet afresh, we put them into referegirator. On the contrary, after putting gas of particles into the bottle, we should heat the bottle to vety high temperature. High speed of particles by high temperature makes half time of the particles in the rocket frame extend so that most of the particles can survive until the rocket arrives to the Earth and they are shared with Earth scientists.

 

[Nugatory]

It would be an interesting exercise to calculate the amount of energy that must be added to a volume of gas to get the particles moving at relativistic speeds.

 

[sweet springs]

Into a paticle at rest we put energy
$$(\gamma-1)mc^2$$
in order to extend half lifettime $\gamma$ times. To double half lifetime, we put the same amount of energy as rest energy. We have to control particle antiparticles pair creation to keep number of particles for $\gamma>3$.

 

[jbriggs444]

Which is more energy-efficient? To heat up the particles so that their half lives are massively time dilated, stuff them in a bottle and lob the bottle at the Earth or to simply shoot the particles toward the Earth at relativistic speeds?

 

[sweet springs]

I am not good at efficiency in sense of engineering but I assume same $\gamma$ requires same energy in the both ways. I might be able to say shooting particle to the Earth requires momentum toward the Earth to input. Heating needs no bulk momentum.

 

[jbriggs444]

I am not good at efficiency in sense of engineering but I assume same $\gamma$ requires same energy in the both ways. I might be able to say shooting particle to the Earth requires momentum toward the Earth to input. Heating needs no bulk momentum.

It's not the same gamma both ways. You need way more gamma in the bottle because the bottle takes so long to arrive.

Despite the slow trip, I think we need to deliver the same momentum either way. The time requirements assure this. Let us try to write down some equations to demonstrate that...

Let $\gamma_{bottle}$ be the required gamma for the bottled particles.
Let $v_{bottle}$ be the bottle velocity
Let $\gamma_{bullet}$ be the required gamma for the particles shot directly.
Let $v_{bullet}$ be the velocity of the particles shot directly.
Let $d$ be the distance to be traversed within proper time limit $t$.

$$t = \frac{d}{v_{bottle} \gamma_{bottle} } = \frac{d} {v_{bullet} \gamma_{bullet} }$$
$$v_{bottle} \gamma_{bottle} = v_{bullet} \gamma_{bullet}$$
$$m v_{bottle} \gamma_{bottle} = m v_{bullet} \gamma_{bullet}$$
$$p_{bottle} = p_{bullet}$$

 

[sweet springs]

It's not the same gamma both ways. You need way more gamma in the bottle because the bottle takes so long to arrive.

Sure. Thanks.

pbottle=pbullet

mvbottleγbottle=mvbulletγbullet

A small correction $$p_{bottle}=m \gamma_{bottle} \frac{v_{bottle}}{\sqrt{1-\frac{v_{bottle}^2}{c^2}}} = m \gamma_{bullet} v_{bullet} = p_{bullet}$$

Though momentum is same, bottle consumes more energy to deliver equally aged particles to the Earth. $M_{bottle} > M_{bullet}$, as for rest energy of particle system.

 

[jbriggs444]

A small correction $$p_{bottle}=m \gamma_{bottle} \frac{v_{bottle}}{\sqrt{1-\frac{v_{bottle}^2}{c^2}}} = m \gamma_{bullet} v_{bullet} = p_{bullet}$$

Good catch. I was assuming a sub-relativistic $v_{bottle}$

 

[pervect]

On a microscopic scale, thermal motion is just motion.

So, to the extent that the particles can be considered to be classical particles, there isn't any conceptual difficulty in the problem, just calculations that need to be performed . I'd suggest considering two idealized cases, one where one has a "thermal" motion that is always in the same direction as the direction of the spaceship, and comparing it to the case where the "thermal" motion is always transverse to the direction of the spaceship. It's probably necessary to specify what frame one mean when one says transverse, I think the frame of most interest for this analysis would be the spaceship frame.

I tend do think that the proper time to reach the Earth would not necessarily be the same for the two cases, so that we'd not see a purely exponential time decay, we'd see a superposition of different decay rates, if we were able to take sufficiently accurate statistics. But I haven't done any detailed calculations, at this point it's just a guess.

In such a classical model, one needs to model the velocity profile of the particles as they bounce off the walls of whatever is containing them. Is it justified to assume that they instantaneously reverse direction at relativistic speeds - the easiest calculation to do - or does one need to account for the details of how long it takes the particles to reverse direction? Doing the later would require a better notion of the interaction of the particles with the wall of whatever box is holding them in.

Quantum effects (such as possible absorbtion and re-emission effects by the walls) won't be covered by this approach. It's a purely classical analysis.

 

[pervect]

Let me add something to my original post. I think some of these issues have a practical impact on the design of NIST-F1, the "cesium fountain clock", that at one time (and may still currently be) our most precise time standard. But I haven't found any specific analysis of thermal effects in the little bit of reading I've done on NIST-F1, for instance https://www.nist.gov/pml/time-and-frequency-division/primary-standard-nist-f1

The clock design is interesting, though, it minizmizes thermal effects by supercooling the cesium atoms. I believe that this implies that it's worth the effort to get rid of the random thermal motions in order to get the most accurate timekeeping, but I didn't find a direct statement of this in what I recently read. I do have a hazy memory of reading something along those lines at one time, though.

 

[Ibix]

The clock design is interesting, though, it minizmizes thermal effects by supercooling the cesium atoms. I believe that this implies that it's worth the effort to get rid of the random thermal motions in order to get the most accurate timekeeping, but I didn't find a direct statement of this in what I recently read

I'm not sure that's due to relativistic effects, though, is it? As far as I recall it's just about eliminating (classical) Doppler broadening of the transition.

 

[pervect]

I'm not sure that's due to relativistic effects, though, is it? As far as I recall it's just about eliminating (classical) Doppler broadening of the transition.

I'm not sure either, to be honest.

 

[A.T.]

[Mentors' note: This thread was split off from another thread about the cosmic ray muon measurements]

The particle at rest in the moving ship is moving against the Earth.
The particle at rest on the Earth is moving against the moving ship.
A scientist in the moving ship measures the life time of the particle at rest in the ship.
Another scientist on the Earth measures the life time of the particle at rest on the Earth.
They get the same half life time of 1 second.

If we catch the particles of half lifetime 1 second in space and would like to bring them to the Earth, how should we do?
In case we would like to keep meet afresh, we put them into referegirator. On the contrary, after putting gas of particles into the bottle, we should heat the bottle to vety high temperature. High speed of particles by high temperature makes half time of the particles in the rocket frame extend so that most of the particles can survive until the rocket arrives to the Earth and they are shared with Earth scientists.

This is related:
https://physics.stackexchange.com/a/400132

 

[sweet springs]

I tend do think that the proper time to reach the Earth would not necessarily be the same for the two cases, so that we'd not see a purely exponential time decay, we'd see a superposition of different decay rates, if we were able to take sufficiently accurate statistics. But I haven't done any detailed calculations, at this point it's just a guess.

Spaceship people observe isotropy in direction of gas molecule motion on its aging. Any other IFR shares this isotropy on aging.

 

[PAllen]

Spaceship people observe isotropy in direction of gas molecule motion on its aging. Any other IFR shares this isotropy on aging.

Isotropy is not frame independent. Consider motion relative to the CMBR, or much more mundane, motion relative to air. The latter has anisotropy of momentum which you feel as wind or pressure. Isotropy of physical laws is a different matter than isotropy of a given system.

 

[sweet springs]

In IFR of spaceship transverse motion particle and parallel motion particle are equally aged, say both a qurter particles alive at one moment.
"In some IFR in a time a half of transverse motion partilces survive and a quarter of parallel motion particle survive, it shows anisotropy on aging", do you want to say like that?
The event or fact in a spaceship IFR "same survival rate for all the direction of motion is observed for partcles in the bottle in a time" should be shared among all the IFRs.

 

[PAllen]

In IFR of spaceship transverse motion particle and parallel motion particle are equally aged, say both a qurter particles alive at one moment.
"In some IFR in a time a half of transverse motion partilces survive and a quarter of parallel motion particle survive, it shows anisotropy on aging", do you want to say like that?
The event or fact in a spaceship IFR "same survival rate of particles for all the direction of motion was observed" can be shared among all the IFRs.

Without getting into frame variance of relativistic statistical mechanics (which I only studied briefly over 40 years ago), consider a simple system of isotropic explosion of some radionuclide in some frame. Decay rates are isotropic in this frame. In a frame in motion relative to this one, forward, backward, and transverse particles will clearly have different observed decay rates.

 

[sweet springs]

Leaving gas in the bottle, going to your explosion case of 1d version, say two same kind of decaying particles of -v, +v flying after explosion.
In IFR where one of them is at rest, they have velocity 0, $\frac{2v}{1+\frac{v^2}{c^2}}$. I ageee that the former dies earlier in that IFR.

Let them put into box or bottle. Particle reflect at the box walll so its rest frame is not fixed. It makes difference, I guess.

 

[sweet springs]

Let them put into box or bottle.

More simple case : a particle moving with speed V in 1d box at rest of length $L_0$.

In Box IFR,
- velosity of particle is V during time interval $\frac{L_0}{V}$, -V during time interval $\frac{L_0}{V}$.
- Time period of above cycle is $\frac{2L_0}{V}$
- Life time spent per cycle is $\frac{2L_0}{V}\sqrt{1-\frac{V^2}{c^2}}$
- Life time spent per unit time is $\sqrt{1-\frac{V^2}{c^2}}$

In IFR moving -V to Box IFR
- velosity of particle is 0 during time interval $\frac{L}{V}$, $\frac{2V}{1+\frac{V^2}{c^2}}$ during time interval
$$\frac{L}{\frac{2V}{1+\frac{V^2}{c^2}}-V}$$.
- Time period of above cycle is
$$\frac{L}{V}+\frac{L}{\frac{2V}{1+\frac{V^2}{c^2}}-V}=\frac{2L}{V}\frac{1}{1-\frac{V^2}{c^2}}$$
- Life time spent per cycle is
$$\frac{L}{V}+\frac{L}{\frac{2V}{1+\frac{V^2}{c^2}}-V}\sqrt{1-\frac{(2V/c)^2}{{(1+\frac{V^2}{c^2})^2}}}=\frac{2L}{V}$$
- Life time spent per unit time is $1-\frac{V^2}{c^2}=\sqrt{1-\frac{V^2}{c^2}}\sqrt{1-\frac{V^2}{c^2}}$

Factor is doubled. We can intepret that one is coming from motion of particle agaist Box as well as in Box IFR. Another comes from motion of Box.

Transverse case, i.e. In IFR moving W in tranvese direction to 1d box
- transverse path during a half cycle is
$$W\frac{L_0}{V}\frac{1}{\sqrt{1-\frac{V^2}{c^2}}}$$
- path length during a half cycle is, from Pythagorean theorem,
$$\sqrt{L_0^2+W^2\frac{L_0^2}{V^2}\frac{1}{1-\frac{V^2}{c^2}}}$$
- speed is
$$\sqrt{\frac{L_0^2+W^2\frac{L_0^2}{V^2}\frac{1}{1-\frac{V^2}{c^2}}}{\frac{L_0^2}{V^2}\frac{1}{1-\frac{V^2}{c^2}}}}=\sqrt{V^2(1-\frac{W^2}{c^2})+W^2}$$
- Life time spent per unit time is $$\sqrt{1-\frac{V^2}{c^2}(1-\frac{W^2}{c^2})-\frac{W^2}{c^2}}=\sqrt{1-\frac{V^2}{c^2}}\sqrt{1-\frac{W^2}{c^2}}$$
Again one factor comes from motion of particle agaist Box as well as in Box IFR. Another comes from motion of Box.

So I may almost conclude that the last formula stands for any direction and magnitude of W. The effect of particle motion against Box and motion of Box in referring IFR is independent and they are multiplied in calculation of lifetime of decaying particles.