Muons hitting Earth half-life explanation in SR

[loislane]

Muons time dilation from the Earth's frame (and length contraction of Earth's atmosphere from muon's frame) is the usual explanation of the fact that muons reach the Earth when they shouldn't just by their rest half-life.
My question is if the explanation based on differential aging(different half-lifes in this case) of a muon at rest on Earth's atmosphere and a cosmic ray traveling muon different path relative to the Earth's muon(a la twin paradox explanation) is considered exactly equivalent to the usual explanation in terms of time dilation and length contraction in different frames(earth's and muon's).

 

[Dale]

Usually on this forum the term "differential aging" is applied to the twins paradox. I am not sure how to interpret it in this context.

 

[loislane]

Usually on this forum the term "differential aging" is applied to the twins paradox. I am not sure how to interpret it in this context.

I wrote between parenthesis "difference of half-life in this case" on behalf of those who might not be able to extrapolate from the concept of differential aging of the twin paradox thought experiment to the case of particles like muons that cannot obviously "age" but that have something called half-life that can be compared much in the way different biological ages can, basically as evidence of different elapsed times. If you need further clarification of the context please specify what you don't understand about the analogy.

So I was wondering if someone could just confirm(or deny) whether the cosmic ray muon time-dilation observed between Earth's and muon's frames is equivalently explained in the same way than the twin paradox (substituting differential aging for different half-life) when analysing it from the pov of a muon at rest on Earth with one half-life and a cosmic ray traveller muon reaching Earth with a different half-life.
Usually the twin paradox is explained in terms of time dilation so I would expect a confirmation but one never knows.

 

[A.T.]

particles like muons that cannot obviously "age" but that have something called half-life that can be compared much in the way different biological ages can, basically as evidence of different elapsed times.

Yes, that's the idea.

 

[Bandersnatch]

I wrote between parenthesis "difference of half-life in this case" on behalf of those who might not be able to extrapolate from the concept of differential aging of the twin paradox thought experiment to the case of particles like muons that cannot obviously "age" but that have something called half-life that can be compared much in the way different biological ages can, basically as evidence of different elapsed times. If you need further clarification of the context please specify what you don't understand about the analogy.

I think that simple difference is not the issue here. In the twin paradox differential ageing comes along as a result of the two twins starting collocated in spacetime, traveling through different paths, and returning to the same spot in spacetime. Those different paths between the same two points is what results in differential ageing.

In the case of muons, you never have two particles traveling between the same two points in spacetime, so you don't get differential ageing.

 

[Dale]

I wrote between parenthesis "difference of half-life in this case" on behalf of those who might not be able to extrapolate from the concept of differential aging of the twin paradox thought experiment to the case of particles like muons that cannot obviously "age" but that have something called half-life that can be compared much in the way different biological ages can, basically as evidence of different elapsed times. If you need further clarification of the context please specify what you don't understand about the analogy.

I have no problem with the analogy of age or lifetime with half life. But as Bandersnatch mentioned "differential aging" usually describes the difference in proper time for a pair of objects which start out together, separate, and then reunite. I just can't see how it applies here.

My guess is that what you are calling "differential aging" is the same as time dilation.

 

[vanhees71]

I think the usual explanation is a very clear example for the consistency of SRT. First of all one should know that particle physicists usually define intrinsic properties of (massive) particles in their restframe. So the life time of the muon is defined as the mean lifetime of the particle in its rest frame, which I call $\tau$. This is by definition a scalar quantity that does not change with Lorentz transformations.

Nevertheless it implies that the mean lifetime of the particle as measured in a reference frame, where the particle moves with a speed $v=\beta c$ is longer by the Lorentz factor,
$$t_{\text{life}}=\gamma \tau=\frac{\tau}{\sqrt{1-\beta^2}}.$$
So for an observer on Earth the muon has a very large speed and thus will life much longer in terms of his time than the nominal proper lifetime $\tau$, and thus it can reach us with a much higher probability than thought when using naively $\tau$ as mean lifetime on earth.

From the point of view of the muon rest frame. This is also consistent, because there its mean life time is indeed $\tau$, but the distance to travel from its creation point to Earth is length contracted, i.e., if $L$ is the distance from Earth to the creation point as seen from the observer at rest on earth, then the distance to be traveled from the point of view of the muon is only
$$L'=\frac{L}{\gamma}=L \sqrt{1-\beta^2},$$
and thus it's likelier from the point of view of the muon restframe to reach the Earth within an average lifetime of $\tau$ as measured in this frame.

So both observers come to the same conclusion concerning the probability for the muon to reach earth. Indeed, for the observer on Earth the muon needs a travel time $t_{\text{travel}}=L/v$ from its point of creation. The survival probatility is
$$P=\exp \left (-\frac{t_{\text{travel}}}{t_{\text{life}}} \right )=\exp \left (-\frac{L \sqrt{1-\beta^2}}{v \tau} \right ).$$

For the observer in the restframe of the muon the calculation is analogous: The travel time from his point of view is $t_{\text{travel}}'=L'/v=L \sqrt{1-\beta^2}/v$ and the mean lifetime $t_{\text{life}}'=\tau$ and thus
$$P'=\exp \left (-\frac{t_{\text{travel}}'}{t_{\text{life}}'} \right )=\exp \left (-\frac{L \sqrt{1-\beta^2}}{v \tau} \right )=P,$$
as it should be.

 

[Mister T]

Usually the twin paradox is explained in terms of time dilation so I would expect a confirmation but one never knows.

If you are an observer on Earth you would use the same time dilation formula to calculate the lifetime of a traveling muon as you would to calculate the aging of your traveling twin. (Making the usual assumption that the traveling twin spends a negligible amount of time switching from his outbound journey to his inbound journey).

The situation for your traveling twin is more complicated, though, because unlike you he must change reference frames (undergo a change in velocity) to be able to turn around and return to you. But even he would use that same time dilation formula for the two parts of his trip (the voyage out and the voyage back) to calculate your aging. He would also have to account for length contraction in the same way as the traveling muon does. But he would also use an additional formula, the one used to account for the relativity of simultaneity, to account for your aging.

 

[loislane]

I think that simple difference is not the issue here. In the twin paradox differential ageing comes along as a result of the two twins starting collocated in spacetime, traveling through different paths, and returning to the same spot in spacetime. Those different paths between the same two points is what results in differential ageing.

In the case of muons, you never have two particles traveling between the same two points in spacetime, so you don't get differential ageing.

Yes, I agree there is this difference between the muon and the twin cases. So I guess an additional assumption is needed for them to be equivalent when it comes to comparing a muon at rest on Earth with the usual half-life of muons and the traveling muon, namely that the clock that measures proper time for the latter(actually for the particles from which it decaya) has gone thru a noninertial path before attaining the veloity with which it enters the atmosphere.
This is a reasonable assumption astrophysically.

 

[vanhees71]

Well, there are also experiments with unstable particles in storage rings, where they are in circular motion. With very high precision also in this case of accelerated motion you find the expected time-dilation factor (Lorentz-$\gamma$).

 

[Mister T]

So I guess an additional assumption is needed for them to be equivalent when it comes to comparing a muon at rest on Earth with the usual half-life of muons and the traveling muon, namely that the clock that measures proper time for the latter(actually for the particles from which it decaya) has gone thru a noninertial path before attaining the veloity with which it enters the atmosphere.

What happened to the muon prior to it being created and (in the process) attaining its speed relative to Earth (a process that happens in Earth's atmosphere, by the way) is not relevant. All that's relevant is what happens afterward. When we say the muon has a half-life of 2.2 µs we mean it as a time measured in the muon's rest frame. In other words it's a proper time $\tau$. Proper time is a so-called relativistic invariant, meaning that all observers agree that in its rest frame this is the value. We would write

$\tau=2.2$ µs.

Likewise, we might say that the traveling twin takes, on average, 15 minutes to eat her lunch. Again, this is a proper time $\tau$ meaning that it's a time measured in her reference frame. Again, this is a relativistic invariant meaning that all observers will agree that in her reference frame this is the value. We would write

$\tau=15$ min.

Let's suppose that both the muon and the traveling twin, relative to Earth, each have a speed of 0.6c, so the relativistic factor $\gamma$ equals 1.25

On Earth we would then calculate that the average lunch time is

$\gamma \tau=(1.25)(15)=18.75$ min,

and the half-life is

$\gamma \tau=(1.25)(2.2)=2.75$ µs.

So, I don't know what you mean when you talk about what needs to be done to make the two scenarios equivalent. To me, all that needs to be done is to either keep the traveling twin forever moving so that she never switches directions, or have the muon turn around and change direction. Either way, we get the same effect both qualitatively and quantitatively.

 

[Mister T]

Let's imagine a different scenario that might help shed light.

Suppose the stay-at-home twin, Alice, works in a lab where she takes 15 minutes to her lunch and she measures the half-lives of muons at rest in her lab to be 2.2 µs. An observer in a space ship, Bob, heads towards Earth at speed 0.6c. Bob measures the time it takes Alice to eat her lunch and gets 18.75 minutes. He measure's the half-life of Alice's muons to be 2.75 µs.

Bob eats his lunch in exactly the same way as Alice, but it takes him only 15 minutes. Bob has muons at rest in his spaceship (meaning they are headed towards Earth at a speed of 0.6c) and he measures their half-life to be only 2.2 µs.

 

[PeterDonis]

So I guess an additional assumption is needed for them to be equivalent when it comes to comparing a muon at rest on Earth with the usual half-life of muons and the traveling muon, namely that the clock that measures proper time for the latter(actually for the particles from which it decaya) has gone thru a noninertial path before attaining the veloity with which it enters the atmosphere.

No, that's not the assumption being made. The assumption being made is that we are using the simultaneity convention of the Earth, not the muons falling through the atmosphere. If we adopt the simultaneity convention of the muons (i.e., if we work in an inertial frame in which they are at rest and the Earth, along with muons at rest on the Earth, is moving towards the muons from the atmosphere at high speed), then the Earth will be greatly time dilated, and all its muons will have much longer half-lives in this frame than the muons from the atmosphere. (Also, of course, the Earth and its atmosphere will be greatly length contracted in this frame, so the muons from the atmosphere will not decay before they intersect the Earth's surface.)

None of the above is inconsistent with what has already been said about things as they are in the Earth frame, because, as has been said already, the two sets of muons (atmosphere and Earth) never pass through the same pair of events in spacetime, so this is not a "twin paradox" scenario. We simply have two different inertial frames and that's it.