• I
• Kairos
N_0 of atoms, then according to the disintegration law, N_A=N_B, which is not what was announced.In summary, the incorrect relativistic decay law was found to be $$N=N_0 \ \exp\left(- \frac{\ln 2 }{t_{1/2}}\int d\tau \right)$$ where ##t_{1/2}## is the half life of the material in the rest frame.

#### Kairos

Replace the twins in the twin experiment by identical radioactive samples containing the same starting number ## N_0 ## of atoms. One (A) stays on Earth while the other (B) makes a round trip at high speed. When back the traveling sample is more radioactive because its half-life ## \Delta t1/2 ## has been dilated, which means that the numbers of radioactive atoms in the two samples are no longer the same (## N_A < N_B ##). This result (confirmed previously in this forum) is expected because it is equivalent to that of the twin experiment in which the traveling twin returns younger than is brother. By writing the disintegration law, for the traveling sample
$$N_A = N_0 \ \exp\left(-\dfrac{\Delta t_A \ln 2}{\Delta t_A 1/2}\right)$$
and for the sample at rest on earth
$$N_B = N_0 \ \exp\left(-\dfrac{\Delta t_B \ln 2}{\Delta t_B 1/2}\right)$$
According to the result of the twins' experiment
$$\Delta t_B = \gamma \Delta t_A$$
and according to the result of the muon half-life
$$\Delta t_B 1/2 = \gamma \Delta t_A 1/2$$
this gives ## N_A = N_B ## contrary to what has been announced. Thank you in advance for pointing out my mistake!

The correct relativistic decay law is $$N=N_0 \ \exp\left(- \frac{\ln 2 }{t_{1/2}}\int d\tau \right)$$ where ##t_{1/2}## is the half life of the material in the rest frame.

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• vanhees71 and Kairos
Kairos said:
and according to the result of the muon half-life
$$\Delta t_B 1/2 = \gamma \Delta t_A 1/2$$
This is wrong. The muon has the same half-life independent of its state of motion. So, you should have:
$$\Delta t_B 1/2 = \Delta t_A 1/2 = \tau$$

• Kairos
As a general rule, adding complications to a scenario seldom aids understanding,

PeroK said:
This is wrong. The muon has the same half-life independent of its state of motion. So, you should have:
$$\Delta t_B 1/2 = \Delta t_A 1/2 = \tau$$
thank you very much! I misinterpreted the sentence "the extended half-life of fast moving muons" mentioned in many posts

• hutchphd
Kairos said:
When back the traveling sample is more radioactive because its half-life has been dilated
Well, not really. Time dilation doesn't change anything. What DOES make the change is the differential aging caused by differing paths through space-time.

A good way to keep this in mind is to realize that right now, you, as you read this, are MASSIVELY time dilated from the frame of reference of an accelerated particle at CERN. Do you feel as though any of your biological functions have slowed down?

• vanhees71
Dale said:
The correct relativistic decay law is $$N=N_0 \ \exp\left( \frac{\ln 2 \ \int d\tau}{t_{1/2}} \right)$$ where ##t_{1/2}## is the half life of the material in the rest frame.

The point here, @Kairos, is that it's seldom a good idea to define a "relativistic anything" (mass, heat capacity, half life,...). So it's better to define half life to be the half life measured in the particles' rest frame, which is what Dale's expression uses. ##\int_0^t d\tau## is the elapsed time for the muon between my coordinate times ##0## and ##t##, which means that ##\int d\tau/t_{1/2}## is the number of half lives it's experienced, and that's why that quantity appears in the exponential decay.

That's conceptually a lot simpler than worrying about a relativistic half life and whether I need to divide by its integral or multiply by the integral of its reciprocal.

• Dale, vanhees71 and PeroK
Ibix said:
Oops, yes. Thanks I fixed it.

• Ibix and vanhees71
phinds said:
right now
Well, there's no beam in the machine right now, but I see your point. Kairos said:
According to the result of the twins' experiment
$$\Delta t_B = \gamma \Delta t_A$$
On top of everything else, this is backwards. If B is the traveling twin, it elapsed less proper time, so it should be:
$$\Delta t_A = \gamma \Delta t_B$$
But note this is a relation in respective proper times. In terms of proper times, half life is always the same.

On the other hand, if you want to use coordinate times in the Earth frame, then the twin times between meetup are the same in Earth coordinate time, else they wouldn't meet. Your second expression for half lives is valid for coorinate time in the Earth frame - traveling muons have longer half life in Earth frame coordinate time.

So you are combining what cannot be combined (as well as getting one relation backwards).

If you use either all proper times, or all Earth frame coordinate times, you will get the expected result either way.

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• vanhees71, Kairos and Ibix
PAllen said:
If B is the traveling twin, it elapsed less proper time, so it should be: ΔtA=γΔtB
oops yes! thank you

PAllen said:
If you use either all proper times, or all Earth frame coordinate times, you will get the expected result either way.
Yes I understand that the formula of post #2 using all proper durations in the numerator and denominator, gives the expected result of the twin experiment, but if A and B are both inertial, this formula would give the same result for both? and unable to predict the extended life of fast moving cosmic muons compared to static muons?

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Kairos said:
but if A and B are both inertial, this formula would always give the same result?
You mean, frame A measures particles at rest in frame B to have a longer halflife than those at rest in A, and frame B measures particles at rest in frame A to have a longer half life than those at rest in B? Sure. That's just the usual symmetry of time dilation.
Kairos said:
and unable to predict the extended life of fast moving cosmic muons compared to static muons?
No, that's exactly what it predicts.

If you are thinking of muons produced in the upper atmosphere, do recall that the atmosphere is moving in the muon rest frame and therefore length contracted. Thus it can pass through the muons before they decay because it is less deep in this frame than it is in its rest frame.

Yes I understand all your comments about time dilation end length contraction, but concerning the formula: if the flight time (numerator) and the half-life (denominator) are both proper durations or both coordinate times, it gives the same result by compensation. To predict the number of surviving muons actually observed, should we not rather put a coordinate duration in the numerator and the proper half-time in the denominator? or conversely the proper duration in the numerator and the extended half-life in the denominator?

This would make the formula to be used different for the twins experiment and the usual symmetry of time dilation..

Kairos said:
To predict the number of surviving muons actually observed, should we not rather put a coordinate duration in the numerator and the proper half-time in the denominator?
We aren't putting a coordinate time in either numerator or denominator. Coordinate times only appear in the limits of the integral, assuming you are still talking about Dale's formula in #2.

I'm not sure what experiment you are thinking about at the moment, since it doesn't seem to be the usual cosmic ray muons one. Are there two groups of muons moving at different speeds? Who is measuring the survival rates and what is their velocity relative to the muons? Are they at some fixed distance from the muon source(s) in their frame? Or if not, what is their criterion for how to count a muon as "survived"?

Kairos said:
Yes I understand all your comments about time dilation end length contraction, but concerning the formula: if the flight time (numerator) and the half-life (denominator) are both proper durations or both coordinate times, it gives the same result by compensation. To predict the number of surviving muons actually observed, should we not rather put a coordinate duration in the numerator and the proper half-time in the denominator? or conversely the proper duration in the numerator and the extended half-life in the denominator?

This would make the formula to be used different for the twins experiment and the usual symmetry of time dilation..
A radioactive sample is a clock. If you do what you say, then you simply cancel out any time dilation factor, restoring absolute time for such a clock. That is Newtonian physics and one possibility.

Experiment shows, however, that our universe has relativistic spacetime, so a radioactive clock does not exhibit absolute time.

• Ibix
Ibix said:
I'm not sure what experiment you are thinking about at the moment
I am thinking of the famous experiment counting the number of muons reaching the ground compared to the number detected at altitude, and the formula allowing to predict the observed result (knowing their number at a given altitude, their speed and their half-life at rest)

Kairos said:
Yes I understand that the formula of post #2 using all proper durations in the numerator and denominator, gives the expected result of the twin experiment, but if A and B are both inertial, this formula would give the same result for both? and unable to predict the extended life of fast moving cosmic muons compared to static muons?
No, the formula in 2 is general. It works for any path, inertial or not, using any coordinates in any spacetime.

Kairos said:
if the flight time (numerator) and the half-life (denominator) are both proper durations or both coordinate times, it gives the same result by compensation.
If you change the formula then it is no longer the same formula. The numerator is proper time and the denominator is the half life that you look up in tables, which is a proper time.

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Kairos said:
Yes I understand all your comments about time dilation end length contraction, but concerning the formula: if the flight time (numerator) and the half-life (denominator) are both proper durations or both coordinate times, it gives the same result by compensation. To predict the number of surviving muons actually observed, should we not rather put a coordinate duration in the numerator and the proper half-time in the denominator? or conversely the proper duration in the numerator and the extended half-life in the denominator?

This would make the formula to be used different for the twins experiment and the usual symmetry of time dilation..
NO, that is nonsense, as I indicated in my earlier post. It only makes physical sense to use proper time (or invariants, more generally) throughout or coordinate values mediated through an associated metric throughout. Any combination is like saying how many apples make up an orange.

Kairos said:
I am thinking of the famous experiment counting the number of muons reaching the ground compared to the number detected at altitude, and the formula allowing to predict the observed result (knowing their number at a given altitude, their speed and their half-life at rest)
A) Using proper times:
The proper time for muons between their creation and arrival at a ground detector is way less than their intrinsic half life, so most reach the ground. This computation can be done in any coordinates, and will come out the same for any coordinates.

B) Using Earth inertial frame
The high speed muons have a longer half life than the intrinsic half life due to time dilation. This is reflected in your formula for half life, which is purely a coordinate dependent formua, in this case valid only in Earth based inertial coordinates. As a result of the lengthening of half life expressed in coordinate time for the high speed muons, most reach the ground.

C) Using inertial frame of created muon
The atmosphere is moving rapidly, and its thickness is such that it passes by way before most muons decay.

Combining these different methods of calculating/explaining the unique physical observation leads to nonsense. Each, alone, is consistent.

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• phinds
Dale said:
No, the formula in 2 is general. It works for any path, inertial or not, using any coordinates in any spacetime.

If you change the formula then it is no longer the same formula. The numerator is proper time and the denominator is the half life that you look up in tables

Thank you. OK The formula of post#2 is general and always works.

PAllen said:
A) Using proper times:
The proper time for muons between their creation and arrival at a ground detector is way less than their intrinsic half life, so most reach the ground. This computation can be done in any coordinates, and will come out the same for any coordinates.

B) Using Earth inertial frame
The muons have a longer half life than the intrinsic half life due to time dilation. This is reflected in your formula for half life, which is purely a coordinate dependent formua, in this case valid only in Earth based inertial coordinates. As a result of the lengthening of half life expressed in coordinate time for the high speed muons, most reach the ground.

C) Using inertial frame of created muon
The atmosphere is moving rapidly, and its thickness is such that it passes by way before most muons decay.

Combining these different methods of calculating/explaining the unique physical observation leads to nonsense. Each, alone, is consistent.

Thank you very much. For concrete verification, could you please tell me if the following exponents are correct, postulating that A are "observer" muons static on the Earth ground and B are observer muons moving relative to the Earth at speed v. The starting numbers of A and B are identical (## N_0 ##) and the final numbers ## N_A ## and ## N_B ## are counted when they meet. The proper mean life time of muons is written ## \tau ##.

For a linear path (experiment of created muons starting from the altitude D and reaching vertically the ground). A and B are inertial.

From the frame of A,
for calculating ## N_A ##, the exponent is ## - D/(v \ \tau) ##
for calculating ## N_B ##: ## - D/(v \ \tau \ \gamma) ##

From the frame of B,
for calculating ## N_A ## (terrestrial muons arriving at B): ## - D/(v \ \tau \ \gamma) ##
for calculating ## N_B ##: ## - D/(v \ \tau) ##

For a circular path of B in a ring of length D. A (inertial) and B (non-inertial) start from the same point and arrive at the same point.

From the frame of A,
for calculating ## N_A ##: ## - D/(v \ \tau) ##
for calculating ## N_B ##: ## - D/(v \ \tau \ \gamma) ##

no calculation from B and just assuming that the result measured from A must be also true for B

PAllen said:
A) Using proper times:
The proper time for muons between their creation and arrival at a ground detector is way less than their intrinsic half life, so most reach the ground. This computation can be done in any coordinates, and will come out the same for any coordinates.
It's the other way round: the lifetime of the muons by definition is the average lifetime as measured in the muons' rest frame, i.e., as measured by their proper time and thus this lifetime is a scalar. The lifetime as measured in any frame, where the muons move is "dilated". In the most simple case, where the muons move with constant velocity, this "coordinate lifetime" is longer by a Lorentz factor ##\gamma##, i.e., ##t_{\mu}=\gamma \tau_{\mu}=\tau_{\mu}/\sqrt{1-v^2/c^2}##.
PAllen said:
B) Using Earth inertial frame
The high speed muons have a longer half life than the intrinsic half life due to time dilation. This is reflected in your formula for half life, which is purely a coordinate dependent formua, in this case valid only in Earth based inertial coordinates. As a result of the lengthening of half life expressed in coordinate time for the high speed muons, most reach the ground.
That's correct! Thus in this frame the muon needs a time ##t=L/v## to reach the detector and its lifetime as measured in this frame is ##t_{\mu}=\gamma \tau_{\mu}##. The probability for being detected thus is
$$\exp(-t/t_{\mu})=\exp[-L/(\gamma v \tau_{\mu})].$$
PAllen said:
C) Using inertial frame of created muon
The atmosphere is moving rapidly, and its thickness is such that it passes by way before most muons decay.

Combining these different methods of calculating/explaining the unique physical observation leads to nonsense. Each, alone, is consistent.
Here the distance between the muon's creation point and the detector is Lorentz contracted by the inverse Lorentz factor and thus the detector reaches the muons after a time ##t'=L'/v=L/(\gamma v)##. The probability to be detected in this frame is ##\exp(-t'/\tau_{\mu})=\exp[-L/(\gamma v \tau_{\mu})]##, which is the same as when observed in the detector-rest frame, as it must be.

vanhees71 said:
It's the other way round: the lifetime of the muons by definition is the average lifetime as measured in the muons' rest frame, i.e., as measured by their proper time and thus this lifetime is a scalar. The lifetime as measured in any frame, where the muons move is "dilated". In the most simple case, where the muons move with constant velocity, this "coordinate lifetime" is longer by a Lorentz factor ##\gamma##, i.e., ##t_{\mu}=\gamma \tau_{\mu}=\tau_{\mu}/\sqrt{1-v^2/c^2}##.

You misread what I wrote. The muon proper time between creation and ground detection is way less than muon half life. That is true, and that is what I wrote.

• vanhees71
Kairos said:
Thank you very much. For concrete verification, could you please tell me if the following exponents are correct, postulating that A are "observer" muons static on the Earth ground and B are observer muons moving relative to the Earth at speed v. The starting numbers of A and B are identical (## N_0 ##) and the final numbers ## N_A ## and ## N_B ## are counted when they meet. The proper mean life time of muons is written ## \tau ##.

For a linear path (experiment of created muons starting from the altitude D and reaching vertically the ground). A and B are inertial.

From the frame of A,
for calculating ## N_A ##, the exponent is ## - D/(v \ \tau) ##
for calculating ## N_B ##: ## - D/(v \ \tau \ \gamma) ##

From the frame of B,
for calculating ## N_A ## (terrestrial muons arriving at B): ## - D/(v \ \tau \ \gamma) ##
for calculating ## N_B ##: ## - D/(v \ \tau) ##

For a circular path of B in a ring of length D. A (inertial) and B (non-inertial) start from the same point and arrive at the same point.

From the frame of A,
for calculating ## N_A ##: ## - D/(v \ \tau) ##
for calculating ## N_B ##: ## - D/(v \ \tau \ \gamma) ##

no calculation from B and just assuming that the result measured from A must be also true for B
The problem for all of this is that proper time is defined along a world line between two events. The event of muon creation and muon ground detection are well define events. However, for a muon at rest on the ground, only the event of meeting an incoming muon is well defined. The event on ground muon's world line corresponding to the atmospheric muon's creation is wholly a matter of convention. So the fact that the muon's reach the ground is invariant. Comparing decay rates of the two batches of muons in the case where their paths have only one intersection has no invariant meaning. It is purely a matter of simultaneity convention.

This is different from comparing a bunch of muons at rest with respect to a muon storage ring, near it, versus muons circulating in the ring. Here, there is a periodic meeting time defining a well defined world line for both muon (batches). So here, invariantly, in any coordinates, you find the circulating muons decay slower, by computing proper times for the two paths between meetings.

• hutchphd, vanhees71, Ibix and 1 other person
Or, in short, the outcome of a "one-way twin paradox" is frame dependent.

PAllen said:
The problem for all of this is that proper time is defined along a world line between two events. The event of muon creation and muon ground detection are well define events. However, for a muon at rest on the ground, only the event of meeting an incoming muon is well defined. The event on ground muon's world line corresponding to the atmospheric muon's creation is wholly a matter of convention. So the fact that the muon's reach the ground is invariant. Comparing decay rates of the two batches of muons in the case where their paths have only one intersection has no invariant meaning. It is purely a matter of simultaneity convention.

This is different from comparing a bunch of muons at rest with respect to a muon storage ring, near it, versus muons circulating in the ring. Here, there is a periodic meeting time defining a well defined world line for both muon (batches). So here, invariantly, in any coordinates, you find the circulating muons decay slower, by computing proper times for the two paths between meetings.
yes indeed concerning the synchronization between A and B.
It was just a calculation of muon survival between any 2 events separated by a proper time of value D/v.

PAllen said:
You misread what I wrote. The muon proper time between creation and ground detection is way less than muon half life. That is true, and that is what I wrote.
Ok sorry, then I really misunderstood what you wrote. That's of course true, and it's true in any reference frame.

PAllen said:
You misread what I wrote. The muon proper time between creation and ground detection is way less than muon half life. That is true, and that is what I wrote.
Actually, the muon half-life is significantly more (by a factor of four or so) than the proper time between creation and ground detection. Most of the muons fail to reach the ground, despite time dilation or length contraction.

It is just that the fraction that fail to reach the ground is not as massively low as the classical prediction (20 or so half lives).

Numbers taken from Hyperphysics site.

jbriggs444 said:
Actually, the muon half-life is significantly more (by a factor of four or so) than the proper time between creation and ground detection. Most of the muons fail to reach the ground, despite time dilation or length contraction.

It is just that the fraction that fail to reach the ground is not as massively low as the classical prediction (20 or so half lives).

Numbers taken from Hyperphysics site.
I’ve always considered that many muons are created at much lower altitude than 10 km, where the atmosphere is denser. However, looking at the density as a function of altitude, it does appear 10 km would be a reasonable median creation altitude. But there is also the question of muon energy. I think many are created with ##\gamma## greater than 5, some much greater (even over 100).

• vanhees71 and jbriggs444
PAllen said:
I’ve always considered that many muons are created at much lower altitude than 10 km, where the atmosphere is denser. However, looking at the density as a function of altitude, it does appear 10 km would be a reasonable median creation altitude. But there is also the question of muon energy. I think many are created with ##\gamma## greater than 5, some much greater (even over 100).
Actually, I just checked and found the average muon energy at creation is well over 4 GeV, which is a ##\gamma## of over 40. Thus, my statement was correct. For some reason, hyperphysics is using a plausible creation altitude, but a way too low energy. There are even substantial numbers of muons created with ##\gamma## of 40,000 or more.

[edit: as a further aside, I note that the cosmic ray muon spectrum is regularly measured out to 100 TEV, which corresponds to a ##\gamma## of 1 million. For such a muon, it would appear only 1 cm of atmosphere passed by before the ground hits.]

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• vanhees71 and jbriggs444
PAllen said:
Actually, I just checked and found the average muon energy at creation is well over 4 GeV, which is a ##\gamma## of over 40. Thus, my statement was correct. For some reason, hyperphysics is using a plausible creation altitude, but a way too low energy. There are even substantial numbers of muons created with ##\gamma## of 40,000 or more.

[edit: as a further aside, I note that the cosmic ray muon spectrum is regularly measured out to 100 TEV, which corresponds to a ##\gamma## of 1 million. For such a muon, it would appear only 1 cm of atmosphere passed by before the ground hits.]
And this page of hyperphysics has the correct number - no idea why the muon experiment page uses misleading numbers:

http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/muonatm.html