Strich.9
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 Posted: Tue Nov 18, 2008 11:51 pm Post subject: What is the Muon Half-Life? |
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[[Mod. note -- This post arrived with some characters garbled.
I have managed to unscramble some, but not all, of these.
It>s safest to confine your postings to plain 7-bit-clean ASCII.
-- jt]]
[[Mod. note -- When pondering special-relativity situations like
these, it>s useful to focus on *events*, rather than time intervals
(which are observer-dependent).
-- jt]]
The MUON Experiment. The standard interpretation of the muon
experiment is as follows. Muons are formed in the upper atmosphere.
They move with a velocity close to light at v=0.995c. The muon has a
postulated half-life of about Th~2M-NM-<s measured by 'stopping' a speeding
muon. But the muon somehow reaches the earth, travelling up to
D=6000m. With this distance and speed, the muon is computed to have a
half-life of:
Th' = D / v = 6000m / 0.995c = 20M-NM-<s
The apparent conclusion is that the muon has been time dilated and
therefore lived 10 times longer.
Let us analyze this in better detail. First let us pick a spot on the
earth E, at the north pole, and let us pick a spot F that is at a
distance D=6000m above the north pole. A muon will form anytime at F
heading towards the north pole. This muon will travel at around
v=0.995c and reach the north pole.
Before we make any computations, let us define the four times we are
going to measure:
Te = native co-inertial measurement
Te' = native trans-inertial measurement
Tm = foreign co-inertial measurement
Tm' = foreign trans-inertial measurement
We define the earth observer as the native observer and the muon
observer as the foreign observer. Obviously the co-inertial
measurements are those measured within the rest frame and the trans-
inertial measurements are those measured in a different frame to the
rest frame.
Now a muon forms at F and the earth observer starts his stopwatch at
time 0. The muon streams down towards the north pole at 0.995c. It
decays just as it reaches the north pole. At the time of impact the
earth stopwatch registers:
Te = D / v = 6000m / 0.995c = 20us.
The earth observer also observes the muon stopwatch start at time 0 at
the beginning of the journey. Upon impact, the muon stopwatch
registers:
Te' = 20us / gamma = 2us.
Note that this kind of direct trans-inertial observation of another
clock has never been performed so we base our measurement using the
Lorentz transformation equation.
This is the standard analyses of the Muon experiment. Note that
experiments have measured a stopped muon and noted a half life of
~2us. Note also that a stopped muon is one that decays within the
detector so it is usually a less energetic muon. But we are not here
to dispute experimental methodology and interpretation so we will
simply leave that as it is.
Note that we have both Te and Te' but not Tm and Tm'. So we continue
with our analysis.
We have the same spot F and the north pole. We use the same muon but
from a different perspective. Our same muon is born at F and starts
his stopwatch at time 0 and right away finds the earth streaming
towards it. The earth just impacts the muon at the same time that the
muon decays, travelling a distance of D=6000m with a velocity of
0.995c. Just as the muon and its observer passes away, he notes that
his stopwatch registered:
Tm = D / v = 6000m / 0.995c = 20us.
Now the muon based observer observes the earth stopwatch at the
beginning of the earth>s journey at time 0, and just as it passes away
notes the earth stopwatch registered:
Tm' = 20us / gamma = 2us.
Note that this kind of direct trans-inertial observation of the earth
clock from a subatomic particle cannot be performed, so we base the
measurement using the Lorentz transformation equation.
Note that both the earth clock measures 20us prior to the decay of the
muon and that the muon clock measure 20us prior its decay.
Now the relativist argument is that there is time dilation since the
earth observer observed the muon clock trans-inertially and measured
it to tick only Te'=2us. However, the same clock was observed co-
inertially by the muon and he saw it tick the full Tm=20us.
Likewise, the muon observer will argue that the earth clock was slow
and ticked Tm'=2us while the earth observer will argue that the earth
stopwatch registered Te=20us.
According to relativity, at the time of impact, we have two clocks,
the earth and the muon clock, each with two different non-reconcilable
readings.
To argue a time dilation in the muon clock would be to argue a time
dilation in the earth clock. Now each second millions of muons are
streaming towards the earth at 0.995c. Does anybody observe their
clock time dilated? Of course not.
Time dilation is a trans-inertial artifact, which the co-inertial
clock will never register.
The final question, what is the resting half life of the muon?
The answer, it was measured by the muon clock co-inertially at
Tm=20us. |
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doug
Guest
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| Posted: Fri Nov 21, 2008 3:21 am Post subject: Re: What is the Muon Half-Life? |
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Strich.9 wrote:
[[Mod. note -- Very-excessively-quoted text snipped. -- jt]]
[quote]Te = native co-inertial measurement
Te' = native trans-inertial measurement
Tm = foreign co-inertial measurement
Tm' = foreign trans-inertial measurement
We define the earth observer as the native observer and the muon
observer as the foreign observer. Obviously the co-inertial
measurements are those measured within the rest frame and the trans-
inertial measurements are those measured in a different frame to the
rest frame.
Now a muon forms at F and the earth observer starts his stopwatch at
time 0. The muon streams down towards the north pole at 0.995c. It
decays just as it reaches the north pole. At the time of impact the
earth stopwatch registers:
Te = D / v = 6000m / 0.995c = 20us.
The earth observer also observes the muon stopwatch start at time 0 at
the beginning of the journey. Upon impact, the muon stopwatch
registers:
Te' = 20us / gamma = 2us.
Note that this kind of direct trans-inertial observation of another
clock has never been performed so we base our measurement using the
Lorentz transformation equation.
[/quote]
[[Mod. note -- Very-excessively-quoted text snipped. -- jt]]
[quote]We have the same spot F and the north pole. We use the same muon but
from a different perspective. Our same muon is born at F and starts
his stopwatch at time 0 and right away finds the earth streaming
towards it. The earth just impacts the muon at the same time that the
muon decays, travelling a distance of D=6000m with a velocity of
0.995c. Just as the muon and its observer passes away, he notes that
his stopwatch registered:
Tm = D / v = 6000m / 0.995c = 20us.
[/quote]
The error here is to assume that the muon sees the distance as
6000m. The muon is created, its clock starts. It times its
approach to the earth. At impact, it reads 2usec, the moun is
traveling near c so it calculates the distance as 2usec x c = 600meters
not the 6000m in the earth frame.
[quote]
Now the muon based observer observes the earth stopwatch at the
beginning of the earth>s journey at time 0, and just as it passes away
notes the earth stopwatch registered:
Tm' = 20us / gamma = 2us.
Note that this kind of direct trans-inertial observation of the earth
clock from a subatomic particle cannot be performed, so we base the
measurement using the Lorentz transformation equation.
Note that both the earth clock measures 20us prior to the decay of the
muon and that the muon clock measure 20us prior its decay.
[/quote]
Wrong. See above and experiments. This means the discussion below is
wrong as well.
[[Mod. note -- Very-excessively-quoted text snipped. -- jt]]
[quote]The final question, what is the resting half life of the muon?
The answer, it was measured by the muon clock co-inertially at
Tm=20us.
[/quote]
No, it was measured here and by other experiments as 2usec. |
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