Math Forum / Mathematics / Mathematical Logic / November 2008

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 supposed resting half-life
of about Th ~ 2μs measured by ʽstoppingʼ 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:

    Thʼ = D / v = 6000m / 0.995c = 20μs

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
Teʼ = native trans-inertial measurement
Tm = foreign co-inertial measurement
Tmʼ = 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.
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 the
beginning of the journey. Upon impact, the muon stopwatch would
register a slower time of:

    Teʼ = 2us.

Note that this kind of direct trans-inertial observation of another
clock has never been performed and is based on a theoretic application
of the Lorentz transformation equation (LTE) relating Te' and Te as
follows: Te' = Te / gamma.  This is not quite correct as the standard
LTE should relate the clock in the muon as measured by the muon
observer (foreign co-inertial) to the clock in the muon as measured by
the earth observer (native trans-inertial).  However, due to some
similarity in measurements this erroneous use of the LTE seem to yield
the correct numbers.

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 Teʼ but not Tm and Tmʼ.  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 right away, finding the earth streaming towards it at
0.995c.  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 also observes the earth stopwatch start at
the beginning of the earthʼs journey, and just as the muon passes away
notes the earth stopwatch registered:

Tmʼ = 2us.

Note that this kind of direct trans-inertial observation of the earth
clock from a subatomic particle cannot be performed.  However it is
computed from the standard LTE relating the earth clock as measured by
the earth observer Te (native co-inertial measurement) to the earth
clock as measured by the muon observer Tm' (foreign trans-inertial
measurement):

Tm' = Te / gamma

In this regard, the time on the muon clock as measured by the earth
observer Te' (native trans-inertial measurement) is also related to
the time on the muon clock as measured by the muon observer Tm
(foreign co-inertial measurement) by the standard LTE;

Te' = Tm / gamma

Note that both the earth clock measures Te=20us prior to the decay of
the muon and that the muon clock measure Tm=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 Teʼ=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
ticking Tmʼ=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: Te=20us and Tm'=2us for the earth clock and
Tm=20us and Te'=2us for the muon clock.

To argue a time dilation in the muon clock would be equivalent to
arguing 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
cannot be observed in the co-inertial clock.  This is the fallacy of
the Hafele-Keating experiment.  The atomic clocks will never measure a
slowing of clock rate in its own frame, so when it is finally brought
back it will still have the same time as the co-inertial time of the
stay at home atomic clock.

Time dilation is a trans-inertial artifact, which the co-inertial
clock will never register.

To answer the original and final question, what is the resting half
life of the muon?  The answer is that it was measured by the muon
clock co-inertially Tm=20us.

The earth observer can also easily compute the actual foreign co-
inertial time in the muon by the LTE:

Te' = Tm / gamma

By solving for Tm we get the inverse LTE:

Tm = Te' * gamma

Thus

Tm = Te' * gamma = 2us x 10 = 20us.

Giving the actual half-life of the atmospheric muon.  And what about
the 2us measured for 'at rest' muons?  Sampling error.  Specifically,
the detector only caught the lower energy muons.  [It is like a cat
concluding that all mice are white, simply because she only caught the
white mice she saw and never even noticed the dark mice go by.]

Strich
Using the Power of IQ

Note that when Strichnine senses he is mistaken he resorts to innuendo.

And what of it?

Your view of the event is entirely impoverished.

End of thread.