Analyzing the Muon Experiment as Proof of Special Relativity

Muons are continually being formed in the upper atmosphere. They live for about 2.26 microseconds. Travelling at speeds close to light, they cover 700 meters as measured in the muon frame, which corresponds to 7000 meters in the Earth frame due to Lorentz contraction, enabling them to reach the lower atmosphere.

In 2.26 microseconds in the earth frame, the same muon can only travel 700 meters as measured in the Earth frame, and not reach the lower atmosphere.

Since the 2.26 microseconds in the muon REST frame is EXACTLY EQUAL* (barring the trivial effects of a gravitational field) to the 2.26 microseconds in the earth REST frame, then we have a contradiction, not a paradox mind you, but a logical contradiction. Thus Special Relativity must be wrong.


[*According to Special Relativity, all clocks and measuring rods at rest in their respective frames keep the same time and length.]

Since we know that muons do reach the lower atmosphere, then we must postulate other explanations. Either the muon has a longer life span, or we are seeing only a tiny fraction of muons formed in the upper atmosphere, about 0.1%, which have life spans several standard deviations above the mean.

If one has difficulty analyzing the above, perhaps the following may help.

Analyzing Multi-Body Special Relativity Problems

Five neutrinos A, B, C, D, E are fired from an accelerator 10km away to a muon at rest in an MIT lab with parameters below.


A B C D E
Velocity (in c)* 0.99 (A) 0.87 (B) 0.60 (C) 0.41 (D) 0 (E)
Velocity (in m/s) 3x10^8 (A) 2.6x10^8 (B) 1.8x10^8 (C) 1.2x10^8 (D) 0 (E)
Gamma 10 (A) 2 (B) 1.25 (C) 1.1 (D) 1 (E)
Distance (m) 10000 (A) 10000 (B) 10000 (C) 10000 (D) 10000 (E)
Length Contraction (m) 1000 (A) 5000 (B) 8000 (C) 9090 (D) 10000 (E)
Muon Lifespan (microsec) 2 (A) 2 (B) 2 (C) 2 (D) 2 (E)
Distance M** (m) 600 (A) 520 (B) 360 (C) 240 (D) 0 (E)
Time Dilation (microsec) 20 (A) 4 (B) 2.5 (C) 2.2 (D) 2 (E)
Distance N*** (m) 6000 (A) 1040 (B) 450 (C) 264 (D) 0 (E)


*velocity relative to muon
**distance “traveled” by muon in its frame prior to disintegration=velocity x time
***distance “traveled” by neutrino in its frame prior to muon disintegration=velocity x dilated time

The M distances are in the muon frame and hence comparable while the N distances are in the neutrino frames which are different and not comparable. Since in our experiment the muon is "motionless", the M distance represents the distance reduction between the muon and a neutrino at the time of its death from its reference frame.

It is seen that the muon covers different distances depending on who is observing it, and when there are five observers, then the muon dies after covering 0, 240, 360, 520, AND 600 meters in its frame, which is obviously IMPOSSIBLE.

M distances are actually correct: these are relative distances between N and M.

However, the N distances are anomalous as they imply superluminal travel, which is obviously IMPOSSIBLE.

The N distances represent the distance the neutrino travels prior to the disintegration of the muon, which is 2 microseconds. If viewed in the muon frame, this is superluminal travel, which is impossible.

We can "cheat" and re-correct the N distances to its frame, making them equal to the M distances.

This reduces the problem to a non-relativistic analyses, wherein the distance reduction between the particles is equal to velocity x time, with no invocation of time dilation and length contraction, which gives consistent results.

If relativistic analyses gives inconsistent results, and non-relativistic analyses gives consistent results, why do we need relativity?

Better still, why is relativity wrong?

these threads always seem to spiral into complexity and the fallacy is lost in the storm. let me distill this issue into a concise 5 step proposition:


PROP 1: Muons formed in the upper atmosphere have a rest life of ~2.26us.

PROP 2: Travelling at speed ~c, they cover 700m as measured in the muon frame prior to disintegration after ~2.26us,

PROP 3: Which corresponds to 7000m in the Earth frame due to Lorentz contraction, enabling them to reach the lower atmosphere.

PROP 4: In 2.26us in the earth frame, any particle with speed ~c, (such as a photon, or a muon in this case) can only travel 700m as measured in the Earth frame, and not reach the lower atmosphere.

PROP 5: Since the 2.26us in the muon REST frame is EXACTLY EQUAL (barring the trivial effects of a gravitational field) to the 2.26us in the earth REST frame, then we have a contradiction, since 700m is not equal to 7000m.

Conclusion: Special Relativity is erroneous.

If this logic is in error, I challenge anybody to pick the erroneous step. This is now only a single multiple choice question.

People who have succumbed to dogma have had their common sense suspended. Arguments with them would have to be very clear to get through years of false logic which they have adapted into their habit of thinking. We will start with a simple case...

THE GREEK THEORY OF RELATIVITY

Everybody knows that a person in the distance looks smaller. If A and B are at distance (d) apart, B looks smaller to A and A looks smaller to B.

There is no absolute point of reference, and Cleveland is as good as Boston, and vice versa, and any other point for that matter. This becomes a relativity problem.

The decrease in size is proportional to the distance (d) in each locus. The greeks have perfected the measurements and equations for this problem, and in the process discovered and developed geometry. We can summarize these in the standard equations of a relativistic theory. Thus,

Let MA be the measure in A and MB be the measure in B. Then:

MA' = f1(d) x MB

MB' = f2(d) x MA

Where d is the distance, f1 and f2 are the Greek transformation equations, MA' and MB' become the relative size measures. Note that since the frames are symmetric, and they have to be in all relativity theories, the f1 and f2 equations are the same, and the frames are equivalent.

Thus:

MA' = f(d) x MB
MB' = f(d) x MA

We could use one equation for both, but I am keeping the two to illustrate a point. If A uses a different measure standard, such as meter, compared to B, such as yard, then the equations will not be the same. However, we will know this right away because they will also use a different distance (d) measure. If we know beforehand that the same (d) units are used, say meter, then this implies that the same units of size measures are operative in both reference frames. B cannot plug in yards into its equation and still come out with meters in its conclusions. We will get back to this later. Right now there is one more very important point here.

Note that obviously the Greek Theory of Relativity is self-consistent and we use their system to this day.

Now suppose that another civilization, say Atlantis, came up with the same theory. Let us call this the Atlantis Theory of Relativity. Now equation for equation the ATR is the same as the GTR. We can even allow that the same units of measure were used. But there is one curious difference. The Atlanteans postulated that when an object is at distance (d), it does not appear smaller. It is actually smaller. This sat well with the magicians of Atlantis, who had always dreamed of shrinking massive objects, like the great pyramids, so they can easily be carried home. Now this did not sit well with some of the intellectual mutants of Atlantis, for they reasoned that an object at a distance still possesses the same weight. This was easily refuted by the orthodoxy, as it held that the weight scale used in the distant object had shrunk as well, and its measurements were correspondingly smaller. The orthodoxy held that if the distant object were weighed on the reference frame of the observer instead of the distant one, then this would prove that its weight had indeed shrunk. As a matter of fact, this was an active area of research in Atlantis. It was also a top secret and top priority project, with its great application in object miniturization. The rest of the story I leave to the reader.

Mathematically, speaking, the ATR is equivalent to the GTR, and is likewise a self-consistent mathematical system. The difference is in the interpretation. The Greeks new that the measure changes were an illusion, while the Atlanteans thought it was reality.

Now the Special Theory of Relativity is a consistent mathematical system. Its error is in the interpretation. If time dilation and length contraction are interpreted not as reality, but as an illusion, then it would be totally correct. That is where the problem lies and that is where Einstein made his crucial mistake.

Let M and N be two inertial frames. There is a resting clock in M and another one in N. Let M use an arbitrary unit of measure of time, say the Msec and let N arbitrarily use the Nsec. There is no means to compare the two units. If M measures the time intervals in N, it will measure these in Msecs, not Nsecs, as M obviously does not have a reference N-clock. Now M measures the relative time intervals in N in Msec units by using transformation equations. By reversing (or inverting) these equations, M can obtain the actual time intervals in N, but still in Msecs.

Likewise, N measures time intervals in N and M using Nsecs.

How does one calibrate the Msec along the Nsec, and vice versa?

While it may appear obvious, it is reiterated and stressed that the units of measure in each frame is invariant. Thus an Msec in M will always represent the same interval of time anywhere and anytime in M; the same goes for the Nsec in N, and any other arbitrary unit of measure Xunit in any other frame X. Note that this applies to any of the basic units of measures for time, distance, mass, charge, etc.

For those who are easily confused, this refers to the units of measure, and not to the actual quantities being measured. For example, a bound quark having less mass than a theoretically unbound quark presupposes a consistent unit of mass (or energy) is being used, without necessarily implying any invariance in the actual mass of the quark.

Note that for the contrary to be true, that the units of measure are not invariant, means that our all observations, measurements, and experiments, become irrelevant, for we lose a standard by which to compare one quantity against another.

Let us go back to our original question, how does one calibrate the Msec along the Nsec, and vice versa?

The simplest method is to let the two frames be at rest relative to one another. Since their relative velocity is zero, then the transformation equations become constants, and now

Nsec = k Msec
Msec = k' Nsec

where k and k' are merely conversion factors from one unit to the other (such as multiplying yards by the factor of 0.914 to get meters). In fact, we can create a new standardized unit in N, callled the mNsec such that,

mNsec = k' Nsec = Msec

or in other words, an identical unit of measure to the Msec in the N frame. Now that we have identical units, we can make measurements in N that are easily referenced in M, and vice versa.

From above: mNsec in N is exactly equal to Msec in M (M and N at relative rest).

Let N move with respect to M, with relative velocity v.

As the clock in M remains at rest in M, its rate does not change in M.
As the clock in N remains at rest in N, its rate does not change in N either.

Since the rates did not change, mNsec in N is still exactly equal to Msec in M (M and N at relative motion).

This allows direct comparison of any interval of time in M to any interval of time in N via a common unit (mNsec = Msec).