# The Time Dilation of the Muon is just Bad Math

 Please note that this paper is a simplification by me of a paper or papers written and copyrighted by Miles Mathis on his site. I have replaced "I" and "my" with "MM" to show that he is talking. All links within the papers, not yet simplified, are linked directly to the Miles Mathis site and will appear in another tab. (It will be clear which of these are Miles Mathis originals because they will be still contain "I" and "my".) The original papers on his site are the ultimate and correct source. All contributions to his papers and ordering of his books should be made on his site. (This paper incorporates Miles Mathis' muon paper and the first section voat paper.) First posted September 20, 2009

Abstract: Miles Mathis will show that the atmospheric muon, now used as proof of time dilation, is not experiencing time dilation at all. Even with MM's corrections to SR shown in An Algebraic Correction to Special Relativity and Refutation of Gamma a muon in approach would actually be time compressed, not time dilated since only objects moving away from us are time dilated. The main problem has never been a problem of Special or General Relativity, it has been a lack of knowledge in applying simple gravitational math, specifically in stacking a velocity and an acceleration in a vector situation, as in the equation v = v0 + at (analyzed below).

Once the basic textbook acceleration equations are rewritten MM will be able to take muon creation up to 270 miles, into the ionosphere where it should be, not in the lower stratosphere (9 miles) where it currently is said to be. Since this problem has nothing to do with Relativity, this paper is neither a refutation of Relativity nor time dilation, both of which are real.

## The equation v = v0 + at

First posted December 30, 2009

The above equation can not be used when we have a particle like a muon with a very high velocity being accelerated by a gravity field. In that case the equation should be:
vf = vo + 2vo2t.

The equation v = v0 + at only works when the acceleration is not a field for example when the acceleration is an internal acceleration, as with a car, but not as a field equation.

This equation in textbooks is derived as follows

a = v/t or v = at
We add the initial velocity and we have: v = v0 + at
This equation is basically a definition of acceleration.

In an example of variable acceleration in MM's paper on the calculus (A Redefinition of the Derivative and the Integral MM showed that the equation can actually be calculated from the time derivative:

v = a[d(t2)]/2 = at

The 2 in the denominator comes from the halved first interval, which we must take into account in any acceleration.

So we have confirmed that part of the equation. The problem must be with the way the initial velocity and the acceleration stack up in a field. Apparently we cannot just add the final velocity from the acceleration to the original velocity. The question is: why is this?

Put simply, the reason is because if we have an initial velocity meeting an acceleration field, we have three velocities. Let us start with just the acceleration field, and no initial velocity. Let us say you have a gravity field, or any other acceleration field that creates a normal or squared acceleration. If you place an object in the field, the object will be given two simultaneous velocities.

That is what an acceleration is, after all. If the object has one velocity, it has no acceleration. If it has an acceleration, it must have more than one velocity over each interval or differential. A squared acceleration, or an acceleration to the power of 2, is composed of two velocities. Therefore an object placed in any acceleration field will be given two velocities.

However, if we take an object that already has a velocity of its own, and we fly it into that acceleration field. It must then have three velocities over each interval. Or, to say it another way, it now has a cubed acceleration. It has three t's in the denominator.

For this not to be the case, we would have to break Newton's first law. If we do not include the initial velocity in each differential of acceleration, that means the acceleration field has somehow negated that velocity during the time of acceleration. The field would have to stop the object, then accelerate it, then give it its original velocity back at the end. Of course all that is impossible and absurd. No, the field would have to keep the initial velocity in every differential of acceleration, meaning that it would be accelerating the velocity, not the object.

The problem with the equation v = v0 + at
is that it does not include a cubed acceleration. The “a” variable stands for a squared acceleration. The equation includes three velocities, but the first velocity has not been integrated into the acceleration. It is only added, as a separate term. But that is not the way the real field would work. The real field would accelerate the initial velocity, not just the object. As it is, the initial velocity is not summed in every differential, as it should be; it is only summed outside the acceleration. That cannot work, either as a matter of calculus or as a matter of physics. The equation is wrong, in any and all cases where a field is involved.

One might ask that since the equation v = v0 + at
clearly works in the simple case of a car, why would it not work in a field? How is a car accelerating different than a field? Thus is because the car is accelerating itself, internally, with its own engines, and the car has no velocity relative to itself. The car and its engines have no way of knowing the car is moving at all, since velocity is relative. The car cannot be its own field, by the definition of field.

Relative to itself, the car always has a velocity of zero. Therefore the acceleration is not a field acceleration. The engines really are accelerating the car, not the velocity of the car. But if we have an external field, the field must be accelerating the velocity itself.

The apparent conflict with MM's new equations is that they show huge final velocities and yet we know that falling objects and objects like meteors entering the atmosphere do not reach velocities like that. The explanation comes from terminal velocity. Most objects will have an appreciable size and mass, so that they are affected strongly by the atmosphere. They cannot reach those velocities simply due to drag. (However, MM new equations do predict that that objects entering our gravity field reach terminal velocity much quicker than previously supposed.)

When we consider very small objects like muons, that are not affected by atmospheric drag, they are not slowed by the atmosphere at. They are either absorbed or deflected by it, or they are not.

If we detect these particles at sea level, then it is because they have avoided collision with molecules in the atmosphere. If they avoided collision, then they avoided "drag". Therefore they can cover these very long distances predicted by MM's field equations.

(For further study see the paper on MM's site on cosmic inflation where he has also developed a simple equation for cubed acceleration.)

The muon is currently used as proof of time dilation, since without it, we are told, the muon would not live long enough to reach the surface of the earth. The muon lives only 2.2 x 10-6 seconds, and travels only about 660m under normal circumstances. But it is thought to be produced at an altitude of 15km. It does reach detectors on the surface of the earth, or at sea level, therefore it must be experiencing time dilation.

The problem is that physicists make this claim without applying the transforms directly to the muons they detect. Since they do not know the altitude of creation, they cannot know the distance or time. Transforms have to be done on data: that is, on measurements that are directly made in experiment. But we only detect the muons, or, in some cases, measure their energy. We cannot and do not measure time or distance, therefore we cannot apply time or distance transforms. The altitude of production is inferred from the energy, as is admitted.

This is a major problem since you cannot apply SR transforms to an inference. You cannot find that a body is dilated without knowing its initial state. We do not know the altitude of production for the muons we detect, therefore we cannot apply transforms. The current theory is circular. We claim that if the muons are dilated as we think they are, then they must come from a certain altitude. And they are able to come from that altitude because they are dilated. The theory has no content and no possible proof. If it cannot be proved, it cannot be proof of anything.

MM does not doubt the theory of muon creation. In fact, he can use muon creation and detection as proof of his own mechanics only because he trusts the experiments. MM believes in the creation and detection, as it is given to us in current journals and texts, but he does not believe in the given time dilation, since there can be no time dilation without a measurement of time or distance.

Muon detection is easy to explain without time dilation, provided you know how to apply the ordinary gravitational field equations. Using gravity in an orthodox way, the muon arrives on the surface of the earth because it is accelerating toward the earth, not just moving at a constant velocity. This must affect the apparent time of the trip.

This simple statement is controversial because it appears to conflict with current theory in several ways. First, because the muon is in freefall, and is not in a gravitational curve, it doesn't appear that GR allows us to find a time differential due to gravity. We can find a time differential due to SR, but not to GR. In fact, it is thought that GR prevents us from accelerating the muon at all: you cannot accelerate something that is already moving at .9996678c (This is the speed the current model believes the atmospheric muon is going, according to the value of gamma of 38.8.), not even in a gravitational field. Or, you could accelerate it a bit, but you would need a much more powerful field and more time.

We can solve this apparent conundrum if we use Einstein's equivalence principle to reverse the vector of gravity as shown in Solving and Arguing General Relativity Problems without the Tensor Calculus (In about 1/100th the time) The reversal makes the surface of the earth move out during each dt, which makes the earth move toward the muon during each dt. The muon's trip is shorter with each passing moment, which makes the time for the trip smaller. The muon doesn't have to go as far, therefore it gets there quicker.

Although many think this is illegal, it is done all the time, even by the mainstream as shown in the following two papers on MM's site: Richard Feynman does it in his Lectures on Gravitation, section 7.2. To calculate a gravitational blue shift, Feynman moves the bottom of his box up toward light that is coming down. Since he is basically repeating the math of the famous Pound-Rebka Experiment of 1959, the math for that experiment must do the same thing. The Earth must be given a velocity toward the light, or no blue shift can possibly be calculated.

Although Einstein's postulate allows for the vector reversal, it seems that in a real situation you cannot move the Earth toward the muon, since that would take the total velocity above c and that would be forbidden. Yes, according to one interpretation of Relativity, one cannot let the Earth move toward the muon while the muon is moving toward the Earth at c, because this would take the total velocity over c. However, this interpretation is wrong. It is wrong because it doesn't take into account the vector situation we have here, and the simple vector math we must do.

If we move the earth toward the muon, we have not increased the total velocity, taking it above c. We have shortened the distance that must be traveled, therefore we have actually decreased the velocity necessary to travel it. If we keep the muon at c, then t will have decreased. It is not the total velocity that has increased, it is the total distance that has decreased. Since velocity is distance over time, the velocity cannot increase with a decrease in distance. They are proportional.

The mistake is in assuming that you add velocities when you approach in this situation, when the opposite is true. Logic says that if a car is coming towards you at 60 miles per hour and you are driving towards it at 50 miles per hour, then the combined velocity is 110 miles per hour, and if we are 110 miles apart to begin with, we will collide in one hour.”

All true, but in that example we are going 110 miles per hour relative to each other, not to the background. To find out how fast we are going relative to the background, we subtract. We subtract because we are in a vector situation. Together, we are going 10 miles an hour relative to the road, since you went 60 miles and I went -50 miles. Velocity is a vector.

If we translate that simple finding to our muon, we find that we are not really interested in how fast the muon is going relative to the earth, since we are not measuring its velocity from the earth. Yes, we are detecting it from the earth, but a detection is not equivalent to a velocity measurement or a time measurement. When we do these velocity and time calculations from a simple detection, what we are calculating is how fast the muon is going relative to the background of the muon+earth. The earth is not the background of the muon; the earth is an object, and both the muon and the earth are objects moving with regard to a background.

If we turn the earth's gravity vector around, this is easy to see. The earth must be accelerating relative to something, since acceleration is a motion and all motion is relative. The earth cannot be accelerating relative to itself. As soon as we have assigned the earth an acceleration, whether it is the old gravitational acceleration or my new expansion, we must assign a third point of reference. In my expansion, the surface is moving relative to the center, so we have the three points of reference: muon, surface, center. A Euclidean coordinate system tied to the earth's center is our background, and both the surface and the muon move relative to it.

Clearly, the surface of the earth does not move out so the vector reversal may be allowed in a thought problem, but not in reality. However, gravity, defined as it is, as an acceleration field arrayed about a central point, creates precisely the same vectors as my explanation in the previous paragraph, with the reversed g. The vectors were reversed to make the concepts clearer, but the concepts and vectors are the same whether we reverse g or not.

If the field of the earth has an acceleration, then this acceleration must be relative to some non-accelerated body or field. Einstein said that all motion was relative, and that applies to accelerations just as much as to velocities. This non-accelerated “body” is a Euclidean system set up around the center of the earth. We require a background to measure or calculate the gravity of the earth: not just how it affects things like muons, but how it defines the acceleration itself. And we require this non-accelerated background whether we reverse g or not. GR contains this background, because in GR the center of the earth acts as this background. The same can be said for Newton.

Historically, we have drawn the vector pointing in to indicate the motion of a particle dropped from rest, but this is not how the vector works in the math or in the field when we have a particle approaching the earth with its own velocity, as with the muon. Since the acceleration acts as an attractive force, it must combine with an incoming velocity in a vector situation that subtracts, not that adds. This is true even if we do not assign any real motion to the surface of the earth. Using conventional gravity theory, the gravity field of the earth must create a vector that combines with the muon velocity in a differential, with a negative sign. We never had any possible conflict with the limit of c here, which makes the modern failure to solve this problem that much more surprising.

If the muons are falling in the earth's gravitational field, as they are, then we cannot simply calculate the muons against the earth, as my critic did when he calculated the two cars' velocities against each other. We have to calculate the muons and earth against a background. Which gives us a vector situation. Which gives one of the velocities a negative sign. Which forces us to subtract velocities in our equation. Which allows us to have both velocities without exceeding c.

You can refuse my vector reversal if you like. It doesn't matter. Your conventional gravitational field, whether it is the field of Newton or Einstein, must work in a vector situation as if the earth is moving toward the muon. This is what the attraction implies, and if you refuse to assign that motion toward the muon to the surface of the earth, you must apply it instead to the field. That is, you can slide your motion off a real object and apply it to the field, but that trick won't help you evade me or my logic. Your field vector and c will still be in opposition, and my point is proved either way.

That explanation was a bit difficult in places, perhaps, but never esoteric. As you see, the explanation is really pretty simple, once you unwind it. It is another failure of contemporary physicists to understand vector math. They can juggle very complex equations, but they cannot seem to comprehend addition and subtraction in vector spaces. We can propose accelerating a particle already moving at c, and it is precisely because the two implied motions are in vector opposition. The muon is moving toward the earth, so the velocities do not add, they subtract. We do not find a velocity above c here, so we do not have to break any of the laws of Relativity. The muon's own velocity never goes above c, and the total velocity of (earth + muon) never goes above c (since the real vector equation is (muon – earth).

You will say, “Even if all that is true, you still cannot explain muons that way, since using the known acceleration of gravity, the surface of the earth will only move 1.1 x 10-11 m in 2.2 x 10-6s. Turning your gravity vector around doesn't explain anything.” That would be true only in the case that we take the initial velocity of the surface of the earth as zero. If the initial velocity is zero, then we can use the equation s = ½at2, which does give us the very small number above. But if the earth does not have an initial velocity of zero, that equation won't work. Well, the surface of the earth does have an acceleration of 9.8m/s2, but it doesn't have an initial velocity of zero. Since these accelerations and velocities are relative to the center of the earth, the velocity at the surface of the earth during a given interval can hardly be zero. The only zero velocity in the field would be expected to be at the center of the earth. The surface of the earth is about 6,378km from the center, so it could not have a velocity of zero relative to that center. If it did, then its acceleration would also be zero. So let us see if we can develop a velocity just from the numbers we already have. If we use the above equation, s = ½at2, and solve for t using 9.8 and 6,378, we get t = 1141. In a field of acceleration of 9.8, the surface of the earth is 1141s away from the center. From that we can develop the velocity: v = 6,378,000/1141 = 5,590m/s. That is just a rough estimate, since the earth's field is not 9.8 at all distances; but it gives you the idea of how things work in a gravity field.

My critic will say, “Even that won't help you, because at that velocity, the earth still goes less than a meter in the given time.” True, but the earth is not moving at a constant velocity, it is accelerating from that speed. When we have the muon moving at the earth, we do not have two velocities being added, we have a velocity of c plus an acceleration of g.

Now to solve. MM developed a velocity for the surface of the earth to prove that it could be done, and to show that, if we reverse the gravity vector, the surface must have an initial velocity. But MM will not use velocity in his solution. Since the vector situation allows us to move the earth toward the muon, even while the muon is at c, let us do the math in the simplest way possible. Let us dispense with the vector reversal and keep the earth steady. This will have the mathematical and mechanical result of giving all our motions to the muon, but it will not contradict all that has shown above. Doing the math this way is a convenience, and implies nothing about the physics involved. Although this will take the total calculated velocity of the muon alone over c, as a matter of addition, it won't concern us. Again it won't concern us, and won't break any of Einstein's rules, because due to the vector situation, the velocities aren't really added, they are subtracted. We apply them all to the muon only to simplify our math, but this in no way implies that the muon is actually going over c.

Everyone else who has looked at this problem has tried to apply the old equation v = v0 + at, which fails to help us here. Even before Einstein forbid us from exceeding c, this equation could not solve problems like the muon problem, where particles approaching the earth at c were supposed to be accelerated by g. It doesn't solve our current problem because it only gives us a tiny correction, one that is frankly counter-intuitive. A particle going c in a gravity field should be accelerated more than that, supposing that it can be accelerated at all. But rather than pull apart that basic equation that was critiqued at the beginning of this paper, showing the equation is false in all cases, let us ask how long it would take to accelerate a particle in freefall from zero to c in the earth's gravitational field. Let us pretend that the earth's field is constant at all distances, and that we can allow bodies to freefall as long as we like. In that case, we lose the v0 in the equation above, and it reduces down to c = at. Solving for t gives us 3.0591067142857 x 107 s. Since the acceleration is constant, the average velocity is c/2, and the distance traveled is therefore 4.58548560580009 x 1015m.

So it takes this hypothetical particle almost a year to go from rest to c. Now, a nice question is, how far would it travel in the next second? In other words, how far would it travel if we continued accelerating it past c, at the same rate of 9.8? Easy, since we just add a second to the time and solve: x = 2.989 x 108m. How far would it travel in the next 2.2 x 10-6 seconds?: x = 617m.

Hopefully, this method will be clear. Rather than assume that the muon is accelerated independently of its velocity, we assume that it is accelerated as if it were already in freefall. We assume the earth cannot tell the difference between a body it has been accelerating for a long time and a body that just arrived or was just created. In other words, the earth accelerates the velocity, not the body. In a gravitational field, the equation v = v0 + at doesn't work. In that equation, the acceleration is independent of the velocity. But we want to accelerate the velocity. To do that without calculus, you do it just like this.

A critic. not doubt say, you have traveled 660m and added another 617m. This would double the “lifetime” of the muon as well, if we give it to the time instead of the distance, but we do not require a doubling, we require a increase of 50x. These muons are still 13,723m short of the surface of the earth.

However, it is not that the muon has traveled another 617 meters on top of the 660, but that the muon was traveling 617 meters while it was traveling 660 meters. It would have traveled 660 meters in no field at all and then that a gravitational field would accelerate it 617 meters, specifically, that the gravitational field would accelerate a particle that far in an interval of that length, assuming the particle never had an independent velocity of its own. Since we accelerated this particle from zero, the particle had no velocity uncaused by the field. Therefore if we want to find how far the muon can travel during that same interval, with both the acceleration of gravity and its own velocity of c, we cannot just add the two numbers. We have to multiply them, giving us 660 x 617 = 407,220m.

Given this, the muon could be over 27 times higher than it is currently proposed to be and still reach the earth, but the math is not complete. If one does not let the acceleration of the earth be 9.8 over the entire trip of the muon, because the acceleration of the earth is not 9.8 at an altitude of 268 miles. It varies over the trip, going from about 8.62 at the start to 9.8 at the end. Giving an estimate of the new number: our average acceleration will be 9.21, which gives us x = 660m, which increases our altitude to 435,600m, putting us way up into the ionosphere.

Take note that the two numbers match. We came into the problem knowing that the muon traveled 660m with no acceleration, then found it was accelerated 660 by the field of the earth. We found that number precisely because the field is accelerating the initial velocity, not the muon. The initial velocity is not added at the end, as in the the naive equation v = v0 + at. The initial velocity is integrated into each and every differential of acceleration, as it must be. Which means the initial distance traveled during each differential is also integrated into the field acceleration. This gives us the equations

vf = v0 + 2v02t
x = v02t2

This allows us to estimate the distance traveled in the field by simply squaring the distance traveled outside the field.

>x0 = v0t
x = x02

Here are the full equations, so that you can see exactly why 660 comes up twice.

c = at0
d0 = ct0/2 = c2/2a
d = vf(t + t0)/2 = a(t + t0)/2
d = a[(c/a) + t]2/2 = [(c2/a) + 2ct + at2]/2
Δd = d - d0 = (c2/2a) + (2ct/2) + (at2/2) - (c2/2a)
Δd = ct + (at2/2)

Because t is very small in this problem and c is very large, the second term is negligible. This makes the distance traveled during each interval due to the acceleration almost equal to the distance traveled due to c. This is why there is no acceleration variable in MM's equations above: it can be ignored when the time period is so small. A field of 9.8 will act pretty much like a field of 1 or a field of 100.

MM could have given us these new equations to begin with, instead of leading us through all that talk of expanding Earths and so on. These new equations show a velocity over c, which is currently forbidden, such as squaring c. So it was necessary to explain precisely why that mathematical manipulation was allowed here, thus the length.

This finding will be of great use to physicists. Their current number for altitude of muon creation is about 9 miles, which is way too low. They have kept it low purposely to blend more easily with this time dilation theory. The higher they create the muons, you see, the more dilation they need. They already have gamma at around 39, which is embarrassing enough. Any additional altitude will just make that number go higher. But they need the muon creation to be much higher than 9 miles, since as it is, they have muon production just above the troposphere, in the lower levels of the stratosphere. That doesn't make any sense. MM's number, which comes out to be about 270 miles, is much better, since we are then in the upper levels of the ionosphere. In the ionosphere, we would expect muon creation. We need those ions for muon creation, and there just aren't enough of them at 9 miles to explain the number of muons we see.

## Conclusion

In the paper Relativity Demystified, MM explains that time compression is indeed equivalent to life extension, and this explains life extension in particle accelerators, where particles are approaching detectors. However, neither time compression nor time dilation has been necessary to explain the detection of the muon at sea level and we still are not measuring the time or distance the muon has gone. Given the muon's known lifespan, MM has been able to take its altitude up to 270 miles, with no discussion of Relativity. We can do a time transform if we like, and yes, it will show time compression and apparent life extension with the muon in approach, but there is no physical reason to do a time transform here.

MM has no need to devise an experiment above in which we are measuring the distance directly such as going up to 270 miles, creating muons, then measuring their arrival at sea level. He merely uses gravity equations to show that the gravity field of the earth can appear to accelerate particles that are already at c, and that it can do it without contradicting Relativity. It can do this because the gravity field acts physically or mechanically as a vector pointing out from the center. It acts to decrease the effective distance the muon must travel, and because it acts like this in the equations, it must act like a vector pointing out.

It is important to note that MM agrees with Relativity, for the most part. Thus, if we do direct measurements on a muon, we will not be able to measure it going over c. Likewise, if the muon measured us, it would not be able to measure a velocity over c. However, that fact does not impact this paper, since in calculating an altitude for the muon and in analyzing its trip, we aren't measuring its velocity. We are calculating a distance, which is not at all the same thing. Although there will still be those who do not accept my analysis, which shows that the vectors in this problem must subtract. They will stick to their interpretation of SR which says we cannot produce any vector pointing at an object going c, since that would create a total velocity above c. However, to answer this, MM simply offers them the blue-shift of light.

To create a blue-shift, you have to have a vector pointing at light, while the light is going c. The waves cannot possibly be shortened unless the receiver has a velocity relative to the source of emission, and the vector of this velocity must be opposite to the c-vector. This is already known and accepted. So we allow vector situations like this all the time. However, we only forbid assigning numbers to the vectors, because that would break a rule of Relativity. It should be clearly absurd to allow the assignment of vectors, but disallow them from having sizes. MM has shown that allowing them sizes does not contradict Relativity, therefore the whole question has been a tempest in a teapot.

Those who scoff at MM's math because the math shows the muon going over c are guilty of gross hypocrisy. They will not allow me to exceed c in any equation, when their equations in this very problem often exceed c. They have their muon going 15,000m in 2.2 x 10-6 seconds, which is a raw velocity of 22.7 times the speed of light. They will explain that in Relativity, velocity is no longer x/t, but that is not true. In fact, 22.7c is the measured velocity in this problem, according to their inferences, and the velocity transform is meant to transform the measured velocity into the real or local velocity. In other words, .9996678c is v' here, and 22.7c is v. They never do a velocity transform for you because they cannot figure out how to use Einstein's velocity transform, but that is the way it should work. They hide all their math, in which c is exceeded all the time, and then attack anyone who creates any equation that exceeds c, even when it is clear that this math is breaking no law of Relativity as will be seen in the second section in this paper

To say it one final time, in MM's equations, the muon is not going over c. The raw velocity is 660c, yes, but no real object is going that fast in any system, neither in its own nor anybody else's system. That number is a result, not a real velocity. It is a result of applying all the motions in the math to the muon, but that is just a convenience. The earth's field also has a motion here, and integrating the two motions gives us a number over c. That is not forbidden by Relativity.

A poor understanding of vector math caused physicists to believe that the muon and other incoming particles could not be accelerated by the earth's gravitational field. They tried to add the motions, taking the total above c. But as a matter of vectors, the motions subtract. The distance between the approaching objects gets smaller, which means the velocities must subtract, not add. Once this is recognized, and the gravity field is fully understood, the muon can be accelerated without any conflict with Relativity or the limit of c. Furthermore, in the gravitational field, the equation v = v0 + gt does not apply. Instead, when we are accelerating particles already at c, we must integrate the acceleration with c, in order to find a total distance traveled. You do not add freefall to the local velocity, you integrate the two.

To see how this affects the gravity field in other ways, see MM's paper on his site on reduced mass, where he shows that accelerations must be integrated as well. Two accelerations must be integrated in the way we have integrated a velocity and an acceleration here.

[A criticism has been raised, "Are you claiming the earth has velocity of 659c here? That is the only other motion we have, right?" No, MM is not claiming that the earth or the earth's field has a velocity of 659c. The earth's field has an acceleration, and you cannot derive a velocity that way from an acceleration. My critic is once again just doing bad math. He is taking a distance traveled by two objects in approach, one of which is accelerating, and then dividing that distance by the time of one of the objects. This number is over c, so a rule has been broken, he says. But no rule has been broken, since you cannot legally assign that velocity to either object or to both of them. Velocity is simply not defined that way. Nothing is actually going 660c, and nothing is being measured to have gone 660c, so no rules of Relativity have come into play. This is easy to see by giving the surface of the earth the acceleration. This shows that the distance in the math is not the real distance. The distance 270 miles is not the distance traveled by either object or both objects, in either system. It is the distance that will seem to be traveled by one object, in the math, if we prevent the other object from moving. But since, physically, we need two separate motions to explain the mechanics and the math, the number 270 miles doesn't apply to any real parameter. The number 270 miles is an integral, but the real motions are causing differentials. The rules of Relativity apply to the real motions and the differentials, not to these integrals.]

## An Analysis of the SR transforms in the Muon Problem

Mainstream Relativists have what they think is an airtight theory for the atmospheric muon, but MM has show already that both the theory and the math are full of holes. MM has shown that the theory was quite poor, beginning with the fact that they have muon creation at 9 miles, way below the ionosphere where it should be. MM also pointed out that, theoretically and experimentally, their use of SR was suspect, to say the least, since they do not have any method for measuring the altitude of muon production. SR is a theory of measurement. The transforms have to be applied to real data, not to inferences. Their theory is circular because they use time dilation to explain the altitude of production, and use the altitude of production to explain the amount of time dilation.

To start with, we will look at the velocity transform. The velocity transform of SR is

V = w + v
1 + wv/c2

We have three velocity variables, as you see. Can you figure out how to apply this to the muon? Our mainstream Relativists will say that this equation is used when you have two observables, like a spaceship firing a rocket. But if we have a muon experiencing time dilation, as they claim, then we must have length contraction as well. You cannot have one without the other, and they go hand in hand. Well, if you have time and length transforms, how can you fail to have a velocity transform? Isn't velocity determined by length and time in Relativity?

Relativists really cannot answer this simple question, and it is because Einstein did not give them a first-degree velocity transform for SR. The velocity transform above is for two-degrees of Relativity, and they seem to understand that. But if you have a velocity transform for two observables, should not you have a velocity transform for one observable, like this muon? If its time and length are different, its velocity must be also, right?

Some of them will say “No, there is no velocity transform for one observable, because the velocity is given.” But MM says, how can the velocity in both systems be given? If the velocity in both systems is given, then we already know times and lengths. If the velocity in both systems is given, we have nothing to seek and no need of transforms. Besides, if the two systems have different x's and t's, they cannot have the same v's, can they?

Also, if the velocity is given for one observable, why is it not given for two? These Relativists are telling us we do not need a velocity transform for a muon, which must mean that the velocity we see and measure is already correct. The velocity is already in our own system, in other words. If so, then by what mechanism has the velocity of two observables gone underground? If what we see with velocity is already good, then how can it fail to be good just because there are two objects rather than one? Does the second object cause our eyes and machines to fail?

Finally, if we measure x and t directly in the muon problem as solved by the mainstream, we get 15,000m and 2.2 x 10-6s. That velocity is way over c. Is that our “given” velocity? No, they will say, our given velocity is .9996678c. But how can that be our given velocity, when that velocity comes from another experiment altogether? According to Einstein's equations, the given velocity is v, not v', which means it is the velocity in our own coordinate system, the velocity we measure directly from our own position, as in a train being measured from the platform. But that means in this problem that v = 22.7c. Only v' = .9996678c, since that is the local velocity of the muon.

The Relativists will then say, “Velocity is already a relative measurement. Our given velocity is the object relative to the observer.” But that is clearly false, since objects can easily measure their own velocity. The Relativists are claiming we can measure the muon's velocity, but that the muon cannot measure its own velocity. When you are in a car, can you not measure your own velocity? Of course you can, and if you were at very high speeds, your measurement would not be the same as a stationary police car measuring you. That is what Relativity is all about, for goodness sakes. There should be a velocity transform for one observable. MM knows this for a fact, because in his corrections to SR (See An Algebraic Correction to Special Relativity and Refutation of Gamma) MM finds a velocity transform for one observable, and it happens to be the very same equation as a frequency transform in optics.

The reason mainstream Relativists do not know this is that Einstein did not know it. They know nothing that he did not tell them. Because his equations are slightly off, he cannot find a velocity transform for one observable. And if you used his equations to find one, it would be the wrong one. In other words, you cannot derive the frequency transform from his equations, but you can from mine.

The next problem is that these mainstream physicists who fudged this time dilation answer for the muon are hiding the length contraction from you. As MM pointed out above, if you have time dilation you should also have length contraction, according to Einstein himself. In fact, you cannot have one without the other. They are hiding this from you because in the muon problem we have a claimed time dilation at the same time that we have a rather obvious length extension. Yes, the x variable has grown by a large margin in this muon problem, since a muon that should have only gone 660m is now claimed to go 15,000m. That is not a length contraction, is it? If you want one of these physicists to go white, just mention this little problem to him or her.

Some of the most dishonest will try to tell you that the length has nothing to do with the distance traveled, but that is simply a lie. All you have to do is go to Einstein's equations, or any of the modern updates of them in any form. You will find that the x transform and the L transform, although slightly different, are not in inverse proportion. As L contracts, x contracts. These physicists would have to show you that as L contracts, x gets longer, and they cannot do that without rewriting history. None of them really want to do that, so they just hide and misdirect. They hope you won't notice that their time-dilated muons should be length-contracted. Meaning, yes, they would live longer if they obeyed current theory, but if they obeyed current theory they would go less than 600 meters, not more.

As MM said earlier in this paper, he believes in Relativity. What is more, he believes our little muons are “living longer”, by the current meaning of that phrase, but they are not time-dilated. Yes, in his corrections, time dilation and length contraction still go together, but they go together for objects that are moving away. Since the muon is moving nearer, it must be time-compressed and length-extended. This immediately explains the increase in x. The muon goes a longer distance, not a shorter, and MM can explain that sensibly whether my muon is in a gravitational field or not.

In the paper Relativity Demystified MM has shown that “longer life” goes with time compression, not time dilation. Time-dilated objects live shorter apparent lives. Time dilation means that we see or measure their clocks to be going slower. Locally, that means that their clocks are going faster. The dilated clock is what we see: it is the t variable in the equations. But the t' variable is the local variable: it is what the object sees itself. In time dilation, t > t'. So the object has the faster clock, locally. Which means it burns out faster than we think it will. We are looking at a clock that is wrong. If all the clocks in your house are ticking too slowly, then you fall behind real life, you do not outrun it.

That being said, MM has also shown that there is a limit on time change and length change. SR by itself cannot show increases of 38.8 times, like these physicists are trying to show with the muon. Time dilation is a phenomenon that comes out of SR, not GR. In the muon problem, current physicists are finding their numbers using SR, not GR. But SR simply cannot find that amount of change. The corrected first-degree transforms have a limit of 2. That is, you cannot do more than double a time or a length or a velocity using SR correctly. An object moving at you at c will only double its apparent length and apparent life. Any increase above that must be explained by GR or by a classical gravitational field or by an acceleration field. Current physicists, both in particle physics and in astronomy, are applying apparent mass increases and distances to SR, when they should be applied to acceleration fields.

As shown in this paper on the muon, physicists do not understand how to accelerate a velocity, and because their equations fail, they must explain the failure in some way. They do this with SR. But MM's solution in the muon paper can be applied to acceleration fields in accelerators as well. In accelerators, they are accelerating velocities, not particles. Like the atmospheric muon, those particles in accelerators always have their own velocities that are already near c. As with the muon, the field is increasing the energy of the particle directly a lot more than they think, so that very little of the mass increase has to be explained with SR. In the accelerator, it is not the velocity that is the problem, it is the mass; and that is because the acceleration field is not in vector opposition to the particle, as it is with the earth and the muon. In the accelerator, the field is stacked on the particle, and the energy cannot go to velocity: it has to go to “mass”. But it is not mainly the energy from SR that causes the mass increase, it is the energy lost to the faulty equations. As in the muon problem, particle physicists are using the wrong acceleration field equations, so they do not understand where their extra energy is coming from.

The muon explanation is an example of how Relativity is being misused all over the place to fill holes in bad equations. SR is a good theory, but it cannot create time increases of 38.8. The specific problem in the current equations is gamma, which is in the wrong form.

1/√[1 – (v2/c2)]

Even Einstein could not justify that square root, as MM has shown. In Einstein's 1905 paper, he simply reduces wrong. There is no square root in his original math. Other physicists have caught that mistake, including Born. In later conceptualizations like the light clock, the Pythagorean theorem is applied to light that passes us on a tangent: the math is applied to a vector we cannot see or measure, to light that never reaches us! Gamma is a ghost, and once MM corrected all the math, he found that gamma never once appeared. The closest thing he found was this, from the mass transforms:

ET = mric2[1 + (v’/2c)]
[1 – (v’2/c2)]

But, as you see, no square root. And the time and distance transforms aren't even as complex as this. The distance transform is just

x = x'/ [1 + (v/c)]

At the limit of c, x is just twice x'.

Relativity is true, but it cannot do all the things it is claimed to do. It is a powerful tool, but it just is not that powerful. And no matter how powerful it is, it should not be used to cover up bad acceleration equations.

This is the reason MM knew he had to correct the field equations in the muon problem. Time and length transforms are simply not capable of making 660m into 15,000m, much less 435,000m. It is the gravity field that is much more powerful than we think it is, and a particle already at c in that field can cover a lot of ground (or air) in a very short time.