return to homepage THE MYSTERIOUS MUON
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 [I have done that now, 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 v in the equation above, and it reduces down to _{0}c = at. Solving for t gives us 3.0591067142857 x 10^{7} s. Since the acceleration is constant, the average velocity is c/2, and the distance traveled is therefore 4.58548560580009 x 10^{15}m.
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 10^{8}m. How far would it travel in the next 2.2 x 10^{-6} seconds?: x = 617m.
Hopefully, you see what I have done. 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 = v 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 as I just did it.
_{0} + atMy critic will say, “Very ingenious, since you doubled the distance traveled by the muon. It would have traveled 660m and you have added another 617m. This would double the “lifetime” of the muon as well, if we gave it to the time instead of the distance. Unfortunately for you, we don't require a doubling, we require a increase of 50x. Your muons are still 13,723m short of the surface of the earth.” Sad that my critic still can't comprehend my numbers, this far into the paper. I didn't just find that the muon traveled another 617 meters on top of the 660; no, I found that the muon was traveling 617 meters while it was traveling 660 meters. Remember, it would travel 660 meters in no field at all. Then I found that a gravitational field would accelerate it 617 meters. I found, 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. Remember, we accelerated this particle from zero, so the particle had no velocity uncaused by the field. So 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 can't just add the two numbers. We have to multiply them, giving us 660 x 617 = 407,220m.
If my math were complete, the muon could be over 27 times higher than it is currently proposed to be and still reach the earth. But my math is not complete. I let the acceleration of the earth be 9.8 over the entire trip of the muon, but 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. I will not do the full math, I will simply estimate a 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. Coincidence? No, of course not. 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 naive equation v = v. 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 _{0} + atv _{f} = v_{0} + 2v_{0}^{2}tx = v _{0}^{2}t^{2}And this allows us to estimate the distance traveled in the field by simply squaring the distance traveled outside the field. x _{0} = v_{0}tx = x _{0}^{2}
Those equation are rough, and some have not been satisfied by this verbal math that I like to do. So here are the full equations, so that you can see exactly why 660 comes up twice. c = at _{0}d _{0} = ct_{0}/2 = c^{2}/2ad = v _{f}(t + t_{0})/2 = a(t + t_{0})/2d = a[(c/a) + t] ^{2}/2 = [(c^{2}/a) + 2ct + at^{2}]/2Δd = d - d _{0} = (c^{2}/2a) + (2ct/2) + (at^{2}/2) - (c^{2}/2a)Δd = ct + (at ^{2}/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 my 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. Some will say, "Why didn't you just give us those new equations to begin with, instead of leading us through all that talk of expanding Earths and so on?" I really wish I could have, but these new equations beg all those questions, unfortunately. By themselves, these new equations show a velocity over c, which is currently forbidden. I am squaring c here, notice. So I had to explain precisely why that mathematical manipulation was allowed here. I had to go to some length to shows its logic and legality. This finding will be of great use to physicists, supposing some few of them dig out long enough to recognize it. 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. My 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.
Some will say, “That all works out pretty well, but didn't you say that the muon was time compressed? In another paper, you say that time compression is indeed equivalent to life extension, and this explains life extension in particle accelerators, where particles are approaching detectors. Shouldn't you have to calculate time compression here?” No, I shouldn't have to calculate time compression here, since I have shown that neither time compression nor time dilation is necessary to explain the detection of the muon at sea level. Given the muon's known lifespan, I have 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, since 1) we don't require it to explain anything, and 2) we still aren't measuring the time or distance the muon has gone. In other words, I haven't devised an experiment above in which we are measuring the distance directly. We are not going up to 270 miles, creating muons, then measuring their arrival at sea level. I am just using 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.
Finally, in closing, I will repeat what I have said in other papers: I agree with Relativity, for the most part. 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. I know some will still 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, I simply offer 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. We only forbid assigning numbers to the vectors, because that would break a rule of Relativity. I hope you see how absurd that is: allowing the assignment of vectors, but disallowing them sizes. I have shown that allowing them sizes does not contradict Relativity, therefore the whole question has been a tempest in a teapot. I would also point out that those who scoff at my 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 isn't 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 can't 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.**To say it one final time, in my 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. My critic will say, "Are you claiming the earth has velocity of 659c here? That is the only other motion we have, right?" No, I am not claiming that the earth or the earth's field has a velocity of 659c. The earth's field has an acceleration, and you can't 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 isn't the real distance. The distance 270 miles isn't 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. [To see an analysis of the SR transforms in the muon problem, you may go to my second paper on the muon.] Conclusion: 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 = v 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. _{0} + gtTo see how this affects the gravity field in other ways, see my new paper on reduced mass, where I show that accelerations must be integrated as well. Two accelerations must be integrated in the way we have integrated a velocity and an acceleration here. *This is the speed the current model believes the atmospheric muon is going, according to the value of gamma of 38.8.**You can read my second paper on the muon to see how current Relativists fudge through their own equations. If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. This will allow me to continue writing these "unpublishable" things. Don't be confused by paying Melisa Smith--that is just one of my many |