© Lloyd1. (THE MYSTERIOUS MUON) http://milesmathis.com/muon.html
2. The muon lives only 2.2 x 10^-6 seconds, and travels only about 660m under normal circumstances.
3. 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.
4. 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, not even in a gravitational field.
5. We can solve this apparent conundrum if we use Einstein's equivalence principle to reverse the vector of gravity, as I have done so many times before.
6. The reversal makes the surface of the earth move out during each dt, which makes the earth move toward the muon during each dt.
7. The muon's trip is shorter with each passing moment, which makes the time for the trip smaller.
8. Richard Feynman does [this] in his Lectures on Gravitation, section 7.2, as I have shown.
9. If we move the earth toward the muon, we have not increased the total velocity, taking it above c.
10. No, we have shortened the distance that must be traveled, therefore we have actually decreased the velocity necessary to travel it.
11. If we keep the muon at c, then t will have decreased.
12. It is not the total velocity that has increased, it is the total distance that has decreased.
13. The mistake is in assuming that you add velocities in approach in this situation, when the opposite is true.
14. If I am driving at you at 60 miles per hour and you are driving at me at 50 miles per hour, then our combined velocity is 110 miles per hour, and if we are 110 miles apart to begin with, we will collide in one hour."
15. All true, but in that example we are going 110 miles per hour relative to eachother, not to the background.
16. To find how fast we are going relative to the background, we subtract.
17. ?>>> Together, we are going 10 miles an hour relative to the road, since you went 60 miles and I went -50 miles.
18. 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, as I said in the opening paragraphs, we are not measuring its velocity from the earth.
19. 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.
20. If we turn the earth's gravity vector around, this is easy to see.
21. The earth must be accelerating relative to something, since acceleration is a motion and all motion is relative.
22. The earth cannot be accelerating relative to itself.
23. 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.
24. In my expansion, the surface is moving relative to the center, so we have the three points of reference: muon, surface, center.
25. 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.
26. I reversed the vector to make the concepts clearer, but the concepts and vectors are the same whether we reverse g or not.
27. If the field of the earth has an acceleration, then this acceleration must be relative to some non-accelerated body or field.
28. Einstein said that all motion was relative, and that applies to accelerations just as much as to velocities.
29. 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.
30. Since the acceleration acts as an attractive force, it must combine with an incoming velocity in a vector situation that subtracts, not that adds.
31. 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.
32. 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 eachother.
33. We have to calculate the muons and earth against a background.
34. Which gives one of the velocities a negative sign.
35. 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.
36. If the initial velocity is zero, then we can use the equation s = ½at2, which does give us the very small number above.
37. But if the earth does not have an initial velocity of zero, that equation won't work.
38. 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.
39. If we use the above equation, s = ½at2, and solve for t using 9.8 and 6,378, we get t = 1141.
40. In a field of acceleration of 9.8, the surface of the earth is 1141s away from the center.
41. From that we can develop the velocity: v = 6,378,000/1141 = 5,590m/s.
42. 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.
43. Let us dispense with the vector reversal and keep the earth steady.
44. This will have the mathematical and mechanical result of giving all our motions to the muon, but it will not contradict all I have shown above.
45. Doing the math this way is a convenience, and implies nothing about the physics involved.
46. 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.
47. In that case, we lose the v0 in the equation above, and it reduces down to c = at.
48. Solving for t gives us 3.0591067142857 x 10^7 s.
49. Since the acceleration is constant, the average velocity is c/2, and the distance traveled is therefore 4.58548560580009 x 10^15m.
50. So it takes this hypothetical particle almost a year to go from rest to c.
51. Now, a nice question is, how far would it travel in the next second?
52. Easy, since we just add a second to the time and solve: x = 2.989 x 10^8m.
53. How far would it travel in the next 2.2 x 10^-6 seconds?: x = 617m.
54. 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.
55. 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.
56. ?>>> In other words, the earth accelerates the velocity, not the body.
57. >>> [Does he mean it accelerates a body with a velocity, rather than a body with no velocity?]
58. In a gravitational field, the equation v = v0 + at doesn't work.
59. In that equation, the acceleration is independent of the velocity.
60. But we want to accelerate the velocity.
61. 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.
62. Remember, it would travel 660 meters in no field at all.
63. Then I found that a gravitational field would accelerate it 617 meters.
64. 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.
65. Remember, we accelerated this particle from zero, so the particle had no velocity uncaused by the field.
66. 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.
67. ?>>> We have to multiply them, giving us 660 x 617 = 407,220m.
68. But my math is not complete.
69. 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.
70. It varies over the trip, going from about 8.62 at the start to 9.8 at the end.
71. I will not do the full math, I will simply estimate a new number.
72. 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.
73. Take note that the two numbers match.
74. 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.
75. Coincidence? No, of course not.
76. We found that number precisely because the field is accelerating the initial velocity, not the muon.
77. The initial velocity is integrated into each and every differential of acceleration, as it must be.
78. Which means the initial distance traveled during each differential is also integrated into the field acceleration.
79. This gives us the equations vf = v0 + 2v0^2t [and] x = v0^2t2
80. And this allows us to estimate the distance traveled in the field by simply squaring the distance traveled outside the field.
81. x0 = v0t; x = x0^2
82. >>> [Does he mean x = v0^2?]
83. So here are the full equations, so that you can see exactly why 660 comes up twice.
84. c = at0
85. d0 = ct0/2 = c2/2a
86. d = vf(t + t0)/2 = a(t + t0)/2
87. d = a[(c/a) + t]^2/2 = [(c^2/a) + 2ct + at^2]/2
88. Δd = d - d0 = (c^2/2a) + (2ct/2) + (at^2/2) - (c^2/2a)
89. Δd = ct + (at^2/2)
90. Because t is very small in this problem and c is very large, the second term is negligible.
91. This makes the distance traveled during each interval due to the acceleration almost equal to the distance traveled due to c.
92. This is why there is no acceleration variable in my equations above: it can be ignored when the time period is so small.
93. My number, which comes out to be about 270 miles, is much better, since we are then in the upper levels of the ionosphere.
94. In the ionosphere, we would expect muon creation.
95. 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.
96. 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.
97. It can do this because the gravity field acts physically or mechanically as a vector pointing out from the center.
98. 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.
99. [Scientists] only forbid assigning numbers to the vectors, because that would break a rule of Relativity.
100. I hope you see how absurd that is: allowing the assignment of vectors, but disallowing them sizes.
101. I have shown that allowing them sizes does not contradict Relativity, therefore the whole question has been a tempest in a teapot.
102. 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.
103. The distance 270 miles isn't the distance traveled by either object or both objects, in either system.
104. It is the distance that will seem to be traveled by one object, in the math, if we prevent the other object from moving.
105. The number 270 miles is an integral, but the real motions are causing differentials.