© Lloyd

THE EQUATION V = V0 + AT IS FALSE as a field equation
http://milesmathis.com/voat.html

1. [] this equation doesn't work when v0 is equal to c.
2. It doesn't work when we have a particle with a very high velocity being accelerated by a gravity field.
3. In that case the equation is vf = v0 + 2v0^2t.
4. [] the equation [v = v0 + at] only works when the acceleration is not a field, [but] an internal acceleration, as with a car.
5. In textbooks, the equation [v = v0 + at] is derived like this: a = v/t; v = at. Then we add the initial velocity [v0].
6. [] The problem must be with the way the initial velocity and the acceleration stack up in a field.
7. Apparently we can't just add the final velocity from the acceleration to the original velocity.
8. Again, why? Put simply, the reason is because if we have an initial velocity meeting an acceleration field, we have three velocities.
9. Let us start with just the acceleration field, and no initial velocity.
10. Let us say you have a gravity field, or any other acceleration field that creates a normal or squared acceleration. >>>[What field, besides gravity, does that?]
11. If you place an object in the field, the object will be given two simultaneous velocities.
12. That is what an acceleration is, after all.
13. If the object has one velocity, it has no acceleration. >>>[But if an object has acceleration, it has a whole range of velocities between the initial and final ones.]
14. If it has an acceleration, it must have more than one velocity over each interval or differential.
15. A squared acceleration, or an acceleration to the power of 2, is composed of two velocities.
16. Therefore an object placed in any acceleration field will be given two velocities. >>>[It has a range of velocities.]
17. OK, but now we take an object that already has a velocity of its own, and we fly it into that acceleration field.
18. It must then have three velocities over each interval.
19. Or, to say it another way, it now has a cubed acceleration.
20. It has three t's in the denominator.
21. For this not to be the case, we would have to break Newton's first law [of motion?].
22. [] 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. >>>[How can a velocity be separate from an object?]
23. [] the problem with the equation in the title [is] It does not include a cubed acceleration.
24. The "a" variable stands for a squared acceleration.
25. The equation includes three velocities, but the first velocity has not been integrated into the acceleration.
26. It is only added, as a separate term.
27. But that is not the way the real field would work.
28. The real field would accelerate the initial velocity, not just the object.
29. As it is, the initial velocity is not summed in every differential, as it should be; it is only summed outside the acceleration.
30. That cannot work, either as a matter of calculus or as a matter of physics.
31. The equation is wrong, in any and all cases where a field is involved.
32. [] The other problem many people have is that my new equations show huge final velocities.
33. And yet we know that falling objects and objects like meteors entering the atmosphere do not reach velocities like that.
34. How can I answer that? I answer it with terminal velocity.
35. Most objects will have an appreciable size and mass, so that they are affected strongly by the atmosphere.
36. They can't reach those velocities simply due to drag.
37. However, I do believe that objects entering our gravity field reach terminal velocity much quicker than previously supposed, and they do so because of my new equations.
38. In addition, when we consider very small objects like muons, we have objects that are not affected by atmospheric drag.
39. Therefore they can cover these very long distances predicted by my field equations.
40. Now, I have developed a simple equation for cubed acceleration in another paper2: v = a[d2(t3)]/2 = 3at
41. But the "a" variable there is the cubed acceleration.
42. [] s = v0t [] is the distance the object would travel with only the initial velocity.
43. Δs = v0t + at2/2 [] is the additional distance it would travel during each interval, due to the acceleration.
44. 2sΔs = 2v0t(v0t + at2/2) = 2v02t2 + av0t3 [] is integrating the three velocities.
45. vf = v0 + 2sΔs/t = v0 + 2v02t + av0t2; vf = v0 (1 + 2v0t + at2)
46. If the time is very small and the initial velocity very large, as in our muon problem, we can ignore the acceleration and estimate the final distance with this equation: s = v02t2
47. Which gave us a distance of about 435,000 m in the muon problem.
48. A distance times a distance in line cannot be a distance squared anymore than a distance plus a distance can be a distance squared.
49. [] I have used algebra here to solve, but notice that if we use integration, we get the same problem: ∫(2x + y)dx = x(x + y).
50. We integrate a sum but get meters squared.
51. This problem ties into the historical problem of velocity squared versus acceleration.
52. A square velocity IS an acceleration, by definition, so if we are getting different dimensions it is by ignoring some mechanics.
53. We see that in this problem very clearly: if we multiply a velocity times a velocity in a field, we should get an acceleration.
54. [] Conclusion: Historically, the equations have failed to properly integrate the initial velocity, but I have corrected that.
55. [] To 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.
56. Two accelerations must be integrated in the way we have integrated a velocity and an acceleration here.