EQUATION V = V0
First posted December 30, 2009
In a recent paper on the muon1, I showed that this equation doesn't work when vo is equal to c. It doesn't work when we have a particle with a very high velocity being accelerated by a gravity field. In that case the equation is vf = vo + 2vo2t. Why?
v = at
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 can't just add the final velocity from the acceleration to the original velocity. Again, why?
OK, but now 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 don't 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.
Now maybe you see the problem with the equation in the title. 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.
Now, some have had difficulty understanding this. They know the equation works in a simple case like a car. Why would it not work in a field? How is a car accelerating different than a field? Well, 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.
problem many people have is that my new equations show huge final
velocities. And yet we know that falling objects and objects like
meteors entering the atmosphere do not reach velocities like that.
How can I answer that? I answer it with terminal velocity. Most
objects will have an appreciable size and mass, so that they are
affected strongly by the atmosphere. They can't reach those
velocities simply due to drag. 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. In addition, when we consider very small objects like
muons, we have objects that are not affected by atmospheric drag.
They aren't slowed by the atmosphere at all. They are either
absorbed or deflected by it, or they are not. If we detect them 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 my field equations.
v = a[d2(t3)]/2 = 3at
But the “a”
variable there is the cubed acceleration. We need to be given that
acceleration in order to find velocities and distances. And that
equation does not integrate an initial velocity either. In the
muon problem, we were given only the initial velocity and the
squared acceleration. That is why we have to use these equations:
That is the
distance the object would travel with only the initial velocity.
That is the
additional distance it would travel during each interval, due to
That is integrating the three velocities.
vf = vo + 2sΔs/t = vo + 2vo2t + avot2
vf = vo (1 + 2vo t + at2)
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 = vo2t2
Which gave us a distance of about 435,000 m in the muon problem.
A reader has pointed out to me that the dimensions appear to be off in that final equation. What causes it is that I am multiplying a distance by a distance, which appears to give us a distance squared. Go back two steps and you will see what I mean. But a distance times a distance is not a distance squared, if they are in line. A distance times a distance is a distance squared only in the case that the two distances are orthogonal, so that we have a square or something. A distance times a distance in line cannot be a distance squared anymore than a distance plus a distance can be a distance squared. This is another mainstream misunderstanding.
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). We integrate a sum but get meters squared.
This problem ties into the historical problem of velocity squared versus acceleration. Physics has never been clear about the mechanical difference. One has meters squared in the numerator and one does not, but are they really different? No. A square velocity IS an acceleration, by definition, so if we are getting different dimensions it is by ignoring some mechanics. We see that in this problem very clearly: if we multiply a velocity times a velocity in a field, we should get an acceleration. Therefore the extra length in the numerator is just that: extra. [You can now read more about this in a separate paper.]
said I can't be right, because if we use c for vo,
we get a final velocity way above c. But I have shown that this
does not break any rules of Relativity, since vf
doesn't really apply to the
object (as with a muon). The final velocity in the equation is a
field result, not a real velocity of any one object. The field
acts like a second object in the math. Relativity does not forbid
a result over c, or a calculation over c, it forbids a measurement
over c. It tells us we will never discover a value over c in our
data, and I do not disagree with that. However, we already know
from accepted data that fields create results over c all the time.
We know this from blueshifts. Blueshifts are impossible without
motion relative to light. Some still think Relativity forbids
motion relative to light, but it doesn't. It only forbids
of motion relative to light. It
cannot forbid calculations relative to light, and we calculate
motion relative to light every time we calculate a blueshift (or a
I have claimed that physicists and mathematicians were hiding away in esoteric equations and arcane problems while they couldn't do basic math and physics, and I have shown yet another shocking proof of that. The highschool physics books are full of false equations, while the top physicists are getting Nobel Prizes for modeling the first split seconds of the Big Bang3. Ask yourself this: is it likely they are getting the right answer for that, when they have the wrong answer for this?
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