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Now the question – why do we believe the weight of a kilometer of pipe equals mg without asking a question as to what precisely is housed within the pipe, and more importantly, what is “it” doing there?

It seems obvious that the weight of one kilometer of pipe equals mg.

Now the question – why do we believe the weight of a kilometer of pipe equals mg without asking a question as to what precisely is housed within the pipe, and more importantly, what is “it” doing there? There may in fact be stones in the pipe, or it may be water, or something else. In calculating weight by mass, we need to know what mass is hidden in the pipe and what it is doing there, don’t we?

If you think that we don’t need to know this, then permit me to inform you that a multitude of spherules are situated in the pipe we are considering, during which the pipe spans the entire earth and is continuous. There is a vacuum in the pipe and the spherules race through the pipe at orbital velocity, like minisatellites. (Fig.1)


Fig.1

They are “halved” – one half of the spherules is hurtling toward the other half; thus, the spherules are flying, not colliding, and thereby ensuring that the total pulse of the spherules inside the pipe equals zero. Therefore, the weight of a kilometer of pipe equals m, does not equal zero, and the total pulse of the filler equals zero due to the opposing motion of the spherules. The pipe itself containing the filler lies on the ground – there is mass, but the weight does not equal mg, since the satellite spherules in flight within the pipe do not put pressure on the ground. The pipe does not put pressure on the ground, but why? Does the weight of this pipe equal zero, or is there a weight of m, but the formula mg does not work?

If, however, the pipe is not continuous, but rather consists, for example, of a one-kilometer segment that is closed at the ends with flanges, then given the absolute elasticity of the spherules and their motion following reflection along the same trajectories, the pipe will put pressure on the Earth (Fig.2).


Fig.2

However, it will not exert pressure due to gravitational force (the minisatellites spherules do not put pressure on the Earth), but rather as a result of the forces applied to the flanges due to a change in the pulses of the spherules striking the flanges. If it is a short pipe with flanges arranged in a strictly radial manner (relative to the Earth) that the spherules are striking, then a resultant force directed at the Earth’s surface will in any event appear. This force will appear due to the fact that the angle between the vectors of the forces applied to the flanges invariably prove to be less that 180 degrees. The pipe will put pressure on the ground with a force that equals mg, but can the force with which the pipe segment acts on the ground be weight? This force is in fact created not by the Earth’s gravitational field, but by the spherules striking the flanges.

Or another example.

Let us consider an elevator that is standing on the ground and that is filled with a multitude of massive spherules (Fig.3).


Fig.3

The weight of the elevator itself is negligible as compared to the weight of the spherules; i.e., the weight of the elevator containing the spherules is practically the same as the total weight of the spherules and equals m. The weight of the elevator containing the spherules equals mg and is made up of the weights of the individual spherules.

By the way, what does “made up” mean?

I forgot to say that the spherules are elastic and are hurtling vertically bottom to top and top to bottom between the ceiling and floor the whole time, rebounding first off the ceiling then off the floor. Therefore, the bulk of the spherules are in a state of flight at a given moment in time and are not exerting a force impact on the elevator. I.e., the vast majority of the spherules are not taking part in the pressure on the ground at a given moment in time. The weight of the elevator nonetheless equals the total weight of the spherules at a given moment in time, the bulk of which are not putting pressure on anything at a given moment in time. The weight of the elevator containing the spherules is not created by the gravitational force acting on all the spherules at a given moment in time, but rather by the difference in the pulses of the spherules striking the elevator’s ceiling and floor at a given moment in time.

For the time being, imagine that an elevator containing spherules that are hurtling between the floor and ceiling is in a state of weightlessness in space. If you now pull on the cable with an acceleration of a, a force of F, equaling ma, will then be applied to the elevator (Fig.4).


Fig.4

In this instance, a force of F will at no point be applied to the basic mass of the spherules, and the formula F = ma works. It is not difficult to comprehend that the force applied to the tractive cable at a given moment in time is created by the difference in the pulses of the spherules striking the elevator’s ceiling and floor at a given moment in time as it accelerates. Indeed, when the elevator accelerates, a spherule returning to the floor following its reflection from the ceiling collides with the floor at a higher velocity than it collided with the ceiling receding from it. There is inert mass, but it is created at a given moment in time not by the inert mass of the vast majority of the spherules, but by the pulses of a negligible portion of the spherules.

The elevator might be filled with a photon gas containing photons that dart between the specular floor and ceiling, and once again, the weight of such photon filling of the elevator standing on the Earth in its gravitational field will be determined by the difference in the pulses of the photons reflected from the floor and from the ceiling rather than by the weight of the photons, which does not naturally exist. The photon velocity near the ceiling and near the floor will be identical, but the wavelengths and pulses of “one and the same” photon on the floor and near the ceiling will be a little bit different. The invariant mass, which has superseded the relativistic mass in major physics, of each photon equals zero, while the invariant mass of the photon gas does not equal zero.

The photon gas also puts up resistance to the elevator’s acceleration in deep space. In this instance, the floor and ceiling photons create the force of inertia at a given moment in time, while the remaining ones do not take part in the creation of this force. But what do the Higgs bosons do in the elevator? Do they work or do they “rest”?

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