Mass Effect Calculation m = f(x,y,z,t) Introduction This webpage proposes a calculation of the mass effect m = f(x,y,z,t). Expression of "m" We have seen that (fig. 1):
On the other hand, we have calculated the Newton Law starting with the following expression:
...where K is an unknown constant having the dimensional quantity of [L/M]. Porting (2) in (1) gives:
or:
Combining equation (20) of the Newton Law webpage and (4) gives the expression of the "mass effect". A parameter has been added, ρ, which is the density of external spacetime in the point of measurement, with a flat spacetime as reference. This parameter increases or decreases the mass effect.
with:
The coefficient of elasticity of spacetime ev can be calculated from spherical particles. However, this calculation is not easy because the only supposed spherical particles known are leptons. To have accurate data, we must know exactly what electrons, muons, and taus are, and their accurate diameter. Nuclei with a null quadripolar moment (magic nuclei) could also be used but we do not know exactly the arrangement of nucleons inside the nucleus, nor the arrangement of quarks inside each nucleon. The space between elements may vary from one nucleon to another, and in this case, may give wrong results. Since the mass effect is calculated from the closed volume, it is necessary to know the latter with accuracy before computing data.As a direct consequence of the proposed theory, it could be possible that a relationship exists between the sphericity of particles and the accuracy of measurements. For example, leptons could be spherical since their mass is known with an excellent accuracy. This is not the case of quarks. Case of a Sphere In the particular case of a sphere, we have V = 4/3 pR3 and S = 4pR2, and expression (5) becomes:
Nuclei Nuclei aren't spherical. Therefore, it is not possible to apply equation (6) to calculate their mass effect since we do not know exactly their shape. Empirical formulae based on A, Z and N, as the Bethe-Weizsäcker formula, give accurate results but it is important to note that the law in A1/3 does not mean that the mass is proportional to the volume. It means that the arrangement of nucleons inside the nucleus follows a A1/3 law. Results are equivalent but the signification is totally different.
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