# Mass and momentum

Fundamental chapter: Mass and inertia

The law of the equivalence of mass and rest energy relates to the electromagnetic field energy ("Mass and inertia", 11): $E=\frac{mc^2}{\sqrt{1-v^2/c^2}}\tag{1}$ According to the model, the particle at a small distance from its center is a kind of the magnetic wave of a length, which is determined by its energy and the Planck constant so: $\lambda_0=\frac{hc}{E}=\frac{h}{mc}\sqrt{1-\frac{v^2}{c^2}}\tag{2}$ The relativistic particles are significantly deformed (see "Mass and inertia").

The condensed associations of several particles (for example, the nuclei) have a common wavelength $$\lambda_0$$, which is determined by the total association energy.

A sphere of radius $$\lambda_0$$ is the actual particle volume, which contains most of its mass.

Particle sizes in comparison with other quantities
Size, m Quantity Comment
1,3·10-15 Proton radius According to the nuclear physics, the proton radius is 0,8…0,9·10-15 m.
2,4·10-12 Electron radius Electron is a large lightweight rarefied cloud.
10-10…10-9 Lattice constant Resolution of the best electron microscope.

The equation of the particle rest energy ("Mass and inertia", 5) is: $E=\frac{A}{4\pi\varepsilon_0}\frac{e^2}{R}=\frac{hc}{\lambda_0}\tag{3}$ The neutron consists of a proton and an electron, which are overlapping each other so that their electrostatic fields are neutralized. The maximum compensation is achieved when the $$R$$ value for the electron is equal to the proton radius $$\lambda_0$$. The proton energy, which is concentrated within its radius $$\lambda_0$$, cannot be compensated by the electron in any case. Thus, $$R$$ and $$\lambda_0$$ are related by an equation with the wavelength ratio of electron and proton (or their mass $$m_e$$ and $$m_p$$ ratio): $R=\frac{m_e}{m_p}\lambda_0\tag{4}$ The equations (3) and (4) give the dimensionless constant $$A\approx 0,47$$.

The wavelength $$\lambda_0$$ is called the Compton wavelength. The particle motion decreases the effective wavelength to the de Broglie wavelength with respect to a motionless point on the particle path, similarly to the Doppler effect: $\lambda=\lambda_0\frac{c}{v}=\frac{h}{mv}\sqrt{1-\frac{v^2}{c^2}}\tag{5}$

A particle, which moves among other particles, excites there an electromagnetic wave with the corresponding de Broglie wavelength. This reduces the particle momentum, so the wave is directed to the particle motion direction. The radiation power depends on the field density, so it increases when the particles approach. Because of this, there are such things as:

• Exchange of momentums in the particles collision (see below).
• Quantum effects in the atoms.
• Diffraction of the particles on the crystal lattice.

The momentum of any system, as a sum of the momentums of its parts, is defined by a force, which acts on it: $\mathbf{\overrightarrow{F}}=\frac{\mathrm{d}\mathbf{\overrightarrow{p}}}{\mathrm{d}t}\tag{6}$ Upon collision of the particles and also the bodies, which are composed of them, the interaction forces are balanced and the total momentum remains constant. The momentum is transferred in a form of the aetheric beam, according to the laws of conservation of momentum and energy.

Substitution of the inertia force ("Mass and inertia", 13) into (6) gives the particle momentum value: $\mathbf{\overrightarrow{p}}=\frac{m\mathbf{\overrightarrow{v}}}{\sqrt{1-v^2/c^2}}\approx m\mathbf{\overrightarrow{v}}\left(1+\frac{v^2}{2c^2}+\frac{3v^4}{8c^4}+\dotsc\right)\tag{7}$ Comparison of (5) and (7) provide the equation for the particle momentum: $p=\frac{h}{\lambda}\tag{8}$