October 10, 2014
Hiroshi Michiwaki
Study introducing parameters that connect physical equations and reducible set
theoretical division by zeroI.

Discussion of photon energy and reducible set theoretical division by zero

By using the Planck constant h, wave-particle photon energy E is expressed as a function ofwavelength λ,Here, if we assume an arbitrary parameter κ asand substitute it in Equation (1), photon energy E can be expressed as

Of course, when λ ≠ 0, the dimensionless coefficient κ/κ is 1; therefore, E(λ) in Equation (3)becomes equivalent to Equation (1). Here, if we assume λ = 0 in Equation (3),is obtained. To the result of Equation (4), if we apply the reducible set theory (March 18,
2014: Reducible set theory), we obtain a quotient of 0 and a remainder of κhc, and energy E
is not emitted. It[A1] also does not simply become 0, showing that for the whole system, theenergy of κhc is conserved.
Here, we should note that κhc is in the original dimension of E. This is a natural interpretation
because hc is divided by λ, and the dimension is the result of the division regardless of thevalue of λ. It would be natural to think that this would be applied the same way in the case of
λ = 0.
II.

Discussions of the Doppler effect and reducible set theoretical division by zero

Using the same method as the above section I, we will consider the Doppler effect. If weassume the frequency of the original sound waves as f, the sound velocity as v, the velocity at
which the sound source is moving as vs, and the velocity at which the observer is moving asvo, the frequency of the sound waves f ́for the observer becomes,

If we assume the arbitrary parameter η asand introduce it into Equation (5), then the observed frequency f ' becomes

Of course, when v − vs ≠ 0, the dimensionless coefficient η/η is 1; thus, the f ' of Equation (7)becomes equivalent to that of Equation (5). Here, in Equation (7), if the velocity at which thesound source is moving vs becomes the same as the sound velocity v; in other words, v − vs =0,

is obtained. If we interpret the result of Equation (8) by applying reducible set theory (March18, 2014, Reducible set theory), it shows that the quotient equals 0, and the remainder isequal to η(v − vo) f. Of course, here, η(v − vo) f is in the dimension of f. In other words, thefrequency perceived by the observer f ' appears to be 0, but it shows that η(v − vo)𝑓 exists as
the conserved amount.
III.

Discussion of special relativity theory and reducible set theoretical division byzero

For the mass m in special relativity theory, if we define m0 as the rest mass, and its movementvelocity as v, it can be expressed as

Where c is the speed of light.
Here, the arbitrary parameter μ as

is introduced into Equation (9). Then, the theoretical mass m in special relativity is expressedasOf course, when v ≠c, the dimensionless coefficient μ/μ is 1; thus, the m in Equation (11)becomes equivalent to that of Equation (9). Here, in Equation (11), if movement velocity v
becomes the same as the speed of light c; in other words, if v = c,

is obtained. If we interpret the result of Equation (12) by applying the reducible set theory(March 18, 2014, subtractable set theory), it shows that the quotient equals 0 and theremainder equals μm0. Of course, μm0 here is in the dimension of m. In other words, mass mappears to be 0 when the movement velocity v reaches the speed of light c, but μm0 exists inthe overall system.
Here, we assume mass m0 as the rest mass of a neutrino, and the flight velocity of a neutrino
as v. If we combine this with the observational results from the Super-Kamiokande, wealmost obtain v = c; thus, the mass m of a neutrino in flight becomes infinitely large if
calculated with Equation (9). This is inconsistent with empirical observation. This resultshows that though a neutrino is extremely small, it has a finite mass that is larger than 0. Thiscan be explained rationally by the reducible set theory and Equation (12).
For example, this means that if an elementary particle, etc., moves at the speed of light, mass
is not discharged. It also means that there is no paradox as described above that would breakthe law of the conservation of energy. In other words, this theory shows that the problem ofdivision by zero where the laws of physics are broken by an essential singularity is solved. It
poses other problems, however. For example, what is light with wavelength zero?
Sound emitted from a sound source that reaches the speed of sound creates a shock wave inthat instant, and subsequently, the wavelength of sound emitted from a sound source thatmaintains the speed of sound continues to be zero. At this time, what does the conserved
remainder term refer to? From such questions, it proposes the problem of a possible
relationship between unsolved dark matter and mass μm0.