How Do Wave Properties Relate to Mass and Energy?

[shintashi]

“wave radius” used for spherical wave patterns, same as wave length.
“Wave density” same as wave frequency.

Gamma wave 1e-11 wavelength multiplied by 3e19 frequency rquals total mass
Total “energy” of gamma wave = 2e-14

Radio wave length 1e-1, multiplied by 3e9 frequency equals total mass
Total “energy” of this wave length = 2e-24

How does this total energy relate to the technical mass ? all of these objects are said to have a technical mass of 3e8, based upon the false belief of light speed. But why would they have different energies ? What is the energy of 1 electron ? certainly the energy of 1 electron compared to the energy of these various energies of EM spectrum will yield the technical frequency of electrons, assuming that 3e8 is the total mass and there is some kind of relationship between the frequency and the total energy.
It was concluded that if you had the same wave length for a visible light wave (e14) frequency) compared to a gamma wave of the same length (3e19) you would have a potential energy 3e5 times greater, which means that if converted into kinetic energy, the wave decompression of a gamma wave would yield 3e5 times more kinetic velocity.
Potential velocity/wave decompression of wave length seems to be exactly equal to the frequency, in some odd respect, while the longer the wave, the more massive the wave, and the more “energy” can be produced by this expansion.

Approximately 4 times more “energy” than a gamma wave, which makes it either 1 octave or 2 (not sure which) up from gamma waves.
8.18622e-14 joules = total energy of electron

gamma wave = e10 times more frequency than radio wave, = e10 times times more “energy”
electron = 4 times more frequency, = 4 times more “energy”
frequency of electron = 1.2e20

wave Wavelength (m) Frequency (Hz) Energy (J)
Radio > 1 x 10-1 < 3 x 109 < 2 x 10-24
Microwave 1 x 10-3 - 1 x 10-1 3 x 109 - 3 x 1011 2 x 10-24- 2 x 10-22
Infrared 7 x 10-7 - 1 x 10-3 3 x 1011 - 4 x 1014 2 x 10-22 - 3 x 10-19
Optical 4 x 10-7 - 7 x 10-7 4 x 1014 - 7.5 x 1014 3 x 10-19 - 5 x 10-19
UV 1 x 10-8 - 4 x 10-7 7.5 x 1014 - 3 x 1016 5 x 10-19 - 2 x 10-17
X-ray 1 x 10-11 - 1 x 10-8 3 x 1016 - 3 x 1019 2 x 10-17 - 2 x 10-14
Gamma-ray < 1 x 10-11 > 3 x 1019 > 2 x 10-14
Electron 2.5 x10-12 1.2 x 1020 8 x 10-14







Mass of electron 9.1e-31 kg
0.511 MeV
5.11e5

0.510 998 918(44) MeV
one eV = 1.602 x 10-19 joules
106 eV = 1 MeV

wave E/wavelength E/Frequency Energy (J)
Radio 2e-23 6.6e-34 < 2 x 10-24
Microwave 1 x 10-3 - 1 x 10-1 6.6e-34 2 x 10-24- 2 x 10-22
Infrared 7 x 10-7 - 1 x 10-3 6.6e-34 2 x 10-22 - 3 x 10-19
Optical 4 x 10-7 - 7 x 10-7 6.6e-34 3 x 10-19 - 5 x 10-19
UV 1 x 10-8 - 4 x 10-7 6.6e-34 5 x 10-19 - 2 x 10-17
X-ray 1 x 10-11 - 1 x 10-8 6.6e-34 2 x 10-17 - 2 x 10-14
Gamma-ray < 1 x 10-11 6.6e-34 > 2 x 10-14
Electron 2.5 x10-12 6.6e-34 8 x 10-14


(1/wavelength) * Frequency = total mass per meter, possibly a new unit.
this value compared to energy in joules should yield some kind of number

wave Wavelength (m) Frequency (Hz) Total Mass (Æ)
Radio > 1 x 10-1 < 3 x 109 3e10
Microwave 1 x 10-3 - 1 x 10-1 3 x 109 - 3 x 1011 3e10-3e14
Infrared 7 x 10-7 - 1 x 10-3 3 x 1011 - 4 x 1014 3e14-2.8e22
Optical 4 x 10-7 - 7 x 10-7 4 x 1014 - 7.5 x 1014 2.8e22-3e22
UV 1 x 10-8 - 4 x 10-7 7.5 x 1014 - 3 x 1016 3e22-3e24
X-ray 1 x 10-11 - 1 x 10-8 3 x 1016 - 3 x 1019 3e24-3e30
Gamma-ray < 1 x 10-11 > 3 x 1019 3e30
Electron 2.5 x10-12 1.2 x 1020 4.8e31

Total mass = Mass per meter…

There seems to be a factor of roughly “2” between the “total mass” of an electron using my system, and the technical mass of an electron as listed in books.
Electron mass
1/total mass = technical mass ?

I'm wondering what exactly this all means, since I did not expect to get 2.08e-32 vs. 9.6e-32 mass ratios, obviously within an order of magnitude

Note a lot of this is built upon my own notes, so it might not make sense, unless it does ^_^

 

[shintashi]

Quote from Petawatt laser Experiment(http://www.llnl.gov/str/MPerry.html)…
“Electron energies greater than about 2 megaelectronvolts produce gamma rays that can be transformed into pairs of electrons and positrons (pair production).”

Note, for people who cannot THINK… remember this quote “Approximately 4 times more “energy” than a gamma wave, which makes it either 1 octave or 2 (not sure which) up from gamma waves.”

0.511 x 4 = 2.022
or another way of putting it, 2 MeV/4 = 0.5 Mev, and poof, you have an electron.

Mass of electron 9.1e-31 kg
0.511 MeV
5.11e5

its funny how an obvious crackpot like me can come up with the same numbers as the high and mighty at Laurence Livermore.. except they probably don't know why.

 

[R0man]

The concept of wave/mass ratios is an interesting one to explore. It seems that you are trying to establish a relationship between the wavelength, frequency, and energy of different types of waves, and how they relate to the mass of an object. Your calculations and comparisons are certainly intriguing, but it's important to keep in mind that these are just theoretical ideas and may not necessarily have a concrete scientific basis.

That being said, the concept of wave radius being the same as wavelength and wave density being the same as frequency is an interesting way to look at it. It's also interesting to consider the potential energy and kinetic energy that can be produced by different types of waves, as well as the potential relationship between frequency and total energy.

However, it's important to note that the concept of technical mass and the belief in the speed of light are not necessarily false. These are principles that have been established through scientific research and experimentation, and while there may be room for further exploration and understanding, it's not accurate to dismiss them as false beliefs.

Overall, your ideas and calculations are thought-provoking and may lead to further exploration and understanding of the relationship between waves, mass, and energy. However, it's important to approach these concepts with a critical and scientific mindset to ensure accuracy and validity.