Quoting: Anonymous Coward 1406242
Doh. Take 2. Quoting: Anonymous Coward 74444
Carefully and completely spell out five different experiments that would specifically falsify and disqualify your theory.
And watch my surprise when you can't.
1)Prove the production of electricity from a photovoltaic cell is not the result of electrons in the form of light being intercepted by the atomic structure of the photovoltaic material and converted to normal low electrons. (By the way, the function of photovoltaic cells is most succinctly explained by my theory)
But the theory that best suits the facts is the current traditional one, and you have proposed no experiment to verify your idea while disqualifying the current theory.
[link to science.nasa.gov
What experiment would you propose that *specifically tests the validity of YOUR model*, and then disqualifies the traditional one? Spell out the experiment!
2)Prove that light frequency energy cannot be conducted by the same materials as electricity in the non EMR form. Radio antennae and wires being used to conduct EMR are in almsot every electronic device, and are conductors of electrons. Repeated experiments mentioned in early debates involving plants producing photosynthesis in total darkness using metal plates and conductors will get in your way though,as well Quoting: Anonymous Astrophysicist
This seems to be the closest you got to actual experimental verification of anything. If I understand you, you are saying that if your explanation is true, and the traditional theory false, that you should be able to put plants in total darkness, and if there are metal plates and conductors, the plants will flourish anyway, proving that light is getting to them through electro-magnetic radiation. You would have, as a control, another identical room without the metal plates and conductors, and the plants should die. Is that an accurate way to describe an experiment that could qualify your theory? And does it really disprove the traditional model? Be specific.
3)Prove that the energy of electromagnetic radiation is not directly related to it's velocity, speed amplitude and frequency, in other words prove that the particle I suggest is not moving the precise distance and velocity to produce the energy it does using the classic 1/2M X V2 formula. (I give the simple math proving this in my theorem) Quoting: Anonymous Astrophysicist
Not sure if I understand you, here. Electromagnetic waves travel through a vacuum at a constant velocity known as the speed of light, c. The relationship between the speed of light, wavelength, and frequency is:
F is the freq in cycles per second, c is light velocity in meters, and (lambda) is the wavelength in meters.
[link to www.1728.org
When light passes through other media, the velocity of light decreases. For a given frequency of light, the wavelength also must decrease. This decrease in velocity is quantitated by the refractive index, n, which is the ratio of c to the velocity of light in another medium, v:
n = c / v
Since the velocity of light is lower in other media than in a vacuum, n is always a number greater than one. which is the refractive index. Refractive index is an intrinsic physical property of a substance, and can be used to monitor purity or the concentration of a solute in a solution. The refractive index of a material is measured with a refractometer, and is usually made versus air. If the precision warrants, the measurements can be corrected for vacuum. Note that the difference between n(air) and n(vacuum) is only significant in the fourth decimal place.
[link to www.chemistry.adelaide.edu.au
So I am not sure how your theory and the traditional theory are at odds here. Again, propose an experiment to VERIFY your idea as the better model. Knocking down the current idea is irrelevant. Giving evidence that SUPPORTS yours is what is relevant, and what you stubbornly refuse to do.
4)Prove that and momentum can exist without mass Quoting: Anonymous Astrophysicist
The use of words can make a lot of confusion. Unfortunately, the word "mass" has been used in two different ways in physics. One was the way Einstein used it in E=mc2, where mass is really just the same thing as energy (E) but measured in different units. This is the same "m" that you multiply velocity by to find momentum, and thus is sometimes called the inertial mass. It's also the mass that provides the source of gravitational effects. Light has this "m" because it has energy. So it is indeed affected by gravity- not just in black holes but in all sorts of less extreme situations too. In fact, the first important *confirmation* of General Relativity came in 1919, when it was found that light from stars bends as it goes by the Sun.
The other way "mass" is often used, especially in recent years, is to mean "rest mass" or "invariant mass", which is sqrt(E2-p2*c2)/c2. This is invariant because it doesn't change when you describe an object at rest or from the point of view of someone who says it's moving. Obviously that's a good type of "mass" to give when you want to make a list of masses of particles. For a light beam traveling in a single direction, E=pc, so this "m" is zero. There is no point of view from which the light is standing still.
However, once you consider light traveling in a variety of directions, the E's from the different parts just add up to give the total E but the vector p's don't. In fact the total p can be zero if there are beams traveling opposite ways. So for many purposes the older definition of m (the inertial mass) is more convenient than the invariant particle mass, since it's the inertial mass that's just the sum of the inertial masses of the parts. For light moving equally in all directions, like the light bouncing around inside a star, total p is zero, so both definitions just give m=E/c2.
[link to van.physics.illinois.edu
Oh, and here's a first-year physics student proving momentum without mass:
"We constructed a torsion pendulum in a vacuum chamber and fired a powerful laser at a mirror attached to the pendulum. The pendulum rotated away from the laser light and we could measure the force of the light by the angle of deflection. By using a second laser to measure the
angle of deflection and a photosensor attached to a computer, we could see the effect of light on the pendulum in real-time, and were able to record precise data about its motion."
[link to www.gamma.nbi.dk
You seem to be sticking with F=G(M1*M2)/R^2 where, m1=mass of heavenly body r=distance f=gravitational force m2=0=mass of light. If you tried to use that force equation, you'd calculate some bending of light in a gravitational field, but it would only be half the observed amount. General Relativity, which describes the distortion of space-time by mass *and* momentum, is needed to get the right answer.
[link to van.physics.illinois.edu
[link to scienceblogs.com
5)Isolate and fully describe the "photon" Quoting: Anonymous Astrophysicist
In physics, the photon is an elementary particle, the quantum of the electromagnetic field and the basic unit of light and all other forms of electromagnetic radiation. It is also the force carrier for the electromagnetic force. This force's easily visible human-scale effects and applications, from sunlight to radiotelephones, are due to the fact that the photon has no mass and thus can produce interactions at long distances. Like all elementary particles, the photon is governed by quantum mechanics and so exhibits wave-particle duality: that is, it exhibits both wave and particle properties.
The photon is massless, has no electric charge,and does not decay spontaneously in empty space. A photon has two possible polarization states and is described by exactly three continuous parameters: the components of its wave vector, which determine its wavelength (lambda); and its direction of propagation.
[link to sawaal.ibibo.com
In 1986, Grangier, Roger, and Aspect performed an elegant
experiment to isolate single photons. Conceptually very simple, their approach was to examine correlations between photodetections at the transmission and reflection outputs of a 50/50 beamsplitter. To quote the experimenters, ‘‘a single photon can only be detected once!’’ Hence, if a single quantum of light is incident on the beamsplitter, it should be detected at the transmission output or at the reflection output, but not both: there should be no coincident detections between the two outputs.
[link to people.whitman.edu
Others have repeated and refined the experiments.
[link to people.whitman.edu
[link to www.tp.physique.usherbrooke.ca
[link to www.nobeliefs.com
[link to www.bourbaphy.fr
[link to www-d0.fnal.gov
[link to www.eng.yale.edu
[link to scienceblogs.com
You see? SOURCES.
But, again, knocking on the current theory does nothing to support yours. Do you have an experiment that would demonstrate your theory models better than experiments involving photons? Be very specific, please, and spell out an experiment that would prove massless photons exist that *you* would accept.
And hey, only 66 days left for yet another experiment to conclude, which will falsify many of your nigh-certain claims... even if you *won't* wager on it.