Quantum Mechanics and Light Intensity

Bill Westmiller asks:

I'm confused on a quasi-technical question you may be able to answer.

The energy of a light wave is directly related to its frequency, as in E = hv.

But, this equation seems to take no account of the intensity of light. For example, red light always has the same frequency ... so one might assume it always has the same energy.

Is there something inherent in E or h that reflects or assumes a particular light intensity standard?


Good question. Quantum mechanics (QM) deals with the “smallest quantity of motion.” Although my blogs have pointed out the numerous philosophical errors in QM, physicists have used Max Planck’s brilliant work to explore the theoretical properties of the smallest microcosms, not without some experimental success. Nonetheless, because of the attendant philosophical obfuscation, QM gets extraordinarily complicated. Let me try to explain what I think it is about in simplified terms.

In neomechanics, we stress the collisions between microcosms. The quantum would be a description of collisions involving the smallest microcosm known, in this case, the photon. In regressive physics, including QM, the photon is the particle that carries light from source to observer. In neomechanics, of course, we consider aether-1 particles to be the constituents of the medium for light. In our theory light is simply wave motion in which aether-1 particles collide with each other as in any other wave motion. The emitted motion from the source travels microcosm-to-microcosm to the observer, as it would in any other medium. As in water waves, the individual aether-1 particles pretty much stay at home. I am not sure, but these aether-1 particles could just as easily be photons. We could call them that, except for the unfortunate connotations given them by Einstein. True, there could be even smaller particles, such as the aether-2 particles that Steve and I hypothesize as the constituents of aether-1 particles.[1] They also would be responsible for the “subquantic” motions hypothesized by physicist David Bohm.[2] It is doubtful, of course, that we will ever be able to detect aether-2 particles, much less hypothesize much about them. For now, let us continue with the smallest of everything, which invariably must use Planck’s constant (h = 6.62606957 × 10-34 m2 kg / s).

From the neomechanical perspective, I speculate that today’s “smallest quantity of motion” would be one “cycle” in the Planck equation you mentioned:

E = hv

The “v” in the equation is the frequency in cycles per second. High frequency light (e.g., blue, with a short wavelength) produces more collisions per second than low frequency light (e.g., red, with a long wavelength). That is about as close as one gets to intensity when working at the photon/aether-1 level of the universal hierarchy.

The next level involves a qualitative change with billions of collisions instead of the individual collisions modeled by QM. Wave-particle duality requires that the photon travel without a medium, bringing its own waves along with it. I must admit that I could never understand that conjecture even though I once believed in the 3 in 1 god stuff. Some of those photons must be gigantic, with some electromagnetic waves being over 10 km long. As a practical matter, we measure light intensity at the next level of the hierarchy in various ways, depending on the discipline (sorry for getting a bit elementary here). Many of these, especially astronomy, assume a spherical light source emitting in all directions. Intensity follows the inverse-square law for light, as it does for gravitation, atomic interactions, and many other phenomena:

Intensity is proportional to:    ______1______

A good illustration of the inverse square law is:[3]

The gist of the inverse square law is that the motion produced by a particular microcosm is transmitted geometrically in the 3-D universe, with its effect being scattered over ever-larger spheres. We see this all the time when practicing archery. If all your arrows hit a 4” circle at 20 yards, expect them to hit a 16” circle at 40 yards. The spread will increase by four times when you double the distance. Similarly, an observer or detector will see only a tiny fraction of the light emitted from a distant body, depending on the subtended angle. Astronomers often indicate this fraction in terms of angular diameter as seen from Earth. Looking straight up and from the western horizon to the eastern horizon would comprise an arc of 180 degrees. The Moon has an angular diameter of less than 34 arc-minutes, while Venus has an angular diameter of less than 66 arc-seconds simply because it is much farther away. If all luminous celestial bodies were identical, it would be a simple matter of estimating their distances by measuring their light intensity. Because they are not identical, much more is involved, but you get the point: intensity is multiple rather than singular. A single water molecule, like the lone fan in the stadium, does not a wave make.

[1] Puetz, S.J., and Borchardt, Glenn, 2011, Universal cycle theory: Neomechanics of the hierarchically infinite universe: Denver, Outskirts Press ( www.universalcycletheory.com ), 626 p.

[2] Bohm, David, 1957, Causality and chance in modern physics: New York, Harper and Brothers, 170 p.


Westmiller said...

It isn't really an answer to my question.
I understand the variable energy inherent in frequency and I understand diminishing wave energy at greater distances. The problem is that there is no distance component - nor wattage of the source - in the equation E=hv.
So, is the equation wrong, or does it just assume some fixed wattage or distance from the light source?

Compare to sound or water waves:

Sound waves have a frequency, but the intensity is a factor of the compression of air particles in the wave crest (or the degree of compression variance between the crest and trough). Taking your aether premise, does that mean that the intensity of light is a greater compression of those particles? But, if so, then the amount of energy in any color of wave is NOT directly related to the frequency. There needs to be another "compression" term to calculate the energy of the wave.

Water waves have a frequency, but the intensity is a factor of the height of the wave (plus some compression factor for water molecules). I don't know how you impute a "height" component to light waves (the wavelength is the distance between crests, not their length), but something beyond frequency would have to account for the variance in intensity to arrive at an energy value.

So, three choices:
1. The equation E=hv is wrong;
2. The equation is right, but only at an assumed intensity; or
3. The equation is right, because h has a fixed term that can be varied to account for intensity.

For Case 2, E=hv is only a relative expression. That is, for any specified intensity, the ratio of energy for different colors is directly related to their frequency. So, it isn't really a calculation of energy, but rather a variation or change in energy (delta v and delta E).

For Case 3, the Planck Constant h is 6.62606957(29)×10−34 J.s, but only for a Joule value of one kilogram times one meter squared per one second squared. In that case, intensity might be considered a variance in mass or area, though that doesn't make any sense to me.

Care to pick an option?

Glenn Borchardt said...


As mentioned early on, I assume Planck’s equation to be indicative of “the smallest unit of motion” (i.e., the collision produced by a single photon or aether particle). Intensity really has nothing to do with it. Like Planck, I assume that this single collision occurs at a velocity of c according to the equation: E = hv. This assumption implies that the E in the above equation is also equivalent to mc^2, which yields a photon or aether mass of 7.362 X 10^-48 g. That result poses serious implications for further speculation.

The known electron mass is 9.109 X 10^-28 g. The known classical electron radius is 2.8179 X 10^-13 cm. Thus, the volume of the electron would be 9.3727 X 10^-38 cm^3. This yields an electron density of 0.97186 X 10^10 g/cm^3, with the number of photons or aether particles being 1.2373 X 10^20 in the aether vortex I hypothesized for the structure of the electron at the end of my E=mc^2 paper published in 2009. All this fits with the extremely small size and high density that Steve and I consider necessary for the aether model we hypothesized in our 2011 book.

Sorry that I was not able to choose one of your three possibilities. I suspect that there is too much evidence supporting Planck’s equation for it to be incorrect. Of course, I disagree with him that all microcosms have wave properties. Once you bring the aether back, all microcosms swim in a macrocosm that produces waves. Just don’t confuse the microcosm with the wave.

Westmiller said...

Glenn: "As mentioned early on, I assume Planck’s equation to be indicative of “the smallest unit of motion” (i.e., the collision produced by a single photon or aether particle)."

That sounds like a variation of B, in that it only applies to the "smallest unit" of light. Assuming a wave theory, it's one cycle. Assuming a particle theory, it's one photon. But, I don't think that's what E=hv represents, since v is cycles PER one second. Therefore, for a particular color, the number of cycles in that second will be higher (or lower) than another frequency: more waves or particles impacting the detector.

However, that means that E=hv doesn't account for intensity. In the wave theory, intensity must be the "density" or compression factor of the media components composing the wave, since the frequency remains constant ... but the energy increases. In a particle theory, intensity must be an increase in the mass of the particles, since the number of particles per second remains constant, but the energy increases.

So, I'm still confused.

Glenn Borchardt said...

As I mentioned, the neomechanical view of Planck’s equation assumes that it pertains only to “the smallest unit of motion.” At present, we have no evidence for aether-2 or other “subquantic” interactions, so that leaves us with the collision produced by a single photon or aether particle. At that level, one must forget about waves. The reason frequency is included in the equation is simply because, one can never measure the energy of a single collision. Thus, if the measurement involves 30 cycles per second, then one needs to divide by 30. If it involves 100 seconds, then one needs to divide by 100. In other words, one would need to do whatever it takes to calculate the energy for one collision.

Like Planck, I assume that this single collision occurs at a velocity of c according to the equation: E = hv. Here is where I think we need to make a change in the analysis. In my previous speculation I assumed that the E in the above equation is also equivalent to mc^2. I now think that the proper equation for a single unidirectional collision is 1/2mc^2, which is simply what one gets from the kinetic energy equation, 1/2mv^2. The reason Einstein’s purloined equation is E=mc^2, is that it describes bidirectional collisions. Such occur when fission results in the transmission of the internal motion of the atom to the aether in all directions (see my E=mc^2 paper). The kinetic energy equation would yield a photon or aether mass of 3.681 X 10^-48 g, which would double the number of photons or aether particles per electron to 2.475 X 10^20.

Note that this analysis does not propose any changes in mass, contrary to your suggestion. To get the intensity factor that you seek, you would have to integrate over all the photon/aether collisions involved. Granted, these usually involve noticeable wave motion, but they don’t have too. Here is an example:

While standing in the water at the ocean, I normally think that my legs are being hit by waves, but that really is not the case. What hit me are trillions of water molecules. The energy calculation would involve the sum of all of them. The motion would be periodic, with the height of the wave furnishing the intensity you wish for. On the other hand, a wall of water, as in a dam collapse, also could hit me. The question of whether that was a wave or not would be moot. The equation for the energy involved would not require a term for periodicity. Hopefully, even quantum mechanists could agree that the water molecules themselves were not also waves.

Glenn Borchardt said...

Glenn: ".... if the measurement involves 30 cycles per second, then one needs to divide by 30 ... one would need to do whatever it takes to calculate the energy for one collision."

BW: To apply an example, the center frequency for green light is 5.66e+14 cycles per second, with an energy of 2.33e-06 Mev. So, dividing energy by frequency, we get ~4.13e-30 electron-volts for one particle/wave, the same result as for any other light frequency.

[GB: Thanks for the calculation. It demonstrates my point that each cycle in quantum mechanics involves only a single photon/aether collision (e.g., Planck’s “smallest unit of motion”)]

BW: I'm not sure whether that gets us past the problem that one electron volt is defined as energy change through one meter of one amp for one second, or one watt of power for one second. [1 joule = 6.24e+18 ev] Can we even talk about ONE particle/wave per second?

[GB: No. In neomechanics, waves are collective properties. One particle does not a wave make. Thus, the “wave” that goes around the stadium must involve more than one waver. Even though he certainly could produce a collision, he could never produce a wave. The “waveitis” that infects regressive physics is simply a logical result of systems philosophy and aether denial. In this regard, I was just referred to this neat quote in Science: "the Planck length is a length at which the smoothness of space breaks down, and space assumes a granular structure" (Pipkin, F.M., and Ritter, R.C., 1983, Precision Measurements and Fundamental Constants: Science, v. 219, no. 4587, p. 915). This is an admission that there is no empty space and that the “granular structure” referred to actually consists of photons/aether particles having the “Planck length”—the average diameter of those same particles.]

BW: Jumping back to my original question. Even if we were to establish that the mass of one photon is 3.681 X 10e-48 g, with a fixed velocity c, then what do we mean by the "intensity" of the source? Green light might be emitted by a 1 watt or 1,000 watt source. Does that mean that *more* photons are being emitted each second ... without affecting the frequency? Or, does it mean that their individual mass increases? Or are we into woo-woo land?

[GB: The “*more* photons is closest to the intensity concept, although I am not sure that “emitted” is the right word. I suspect that individual photons/aether particles always travel at greater than c within the medium, just like nitrogen molecules travel at greater than the velocity of sound in air. The “emission” could just as well be a ricochet from such high-speed collisions with a particular microcosm. The upshot is that light intensity should be related to wave amplitude and not frequency.]

Glenn: "... On the other hand, a wall of water, as in a dam collapse, also could hit me. The question of whether that was a wave or not would be moot. The equation for the energy involved would not require a term for periodicity."

BW: Periodic or not, a singular "wall" is still a wave of water particles with their compression/height/width determining their energy. What are the equivalent factors when determining the intensity of a singular photon? I don't think it's compressed, higher, or wider ... though it would presumably have more energy.

[GB: Again, a single photon/aether particle cannot have an “intensity,” just as a single water molecule cannot have an intensity. Intensity is the property of waves, not of the particles that are the ingredients of the waves. BTW: This is a good example of the old dialectical materialist law of the transformation of quantity into quality and vice versa (TSW, p. 342).]

Post a Comment

Thanks so much for your comment. Be sure to hit "Preview" to see if it will publish correctly. Then hit "Publish". Include your email address if you wish to receive copies of your comment as well as all other published comments to this Blog.

For those having trouble getting this comment section to work:

Nitecruzr writes:

[FAQ] Why can't people post comments on my blog?

The Blogger / Google login status, and the ability to post comments, is sensitive to both cookie and script filters. Your readers may need to enable (stop filtering) "third party cookies", in their browser and on their computer. The effects of the newly unavoidable CAPTCHA, and the Google "One account" login, requires third party cookies, even more than before.




Third party cookies filtering, in a browser setting, is the most common solution, overall - but your readers may have to search for other filter(s) that affect their use of Blogger / Google.

Any filters are subject to update, by the creator. If the problem started a few days ago, your readers may have to look on their computers, and find out what product or accessory was updated, a few days ago.