7 Reasons to Abandon Quantum Mechanics-And embrace this New Theory

The Bohr model is absolutely incorrect, yet it provides a model which accurately reflects the experimental evidence Planck was wrestling with.

It doesn't seem unreasonable that certain aspects of QM would also be incorrect, yet sound mathematical models. But this, sir, is not a theory until it is tested. Right now it is hypothesis, and a decent one I'll give you that.

I encourage you to find some means to investigate these predictions. Best of luck to you.

If it turns out you are completely, 100% wrong, then you have done good science, my friend.
 
AndrewGray said:
Again, I suspect that the lampblack materials used to find the "ultraviolet catastrophe" are actually reflective of UV radiation, since they do not emit UV when heated, and because Lampblack coatings make excellent "UV protectants".
Given that sand is both a UV protector and a UV absorber, and that there is no apparent mechanism for amorphous carbon to be degraded by UV, can you articulate further why you think that "UV Protectant" = "UV reflector"?

Have you seriously attempted to determine (through experiment or literature search) whether lampblack is a UV reflector or not?
An experiment would be neither difficult nor expensive.
 
Mr. Gray, in spite of the fact that I do think that a lot of scientists treat science as if it is a religion and they are the rather inept priesthood, I have to object to this. Your thesis has a huge hole in it at the very start and I can see why scientists simply don't want it to appear in their magazine and don't want to deal with it. It is illucid.

I do respect someone who has spent 3 years working with John Wheeler, who is a legend, and I am sorry to say that I would not have absorbed much working with him myself because of my own personality disorder. However, your very first message shows that you have evidently confused force with momentum. Your P(UV) is in newtons. That's your kg m/s. I just looked it up. A force is not comparable to a force without the measure of the distance that the force acts through. This is why a little old lady can lift a car using a bottle jack.

In other words, a newton does not convert to a joule unless it acts through a distance. You already had a measure of the momentums of the photon and the electron before you did the math. Joules are valid units to use to measure momentum, equal to the force of one newton acting through the distance of one meter. I'm sorry, but P(e)=1.12/1.6=0.7P(uv).

With the Bremsstrahlung problem, we are talking about hitting a target that is made of a fairly heavy metal with a large electron cloud and a massive nucleus. This is an adequate backstop for the sudden braking that has to occur. You also seem to forget the basics of elastic collisions.

Richard Feynman admitted that he never really learned some of the basics of calculus, something like a definitive integral or something, in his biography. I am not sure that he ever understood something that I took a long time to understand, that the numbers that are discarded, those "infinities", probably "infinitesimals" that you quoted him on, are actually zero. This is because the length of the interval over which you derive things like the slope of the equation is actually zero. It's a long story. It is what makes an integrated or differentiated equation the exact equation for the given point. All errors have been reduced to zero, and I think that because Feynman didn't understand this he didn't understand renormalization.

The "frequency" of the photon is dependent only on its packet energy and not on any resonant frequencies of anything that it comes from or touches or is affected by. Any individual photon does not actually have a "frequency" but it has a wavelength, and the frequency is derived from that wavelength. Also, it's pretty bogus to try to promote the development of a theory that gives a different answer for the spectrum of black body radiation because that is a measurable and testable reality.

Anyway, I'm afraid that your work here is not even wrong.
 
This New Theory has to be almost completely correct before submission, in my opinion, or it will not be succeed. It is getting there, though, because when I ask for flaws, nothing much really comes back.

Or, is your theory filled with so much nonsense, it gets ignored?
 
QUOTE:
"Given that sand is both a UV protector and a UV absorber, and that there is no apparent mechanism for amorphous carbon to be degraded by UV, can you articulate further why you think that "UV Protectant" = "UV reflector"?"

"Have you seriously attempted to determine (through experiment or literature search) whether lampblack is a UV reflector or not? An experiment would be neither difficult nor expensive."


Pete,

I understand why you say that a "UV absorber" might be a UV protectant. Just put a thick enough coat on there so the UV can't penetrate to the stuff you are trying to protect. A UV reflector also works as a protector.

So I just suspect that Lampblack is a UV reflector. (or perhaps it's transparent at really short wavelengths near soft x-rays). The point is, somewhere in the shorter wavelengths, Lampblack will probably stop absorbing radiation like a blackbody. Do you think that it will absorb 90% of soft x-rays that are incident? Of course not. Then right next to soft x-rays is 1 nm "deep" UV. I just suspect that Lampblack will not absorb 90% of 1 nm deep UV. That's all.


And if it is not an absorber, then it is not a thermal emitter according to this New Theory. Why is this? It has to do with the phase shift that the wave experiences upon incidence.

Consider a wave incident on some material. Part of the wave is transmitted, and part is reflected. If the transmitted wave tends to have a large phase shift near 180[sup]o[/sup] when it encounters the atoms in the material, then the forward wave will tend be canceled in the material. The reflected wave is just the opposite. If the phase shift of the reflected wave is 180[sup]o[/sup], then this makes the material a good reflector, since a mirrored 180[sup]o[/sup] phase shift is additive.

So a metal has free electrons, for example. There will tend to be minimal phase shift between the incident radiation and the electron acceleration. This will cause an approximate 180[sup]o[/sup] phase shift in both the forward wave and the reflected wave, since a charge accelerated upwards re-emits downwards radiation. Hence, the reflected wave will be strong, and the transmitted wave will be non-existent: a reflective material.

However, if the electrons are bound and have a resonant frequency of some sort, the phase shift will not be 180[sup]o[/sup] for the reflected wave. If the resonance oscillation adds another 180[sup]o[/sup] to the phase shift, then the transmitted wave will be strongest, and the reflected wave will be minimal: an absorbing material.

Finally, if the material has electron resonant frequencies in its atoms, when you "bang on them with heat", they will tend to emit these resonant frequencies.

Pete, I agree that an experiment to measure the reflectivity of Lampblack all the way down to soft x-rays would not be difficult.

Inexpensive though? I don't know about that. I do not have access to an x-ray/UV/visible source and spectrometer at my company or my home. Suggestions?

I am open to try with some clever suggestions.



Andrew A. Gray
 
Hi Andrew,
Why don't you stick to the wavelengths that you can test easily and cheaply?

You said earlier that a substance that absorbs UV will radiate UV strongly when heated. You can test that notion easily and inexpensively - you don't need to go to ultra-short wavelengths.

You suspect that lampblack is a UV reflector. That's fine.
I suspect that you are so strongly committed to your idea that you are unable to objectively assess it. You seem to have an almost religious devotion to it. Discussing it with you is similar to discussing origins with creationists - they too mutter darkly about closed-minded scientists, misunderstood experiments, and they also back off when confronted with simple experiments.

So:
Do you maintain that a UV-absorbing substance will radiate UV more strongly when heated than the standard blackbody model suggests?
Why don't you test that idea?
 
QUOTE:
"Your P(UV) is in newtons. That's your kg m/s. I just looked it up. A force is not comparable to a force without the measure of the distance that the force acts through."

"In other words, a newton does not convert to a joule unless it acts through a distance. . . . I'm sorry, but P(e)=1.12/1.6=0.7P(uv). "


MetaKron,

I think that you are a little confused. Momentum is
mv, or (mass) x (velocity). Mass ≡ kg and Velocity ≡ meter/s in metric units. That makes


Momentum ≡ $$(kg)(meter/sec)$$

Now

Energy ≡ Force x distance. Force is (mass)(acceleration), distance is (meter) . So we have

Energy ≡ (kg)(meter/s²) x (meter) = (kg)(meter²/s²) ≡ Joules

For light we have that P[sub]light[/sub] = E[sub]light][/sub]/c. so we have

Momentum ≡ $$ \frac {\frac{(kg) (meter^2)}{s^2}} {meter /s } = (kg)(meter/s)
$$


So for UV,

$$P_{UV} = \frac {E_{UV}}{c} = \frac{(10eV)(1.6 \times 10^{-19} \frac {J}{eV})}{3 \times 10^{8}m/s} = 5.3 \times 10^{-27} (kg)(m/s)
$$

Now for the electron,

$$E_e = \frac{1}{2}m v^2 = \frac {P_e^2}{2m_e}
$$

So

$$
P_e = \sqrt{2m_eE_e} = \sqrt{2 (9.1\times 10^{-31}kg)(10eV-3eV)(1.6 \times 10^{-19}\frac{J}{eV})} = 1.43 \times 10^{-24} ( kg )( m/s).
$$

So

$$
|P_e| = 270|P_{UV}|
$$

 
Pete,

I apologize if I seem illogical like creationists.

I am open to checking some easily obtainable frequencies. So let me think out loud here. Help me out.

I would need:

1) An oven capable of 1000[sup]o[/sup]-1500[sup]o[/sup] K.
2) Some way to accurately measure the temperature.
3) Some UV absorber that can tolerate these temperatures.
4) A cavity in some kind of material.
5) Spectrometer.
6) Intensity meter.


Can you think of any details to fill in about these items?
Are you familiar with the least expensive technology for these?
For example, perhaps a pottery kiln for the oven.
Perhaps I will go this weekend and look up the original experiments.

What do you think about this material:

http://www.modelofreality.org/UVabsorber.gif

It is definitely "black" in the UV range.


Andrew A. Gray

P.S. I do maintain that a good UV-absorbing substance will radiate more UV than Planck predicts.
 
Last edited:
Pete,

I apologize if I seem illogical like creationists.

I am open to checking some easily obtainable frequencies. So let me think out loud here. Help me out.

I would need:

1) An oven capable of 1000[sup]o[/sup]-1500[sup]o[/sup] K.
2) Some way to accurately measure the temperature.
3) Some UV absorber that can tolerate these temperatures.
4) A cavity in some kind of material.
5) Spectrometer.
6) Intensity meter.


Can you think of any details to fill in about these items?
Are you familiar with the least expensive technology for these?
For example, perhaps a pottery kiln for the oven.
Perhaps I will go this weekend and look up the original experiments.


Andrew A. Gray

P.S. I do maintain that a good UV-absorbing substance will radiate more UV than Planck predicts.

Hi Andrew,
An engineer or applied scientist of some description would probably be more helpful, particularly with controlling errors - it might be hard to get great accuracy doing this at home - but here are some ideas:
1) An oven capable of 1000[sup]o[/sup]-1500[sup]o[/sup] K.
Maybe a carbon-arc setup? Preferably in an inert gas bath or vaccuum, I think.
2) Some way to accurately measure the temperature.
A high temperature thermocouple.
3) Some UV absorber that can tolerate these temperatures.
A carbon rod or two.
4) A cavity in some kind of material.
Not strictly necessary, but a pit in the carbon rod would do.
5) Spectrometer.
6) Intensity meter.

These I'm not so sure about. Google tells me you can get a brand new UV spectrometer with software for under US$2000 (http://www.photon-control.com/spectroscopy.html)
You might also find something on e-bay (EG&G Digital Triple-Grating Spectrograph model 1235.

Or, you might get good results if you buy a graduate optics student a case of beer :).
 
Why don't you look up the black body radiation spectrums that were made by other researchers?
 
QUOTE:
"Your P(UV) is in newtons. That's your kg m/s. I just looked it up. A force is not comparable to a force without the measure of the distance that the force acts through."

"In other words, a newton does not convert to a joule unless it acts through a distance. . . . I'm sorry, but P(e)=1.12/1.6=0.7P(uv). "


MetaKron,

I think that you are a little confused. Momentum is
mv, or (mass) x (velocity). Mass ≡ kg and Velocity ≡ meter/s in metric units. That makes


Momentum ≡ $$(kg)(meter/sec)$$

Now

Energy ≡ Force x distance. Force is (mass)(acceleration), distance is (meter) . So we have

Energy ≡ (kg)(meter/s²) x (meter) = (kg)(meter²/s²) ≡ Joules

For light we have that P[sub]light[/sub] = E[sub]light][/sub]/c. so we have

Momentum ≡ $$ \frac {\frac{(kg) (meter^2)}{s^2}} {meter /s } = (kg)(meter/s)
$$


So for UV,

$$P_{UV} = \frac {E_{UV}}{c} = \frac{(10eV)(1.6 \times 10^{-19} \frac {J}{eV})}{3 \times 10^{8}m/s} = 5.3 \times 10^{-27} (kg)(m/s)
$$

Now for the electron,

$$E_e = \frac{1}{2}m v^2 = \frac {P_e^2}{2m_e}
$$

So

$$
P_e = \sqrt{2m_eE_e} = \sqrt{2 (9.1\times 10^{-31}kg)(10eV-3eV)(1.6 \times 10^{-19}\frac{J}{eV})} = 1.43 \times 10^{-24} ( kg )( m/s).
$$

So

$$
|P_e| = 270|P_{UV}|
$$


Andrew, this proves that you don't even have the basics down. Momentum and energy are exactly the same thing. I could simply say that you've gone through the trouble of converting two comparable measures into two measures that cannot be directly compared to each other, but I'm going to go into a little more detail than that. I'm not sure why.

Your mathematical definition of momentum is wrong:

[/color]
Momentum ≡ $$(kg)(meter/sec)$$


As you are no doubt well aware, a number squared divided by another number squared is not the same as the first number divided by the second number. Also, we already have a well established equation:

Momentum=$$(mass)(velocity^2)/2$$

Or: Momentum=$$(mass)(meters^2/sec^2)/2$$ to relate it to your equation above.

That is also: $$E_k=\frac {mv^2}{2}$$

Velocity is meters per second, distance over time, so velocity squared equals $$(meters^2/sec^2)$$

Since you already started with the measures in joules, you needed no conversions at all, and when you did the conversions, you made very obvious errors in equations that you should have learned the first year in school. How much of this does it take to get you to go back and think this through?

The spectrums of black body radiation versus temperature are among the most established areas of scientific investigation because these can be more or less directly observed with 19th century equipment, with great precision. Why go on about what those should show when you can find out what they do show?
 
As you are no doubt well aware, a number squared divided by another number squared is not the same as the first number divided by the second number. Also, we already have a well established equation:

Momentum=$$(mass)(velocity^2)/2$$

Hi Metakron,
That's Kinetic Energy, not Momentum.
Kinetic Energy
Momentum
 
Thank you, Pete. I guess Andrew's work is better than it looked. The problem is that now we have the fact that two objects can have the same kinetic energy but different momentum. Half the mass times twice the speed gives you the same momentum but twice the kinetic energy.

I knew it was wrong but didn't quite get the why right. The same momentum does not mean the same kinetic energy and momentum is neither a measure of energy nor a measure of force.
 
Metakron:

Here's an easy way to keep momentum and energy straight.

This was a question asked during the "orals" of a Ph.D. candidate in chemistry (who failed).

"If you have an object that has some internal potential energy, and it separates into two smaller objects going in opposite directions, which of the two carries away the greater amount of kinetic energy; the smaller (less massive) object or the bigger (more massive) object?"

The answer is not always intuitively obvious to most people, I've found.

This can be likened to the nucleus of a large atom (say, Uranium-238) that separates into two smaller nuclei (Thorium-234 and Helium-4), with the Helium-4 nucleus (an alpha particle) travelling away at very high speed, and the Uranium 'recoil' nucleus travelling away at very low speed and opposite direction, comparatively. Both of the atoms formed from the original parent Uranium atom would have equal but opposite momentums (opposite directions, equal massXvelocity component). However, the alpha particle carries away a tremendous amount of energy, which goes as the SQUARE of the speed (neglecting relativistic speeds). The alpha speed has to be much much higher than the parent-nucleus recoil speed, by the ratio of their masses. Thus, even though the momentums are the same, the alpha particle has much greater energy.

An even easier way is to imagine a loaded rifle hanging on a string. If the bullet goes off, the rifle recoils (kick to your shoulder in normal usage) in one direction, and the bullet goes in the opposite direction but with equal momentum. Now, at which end of the rifle would be prefer to be standing when the bullet goes off? Clearly, the rifle kick can break a shoulder if not properly braced, but the rifle bullet carries far greater energy, even though it has only the same momentum as the recoiling rifle.

An interesting outgrowth of this, not yet recognized in the medical physics field, is the erroneous interpretation of results for the "Relative Biological Effectiveness (RBE)" for alpha particles used in government regulations for determining how much more effective alpha particles are at causing cancer compared to electrons or photons that deposit the same amount of energy total (lots and lots of particles).

The experimental set-up for determining those values (which results in regulations fixing the values at 1 for electrons and photons, 10 for neutrons, and 20 for alphas) utilized petri dishes of different types of cells that were then exposed to electrons, photons, neutrons and alphas. This resulted in measureable types of cell damage for various amounts of energy deposition. The damage was fixed at "1" for photons, and electrons then came back with the same value, i.e. they were no more effective at causing cell damage than photons, for equivalent total energy delivered to the cell. However, neutrons had values ranging from 4-8, depending on whether they were primitive or advanced cells, and so the regulations fixed the value, to be safe, at 10.

Likewise, the alphas had values ranging from 8-12, compared to photons depositing the same amount of energy, so to be safe, the regulations fixed the valued at 20.

What was overlooked is that the experimental set-up, though valid for photons, electrons and neutrons (in which the ionizations of the cellular damage take place hundreds of Angstroms apart), is not valid for alphas.

That's because the recoil nucleus of the alpha emitter, though it does have a much smaller energy than the alpha, does in fact carry away some energy, being simply the ratio of the masses smaller than the alpha's energy (which is typically in the range of 4-6 MeV, depending on the emitter). HOWEVER, the alpha particle will travel about one cell diameter while depositing its kinetic energy in the cell, and that was what was being measured when the experimenters used external alpha sources to irradiate various types of cells.

In real life, alpha emitters only cause damage if they are INSIDE of the cell (which is where regulations arise from to limit the amount of alpha emitter one may absorb into one's body). While far more energy is deposited by the alpha compared to its recoil nucleus, the recoil nucleus, being a heavy metal, is almost always located on the DNA in the nucleus of the cell, and because of its high mass, it travels only a short distance in causing its ionizations, all of which end up clustered on the DNA, rather than mostly harmlessly dispersed in the cellular waters, as is the case for the alpha particle. The real RBE for alphas is thus much closer to 1,000 than the "20" figure of the regulations, but you won't read about that anywhere else yet, except here.

Anyway, enough thread hijacking for today!

Do a few practice problems with momentum (mv) and kinetic energy (1/2 vmv), and you'll get the feel for it.

Regards,



Walter
 
Thank you for the info, Walter. It doesn't sound to me, by the way, as if it is healthy to have heavy metals in the nuclear DNA even if they don't emit alpha particles.
 
Express kinetik energy in terms of momentum:

$$E = \frac{p^2}{2m}$$

and it shows clearly that the smaller mass will have more kinetik energy if the momentum is constant.
 
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