Sub-photon Radio wave experiment - Your predictions

[Theoretical]

Question: Experiments at 40 MHz are unnecessary because we know x-ray experiments show the photon is quantized, right?

 

Answer: The electric charge is accelerated longitudinally (for the most part) in the x-ray spectrum. Take special note that at radio frequencies it's a transverse effect where electrons move perpendicular to the direction the electromagnetic wave is traveling. That should be a major hint that radio wave experiments are extremely important.

 

 

Question: Will the simple version, experiment 1, reveal if the photons don't emit during each pulse cycle? It's been brought to my attention that if 0.1 photon amount of energy is pulsed to the antenna during each pulse cycle, that one photon will be emitted once every ten pulse cycles on average.

 

Answer: Yes, even experiment 1 will reveal if the antenna does not receive a pulse signal every pulse cycle. Furthermore, it will know if the received signals are not being received at the expected time. After doing the experiment I thought of a very simple clear cut method to reveal this information for experiment 1. To explain this requires some documentation. I'll detail all of this when the experiment is documented in a video.

[Theoretical]

 

 

I'll take some time to go over a few things such as the simple straightforward math of how experiment 1 can tell if there are missing pulse signals when the sub-photon amount of energy goes to the transmitting antenna every pulse.

 

So the prediction I've received from most mainstream physicists is that if the average transmitting antenna pulse falls below h*f joules, then a photon is not emitted every pulse. For example if the average energy per pulse is 1/100th h*f, then on average mainstream physicists expect one photon to emit every 100 pulses on average. To understand what the software would receive, we can do a simple test by hardwiring one pulse to the receiving circuit every millisecond. The signal amplitude average is y volts. Next, we only send one pulse every other time, which means the circuit is getting a pulse every 2 ms. Essentially were sending a pulse half the time. The amplitude average now becomes y/2. So the received signal voltage is relative to how often a photon is received, which means it's relative to the transmitted energy per pulse. However, if on rare occasion the pulses overlap, then it's a bit more complex to compute. Therefore, to simplify things well set the energy per pulse significantly less than h*f so the probability of there being two photons per pulse is low. The received signal voltage is therefore relative to the transmitted energy per pulse:

 

Energy

 

 

Now hypothetically speaking let's see what happens if were not receiving photons, but merely a non-quantized electromagnetic field. We know that the measured peak voltage of an exponentially decaying pulse into a load is relative to the square root of the energy of the pulse. In other words, if we quadruple the energy per pulse, the voltage doubles.

Voltage = sqrt(Power * Resistance)

Power = Energy / time

And hence:

Voltage = sqrt( (Energy / time) * Resistance)

Therefore, the received signal voltage is relative to:

 

sqrt(Energy)

 

 

So as you can see, the two models show different results. In the photon model (where transmitted energy is significantly less than h*f), the received voltage signal is relative to the energy per pulse, while the EM model is relative to the sqrt(energy) per pulse.

 

Here's a simple way to understanding why the two models show different results. If you place a battery across a load, the power is V^2 / R. Now if we place two batteries in-series, then obviously the power quadruples. Very basic electronics. However, if we repeat the experiment by replacing the batteries with pulsing signal generators such that the pulses do not overlap (they pulse at different times), we see that the averaged power does not quadruple, but only doubles. It's only when the pulses overlap that the averaged power quadruples. In other words, if we have one signal generator, and then add another signal generator such that the pulses do not overlap, then the averaged power only doubles.

 

This is a great technique to see which model is correct at low radio frequencies. If you replicate the simple experiment #1, remember that detected radio signals are ~ transverse to the direction of the propagating wave, while at higher frequencies such as IR, visible, UV, x-rays, gamma rays the charge is accelerated longitudinally.

 

I will be releasing a video detailing the performed experiments at the *appropriate* time.

[John Cuthber]

 

Answer: The electric charge is accelerated longitudinally (for the most part) in the x-ray spectrum.

Not really.

The x ray is emitted with a polarisation along the direction of the acceleration

And the Xrays emitted from most sources (other than things like synchrotrons) are emitted in all directions- so we know that the acceleration is not longitudinal.

That's because many of them are produced by this process

https://en.wikipedia.org/wiki/Auger_effect

it's not meaningful to define the "direction" of an electron hopping between two energy levels deep in an atom.

 

Of course, if we look at synchrotron radiation we know exactly what the acceleration of the electrons is- because we carefull designed it.

They are accelerated towards the centre of the circle.

And the Xrays come off tangentially- just the way we expect.

 

So we actually know that the emission process is pretty much the same for radio waves and Xrays.

 

So, fundamentally the answer to your question "Question: Experiments at 40 MHz are unnecessary because we know x-ray experiments show the photon is quantized, right?" is yes- the 40MHz experiment is unnecessary.

[Theoretical]

Not really.

The x ray is emitted with a polarisation along the direction of the acceleration

And the Xrays emitted from most sources (other than things like synchrotrons) are emitted in all directions- so we know that the acceleration is not longitudinal.

That's because many of them are produced by this process

https://en.wikipedia.org/wiki/Auger_effect

it's not meaningful to define the "direction" of an electron hopping between two energy levels deep in an atom.

 

Of course, if we look at synchrotron radiation we know exactly what the acceleration of the electrons is- because we carefull designed it.

They are accelerated towards the centre of the circle.

And the Xrays come off tangentially- just the way we expect.

 

So we actually know that the emission process is pretty much the same for radio waves and Xrays.

 

So, fundamentally the answer to your question "Question: Experiments at 40 MHz are unnecessary because we know x-ray experiments show the photon is quantized, right?" is yes- the 40MHz experiment is unnecessary.

You're confusing emission with absorption. As stated, I was talking about absorption.

 

Synchrotron radiation is the emission of radiation, not electrons.

 

Let's skip to the chase and look at the math that determines the photon and electron angle, energy, as transfer ration as shown in the Compton scattering equation.

 

Energy transfer ratio tells how much energy the photon transfers to the electron. A ratio of 1 means it transfers all. Numbers are rounded to 2 significant figures. So 1.0 does not mean 1.000000.

Photon angel is the angle the photon is scattered.

Electron angle is the angle the electron is scattered. The following example is for a gamma ray with a wavelength of 1e-20 meters. The scattered photon is long wavelength (low energy) since nearly all of the energy is transferred to the electron.)

Zero degrees means the particle goes forward.

 

photon angle: 15 deg.

electron angle: -2.8e-2 deg

energy transfer ratio: 1.0

 

photon angle: 60 deg.

electron angle: -6.4e-3 deg.

energy transfer ratio: 1.0

 

photon angle: 90 deg.

electron angle: -3.7e-3 deg.

energy transfer ratio: 1.0

 

photon angle: 135 deg.

electron angle: -1.5e-3 deg.

energy transfer ratio: 1.0

 

photon angle: 180 deg.

electron angle: -2.5e-12 deg.

energy transfer ratio: 1.0

 

 

As you can see, for such high energy gammas, the electron is accelerated forward regardless.

 

 

If you don't believe Compton scattering equation, then take a look at various photon momentum experiments.

 

 

Btw, polarization requires two axis and therefore cannot be longitudinally. This is seen in radio antennas.

 

 

This might help you. Here's a cloud chamber image that caught a gamma ray (top of image) that creates an electron and positron, both of which go shooting forward in the same direction of the photon. Photons have momentum.

 

http://www.nuclear-power.net/wp-content/uploads/2015/03/Pair-production-in-chamber.jpg?11abca

 

...

 

 

This is the nail in the coffin. Quote from Wikipedia:

 

"Since p points in the direction of the photon's propagation, the magnitude of the momentum is"

 

https://en.wikipedia.org/wiki/Photon#Physical_properties

 

Wikipedia states a bit earlier that p is the photon momentum.

Edited August 24, 2016 by Theoretical

[John Cuthber]

You're confusing emission with absorption. As stated, I was talking about absorption.

 

Synchrotron radiation is the emission of radiation, not electrons.

 

The two processes look the same, apart from the direction of time.

Nobody said anything different.

[Theoretical]

Regarding the IR communication system mentioned in both experiments that's used to sync the oscilloscope with the transmitting antenna. I found visible light components work better than IR, and it's cheaper. For about $10 I found a 30 foot new fiber optic cable on eBay. Normally these brand new cables sell for only a bit more. To match this I bought a HFBR2416TCK fiber optic receiver for about $5. I would have bought a fiber optic transmitter as well, but I had a bunch of old DVD players laying around doing nothing, which have fiber optic transmitters used for digital audio. So I removed one. Works great with my HFBR2416TCK. Fast switching speeds in the nano seconds.

 

So the $10 fiber optic cable has the same type of connector. Nice fit. The 30 foot fiber optic cable separates the transmitting and receiving antennas. If you want more separation, then for practically no amount of money you can be an extension part and another 30 foot fiber optic cable on eBay.

 

 

Another topic of interest is that someone had concerns with low end oscilloscopes being able to reliably sync enough well enough. As mentioned before, yes that would be some concerns if we were analyzing each signal alone, but these experiments are about taking thousands of averages. Of course there's always some jitter in such equipment. This is by no means a problem when averaging.

 

By the way, for these experiments your software will be converting the signal to spectrum, and you'll be analyzing a specific frequency. If you're interested in seeing the final averaged signal, then you'll need to filter out some of the low end frequencies, unless perhaps you happen to live far away from noisy radio stations.

 

But anyhow, if you get an oscilloscope, then make sure they offer PC/Mac source code so you can write custom software, or if the oscilloscope itself can be manually programmed in from the panel. Anyhow, send me a private message if you want.

 

Best wishes in the biggest shock of your entire life!!

[swansont]

Is there a point to posting all this new information, when you have so many questions still awaiting answers from before?

[Theoretical]

Is there a point to posting all this new information, when you have so many questions still awaiting answers from before?

You saying so doesn't make it true. I find your comment to be dishonest because you know darn well I've addressed every question. You may disagree with my answers, but I know for fact that stick with well established math, while the few opposing questions are not founded upon math.

 

If you have a question that you feel wasn't addressed enough to your liking, then posting.

 

 

ps, a friendly reminder to everyone that of the fact that photon's have forward momentum, while the forces at radio wave frequencies are perpendicular to propagation.

 

 

pss, the problem is probably that you're assuming I haven't addressed all the issues, because heaven forbid if I'm correct lol. That couldn't possibly be true lol. I think you need to study the entire thread.

[swansont]

You say you're integrating your signal. How will you distinguish between e.g. a half a photon's worth of energy detected every measurement cycle and one photon being detected every other cycle?

 

Where did you answer this?

 

 

 

ps, a friendly reminder to everyone that of the fact that photon's have forward momentum, while the forces at radio wave frequencies are perpendicular to propagation.

Can you show the classical EM theory that predicts this?

[Theoretical]

You say you're integrating your signal. How will you distinguish between e.g. a half a photon's worth of energy detected every measurement cycle and one photon being detected every other cycle?

 

Where did you answer this?

 

I've detailed this and provided the math. Receiver has an amplifier that feeds into the scope, which averages and converts to a spectrum. If the receieved signal decreases by half the voltage, then the amp output drops by half, which is reflected in the software.

 

That is well documented in the thread. Please read it.

 

 

Can you show the classical EM theory that predicts this?

Didn't you see my posts in this thread that give the math? It's there.

 

 

You know, to save time, why not read the thread and repost what's incorrect. Let's stick to facts. If I've made a math error then post it. If my circuit doesn't work as claimed then post a circuit that's basically the same the shows my circuit doesn't work. If I was on my desktop I could post hundreds of Spice circuits from LTspice that clearly predicts it would work as claimed. No more vague endless chit chat.

Please!

 

[edit: spelling correction]

Edited August 25, 2016 by Theoretical

[swansont]

I've detailed this and provided the math. Receiver has an amplifier that feeds into the scope, which averages and converts to a spectrum. If the receieved signal decreases by half the voltage, then the amp output drops by half, which is reflected in the software.

 

That is well documented in the thread. Please read it.

Where? The answer you give here doesn't answer the question. A signal dropping by half doesn't tell you anything about whether you have a single photon or not.

 

Didn't you see my posts in this thread that give the math? It's there.

 

You know, to save time, why not read the thread and repost what's incorrect. Let's stick to facts. If I've made a math error then post it. If my circuit doesn't work as claimed then post a circuit that's basically the same the shows my circuit doesn't work. If I was on my desktop I could post hundreds of Spice circuits from LTspice that clearly predicts it would work as claimed. No more vague endless chit chat.

Please!

 

[edit: spelling correction]

I didn't ask you about your circuit. I don't really give a damn about your circuit. It's moot if the underlying physics is wrong.

 

It should be no trouble for you to identify the post where you showed this math. But I don't see a post with a Poynting vector equation or anything resembling Maxwell's equations. Which one is it?

[Theoretical]

swansont, I'm not here to teach anyone classical mechanics. Are you seriously question how CM shows electrical current is produced by electromagnetism?? If you want to know how the software detects sub-photons, then that falls into the region of electrical engineering in conjunction with the QM h*f equation. If memory holds true the amp at the frequency is around 110 to 140 ohms. The software converts the signal to a spectrum via FFT. So it knows the voltage and the spectrum time frame. Energy = Voltage^2 * time / resistance.

Btw the exponentially decaying pulse bandwidth is clearly seen in the FFT. So it's easy to include the entire pulse in the software's energy equation.

[swansont]

swansont, I'm not here to teach anyone classical mechanics.

 

 

You accused me of being dishonest because you've "answered every question", and yet won't do me the courtesy of pointing out where you answered one question, and now refuse to answer another, which would support a rather bold claim that the momentum of RF radiation is perpendicular to the direction of motion. I don't need (or particularly want) you to teach me anything. I just want you to straight up answer some questions about your conjecture. At what point do we take "won't answer" to mean "can't answer"?

 

(and your response strongly suggests that the post where you say you showedthis math doesn't exist. You doth protest to much, methinks)

[Theoretical]

 

 

You accused me of being dishonest because you've "answered every question", and yet won't do me the courtesy of pointing out where you answered one question, and now refuse to answer another, which would support a rather bold claim that the momentum of RF radiation is perpendicular to the direction of motion. I don't need (or particularly want) you to teach me anything. I just want you to straight up answer some questions about your conjecture. At what point do we take "won't answer" to mean "can't answer"?

 

(and your response strongly suggests that the post where you say you showedthis math doesn't exist. You doth protest to much, methinks)

Because such a question has to do with question how CM math predicts how radio wave antennas works lol. Do you know how ridiculous and desperate that makes you?

 

If you want to know such equations, which btw the amplifier circuit obviously doesn't need to understand, then start here:

https://en.m.wikipedia.org/wiki/Antenna_(radio)

There! That will answer your questions. Your ridiculous question has been answered. Click on the Wikipedia links for further details. I am not going to teach you classical electrodynamics!

 

Now what desperate side distraction will you come up with? Do you actually think I can't teach you classical electrodynamics?? smh

 

 

... I came back to say how it's mind boggling that you actually don't know basic antenna theory. It is a fact, dear swanson, that the electric field is perpendicular at radio frequencies. If you honestly do not know that, then please take some classes or buy some books. Or download a free copy of 4nec2 that's based on the NEC2 electrodynamics engine where you can see till your heart is content that the electric field and electrical current in the dipole antenna is perpendicular to the traversing wave. ....... Good grief!

 

 

ps, The Wikipedia article I referenced above contains a nice animated dipole antenna showing you the induced voltage is perpendicular to the propagating wave.

[John Cuthber]

As stated, I was talking about absorption.

 

Let's skip to the chase and look at the math that determines the photon and electron angle, energy, as transfer ration as shown in the Compton scattering equation.

 

This is the nail in the coffin

Since absorption is not the same as scattering the wiki page about Compton scattering has nothing to do with it.

It certainly isn't the "nail in the coffin" of anything except your credibility.

[Theoretical]

Since absorption is not the same as scattering the wiki page about Compton scattering has nothing to do with it.

It certainly isn't the "nail in the coffin" of anything except your credibility.

Lmao you guys are soooo obvious in your attempts that you you're dishonest. Here's my direct quote regarding the nail in the coffin:

--

This is the nail in the coffin. Quote from Wikipedia:

 

"Since p points in the direction of the photon's propagation, the magnitude of the momentum is"

--

 

As you can clearly see it was the Wikipedia quote that was the nail in the coffin. Clearly you don't understand basic mainstream physics. Photon momentum is the direction of propagation. While in antennas at radio frequencies it's a transverse effect.

 

ps you still think the polarization longitudinally lol?

 

 

You know what... Forget you people. What a shame that a few people at this website can mess up a thread and keep legitimate researchers away from this vitally important experiment. I'm out of here. No time for this.

[John Cuthber]

Lmao you guys are soooo obvious in your attempts that you you're dishonest. Here's my direct quote regarding the nail in the coffin:

--

This is the nail in the coffin. Quote from Wikipedia:

 

"Since p points in the direction of the photon's propagation, the magnitude of the momentum is"

--

 

As you can clearly see it was the Wikipedia quote that was the nail in the coffin. Clearly you don't understand basic mainstream physics. Photon momentum is the direction of propagation. While in antennas at radio frequencies it's a transverse effect.

 

ps you still think the polarization longitudinally lol?

 

 

You know what... Forget you people. What a shame that a few people at this website can mess up a thread and keep legitimate researchers away from this vitally important experiment. I'm out of here. No time for this.

Lmao you guys are soooo obvious in your attempts that you you're dishonest.

"As stated, I was talking about absorption.

Let's skip to the chase and look at the math that determines the photon and electron angle, energy, as transfer ration as shown in the Compton scattering equation."

 

"ps you still think the polarization longitudinally lol?"

Longitudinal WRT what?

If I set up a dipole antenna and hook it up to a radio transmitter the electrons move up and down the dipole. The radio waves are emitted from the sides of that dipole.

However the polarisation of the radio waves is vertical- just like the motion of the electrons.

So, WRT the acceleration of the electrons (and that's what you were actually talking about- you said "The electric charge is accelerated longitudinally (for the most part) in the x-ray spectrum. " the polarisation is along the longitudinal direction.

 

Did you think you had made some sort of point?

[swansont]

Because such a question has to do with question how CM math predicts how radio wave antennas works lol. Do you know how ridiculous and desperate that makes you?

If you want to know such equations, which btw the amplifier circuit obviously doesn't need to understand, then start here:https://en.m.wikipedia.org/wiki/Antenna_(radio)

There! That will answer your questions. Your ridiculous question has been answered. Click on the Wikipedia links for further details. I am not going to teach you classical electrodynamics!

Now what desperate side distraction will you come up with? Do you actually think I can't teach you classical electrodynamics?? smh

... I came back to say how it's mind boggling that you actually don't know basic antenna theory. It is a fact, dear swanson, that the electric field is perpendicular at radio frequencies. If you honestly do not know that, then please take some classes or buy some books. Or download a free copy of 4nec2 that's based on the NEC2 electrodynamics engine where you can see till your heart is content that the electric field and electrical current in the dipole antenna is perpendicular to the traversing wave. ....... Good grief!

ps, The Wikipedia article I referenced above contains a nice animated dipole antenna showing you the induced voltage is perpendicular to the propagating wave.

What does that have to do with the momentum of the EM wave? Induced voltage in the antenna is not momentum.

[imatfaal]

!

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