Experiment: QM fails, CM succeeds

[Theoretical]

 

Not necessarily. Particularly with reflected power, never seen an antenna lose transmitting power? Through impedance mismatch?

Plenty of times but the equation is for a specific amount of energy absorption.

You should study what QM states and what the mathematics actually covers before claiming to debunk it

Single quantized photons don't exist. How's that for a start. Packets of electromagnetic waves exist.

[Mordred]

Plenty of times but the equation is for a specific amount of energy absorption.

Single quantized photons don't exist. How's that for a start. Packets of electromagnetic waves exist.

So what? I don't care if you use packets in discrete units of quanta instead of photons or not. This doesn't make any difference in the formulas I provided in any of the articles I posted.

 

This brings up the key issue

 

How do you expect a macro experiment to discount the need for the Planck constant?

 

Your test has no where near the level of precision for that

You haven't provided a single solution to particle-wave duality. As far as I know no such solution exists to seperate the two.

Edited August 28, 2015 by Mordred

[Theoretical]

So what? I don't care if you use packets in discrete units of quanta instead of photons or not. This doesn't make any difference in the formulas I provided in any of the articles I posted.

 

This brings up the key issue

 

How do you expect a macro experiment to discount the need for the Planck constant?

 

Your test has no where near the level of precision for that

You haven't provided a single solution to particle-wave duality. As far as I know no such solution exists to seperate the two.

The DC magnetic loop experiment shows emr being emitted from empty space. Nobody has been able to explain that through QM, including yourself. You miss the fact that you need to use DC to explain it.

 

But that's one of many experiments I've done. I've tested for the single photon at radio wavelengths. And don't tell me it can't be done because I did it. You just need to be clever enough. Also I've done experiments at visible light that show nothing but a continuous sine wave where the emr intensity is well below one hf per billion wavelengths. Where's your photon? Where's your packet? The probes is you people go far out of your way to emit packets of photons lol. For once why don't you people get off your high and mighty throne and listen to me by doing my simple experiment. Emit at least 2mA from a non-focused LED, view the spectrum far far away with a linear detector. View the spectrum. If there are packets then it will show up in the spectrum. It's not there. You will *never* find your photon. Try it!

[Mordred]

Why don't you get off your high horse.

 

I provided you the tools to improve your model. Langrene and Hamilton's can be done strictly classical.

 

I mentioned before your mathematics is not up to a sufficient level to overthrow QM.

 

Go ahead waste your time. Make your video, without the proper level of math no professional will even consider it. Considering there is hundreds of varying experiments in fine tuning the Planck constant. It's part of the university curriculum. I've done 2 variations of those tests myself.

 

Is it possible to mathematically eliminate the Planck constant.

 

Yes it is. However with your attitude I don't feel inclined to show you how to use the Langrene and Hamilton's to do so.

Best part is I even know the specific peer reviewed professional paper, that covers this in the numerous related earlier experiments relating to fine tuning the Planck constant. They then mathematically eliminate h.

 

Your loss.

 

( Your far too insulting to waste my time trying to help, particularly since you refuse to understand the tools I've provided)

You claim photons don't exist. Yet you expect to prove this via one type of experiment. Roflmao. Good luck with that.

 

Be well

 

here is a counter experiment for you.

 

 

http://www.iflscience.com/physics/researchers-image-wave-particle-duality-light-first-time-ever

 

Let's see pros and cons between the two experiments.

 

Yours doesn't conform to other experiments, done by thousands of PH.Ds

-mathematics sub par.

 

Their experiment does conform to numerous other tests, has the related detailed mathematics. Done by professionals.

 

Who do you think will get the most attention?

 

Get the picture?

 

But I guess that's just us being on our high horse right? Got it have a good day

Edited August 28, 2015 by Mordred

[Sensei]

I already said why the equation uses h and said it doesn't have to. It could easily be in units of 1 joule instead of hf.

Ridiculous. It must have Planck constant. As it would ruin plentiful in either classic and quantum physics.

 

If you have light source which is emitting uniform wavelength photons with power P from point source, in the all directions (inverse-square law).

 

Single photon has energy:

E=h*f

or

E=h*c/wavelength

 

Power P [W] is E/t, energy by time.

 

If you sum all photons energies, multiply h*f by their quantity, you will receive total energy emitted in time unit.

 

Etotal = h*c/wavelength * quantity of photons

 

Inverse-square law E=Etotal/(4*PI*r^2)

so finally:

E=h*c/wavelength * quantity / (4*PI*r^2)

 

These photons deliver energy from source to our detector. f.e. Sun is heating Earth by photons.

If you would replace h*f by some other value than 6.62607*10^-34 J*s * frequency, energy received in Joules would be different.

Which would be easily detectable as different temperature.

 

Actually that's how it has been detected that different wavelength photons have different energies, by splitting white light on prism, and then heating few samples of water (for red, green, blue light) and reading temperature increase.

Edited August 28, 2015 by Sensei

[Strange]

Wrong. Feeding hf or h/wavelength into an equation absolutely does not suggest quantization. hf is energy.

 

Energy of what?

[Mordred]

Here is some tests you can do at home.

http://www.google.ca/url?sa=t&source=web&cd=2&ved=0CB8QFjABahUKEwi-4uvDmcvHAhUBlYgKHUuHC6I&url=http%3A%2F%2Fjsharkey.org%2Fprojects%2Fphysics%2Flab2.pdf&rct=j&q=Planck%20constant%20lab%20tests&ei=UgbgVb7xKIGqogTLjq6QCg&usg=AFQjCNGVgi0EDlaRwB2aVZNOdSC76usO5g&sig2=zpy32IPekIT1oDkWgJVGxg

 

http://www.foothill.edu/~marasco/4dlabs/4dlab4.html

 

http://demoweb.physics.ucla.edu/content/experiment-6-photoelectric-effect

 

http://www.google.ca/url?sa=t&source=web&cd=5&ved=0CCUQFjAEahUKEwiP37z6msvHAhUPNIgKHTmyD20&url=http%3A%2F%2Fweb.mit.edu%2Flululiu%2FPublic%2Fpixx%2Fnot-pixx%2Fphotoelectric.pdf&rct=j&q=Planck%20constant%20lab%20tests&ei=0QfgVc-yKY_ooAS55L7oBg&usg=AFQjCNF0r6Dp5SCXCEGPCc_G1SsI249UPw&sig2=3AJTR40TLJ9S4S8eiyxLhw

 

keep in mind these are basic setups.

[Sensei]

Here is some tests you can do at home.

http://www.google.ca/url?sa=t&source=web&cd=2&ved=0CB8QFjABahUKEwi-4uvDmcvHAhUBlYgKHUuHC6I&url=http%3A%2F%2Fjsharkey.org%2Fprojects%2Fphysics%2Flab2.pdf&rct=j&q=Planck%20constant%20lab%20tests&ei=UgbgVb7xKIGqogTLjq6QCg&usg=AFQjCNGVgi0EDlaRwB2aVZNOdSC76usO5g&sig2=zpy32IPekIT1oDkWgJVGxg

It has error, page 7

"For Plank's constant h, the accepted value is 6.626×10−39 J·s"

It should be -34

[ajb]

By Joe I've spotted the problem lol. If Albert Einstein was alive today, would he be hanging out at public science forms? No! Einstein exchanged letters with selective people. Wow what a vast spectrum of minds in the science community with only a few in the gamma region, unfortunately.

This is irrelevant!

 

Do you really not think that those who are professionals here do not talk to other experts in our respective fields? Today it is usually via email or Skype.

 

 

... derived from classical mechanics, again in units of hf. Again, the reason for using hf is for your convenience so that you can see the classical mechanics gets the same answer as quantum mechanics.

This seems to be the 'fudge factor'. You cannot classically argue that hf has anything to do with anything. Assuming E= hf is using quantum theory. Using hf is not for our convenience, it is to ensure that we get the correct answer.

 

It is not a mathematical problem as such, but a physics one.

Edited August 28, 2015 by ajb

[Mordred]

It has error, page 7

"For Plank's constant h, the accepted value is 6.626×10−39 J·s"

It should be -34

Good catch, I posted it as an example setup, it's def not a professional paper. More likely some lab student.

 

It was the test setup I wanted primarily to show, something that's also not expensive.

Edited August 28, 2015 by Mordred

[ajb]

Just having another quick look....

 

F=q*v*B

v is speed of light:

F=q*c*B

This is not physical. You have used the Lorentz force (you have set it up so that there is no E, but okay), in the expression v is the speed of some test charge in that background B field. How can it be equal to the speed of light? At best you have some limit here, so lets go with that...

 

calculate the charges traverse peak velocity from one h*f amount of energy:

E=(m*v^2)/2=h*f

As noticed by others, the energy of what is equal to what?

 

It looks like you have put the kinetic energy of the test particle equal to a photon. Why and how is this not quantum?

 

Again, none of this is necessarily a mathematics problem, it may well all be mathematically consistent, but physically the meaning is very unclear.

[imatfaal]

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[Klaynos]

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