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Quantum Physcis and the third force.

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posted on Feb, 1 2016 @ 12:41 AM
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I have been researching the existence of what amounts to the quantum of a third force maybe even a fourth, but i want to focus on the third quantum and how it relates to the macro and micro effect.

This third force that i call the midro in relation to the macro and micro bears out to what many would call simply the middle, however it does not constitute a true middling which would make it a third force in opposition to the macro and micro effect, or the positive and negative, in relation to the positive and negative i call the midro force the miditive.

Let me explain, even in the midro there bears a resistance to both the higher and lower laws of the quantum, for instance try this test run warm water in a bathtub fill it up about two to three feet and let the warm water soak in the tub aspect, then go into the tub put your feet in the warm water and run the shower on your body at either cold or hot, you will notice that either the colder water from the shower feels warmer with the warm water touching your feet or the warm water touching your feet feels colder, you will also feel the same if you chose to run hot water out of the shower and will feel something different in the warm water at your feet or the hot water out of the shower, this proves a resistance and thus a third quantum factor.

I would label the midro gray to the naked eye beings it seems to be the middle of the micro and macro effect however since it proves a resistance it would only be gray to the naked eye and thus i would say the color would resembles something of a peach type color and not gray because it has a resistance force.

Other theories involving my findings would also be in direct contrast to the duality effect, which means the ying and yang to some however my findings have shown that there is no true duality in the realm of existence, just take the meaning of a word often times there is way more then two meanings, thus labeling the duality effect in all flawed, then take this hypothesis to the quantum realm and you will see we are limiting science to a point where there needs no limitation....




posted on Feb, 1 2016 @ 02:16 AM
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Does your midro theory interact on a constant basis?

Does it have certain defining points?



posted on Feb, 1 2016 @ 06:06 AM
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To my understanding, that's the thing with quantum physics, there is no absolute zero. Isn't that what the pi equation is all about?



posted on Feb, 2 2016 @ 04:13 AM
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a reply to: King Seesar

There are already four forces: Gravitational, Electromagnetic, Strong Nuclear and Weak Nuclear.

Color in quantum physics also does not describe optical wavelength (color as our eyes see it).

You have mentioned that the meanings of words (semantics) can be a point of confusion. That is why physicists usually define things using mathematical language, for example, equations such as E=mc^2.

Also, understandings of things in physics lead us to be able to determine outcomes in particular scenarios. Things such as velocities, accelerations, energies, temperatures and work done.

As clear as it may be in your mind, I cannot see how your concept actually tells us anything. Let alone gives actual values of things.

Perhaps you could try and re-state your idea by carefully using language that does not have subjective or double meaning. Mathematics is such a way of describing things.



posted on Feb, 2 2016 @ 12:38 PM
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originally posted by: Vector99
Does your midro theory interact on a constant basis?

Does it have certain defining points?



Hi thanks for the question, yes as the macro and micro interact on a constant basis so does the midro on the quantum scale all resonating in harmony.

As far as the defining points about 10 to 25 percent of the middling of the positive and negative diagram, let me elaborate, for instance take a charge of the nuclear in opposing contrast of the P (positive) and N (negative) and examine the 10 to 25 electro pulse this will equate as the touching of the contrast.



posted on Feb, 2 2016 @ 12:42 PM
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originally posted by: chr0naut
a reply to: King Seesar

There are already four forces: Gravitational, Electromagnetic, Strong Nuclear and Weak Nuclear.

Color in quantum physics also does not describe optical wavelength (color as our eyes see it).

You have mentioned that the meanings of words (semantics) can be a point of confusion. That is why physicists usually define things using mathematical language, for example, equations such as E=mc^2.

Also, understandings of things in physics lead us to be able to determine outcomes in particular scenarios. Things such as velocities, accelerations, energies, temperatures and work done.

As clear as it may be in your mind, I cannot see how your concept actually tells us anything. Let alone gives actual values of things.

Perhaps you could try and re-state your idea by carefully using language that does not have subjective or double meaning. Mathematics is such a way of describing things.


You are correct about the four forces, the more i study i feel there may even be more, but no one tries to relate such a equation to the quantum which is foolish unto it's self, i was using color as a defining point to prove that this midro is only gray to the naked eye and thus gave it a different color because of the resistance it offers the other molecular level in contrast.

If i were to a sign a number which i already have just did not mention it above, it would be 3, i a signed the macro 1 the micro 2 and thus the midro 3...

Thanks for the question.



posted on Feb, 2 2016 @ 12:46 PM
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originally posted by: amsterdamn87
To my understanding, that's the thing with quantum physics, there is no absolute zero. Isn't that what the pi equation is all about?


Correct that is how i figured out that at least three or four are within realm, and i am also looking into more exists in the pi equation of the level space of the quantum.
edit on 2-2-2016 by King Seesar because: (no reason given)



posted on Feb, 3 2016 @ 04:40 PM
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originally posted by: King Seesar

originally posted by: chr0naut
a reply to: King Seesar

There are already four forces: Gravitational, Electromagnetic, Strong Nuclear and Weak Nuclear.

Color in quantum physics also does not describe optical wavelength (color as our eyes see it).

You have mentioned that the meanings of words (semantics) can be a point of confusion. That is why physicists usually define things using mathematical language, for example, equations such as E=mc^2.

Also, understandings of things in physics lead us to be able to determine outcomes in particular scenarios. Things such as velocities, accelerations, energies, temperatures and work done.

As clear as it may be in your mind, I cannot see how your concept actually tells us anything. Let alone gives actual values of things.

Perhaps you could try and re-state your idea by carefully using language that does not have subjective or double meaning. Mathematics is such a way of describing things.


You are correct about the four forces, the more i study i feel there may even be more, but no one tries to relate such a equation to the quantum which is foolish unto it's self, i was using color as a defining point to prove that this midro is only gray to the naked eye and thus gave it a different color because of the resistance it offers the other molecular level in contrast.

If i were to a sign a number which i already have just did not mention it above, it would be 3, i a signed the macro 1 the micro 2 and thus the midro 3...

Thanks for the question.


Yes but electromagnetic wavelengths are ginormous from a quantum scale. They are simply not reasonable as quantum sized objects.

E=mc^2, arguably, does apply at quantum scales (in many theories), but usually in its more complete form: E^2 = (mc^2)^2+(pc)^2.

There are other equations determining quantum values that must be factored in to the equation to 'flesh-out' its applicability.



posted on Feb, 3 2016 @ 09:01 PM
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a reply to: chr0naut

Quote"Yes but electromagnetic wavelengths are ginormous from a quantum scale. They are simply not reasonable as quantum sized objects.
E=mc^2, arguably, does apply at quantum scales (in many theories), but usually in its more complete form: E^2 = (mc^2)^2+(pc)^2. "End Quote.


I think you did your own math, this is sort of what i was getting at without the pi equation... E^3 = (mc^3)^3+(pc)^3.

Right now i am in the test stages of this application, measuring the wave length tests, you gave me a idea though, i want to see how far in and out the midro connects before it is consumed with the macro and micro i had a equation that was about right on point but i need the exact even though this will take sometime, i will get back to you, thanks.
edit on 3-2-2016 by King Seesar because: (no reason given)



posted on Feb, 4 2016 @ 03:26 PM
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Light Side of the Force, Dark Side of the Force, and the Brute Side of the Force.



posted on Feb, 4 2016 @ 06:00 PM
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originally posted by: King Seesar
a reply to: chr0naut

Quote"Yes but electromagnetic wavelengths are ginormous from a quantum scale. They are simply not reasonable as quantum sized objects.
E=mc^2, arguably, does apply at quantum scales (in many theories), but usually in its more complete form: E^2 = (mc^2)^2+(pc)^2. "End Quote.


I think you did your own math, this is sort of what i was getting at without the pi equation... E^3 = (mc^3)^3+(pc)^3.

Right now i am in the test stages of this application, measuring the wave length tests, you gave me a idea though, i want to see how far in and out the midro connects before it is consumed with the macro and micro i had a equation that was about right on point but i need the exact even though this will take sometime, i will get back to you, thanks.


Mathematically, the equation: E^3 = (mc^3)^3+(pc)^3 does not give the same answers as the equation E^2 = (mc^2)^2+(pc)^2.

If you try substituting values for 'm' & 'p' to calculate the 'E', in both cases, the difference in the answers becomes obvious ('c', of course, is a constant).

The cubed formula is not valid in respect to observed phenomena.


edit on 4/2/2016 by chr0naut because: (no reason given)



posted on Feb, 5 2016 @ 09:50 AM
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a reply to: chr0naut

Yea i agree, the more i study this equation the subtle of 3 in that math equation is not applicable to my findings, it would be more along the lines of E^1 and a half = (mc^1 and a half)^1 and a half+(pc)^1 and half...more correctly...

However that wold be semantics to me at this point and i am trying to apply it to other theories at first, i tried a test and i am waiting on results, i used a device called a radionic device which is based on Tesla scalar wave technology, to have two opposing waves collide and to get a desired effect, but i set the trend of the device to have both the macro (positive) up against my theory of the midro (miditive), to see if i get the out come i am looking for...

On the applicable design of the unit, it has 7 dials with numbers between 1 and 10, right after the number 4 and right before the number 7 is where i have the collapsed wave registered as the midro, i will let you know if i am successesful in my findings, thanks for the question...


edit on 5-2-2016 by King Seesar because: (no reason given)



posted on Feb, 5 2016 @ 09:52 AM
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originally posted by: yosako
Light Side of the Force, Dark Side of the Force, and the Brute Side of the Force.


Yes i would agree that this could be looked into as a Star Wars meme for sure, that movie does explain certain quantum in the universe, but i feel there may even be more then three, but right now i am studying this one.

Thanks for the reply....



posted on Feb, 5 2016 @ 08:47 PM
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a reply to: chr0naut

Hey, yes i used my experiment of a Tesla theory of a collapsed wave with the macro and miditive (my theory) i used numbers i played in a 3 digit number as the target of the test, i had the numbers 555 and 111, and in the day number it came out 550 and at night 116, i did not hit the number but my test's got me the first two numbers correct, which i feel proves my theory has merits, i will try this test again to see what other results i get with other applications....
edit on 5-2-2016 by King Seesar because: (no reason given)



posted on Feb, 14 2016 @ 11:44 PM
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So uhh, I'm pretty interested in following this thread , but can't really understand the equations, ( just a simple layman) maybe someone could point me to some reading material that might help.



posted on Feb, 19 2016 @ 09:31 PM
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a reply to: amsterdamn87

I would recommend any book on Quantum Psychics, also a good read but a little off topic is Radionics and Radiesthesia by Jane E Hartman , it is a book on radionics but it explains seven forces that i quantum as mini forces and not the scalar whole.



posted on Mar, 5 2016 @ 11:10 PM
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originally posted by: chr0naut

Yes but electromagnetic wavelengths are ginormous from a quantum scale. They are simply not reasonable as quantum sized objects.

E=mc^2, arguably, does apply at quantum scales (in many theories), but usually in its more complete form: E^2 = (mc^2)^2+(pc)^2.

There are other equations determining quantum values that must be factored in to the equation to 'flesh-out' its applicability.



This is all wrong.



posted on Mar, 6 2016 @ 01:01 PM
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originally posted by: joelr

originally posted by: chr0naut

Yes but electromagnetic wavelengths are ginormous from a quantum scale. They are simply not reasonable as quantum sized objects.

E=mc^2, arguably, does apply at quantum scales (in many theories), but usually in its more complete form: E^2 = (mc^2)^2+(pc)^2.

There are other equations determining quantum values that must be factored in to the equation to 'flesh-out' its applicability.



This is all wrong.



If it is all wrong, then explain why.

I was basing this off the radius of charge of a photon, roughly 0.5 x 10^-15 m (or half a Fermi). Above this size, the distinction of individual photons ceases to be evident and therefore quantum effects are less noticeable. This is the distinction of the quantum realm vs classical mechanics. The quantum effects are still operational but become smeared out into a measurement of the whole data set. The lower bound for a wavelength of electromagnetic radiation is therefore the diameter of the charge of a photon (a Fermi). Practically, however, the shortest wavelength of electromagnetic radiation that we can achieve is that of Gamma radiation, which is about 10^-11 m.

Since we are talking usually talking about photons when measuring quantum effects, it makes no sense to use E=Mc^2 because photons have no rest mass and the energy (from that equation) is therefore zero. But photons do have energy because they have momentum and if we use the full equation, we get sensible values.

I stand by what I said in my previous post.

edit on 6/3/2016 by chr0naut because: (no reason given)



posted on Mar, 7 2016 @ 09:44 PM
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originally posted by: chr0naut

If it is all wrong, then explain why.

I was basing this off the radius of charge of a photon, roughly 0.5 x 10^-15 m (or half a Fermi). Above this size, the distinction of individual photons ceases to be evident and therefore quantum effects are less noticeable. This is the distinction of the quantum realm vs classical mechanics. The quantum effects are still operational but become smeared out into a measurement of the whole data set. The lower bound for a wavelength of electromagnetic radiation is therefore the diameter of the charge of a photon (a Fermi). Practically, however, the shortest wavelength of electromagnetic radiation that we can achieve is that of Gamma radiation, which is about 10^-11 m.

Since we are talking usually talking about photons when measuring quantum effects, it makes no sense to use E=Mc^2 because photons have no rest mass and the energy (from that equation) is therefore zero. But photons do have energy because they have momentum and if we use the full equation, we get sensible values.

I stand by what I said in my previous post.



If you're talking about a photon then why not E = pc?



posted on Mar, 10 2016 @ 12:19 AM
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It's what makes quantum theory so interesting imo, there's still so much we can learn and how it effects our universe, i still am doing some tests with my theory on the 3rd force, i will get back to this thread when i have more in terms of what i am looking for...




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