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originally posted by: Vector99
Does your midro theory interact on a constant basis?
Does it have certain defining points?
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.
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?
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.
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.
originally posted by: yosako
Light Side of the Force, Dark Side of the Force, and the Brute Side of the Force.
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.
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.
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.