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New Equivalence Principles?


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There was very little there that made sense... I think, if you want to talk in private that's fine. But as far as it goes here I won't say any more, because I really don't want to hurt your feelings. You seem younger than me, and hurting peoples feelings is never my intention.

I will need a math tutor whilst continuing my education. That's what I might contact you about.

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I do have to pickle you though... a black hole temperature, really is a true temperature emission. A black hole does not absorb all the light, remember that one, because it led Hawking to state, ''well actually, black holes are not really black''.... plus, there is no other temperature gradient other than through the Hawking process.

 

Just a nit pick.

 

You are correct, however the distinction lies, in a blackbody first absorbs the incident radiation. This causes the temperature to raise, thus increasing the temperature radiation. The thermodynamic state is determined by the absorption and emission properties. In the other case, example a closed system the temperature lies behind the barrier and is due to internal processes.  You can see how this will affect how one treats the thermodynamic equations in terms of closed vs open systems and the barrier. An idealized blackbody having 100% absorption.

Edited by Shustaire
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You are correct, however the distinction lies, in a blackbody first absorbs the incident radiation. This causes the temperature to raise, thus increasing the temperature radiation. The thermodynamic state is determined by the absorption and emission properties. In the other case, example a closed system the temperature lies behind the barrier and is due to internal processes.  You can see how this will affect how one treats the thermodynamic equations in terms of closed vs open systems and the barrier. An idealized blackbody having 100% absorption.

 

 

Certainly I mean... the black hole is certainly a 'near perfect black body.' No quarrels at all. 

 

A black hole does not however need to first asborb any incident radiation, to give off radiation. If a black hole is isolated from any source of matter of energy, it has a specific decay time. It doesn't actually need to absorb to give off radiation. If I understand the transformation correctly, the most agreed upon interpretation is that pair particles are created at the black hole event horizon. One flings off back into the black hole and the other drifts off into space and this is what led scientists to speculate on what happens to entanglement between the particles because of this.  

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Look again at the Unruh process they are very specific to the surface area why? Does not Hawking radiation prove that black-holes are not perfect blackbodies but are instead a grey body ? The definition of a true blackbody still applies the emitted radiation simply proves that the surface boundary is not a true blackbody as well as not being a true ground state. As all energy is observer dependent so is the associated vacuum states of a field. This is where most papers usually discuss killing vectors of a field lol but lets gloss over that for simplicities sake....

 

 With both Unruh and Hawking the effect it is describing how the surface boundary described under observer treatment and in Hawkings case the time reversal symmetries and the subsequent negative energy states of the Rindler-Fock vaccum. The internal workings are not even involved except to generate the mass term.

Edited by Shustaire
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Some thoughts I had t construct... haven't been to bed because of it yet  :)

 

Ok, the suggested equations i have given was:

 

[math]\Delta \mathbf{S} = 2T \sqrt{1 - \frac{v^2}{c^2}} \frac{1}{1 - \frac{v^2}{c^2}} = \frac{2T}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma^{-1}T[/math]

 

how do you transform between Ott's law and Einstein's? Well, you can see the above as already satisfying Einstein's version. In which case, to get Ott's law you just multiply through by [math]\gamma^{-2}[/math]

 

[math] \Delta \mathbf{S} =  2T \sqrt{1 - \frac{v^2}{c^2}} = \gamma^{-2} \frac{2T}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma T[/math]

 

............................................

 

From Lorentz contractions, the transverse and longitudinal direction in time is given as:
 
 
[math]t_l = \frac{2L \sqrt{1 - \frac{v^2}{c^2}}}{c} \frac{1}{1 - \frac{v^2}{c^2}} = \frac{2L}{c}\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = t_t[/math]
 
A relativistic form of energy has the appearance of 
 
[math]E = W = Fvt[/math]
 
From the work-energy theorem you can obtain the miiddle term which replaced energy for the definition of mechanical work. 
 

[math]Fvt_l = Fv \frac{2L \sqrt{1 - \frac{v^2}{c^2}}}{c} \frac{1}{1 - \frac{v^2}{c^2}} = Fv \frac{2L}{c}\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = Fv t_t[/math]

 

For mechanical work with a transverse time equal to a longitudinal time.

Edited by Dubbelosix
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That's NOT a "photon compactment", when I say photon compactment I mean a particle condensate 006 is already describing some of what I've fathomed. You'd mathematica to code it construct an evolving plot of illustrating the negative to positive photon charge bounce & Gareth has the calc. vocab to code it I don't know that Newtonian shambolic language of xyz equations

 

 

the dagger [math]A^\dagger[/math] is the adjoint operator.

 

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

 

see link for the bounded operators in context of the adjoint relations

 

 

 

 
Where [math]\mathbf{n}[/math] is the particle number density
 
[math]a|n>\ = \sqrt{n_k + 1}|n_k + 1>[/math]
 
Keep in mind also that 
 
[math][a,a^{\dagger}] = aa^{\dagger} - a^{\dagger}a = 1[/math]
 
[math]aa^{\dagger} = a^{\dagger}a + 1[/math] 

 

You're not fooling anyone Shoester the cosmos & ESP tachyon signaling involves Diagravitoelectromagnetism not Adiagravitoelectromagnetism - a matter dominated universe lies within a superverse positron orbiting around a superverse anti-proton. An antimatter dominated universe made of said aforementioned anti-hydrogen atoms lies within an electron orbiting a proton etc etc ad infinitum.

Edited by Super Polymath
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You can imagine any fallacies you wish Poly, the utter word salad above makes absolutely no sense please quit hijacking threads with your utter nonsense. Your personal beliefs in how the universe is just that unfounded personal belief with utterly no basis of proof.

 

 The equations you quoted of Dubbleosix are the creation/annihilation operators under QFT. This is part of mainstream physics despite your lack of faith in anything mainstream.

 

You wouldn't understand a path integral under Feymann to even begin to show you how those equations apply.

Edited by Shustaire
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You can imagine any fallacies you wish Poly, the utter word salad above makes absolutely no sense please quit hijacking threads with your utter nonsense. Your personal beliefs in how the universe is just that unfounded personal belief with utterly no basis of proof.

Well technically you cannot say a mathematical conjecture is either correct nor incorrect without a plot of said conjecture, that provides either validation or invalidation of said conjecture. Then, provided the former, you can move on to experimentation - which in this case would involve a tachyon processor.

Edited by Super Polymath
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Well technically you cannot say a mathematical conjecture is either correct nor incorrect without a plot of said conjecture, that provides either validation or invalidation of said conjecture. Then, provided the former, you can move on to experimentation - which in this case would involve a tachyon processor.

 

 

I don't need to, I can back any statement I make with professional peer review backing. That backing contains the experimental evidence, I simply need to understand the bulk of the mathematics to understand what those papers truly state without requiring the words written in those papers. The mathematics you shun are representations of experimental evidence, they describe the processes of measurements in a predictability and testable manner. If a theory cannot describe a process sufficient too make predictions of how and when said process occurs then it is useless.

 

Here is a little hint if you ever decide to truly understand physics, String theory, QFT, relativity all apply the same classical physics in so far as they must show a correspondence to everyday observable phenomena. Mathematically they have distinctions , however they have for more mathematical similarities than many realize.

 

Here is a really good trick, every major model uses scalar, vector, spinors. As well as tensors under gauge groups, study innr/outer exterior products and you will have the tools to understand any gauge group once you also learn matrix mathematics.

Edited by Shustaire
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She has a point once you learn to read math it is a simple process to know what a statement says unfortunately most people cannot read math. It is not something you are born knowing just like any verbal language. Though, it is always quite evident that Shustaire knows nothing about physics  She always understands the math but never the dynamics of what it describes, it is very noticeable that Shustaire was trained in pure math and not so much physics. She reminds me of a old friend of mine Scarlet, Scarlet was a master math person but did not understand the meaning behind the math when it was applied to physics, that is something that Shustaire needs to learn, just as there is a skill to read math, there is a skill to read physics.

 

There have been several times I have put forward widely used physics equations and shustaire has not understood them while when I use widely used math equation she always understands, physical systems do not always match the mathematical constructs that mathematicians have put forward over time this has created a gap between Math Math and Physics Math. She even does this to dubbel sometimes, I have watched this happen her not understand a physics equation and say that math math finds it unsuitable or not understood the material parts of the system.

 

That is why shustaire thinks you are crazy polymath, you on the other hand never use math math and always use purely physics math and physics talk, you have to remember to Math Math people this sounds like gibberish.  What do you you mean it has to be that way because that is all that experimentally is pointed to as the system's function ...... Nonsense. -Math Math person's thinking to a physics math person not so much.

 

For instance, I bet this looks like nonsense to shustaire but I am sure that dubbel understands it for a solution to the Generalized BB state as it is written in physics math and not math math.

 

Vsphere0rE= (4/3)πru3 0rENucleon Binding Energy + (4/3)πru3 0rELorrentz Force + (4/3)πru3 0rERest Mass +(4/3)πru3  0rERelativistic Newton gravity + (4/3)πru3 0rETemperature avg. + (4/3)πru3 0rErelativistic kinetic + (4/3)πru3 0rERelativistic momentum linear + (4/3)πru3 0rEDark Energy + (4/3)πru3 0rERelativisitic momentum Angular

 

On the other hand, I bet that both dubbel and shustaire get this one, as destructive interference equation being physics in Math Math.

 

Q1cos(ΦQωQ1t) + Q2cos(ΦQωQ2t 0

 

-V.M. and Scarlet

Edited by VictorMedvil
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Your definitely right I do not understand what you are trying to do with the first equation. You have too many terms that would be far simpler to describe under gauge symmetries that the RHS would be far simpler in the long run. If you did a parts by integration you would discover many of the RHS terms are reducible to a few simple relations under symmetry and that you have numerous terms being repeated in terms of energy of a system state.

 Don't make the mistake of trying to separate physics from math, math is the language physics uses, it follows every math rule there is no separate group physics math or math math. Physics applies math.

 

The physics method to describe everything you have in the first equation is to use the Langrangian for each degree of freedom in a state. However physics already does this under the SM qauge symmetries.

 

something along the lines (without filling out the actual langrangian.

 

[math]\mathcal{L}=\mathcal{L}_{relativity}+\mathcal{L}_{Higg's}+\mathcal{L}_{Yukawa}+\mathcal{L}_{Dirac}+\mathcal{L}_{ghosts}[/math]

 

All of the above already use Euclidean as the inner product spaces of each term applies under Euclid coordinate basis. Voila this is the action under the standard model which accounts for rest mass and energy already. Other than the renormalization problem for gravity describes a GUT.

 

Why does every term you have in the first equation describing a central potential system denoted with your gradient operators ? The first rule of cosmology is the cosmological principle (homogeneous and isotropic ) which is not a central potential.

 

For the record every equation or post I have ever applied is the mainstream physics under mathematical terminology. Physics terminology are all describable under mathematical basis. mainly (calculus, differential geometry and linear/nonlinear algebra) for the main math studies.

 

 You can also simplify your equation by recognizing that inner products apply to linear momentum while the outer products of a group describe the angular momentum terms. For example the Minkowskii tensor is an inner product space denoted

 

[math]\mu\cdot\nu=\nu\cdot\mu[/math]

 

LOL you guys should really read 100 Roads to reality by Sir Roger Penrose, it describes the bulk of physics in one book over a 1000 pages, but its all done including the mathematics. He is one of the leading mathematicians (though I laughed my head off when I read his Zing zang EM model, mathematically it is well described)

Edited by Shustaire
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