# New Equivalence Principles?

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### #18 VictorMedvil

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Posted 13 July 2018 - 06:31 PM

Cool, the angular frequency of spinning black-hole from temperature

### #19 Super Polymath

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Posted 14 July 2018 - 02:27 AM

Matter inside a black hole can be charged, which gives a black hole its charge as I am only aware of stagnent black holes (non-rotating) as non-charged. This may translate to mean that black hole charge increases as mass increases. Detection of a black hole charge would be difficult it is assumed by scientists.

There may be a relationship between the angular momentum and the charge. We have already explained that there has been speculation that angular momentum changes with mass as well with only some of the largest black holes spinning near the speed of light.

The Luminosity is

$L = 4 \pi R^2\sigma T^4$

And the Bolometric flux density is

$f = \frac{L}{4 \pi R^2(1 + z)^2}$

Plugging the first equation into the last gives

$f = \frac{4 \pi R^2 \sigma T^4}{4 \pi R^2(1 + z)^2} = \frac{\sigma T^4}{(1 + z)^2}$

In which our denominator features a dimensionless redshift term. The power emitted is

$P = \frac{Aj}{(1 + z)^2} = \frac{A \epsilon_0 \sigma T^4}{(1 + z)^2}$

Solving for temperature yields

$T = ^4\sqrt{\frac{j(1 + z)^2}{\epsilon_0 \sigma}}$

Nice to just cover that. Anyway, I said earlier the transformation could be done on the Hawking black body energy in the Planck spectrum is

$E = k (\frac{T}{\gamma}) = k(\frac{T_0}{(1 - \frac{v^2}{c^2})}) = \frac{\hbar c^3}{8 \pi Gm(1 - \frac{v^2}{c^2})}$

In this case you may notice I was somewhat forced to consider the alternative form for the inverse since the last part of the equation only has constants in the numerator with the only variable, mass in the denominator.

That inverse barely-causality is the most convoluted model of the black "shell" in regards to the practicality of time travel, considering apart of the past of our very continuum keeps getting erased as time dilates, altering the present with a domino effect reverberating from a change in our past that's a paradox because it's the truer past even after replacing the previous one...

Retrocausality is more bizarre than alternate timelines, but as bizarre as the other side of the EH in this model as it's not just time that runs in reverse, but space & energy are inside out as well.

Edited by Super Polymath, 14 July 2018 - 02:33 AM.

### #20 Super Polymath

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Posted 14 July 2018 - 11:16 AM

The incredible

if an electron was a black hole,

Quintillions of them, within billions of locations within the electron creating a dynamic coordinate drag appearing as one negatively charged particle. It's not a singular coordinate drag until you get to the anti-proton, the strangelets/quark star (which are stuck in a non-bh positive charge), the planck particle, or a black hole stellar mass to supermassive & beyond.

That's the incredible thing you can do with this model.

### #21 VictorMedvil

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Posted 14 July 2018 - 06:22 PM

I dabbled around because I could see that I could form a Gaussian curvature definition but required a few steps, which led me formulate a Gauss-Bonnet type of description for the black hole.

The Gaussian curvature is

$8 \pi \rho (\frac{G}{c^2}) = 6(\frac{\hbar}{Mc})^{-2}$

and the equation in question is:

$Q^2 + \frac{c^2}{G} (\frac{J}{m})^2 \leq Gm^2$

We take the inverse

$\frac{1}{Q^2} + \frac{G}{c^2}(\frac{m}{J})^2 \leq \frac{1}{Gm^2}$

And distributed $8 \pi \rho$,

$\frac{8 \pi \rho}{Q^2} + 8 \pi \rho \frac{G}{c^2}(\frac{m}{J})^2 \leq \frac{8 \pi \rho}{Gm^2}$

Ignore inequality and flip equation then rearrange:

$\frac{8 \pi \rho}{Gm^2} - \frac{8 \pi \rho}{Q^2} = 8 \pi \rho \frac{G}{c^2}(\frac{m}{J})^2$

Write in the Gaussian curvature and multiply through by $(\frac{J}{m})^2$ which gives ~

$\frac{J^2}{m^2}(\frac{8 \pi \rho}{Gm^2} - \frac{8 \pi \rho}{Q^2}) = K$

and pull the constants out, gives,

$8 \pi \rho \frac{J^2}{m^2}(\frac{1}{Gm^2} - \frac{1}{Q^2}) = \frac{1}{k_1k_2} = K$

With $(k_1,k_2)$ as the principle curvatures each with inverse of length. For a black hole we would integrate with a boundary associated to the horizon yielding a Gauss-Bonnet equation:

$\int_H 8 \pi \rho \frac{J^2}{m^2}(\frac{1}{Gm^2} - \frac{1}{Q^2})\ dS = 2 \pi \chi (H)$

With $\chi(H)$ playing the role of the Euler characteristic of $H$.

Ya, all the constants looks correct along with the Einstein Metric brought across the equation over density, you have the Density per Electromagnetic fields and gravitational field which has the Einstein Metric curvature. After that you have J which I assume is the J from Maxwell's equations which is magnetic with electric then gravitation field density equals Gaussian curvature K. All of this equals a Special Unitary group. SO(8 )(J) (M,M') (q, B )(X,Y,Z)

Edited by VictorMedvil, 14 July 2018 - 06:29 PM.

### #22 VictorMedvil

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Posted 17 July 2018 - 04:28 AM

Yeah, a big stipulation was made by the poster ''the third law is not useful for quantum mechanics,'' which tells me this person really doesn't understand the implication of the law or its universal role in quantum mechanical zero point fields...

Regardless, I'll still be open that they may be right the Von Neumann entropy allows a non-zero entropy at [actual zero point temperatures] because a casual glance through the internet hasn't revealed for me yet any corroborating evidence.

It couldn't possibly have a zero temperature and still emit hawking radiation, there has to be some level of atomic or molecular movement to generate hawking radiation in the first place dubbel. Otherwise, the hawking radiation has no source to come from without it, itself moving away from the BH and would not be there without some amount of temperature driving it, some level of entropy as well this implies, even if it is just quantum movement and quantum vibration there is some temperature within the system absolute zero temperature is still impossible even for a BH, especially with all the gravitational potential energy crushing the Energy-mass together. The level of energy that a BH outputs is that of our sun in a year in an hour or more, there has to be especially in the photon sphere much Atomic,Quantum and Molecular Movement.

Edited by VictorMedvil, 17 July 2018 - 04:36 AM.

### #23 VictorMedvil

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Posted 17 July 2018 - 03:11 PM

Let the math checkers look over this while I go for an appointment, just some scribbles this morning:

To get the acceleration came from the modified equation to balance the units of charge satisfying a black hole with mass and angular momentum: We'll work in cgs units $(4 \pi \epsilon_0)^{-1} = 1$ and the power emitted like a charge is

$\frac{Q^2a^2}{ c^3} + \frac{c^2a^2}{Gc^3}(\frac{J}{m})^2 \leq \frac{a^2}{c^3}Gm^2$

$\frac{Q^2a^2}{ c^3} + \frac{a^2}{Gc}(\frac{J}{m})^2 \leq \frac{a^2}{c^3}Gm^2$

$\frac{Q^2a^2}{ c^3} + \frac{a^2}{Gc}(\frac{J}{m})^2 \leq \frac{a^2mr_g}{c}$

Notice this last term in the inequality is formally identical to the classical equation describing the radiation for a electron, with the exception that neither the mass or radius are associated to the electron but instead the black hole:

$P = \frac{2}{3}\frac{a^2m_er_e}{c}$

It is theoretically possible, to justify the gyromagnetic ratio since our system incorporates the angular momentum $J$. We will also change both acceleration terms on the left for their equivalent terms:

$\frac{Q^2}{m^2 c^3}(\frac{da}{dt})^2 + \frac{1}{Gm^2c}(\frac{J}{m} \frac{dp}{dt})^2 \leq \frac{a^2mr_g}{c}$

Which means there has to be the adjustable parameter of two since $\gamma = \frac{Q}{2m}$. If there is a usual analogue of $\frac{2}{3}$ for the charge we would get

$\gamma \frac{Q}{m c^3}(\frac{da}{dt})^2 + \frac{a^2}{Gc}(\frac{J}{m})^2 \leq \frac{a^2mr_g}{c}$

The energy of a system is related to the voltage and the charge via

$E = V Q$

In which case we may notice that rearranging we get

$\frac{mc^2}{Q} = V$

which is the inverse of one such term and we have a speed of light on the other terms so we can simplify:

$\frac{1}{2V}(\frac{da}{dt})^2 + \frac{1}{Gm^2}(\frac{J}{m})^2 \leq a^2mr_g$

This would be equivalent to have set the previous equation to units of $c = 1$. We do get an inverse gravitational charge squared through this derivation $Gm^2$. Let's remove acceleration description from the equation entirely now, you get:

$\frac{1}{2V}(\frac{da}{dt})^2 + \frac{1}{Gm^2}(\frac{J}{m})^2 \leq \frac{r_g}{m}(\frac{dp}{dt})^2$

Great thing is, we can simplify even further:

$\frac{r_g}{m}(\frac{dp}{dt})^2 \rightarrow \frac{r_g}{m}\frac{dmv}{dt}(\frac{dp}{dt}) \rightarrow r_g\frac{dv}{dt}(\frac{dp}{dt})$

$\frac{dv}{dt}$ is simply an acceleration term which has popped back up. The acceleration and gravitational radius may further be interpreted from the appropriate formula for acceleration and the radius through the normal equations of motion:

$v^2 = v^2_0 + 2a r_g$

flip the equation

$2ar_g + v^2_0 = v^2$

and rearrange

$ar_g = \frac{v^2 - v^2_0}{2}$

So we get to form something entirely different:

$\frac{1}{2V}(\frac{dp}{dt})^2 + \frac{1}{Gm^2}(\frac{J}{m} \frac{dp}{dt})^2 \leq \frac{v^2 - v^2_0}{2}(\frac{dp}{dt})$

In that form we have it as a function of velocity the velocity must be greater than the gravitational and electric attraction for it to escape that lends to the idea that hawking radiation is actually escaping from beyond the Schwarzchild Radius of the BH as some sort of particle, the density with velocity pushes the hawking radiation out of the BH, which says it is absolutely generated by gravity crushing the particles and pushing them back out along with charge with a temperature or kinetic energy that releasing the hawking radiation from the BH. There is an escape density and velocity for hawking radiation or the precursor particles.

Edited by VictorMedvil, 17 July 2018 - 03:19 PM.

### #24 VictorMedvil

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Posted 17 July 2018 - 03:25 PM

you have the wrong function for velocity - the velocity is not an escape velocity, it's the velocity difference of the black hole. Hence, acceleration may well be implied, but the voltage implies a difference in the electric charge. Which may in fact imply charge is invariant with motion for the black hole.

But I need to hold my horses, I am speculating before investigating.

I hope charge is invariant in a BH my Blackhole Weapon Dyson sphere depends on there being some level of charge generating a magnetic field throughout the BH, otherwise I have been lying to people on the internet.....

### #25 VictorMedvil

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Posted 17 July 2018 - 03:30 PM

Victor I have seen a big improvement, with the clarity of your posts and acceptance to keep within the boundary of what is acceptable. Keep this up.

It is because being around science people all the time has made be even better at this stuff, I was a bit rusty when we first met being around laymen that didn't understand what the hell i was talking about half the time. I had to explain it simply back then.

Edited by VictorMedvil, 17 July 2018 - 03:32 PM.

### #26 VictorMedvil

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Posted 17 July 2018 - 03:39 PM

Besides, don't let me down. I see the name victor, and I think of this badass, Viktor

Don't get too ahead of yourself, you have a long way to go and I have only known the best part of 2 years with you and it wasn't good before. But.... you have been more willing to ask sensible questions when you understand the physics. I don't know is this is about being around science people who explain it in a way that compels you to want to ask sensible questions, but this is still a remarkable improvement. I can only hope, Poly and Moronium follow example.

Lol, I have been around people like polymath and moronium my entire life they are just hopeless speculators. They always want to defy the norm and find problems when the current models to be like Einstein deep down, I on the other hand just want to make the UFT it has been my dream since I was 11 and watched Brain Greene's Movie, Medvil only asks one question, Does it work and how do I apply it to my models?

Edited by VictorMedvil, 17 July 2018 - 03:40 PM.

### #27 Shustaire

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Posted 17 July 2018 - 03:53 PM

Try living with a physicist for over 15 years lol

### #28 VictorMedvil

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Posted 17 July 2018 - 03:57 PM

Keep it for six months more and I promise you can write a paper with me. But.... this means, absorb what you can, learn and deviate away from what you learned. All novices have ... distorted if not plainly wrong views. The difference between the master and the student is that the master has lost far more times than the student has ever won. This holds true, its all about experience, so learn as much as you can,... from me, and a select few others here.

I look forward to writing a paper with you dubbel, we have had so much fun and discovered so much when we worked together sometimes. You helped so much on my Invariant Gravity, I feel that I owe you one, which stands as a E8 partially because of you and Mordred and even Polymath got an equation in.

Edited by VictorMedvil, 17 July 2018 - 03:58 PM.

### #29 Shustaire

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Posted 17 July 2018 - 03:59 PM

? How can you have invariant gravity when only thee geometry of the worldline between two events is invariant ? NVM best left for a different thread

Edited by Shustaire, 17 July 2018 - 04:00 PM.

### #30 VictorMedvil

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Posted 17 July 2018 - 04:00 PM

? How can you have invariant gravity when only thee geometry of the worldline between two events is invariant ? NVM bet left for a different thread

Making it of constants and planck units helps, it never changes because it is constructed of constants. It is true throughout the entire universe with one equation.

Edited by VictorMedvil, 17 July 2018 - 04:02 PM.

### #31 Shustaire

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Posted 17 July 2018 - 04:02 PM

I still fail to see how thaat makes sense if you treat gravity as the spacetime geometry ie the curvature itself. Its not a force but an effect of spacetime curvature in essence.

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### #32 VictorMedvil

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Posted 17 July 2018 - 04:03 PM

Abandon all that.  I know it is hard to do... but when your physics starts working, your model will appear largely nonsense, but you will find parts of it may hold. You'll see in time... but hold to me your promise of what I asked for six months, then how can I deny you this opportunity?

Probably that has happened to other models of mine

I still fail to see how thaat makes sense if you treat gravity as the spacetime geometry ie the curvature itself. Its not a force but an effect of spacetime curvature in essence.

It takes it as curvature from all the forces or fields of nature and puts them as causes of gravity or curvature in a sort of lattice structure.

Edited by VictorMedvil, 17 July 2018 - 04:04 PM.

### #33 VictorMedvil

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Posted 17 July 2018 - 04:10 PM

Indeed. In fact, people know I have strongly protested against field theorist idea's about gravity leading to gravitons. They have been part of Victors imagination for  a while, but old habits break hard if he wants to learn about the real facets of relativity and why they do not translate to mediator particles. In fact, by first principles, they never could. Which is a wonder how many scientists have been duped into this idea that psuedoforces suddenly require mediator particles.

Alright, the reason I think there is a graviton mediators is because of gluons, spacetime must have a particle like a gluon, if you want to break me of this explain how there is not a gluon for gravitational attraction.

Edited by VictorMedvil, 17 July 2018 - 04:10 PM.

### #34 Super Polymath

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Posted 17 July 2018 - 04:10 PM

I am speculating before investigating.

Investigate deeply enough & you'll find my model to be the winner.