Singularity Mechanics? Blackhole Spin

[TWest]

Singularity Spin Mechanics

Okay I within the last week too a look at a article about black hole spin and in this article it was said by a scientist that black hole's spin near the speed of light that are supermassive blazars I was thinking if a object is spinning wouldn't that make this massive object subject to angular momentum laws and found I think a equation that accurately explains this phenomena if this is the case. what happens is simple The black hole spins and opens a hole in event horizon allowing matter to escape where the spin counteracts the gravitational pull which I call a "Void zone or spot where the jet escapes" which makes a coordinate system. dx'∇ = dx(1-(∇Δv^2/C^2))^(1/2) which shows external spin pressure is making the singularity expand in a direction. Δv^2 = (V_g-V_s)^2 , V_g^2 = 2M_bG / R_s, V_s^2 = E_s/M_b

Final Equation

∇'(x',y',z') = ∇(1-(((2M_bG / R_s) - (E_s/M_b))^2/C^2))^(1/2)
Blazar means the number is Less than C at a point or more, and Supermassive black hole means all equal to C.
Does this seem possible to you guys?

Tory Cortland Weston 6/29/2013

Proof

Coords = http://www.wolframalpha.com/input/?i=dx'∇+=+dx(1-(∇Δv^2/C^2))^(1/2)+

Part I = http://www.wolframalpha.com/input/?i=2*M*G+/+R

part II = http://www.wolframalpha.com/input/?i=E/M

Article = http://www.gizmag.com/supermassive-black-hole-spin-nustar-Newton/26543/

 

[Simon Bridge]

if a object is spinning wouldn't that make this massive object subject to angular momentum laws

Yep - see how angular momentum works in GR.

the spin happens in multiple directions

... the spin creates a coordinate system.

The black hole spins and opens a void zone in the black hole where the forces are balanced pulling all this matter to the middle then it escaping out of the hole in the event horizon.

AFAICT this sentence has no meaning.

I think you need to clean up the math.

 

[TWest]

Final Equation

∇'(x',y',z') = ∇(1-(((2M_bG / R_s) - (E_s/M_b))^2/C^2))^(1/2)


Not Quite it would be like this wouldn't it

∇'(x',y',z') = ∇(1-(((2M_bG / R_s) - (I_sw_s^2/2M_b))^2/C^2))^(1/2)

 

[Simon Bridge]

∇'(x',y',z') = ∇(1-(((2MbG / Rs) - (Isws^2/2Mb))^2/C^2))^(1/2)

Well... while we are at it - more like:

$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{I_s\omega_s^2}{2M_b} \right )^2 \right ]^{1/2}
$$

... ??

 

[TWest]

I guess?

 

[TWest]

∇'(x',y',z') = ∇(1-(((2MbG / Rs) - (Isws^2/2Mb))^2/C^2))^(1/2)

Well... while we are at it - more like:

$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{I_s\omega_s^2}{2M_b} \right )^2 \right ]^{1/2}
$$

... ??

$$
i\hbar \frac{\partial}{\partial t}\Psi(\mathbf{r},\,t)=-\frac{\hbar^2}{2m}\nabla^2\Psi(\mathbf{r},\,t) + V(\mathbf{r})\Psi(\mathbf{r},\,t).
$$
which makes Quantum Relativity for a black hole

$$
i\hbar \frac{\partial}{\partial t}\Psi(\mathbf{r},\,t')=-\frac{\hbar^2}{2m}\nabla'^2 (x',y',z')\Psi(\mathbf{r},\,t) + V(\mathbf{r})\Psi(\mathbf{r},\,t).
$$

?

 

[TWest]

$$
i\hbar \frac{\partial}{\partial t}\Psi(\mathbf{r},\,t)=-\frac{\hbar^2}{2m}\nabla^2\Psi(\mathbf{r},\,t) + V(\mathbf{r})\Psi(\mathbf{r},\,t).
$$
which makes Quantum Relativity for a black hole

$$
i\hbar \frac{\partial}{\partial t}\Psi(\mathbf{r},\,t')=-\frac{\hbar^2}{2m}\nabla'^2 (x',y',z')\Psi(\mathbf{r},\,t) + V(\mathbf{r})\Psi(\mathbf{r},\,t).
$$

?

Or more Correctly

$$
i\hbar \frac{\partial}{\partial t'}\Psi(\mathbf{r},\,t')=-\frac{\hbar^2}{2m}\nabla'^2 (x',y',z')\Psi(\mathbf{r},\,t) + V(\mathbf{r})\Psi(\mathbf{r},\,t).
$$

 

[TWest]

$$
i\hbar \frac{\partial}{\partial t}\Psi(\mathbf{r},\,t)=-\frac{\hbar^2}{2m}\nabla^2\Psi(\mathbf{r},\,t) + V(\mathbf{r})\Psi(\mathbf{r},\,t).
$$
which makes Quantum Relativity for a black hole

$$
i\hbar \frac{\partial}{\partial t}\Psi(\mathbf{r},\,t')=-\frac{\hbar^2}{2m}\nabla'^2 (x',y',z')\Psi(\mathbf{r},\,t) + V(\mathbf{r})\Psi(\mathbf{r},\,t).
$$



Or more maybe more correctly.

$$
i\hbar \frac{\partial}{\partial t'}\Psi(\mathbf{r},\,t)=-\frac{\hbar^2}{2m}\nabla'^2 (x',y',z')\Psi(\mathbf{r},\,t) + V(\mathbf{r})\Psi(\mathbf{r},\,t).
$$

 

[TWest]

Which makes time the spin of the universe and a black hole's Quantum tunneling events just like photons leaving a Quantum Field, time is non reversible, fractal until the moment happens, and there are others .

If I am thinking about this correctly.

 

[Simon Bridge]

I think that's unlikely - I don't see how you've done anything other than put stuff arbitrarily into formulas without reference to what they describe.

Have you done any GR at all?

 

[TWest]

Well, I am just thinking aloud, It could be could not I would have to solve it to find out if I was right but the visual translation looks promising.

 

[TWest]

Lets See first I subbed in the Schwarzschild radius Singularity Equation that explains the event horizon of a singularity then I took it against the angular momentum of the black hole pushing outward then tossed that into Lorentz Contraction and made it three dimensional with the Laplacian then subbed in the relativistic Laplacian for the standard in 3-D Schrödinger equation.

why wouldn't this work?

I think I have http://www.physics.fsu.edu/courses/spring98/ast3033/Relativity/GeneralRelativity.htm

 

[TWest]

$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{I_s\omega_s^2}{2M_b} \right )^2 \right ]^{1/2}
$$

Which Makes a mass field in a direction
$$
I_s = ∑i m_i r_i = m_1 r_1 + m_2 r_2 + ... + m_nr_n
$$

which is the expansion rate of time
$$
ω_s^2 = C^2
$$


$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_uG}{R_u} - \frac{(∑im_i r_i = m_1 r_1 + m_2 r_2 + ... + m_nr_n) C^2}{2M_u} \right )^2 \right ]^{1/2}
$$


or i have i just taken this way to far?

 

[TWest]

$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{I_s\omega_s^2}{2M_b} \right )^2 \right ]^{1/2}
$$

Which Makes a mass field in a direction
$$
I_s = ∑i m_i r_i = m_1 r_1 + m_2 r_2 + ... + m_nr_n
$$

which is the expansion rate of time
$$
ω_s^2 = C^2
$$


$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_uG}{R_u} - \frac{(∑im_i r_i = m_1 r_1 + m_2 r_2 + ... + m_nr_n) C^2}{2M_u} \right )^2 \right ]^{1/2}
$$


or i have i just taken this way to far?

$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_i r_i = m_1 r_1 + m_2 r_2 + ... + m_ur_s) C^2}{2M_b} \right )^2 \right ]^{1/2}
$$

making the limit of the series Schwarzschild Radius

 

[TWest]

$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_i r_i = m_1 r_1 + m_2 r_2 + ... + m_ur_s) C^2}{2M_b} \right )^2 \right ]^{1/2}
$$

making the limit of the series Schwarzschild Radius

Which makes them all $$M_G$$ or Mass Graviton I think

$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_G r_i = m_G r_1 + m_G r_2 + ... + m_Gr_s) C^2}{2M_b} \right )^2 \right ]^{1/2}
$$

 

[TWest]

Which makes them all $$M_G$$ or Mass Graviton I think

$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_G r_i = m_G r_1 + m_G r_2 + ... + m_Gr_s) C^2}{2M_b} \right )^2 \right ]^{1/2}
$$

Lets take a step back

$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_G r_i = m_G r_1 + m_G r_2 + ... + m_Gr_s) ω^2}{2M_b} \right )^2 \right ]^{1/2}
$$
$$
ω = √(k/m)
$$

Then

$$

\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_G r_i = m_G r_1 + m_G r_2 + ... + m_Gr_s) (k/m)}{2M_b} \right )^2 \right ]^{1/2}
$$

 

[TWest]

Lets take a step back

$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_G r_i = m_G r_1 + m_G r_2 + ... + m_Gr_s) ω^2}{2M_b} \right )^2 \right ]^{1/2}
$$
$$
ω = √(k/m)
$$

Then

$$

\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_G r_i = m_G r_1 + m_G r_2 + ... + m_Gr_s) (k/m)}{2M_b} \right )^2 \right ]^{1/2}
$$

Which shows the Construction of a black hole...Boson.

 

[TWest]

Lets take a step back

$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_G r_i = m_G r_1 + m_G r_2 + ... + m_Gr_s) ω^2}{2M_b} \right )^2 \right ]^{1/2}
$$
$$
ω = √(k/m)
$$

Then

$$

\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_G r_i = m_G r_1 + m_G r_2 + ... + m_Gr_s) (k/m)}{2M_b} \right )^2 \right ]^{1/2}
$$

which finishes to
$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_G r_s = m_G r_1 + m_G r_2 + ... + m_Gr_s) k}{2M_b^2} \right )^2 \right ]^{1/2}
$$

making
$$
k = 1/(1-ε^2)^{1/2}
$$

 

[TWest]

which finishes to
$$
\nabla'(x',y',z') = \nabla \left [ 1-\frac{1}{C^2}\left ( \frac{2M_bG}{R_s} - \frac{(∑im_G r_s = m_G r_1 + m_G r_2 + ... + m_Gr_s) k}{2M_b^2} \right )^2 \right ]^{1/2}
$$

making
$$
k = 1/(1-ε^2)^{1/2}
$$

$$
ε ^2 = Δt_u^2
$$

 

[TWest]

$$
ε ^2 = Δt_u^2
$$

$$
Δt_u ^2= 0
$$
maybe?

 

[TWest]

Code Numbers one thru four if right.