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Re: "Unified Revolution" new book by Espen Haug

May 11th, 2021, 11:54 am

as u can hear from the record above, all quantum probabilities P_n and P_c between 0 and 1.

The Hubble constant is simply 1 divided by collision-time of the Hubble sphere. Okay re-deriving Freidemann eq. now, and will show what happen if include relativistic mass (that Albert abandoned).
 
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Re: "Unified Revolution" new book by Espen Haug

May 11th, 2021, 12:42 pm

as u can hear from the record above, all quantum probabilities P_n and P_c between 0 and 1.

The Hubble constant is simply 1 divided by collision-time of the Hubble sphere. Okay re-deriving Freidemann eq. now, and will show what happen if include relativistic mass (that Albert abandoned).
So, negative probabilities are ausgeschlossen?
"Compatibility means deliberately repeating other people's mistakes."
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http://www.datasim.nl
 
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Re: "Unified Revolution" new book by Espen Haug

May 11th, 2021, 1:05 pm

yes!

to write on litho-stone u have to write the mirror reflection, so I had a small mirror when writing.

The stone tablet broke in two on first print, can hardly be seen in the print, but yes that is what happen when the equations have power!

What will they do when they find the two pieces of the stone tablet in the future (or in the past), will they think it is some magic runes? Or will they understand these are the laws of the universe that not can be broken?
Screen Shot 2021-05-11 at 3.03.32 PM.png
(ps: the printing press used has been standing in Munchs work place)
 
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Re: "Unified Revolution" new book by Espen Haug

May 15th, 2021, 7:46 pm

The Friedmann equation for the critical universe can be derived also from GR escape velocity, this I have not seen before (but possibly out there), this give same as Friedmann. No big surprise as the escape velocity is also derived from GR. All we need to do is set the escape velocity to c and the radius to the Hubble radius (the Hubble constant is yes observable so we know these)
\begin{eqnarray}
v_e&=&\sqrt{\frac{2GM}{R}} \nonumber \\
c&=&\sqrt{\frac{2GM}{R_H}} \nonumber \\
c&=&\sqrt{\frac{2 G4\pi \frac{M}{\frac{4}{3}\pi R_H^3} R_H^2}{3}} \nonumber \\
c&=&\sqrt{\frac{8\pi G\rho R_H^2}{3}} \nonumber \\
c^2&=&\frac{8\pi G\rho R_H^2}{3} \nonumber \\
\frac{c^2}{R_H^2}&=&\frac{8\pi G\rho }{3} 
\end{eqnarray}
And since \(H_0=\frac{c}{R_H}\), we get

\begin{eqnarray}
H_0^2&=&\frac{8\pi G\rho }{3}  
\end{eqnarray}
Which is the well known Friedmann equation for a critical universe. But how can the GR escape velocity be IDENTICAL to that of Newton, a well known fact. The Newton escape velocity is definitely only a decent approximation for \(v<<c\). I suspect Albert ignored the relativistic mass that he derived incorrectly in 1905, so he abandoned it later on. Lorentz had it correct in 1904.

So I years ago derived a full relativistic escape velocity, Now I understood I can derive an alternative to the Friedmann critical universe from it. 

\begin{eqnarray}
v_e&=&\sqrt{\frac{2GM}{R}-\frac{G^2M^2}{c^2R^2}} \nonumber \\
c&=&\sqrt{\frac{2GM_u}{R_H}-\frac{G^2M_u^2}{c^2R_H^2}} \nonumber \\
c&=&\sqrt{\frac{8\pi G\rho R_H^2}{3}-\frac{16\pi^2G^2\rho R_H^6}{9c^2R_H^2}} \nonumber \\
c&=&\sqrt{\frac{8\pi G\frac{M}{\frac{4}{3}\pi R_H^3} R_H^2}{3}-\frac{\frac{16}{9}\pi^2 G^2\frac{M^2}{\frac{16}{9}\pi^2 R^6}R^6}{c^2R_h^2}} \nonumber \\
c&=&\sqrt{\frac{8\pi G\rho R_H^2}{3}-\frac{16\pi^2G^2\rho R_H^4}{9c^2}} \nonumber \\
c^2&=&\frac{8\pi G\rho R_H^2}{3}-\frac{16\pi^2G^2\rho R_H^4}{9c^2} \nonumber \\
c^2&=&\frac{8\pi G\rho R_H^2}{3}-\frac{16\pi^2G^2\rho R_H^4}{9c^2} \nonumber \\
\frac{c^2}{R_H^2}&=&\frac{8\pi G\rho }{3}-\frac{16\pi^2G^2\rho R_H^2}{9c^2} \nonumber \\
H_0^2&=&\frac{8\pi G\rho }{3}-\frac{16\pi^2G^2\rho^2 }{9H_0^2}  
\end{eqnarray}

This can be simplified further to

\begin{equation}
H_0^2=\frac{4\pi G\rho}{3} 
\end{equation}

The difference is just I get \(4\pi\) rather than \(8\pi \) as in GR and Friedmann. But this means a big deal, it means twice the mass density in the Hubble sphere, it means it takes the Hubble radius is identical to the information horizon, not that it is different as in Friedmann solution. Expansion of space is no longer needed it looks like, the Big Bang is Big Busted. 

Good luck with the Big Bang Relgion, it is likely nothing but a religion based on an incomplete equation!

GR also do not matches the different Planck units at the same time. For example use the GR escape velocity formula to calculate the escape velocity for the Planck mass with a radius equal to the Planck length, one will get a speed above c, one could try to fix this incomplete equation also here by hypothetically suggesting space expansion so above c was allowed. One did not do that, instead one reduced the Planck mass to get it to fit GR solution for micro black holes. Hawking said the micro black hole was approximately equal to the Planck mass, not exactly equal to the Planck mass, this he likely did on purpose as there is no way to get GR to match the Planck units for micro black hole idea completely, one need to alter c, or the Planck length or the Planck mass. The full relativistic escape velocity gives perfect match to all Planck units the same time, and it is nothing more than a collision between two indivisible particles.
 
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Re: "Unified Revolution" new book by Espen Haug

May 16th, 2021, 11:34 am

There is no more to say:
Screen Shot 2021-05-16 at 1.47.32 PM.png
 
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Re: "Unified Revolution" new book by Espen Haug

May 16th, 2021, 11:50 am

Interesting and maybe definitive (nothing more to say), but the annoying liberal arts eye sees two s in absurd and Freid in Friedman.

just tiny details though - Must leave the true elegance to the tailors, as Einstein said! : )
The spectacle is not a collection of images, but a social relation among people, mediated by images. - Guy Debord
 
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Re: "Unified Revolution" new book by Espen Haug

May 17th, 2021, 10:33 pm

So many kinds of black holes these days... Soon everything spherical in shape will be pronounced a black hole! BH?! The Norwegians knew it all along! :O
 
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Re: "Unified Revolution" new book by Espen Haug

May 21st, 2021, 12:29 am

Friedmann "standard" way, + proof we have steady state in full relativistic solution:


 \begin{equation}
U=T+V=\frac{1}{2}m\dot{R}^2-\frac{GMm}{R}=\frac{1}{2}m\dot{R}^2-\frac{4\pi}{3}G\rho R^2m
\end{equation}

 Assume \(\vec R(t)=a(t) \vec x\), and substitute this in the equation above and we get
 
  \begin{equation}
U=\frac{1}{2}m\dot{a}^2x^2-\frac{4\pi}{3}G\rho a^2x^2m
\end{equation}
we can re-arrange this and we get

  \begin{eqnarray}
\frac{\dot{a}^2}{a^2} &=& \frac{8\pi}{3}G\rho -\frac{kc^2}{a^2} \nonumber \\
H_0^2 &=& \frac{8\pi}{3}G\rho -\frac{kc^2}{a^2}
\end{eqnarray}
where \(kc^2=-\frac{U}{x^2m}\). The equation above is the Friedmann equation including the constant k. When setting k=0 we get the critical universe solution that we also got from the escape velocity formula. 

If we take into account relativistic energy as well as relativistic mass (Albert gave up on relativistic mass to force it into four-vector framework = Minkowski, my theory needs only three vector framework) we get

\begin{eqnarray}
U&=&T+V=m\dot{R}^2\gamma-m\dot{R}^2- \frac{GMm\gamma}{R}=m\dot{R}^2\gamma-m\dot{R}^2-\frac{4\pi}{3}G\rho R^2m\gamma \nonumber \\
U&=&m\gamma\dot{a}^2x^2-m\dot{a}^2x^2-\frac{4\pi}{3}G\rho a^2x^2m\gamma \nonumber \\
\gamma\frac{\dot{a}^2}{a^2}-\frac{\dot{a}^2}{a^2}&=&\frac{4\pi}{3}G\rho\gamma -\frac{kc^2}{a^2}
\end{eqnarray}
where \(\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\) and v is the velocity of m, and where \(kc^2=-\frac{U}{x^2m}\).  Divide by \(\gamma\) on both sides and we get


\begin{eqnarray}
\frac{\dot{a}^2}{a^2}-\frac{\dot{a}^2}{a^2}
\sqrt{1-\frac{v^2}{c^2}}&=&\frac{4\pi}{3}G\rho -\frac{kc^2}{a^2}\sqrt{1-\frac{v^2}{c^2}}
\end{eqnarray}
when \(v_e=v=c\), that is when we are at the Hubble radius, we get 


\begin{eqnarray}
\frac{\dot{a}^2}{a^2}&=&\frac{4\pi G\rho}{3}\nonumber \\
H_0^2&=&\frac{4\pi G\rho}{3}
\end{eqnarray}

This is the same solution we got from the escape velocity formula. That is unlike in the Friedemann solution where it in the formula is an open question what to set k to when solved this way. In our full relativistic framework it seems we have no choice other than that the constant k has no impact on our universe when our theory is linked to the Hubble scale. This proves the universe not can be expanding in a full relativistic Newton mechanics. 

The Big Bang theory is just a hypothesis that now has got a competitor! (The Big Bang Big Bust is on its way!)
 
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Re: "Unified Revolution" new book by Espen Haug

May 21st, 2021, 10:11 am

"Data from the Wilkinson Microwave Anisotropy Probe (measuring CMB anisotropies) combined with that from the Sloan Digital Sky Survey and observations of type-Ia supernovae constrain Ω_0 to be 1 within 1%" In other words it in reality support k=0, or that k falls out (same thing basically).

 Euclidean geometry holds, looks like no need for curvature, hyperbolic or spherical curvature of space of no need in the universe just described (created ;-).
 
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Re: "Unified Revolution" new book by Espen Haug

May 21st, 2021, 10:28 am

I’m a bit confused here, collector, what are you trying to do or understand?
Also guys, this is introductory physics and a bit of quantum right?
CBCA->CMSA->FMVA->Ba.Sc.Math-stat->CFA1,2,3->CISI->FRM->CQF->IFoA->than what?
The goal is to write all three and four letter combinations and try to make sense of it.
 
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Re: "Unified Revolution" new book by Espen Haug

May 21st, 2021, 11:23 am

The Friedmann derivation is indeed introductory physics today. I am happy to see if someone have derived it also with relativistic mass as I have done here, second derivation. Einstein, Wheeler, Adler etc. where or are all allergic to relativistic mass, this is well documented in literature. Lorentz got the relativistic mass correct already in 1904. Einstein could just have divided his relativistic energy by c^2, but instead derived two relativistic mass equations in end of his paper, these solutions are today known to be wrong, but hardly talked about as Einstein simply abandoned relativistic mass concept.

They can not get their theory to match all the Planck units same time (for what they interpret as micro black holes), and they get strange interpretations at Hubble scale...

My just updated draft here

The Collision Time of the Observable Universe is 13.8 Billion Years per Planck time: A New Understanding of the Cosmos based on Collision Space-Time

basic math, but a small difference in foundation can give massive differences in interpretations...

the paper still need some re-writing to clarify thoughts, but it is basically there already. My theory for example matches all  Planck units at the same time when linked to escape velocity c, unlike GR.

By re-introducing relativistic mass we also get perfect supernova fit without need for any dark energy, yes it is that simple. This naturally support my universe equation where no cosmological constants needed except the observed Hubble constant, which is nothing more than one divided by the collision-time inside a sphere with escape velocity c.

In one sweep (re-introducing Lorentz 1904 relativistic mass) one get rid of curvature of space, and no need for dark energy. Tested against data already!

No need for dark energy when re-intorducing relativistic mass

Einstein was right, his cosmological constant was a blunder, he should instead have picked up on Lorentz relativistic mass, and simply divided his relativistic energy by c^2 and said it is fully compatible. He followed on a different path...that has lead us to dark energy that never have been detected, but makes up most of today's hypotetical universe. I am not buying into it, even if it makes me unpopular.
 
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Re: "Unified Revolution" new book by Espen Haug

May 21st, 2021, 11:57 am

But how do you than propose to capture the bending of light by massive objects and electromagnetic phenomena caused by the charge of particles?
CBCA->CMSA->FMVA->Ba.Sc.Math-stat->CFA1,2,3->CISI->FRM->CQF->IFoA->than what?
The goal is to write all three and four letter combinations and try to make sense of it.
 
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Re: "Unified Revolution" new book by Espen Haug

May 21st, 2021, 12:56 pm

Inertia of light.
 
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Re: "Unified Revolution" new book by Espen Haug

May 21st, 2021, 1:14 pm

Friedmann "standard" way, + proof we have steady state in full relativistic solution:
 \begin{equation}
U=T+V=\frac{1}{2}m\dot{R}^2-\frac{GMm}{R}=\frac{1}{2}m\dot{R}^2-\frac{4\pi}{3}G\rho R^2m
\end{equation}
 Assume \(\vec R(t)=a(t) \vec x\), and substitute this in the equation above and we get
  \begin{equation}
U=\frac{1}{2}m\dot{a}^2x^2-\frac{4\pi}{3}G\rho a^2x^2m
\end{equation}
we can re-arrange this and we get
\begin{eqnarray}
\frac{\dot{a}^2}{a^2} &=& \frac{8\pi}{3}G\rho -\frac{kc^2}{a^2} \nonumber \\
H_0^2 &=& \frac{8\pi}{3}G\rho -\frac{kc^2}{a^2}
\end{eqnarray}
where \(kc^2=-\frac{U}{x^2m}\). The equation above is the Friedmann equation including the constant k. When setting k=0 we get the critical universe solution that we also got from the escape velocity formula. 
If we take into account relativistic energy as well as relativistic mass (Albert gave up on relativistic mass to force it into four-vector framework = Minkowski, my theory needs only three vector framework) we get
\begin{eqnarray}
U&=&T+V=m\dot{R}^2\gamma-m\dot{R}^2- \frac{GMm\gamma}{R}=m\dot{R}^2\gamma-m\dot{R}^2-\frac{4\pi}{3}G\rho R^2m\gamma \nonumber \\
U&=&m\gamma\dot{a}^2x^2-m\dot{a}^2x^2-\frac{4\pi}{3}G\rho a^2x^2m\gamma \nonumber \\
\gamma\frac{\dot{a}^2}{a^2}-\frac{\dot{a}^2}{a^2}&=&\frac{4\pi}{3}G\rho\gamma -\frac{kc^2}{a^2}
\end{eqnarray}
where \(\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\) and v is the velocity of m, and where \(kc^2=-\frac{U}{x^2m}\).  Divide by \(\gamma\) on both sides and we get
\begin{eqnarray}
\frac{\dot{a}^2}{a^2}-\frac{\dot{a}^2}{a^2}
\sqrt{1-\frac{v^2}{c^2}}&=&\frac{4\pi}{3}G\rho -\frac{kc^2}{a^2}\sqrt{1-\frac{v^2}{c^2}}
\end{eqnarray}
when \(v_e=v=c\), that is when we are at the Hubble radius, we get 
\begin{eqnarray}
\frac{\dot{a}^2}{a^2}&=&\frac{4\pi G\rho}{3}\nonumber \\
H_0^2&=&\frac{4\pi G\rho}{3}
\end{eqnarray}

This is the same solution we got from the escape velocity formula. That is unlike in the Friedemann solution where it in the formula is an open question what to set k to when solved this way. In our full relativistic framework it seems we have no choice other than that the constant k has no impact on our universe when our theory is linked to the Hubble scale. This proves the universe not can be expanding in a full relativistic Newton mechanics. 
The Big Bang theory is just a hypothesis that now has got a competitor! (The Big Bang Big Bust is on its way!)
I wish I were 20 years younger to study physics under your direction in your university in Norway. It certainly would have been a pleasure having you as a professor. I really would had loved it. Science is a pasion no doubt
Last edited by ExSan on May 21st, 2021, 1:27 pm, edited 1 time in total.
 
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Re: "Unified Revolution" new book by Espen Haug

May 21st, 2021, 1:18 pm

Katastrofa, is that his proposal or are you joking?
CBCA->CMSA->FMVA->Ba.Sc.Math-stat->CFA1,2,3->CISI->FRM->CQF->IFoA->than what?
The goal is to write all three and four letter combinations and try to make sense of it.