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Collector
Posts: 4722
Joined: August 21st, 2001, 12:37 pm

### Re: "Unified Revolution" new book by Espen Haug

"How do you measure the Earth radius at the experimental location?"

Jump in the ship (if not already there) and travel east until I am outside the same island. Then I know the circumference of the Earth. (my equator air-ship)...or if lazy, study the shadows!

"How to Determine the Earth’s Circumference from a Shadow for Dummies"

Another ancient method known to few: travel from where you find polar bears to where you find Penguins in a straightest possible line, and the distance is 1/2 the circumference of the Earth. Before the Earth got messed up with Zoos this method was quite fool proof!

Or use a Celtic Cross!

series of typos that I am working on fixing now before it goes in PRINT!

katastrofa
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Location: Alpha Centauri

### Re: "Unified Revolution" new book by Espen Haug

I was referring to your owns enlightened words:
"R is the radius of the gravity object, that is, from the center of the gravity object to where the measurement is performed. If it is done on the surface of the Earth, this is simply the Earth’s radius."

The Earth radius differs from place to place. You need to know it with high precision for your experiments, I guess - ?

Collector
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Joined: August 21st, 2001, 12:37 pm

### Re: "Unified Revolution" new book by Espen Haug

not sure how much the radius vary,  less than half percent it seems? more than good enough for this kind of measurements... and if we know where we are on earth we can make much better than half a percent. This is not about finding a more precise value of l_p than today, but mostly about constant reduction. Even if the error in l_p (percent wise) is exactly half of what it is in G (if from same measurement), this is not important here.

G and h can be replaced with l_p and also one can unify and one understand what mass is.

Standard physics uses two different mass definitions (indirectly) without even knowing about it

katastrofa
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Location: Alpha Centauri

### Re: "Unified Revolution" new book by Espen Haug

Actually you can perform the measurement of the local radius quite easily near Oslo. You'll need a torch producing a narrow beam (something small or with diodes instead of a bulb), a tripod, an A4 sheet of paper, running shoes and a local map. Go at night to the lake. Place the torch in the tripod at the shore in the direction of the furthest opposite shore. Run to the opposite shore. When you are there, roll the paper and look through it straight in the torch light at the opposite shore. Start squatting down slowly until the light disappears from your sight line (you'll probably need to lie down on the ground). If you still see it, run back to the tripod, lower the torch in it, and run to the opposite shore again. Repeat until you finally find the level at which you no longer see the light, h1. Measure this height and measure the height of the torch on the tripod, h2. Ops, you didn't do that? then run again to the torch. You left the map at the opposite shore? Run back for it. Check on the map the distance d between the two points. The radius of Earth at the torch location is $d^2/(2\sqrt{h_1} + 2\sqrt{h_2})^2$. (Don't forget to run back for the torch and tripod.)
If it doesn't work, you'll at least have a nice jog.

Collector
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Joined: August 21st, 2001, 12:37 pm

### Re: "Unified Revolution" new book by Espen Haug

a fast growing percentage of the world population will not agree on that method

Cuchulainn
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### Re: "Unified Revolution" new book by Espen Haug

a fast growing percentage of the world population will not agree on that method

Of course it's flat. The Netherlands is flat and hence by extrapolation, the Earth.
And 90% of Republican believe the Earth is 6,000 years old. So wrong .. it's 8.000 years old, duh.
Step over the gap, not into it. Watch the space between platform and train.
http://www.datasimfinancial.com
http://www.datasim.nl

Collector
Posts: 4722
Joined: August 21st, 2001, 12:37 pm

### Re: "Unified Revolution" new book by Espen Haug

a fast growing percentage of the world population will not agree on that method

Of course it's flat. The Netherlands is flat and hence by extrapolation, the Earth.
And 90% of Republican believe the Earth is 6,000 years old. So wrong .. it's 8.000 years old, duh.
yes the Republicans of Netherland is in a class of their own! 90% think flat? well that is what happens when u not maintain your mountains!

Collector
Posts: 4722
Joined: August 21st, 2001, 12:37 pm

### Re: "Unified Revolution" new book by Espen Haug

Back in 2018 (dec 28 vixra) I put out my first attempt to make a new relativistic wave equation rooted in what I call the (kinetic) Compton momentum, rather than the standard (kinetic) momentum, that from a quantum perspective is rooted in the de Broglie wave (which is a mathematical derivative of the true matter wave the Compton wave.). My first wave equation contained i, and as I am allergic to imaginary numbers I quickly abandoned it (the day after) as I could see also one plausible solution based on total Compton momentum (instead of kinetic Compton momentum), that got rid of i in final result (canceled out). It was likely a mistake to abandon it, so now I am back to it, and this possibly is more promising than my second wave equation that I have posted here before.
Comments welcome, as paper in progress

Deeper insight on Existing and New Wave Equations in Quantum Mechanics

$i\frac{\partial\psi}{\partial t}=\left(-ic\nabla+\frac{mc^2}{\hbar}\right)\psi$
that also can be re-written as (because $\bar{\lambda}=\frac{\hbar}{mc\gamma}$)
$i\frac{\partial \psi }{\partial t} = \left( -ic\nabla +\frac{c}{\bar{\lambda}}\right)\psi$

(please do not check against standard relativistic energy momentum relation, as that is the relativistic energy de Broglie momentum relation, and here one need the relativistic energy Compton momentum relation that is the foundation of these new wave equations. They are related, but not the same! )

relativistic, first order in time and spatial dimension, fully rooted in Compton, so valid also for v=0, not linked to a mathematical wave that converge to infinite as v approaches 0, rock solid fundament (in my 2018 version, never posted here, I also had a sign error due to error in sign of energy operator). I suspect standard  QM wave equations give predictions very hard to understand as they not have dive deep enough into the rabit hole. That said I still do not understand my own equation, but then they have used decades on understanding established QM...

Collector
Posts: 4722
Joined: August 21st, 2001, 12:37 pm

### Re: "Unified Revolution" new book by Espen Haug

I continue my moon-log

Cool, it looks like a wave equation that satisfies the relativistic energy Compton momentum relation will, in addition automatically satisfy the standard relativistic energy momentum relation (de Broglie, not strange as de Broglie is derivative of Compton).  The relativistic energy momentum relation and the relativistic energy Compton momentum relation are effectively two sides of the same coin, where the the relativistic energy Compton momentum relation is much closer to what is going on at the deepest level of reality, while the standard energy momentum relation is an over complicated way to model quantum mechanics through the de Broglie momentum, p, and therefore indirectly the de Broglie wavelength.

It is useful for the following derivation to use the notation

$p=mv\gamma$, $p_0=mv$, and $p_t=mc\gamma$ (Compton momentum) and  $p_k=mc\gamma-mc$ (kinetic Compton momentum)  so we have

(with disclaimer as just did this calculation, trivial, but tired after boxing half day)

\begin{eqnarray}
E^2&=&p^2c^2+m^2c^4\nonumber \\
E^2&=&p_0^2c^2\gamma^2+m^2c^4\nonumber \\
E^2&=&m^2v^2c^2\gamma^2+m^2c^4\nonumber \\
E^2&=&m^2c^4v^2/c^2\gamma^2+m^2c^4\nonumber \\
E^2&=&m^2c^4\left(\frac{v^2}{c^2}-1\right)\gamma^2+m^2c^4\gamma^2+m^2c^4\nonumber \\
E^2&=&-m^2c^4+m^2c^4\gamma^2+m^2c^4 \nonumber \\
E^2&=&m^2c^4\gamma^2 \nonumber \\
E&=&mc^2\gamma \nonumber \\
E&=&p_tc=p_kc+mc^2 \nonumber \\
\end{eqnarray}
setting up a wave equation from the first line require operator to work on standard momentum (indirectly de Broglie), while setting up wave equation based on last line require operator on Compton momentum, indirectly Compton wave, last line make all easier as nothing squared, and is closer linked to depth of reality.

todays physics is top down, and rooted in old definitions that have gone out on date

by the way no one have ever observed standard momentum. Assume you have an apple, you can observe its relative mass, by comparing its weight to one kg mass. You can measure its velocity, but you can never observe an apple*v =mv, not directly not from impact, only by mathematical manipulation of observed entities.

what you can observe is its impact energy, which is a function of v^2 (when v<<c) and not of v. The Compton momentum indeed is a function of v^2 when (v<<c)
Last edited by Collector on August 4th, 2020, 4:38 pm, edited 4 times in total.

Cuchulainn
Topic Author
Posts: 62887
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: "Unified Revolution" new book by Espen Haug

Back in 2018 (dec 28 vixra) I put out my first attempt to make a new relativistic wave equation rooted in what I call the (kinetic) Compton momentum, rather than the standard (kinetic) momentum, that from a quantum perspective is rooted in the de Broglie wave (which is a mathematical derivative of the true matter wave the Compton wave.). My first wave equation contained i, and as I am allergic to imaginary numbers I quickly abandoned it (the day after) as I could see also one plausible solution based on total Compton momentum (instead of kinetic Compton momentum), that got rid of i in final result (canceled out). It was likely a mistake to abandon it, so now I am back to it, and this possibly is more promising than my second wave equation that I have posted here before.

Screen Shot 2020-08-03 at 4.06.25 PM.png

Comments welcome, as paper in progress

Deeper insight on Existing and New Wave Equations in Quantum Mechanics

$i\frac{\partial\psi}{\partial t}=\left(-ic\nabla+\frac{mc^2}{\hbar}\right)\psi$
that also can be re-written as (because $\bar{\lambda}=\frac{\hbar}{mc\gamma}$)
$i\frac{\partial \psi }{\partial t} = \left( -ic\nabla +\frac{c}{\bar{\lambda}}\right)\psi$

(please do not check against standard relativistic energy momentum relation, as that is the relativistic energy de Broglie momentum relation, and here one need the relativistic energy Compton momentum relation that is the foundation of these new wave equations. They are related, but not the same! )

relativistic, first order in time and spatial dimension, fully rooted in Compton, so valid also for v=0, not linked to a mathematical wave that converge to infinite as v approaches 0, rock solid fundament (in my 2018 version, never posted here, I also had a sign error due to error in sign of energy operator). I suspect standard  QM wave equations give predictions very hard to understand as they not have dive deep enough into the rabit hole. That said I still do not understand my own equation, but then they have used decades on understanding established QM...
Is this 'new' equation rigorous? What is the motivation? Where does it come from?
In its current form , it has a \nabla making it hyperbolic..which makes the whole story into a completely new ball game IMO.

Ladder operator? (just a guess, it's been 50 years since I did QM))
https://eng.libretexts.org/Bookshelves/ ... _Operators
Step over the gap, not into it. Watch the space between platform and train.
http://www.datasimfinancial.com
http://www.datasim.nl

Collector
Posts: 4722
Joined: August 21st, 2001, 12:37 pm

### Re: "Unified Revolution" new book by Espen Haug

"Is this 'new' equation rigorous? "  as far as a farm boy can see yes!

well so yes I assume the Compton wave is the real matter wave and not de Broglie wave, the standard momentum is linked to de Broglie, so I define a new momentum based on Compton wave, it is all rigoroouuusis, and if wave equation satisfy my new relativistic energy Compton momentum relation it automatically satisfy standard energy momentum relation...

compared to Schrodinger I am getting time and spatial dimension on same order, first order derivatives (schrodinger time first, spatial second order). Klein-Gordon is all second-order...and Dirac (all first order) is very complicated for a farmer.

I need to find out what spin my equations are valid for

katastrofa
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Location: Alpha Centauri

### Re: "Unified Revolution" new book by Espen Haug

I think you ate $\gamma^2$ when you put $\frac{v^2}{c^2} \approx0$:
\begin{eqnarray}
E^2&=&m^2c^4\left(\frac{v^2}{c^2}-1\right)\gamma^2+m^2c^4\gamma^2+m^2c^4\nonumber \\
E^2&=&-m^2c^4\gamma^2 + m^2c^4\gamma^2+m^2c^4 \nonumber \\\\
E^2&=&m^2c^4 \nonumber \\
E&=&mc^2\nonumber \\
\end{eqnarray}

Sorry, but you're just Einstein.

bearish
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Joined: February 3rd, 2011, 2:19 pm

### Re: "Unified Revolution" new book by Espen Haug

You’re damning him with faint praise!

Collector
Posts: 4722
Joined: August 21st, 2001, 12:37 pm

### Re: "Unified Revolution" new book by Espen Haug

I think you ate $\gamma^2$ when you put $\frac{v^2}{c^2} \approx0$:
\begin{eqnarray}
E^2&=&m^2c^4\left(\frac{v^2}{c^2}-1\right)\gamma^2+m^2c^4\gamma^2+m^2c^4\nonumber \\
E^2&=&-m^2c^4\gamma^2 + m^2c^4\gamma^2+m^2c^4 \nonumber \\\\
E^2&=&m^2c^4 \nonumber \\
E&=&mc^2\nonumber \\
\end{eqnarray}

Sorry, but you're just Einstein.
Missus Higgs, thanks for the comment, but why would I put $\frac{v^2}{c^2} \approx0$ when working on getting a deeper foundations for relativistic QM, I am not Schr\{"o}dinger that is happy with just approximations that only holds for v<<c, my standards are higher, or more precisely faster:

\begin{eqnarray}
E^2&=&m^2c^4\left(\frac{v^2}{c^2}-1\right)\gamma^2+m^2c^4\gamma^2+m^2c^4\nonumber \\
E^2&=&\frac{m^2c^4\left(\frac{v^2}{c^2}-1\right)}{1-\frac{v^2}{c^2}}+m^2c^4\gamma^2+m^2c^4\nonumber \\
E^2&=&\frac{-1\times m^2c^4\left(\frac{v^2}{c^2}-1\right)}{-1\times \left(1-\frac{v^2}{c^2}\right)}+m^2c^4\gamma^2+m^2c^4\nonumber \\
E^2&=&-m^2c^4+m^2c^4\gamma^2+m^2c^4\nonumber \\
E^2&=&m^2c^4\gamma^2\nonumber \\
E&=&mc^2\gamma\nonumber \\
E&=&p_tc=p_kc+mc^2
\end{eqnarray}

But it is correct, that if you set $\frac{v^2}{c^2} \approx0$ you eat the gamma also here, but I am not interested, try Schr\{"o}dinger, I heard rumor his focus is on non-relativistic QM (after giving up on relativistic). So he would love to eat $\gamma$

katastrofa
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### Re: "Unified Revolution" new book by Espen Haug

That's too fast for me, unless I'm driving a manual: