Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.

Astro · 2010-06-04 18:49:28

Since the Observer and the Observed can be given a position in the wave function of spaces $|\psi>$ , then it has a translation operator which can be expressed in matrix form:

Iff $T(a)_i \in \alpha_i (\beta_i)$ relation according to $\alpha_i(\beta_i) \in R^4$ where the observer measures the time coordinates in linearity, then there is a special condition in the translation where we are given a specific direction; this could be seen arbitrarily as a manifold:

$T(a)_{j}=\begin{pmatrix} 1 & 0 & 0 & x \\ 0 & 1 & 0 & y \\ 0 & 0 & 1 & z \end{pmatrix}$

proportional to $\beta \in R^3$

Notice how it’s in three dimensions – this is because our observer is not acting on measuring the object, which is why we have reintroduced the subscript $j$ . As soon as we apply the ''reference principle'', we can have a four dimensional manifold, because the mind plays the role of the linearity of time:

$\box T(a)_{j}=\begin{pmatrix} 1 & 0 & 0 & x \\ 0 & 1 & 0 & y \\ 0 & 0 & 1 & z \\ 0 & 0 & 0 & t \end{pmatrix}$

proportional to $\alpha_i(\beta_i) \in R^4$

Last edited by Astro (2010-06-19 19:55:27)

Astro · 2010-06-08 16:18:06

I have decided to now introduce the reason for a fourth dimension of space. It's actually a requirement according the renown physicist and prof. Barrs http://www.telegraph.co.uk/science/larg … -time.html who has been working on relativity for many years now, that if you introduce a new dimension of time to something like a vacuum, there must be another space dimension to accomodate this dimensions existence. Since the observer is shown to be unique to have $\alpha_i(\beta_i) \in R^4$ this four dimensional holograph/representation of the outside world, then by some elegence we can refer to the relativistic accounts of adding a spatial dimension to the external world.

So a postulation will be forwarded suggeting that only when the reference principle is $\alpha_i(\beta_i)$ satisfied, can $\beta \in R^3$ now be seen as $\beta \in R^4$ - a static dimension of time which runs relative to the observers temporal subjective but yet quite real experience of time.

This static dimension existing could have massive implications when the role of the observer is involved; one i have just recently though of as ''what if this new additional static dimension is the representation of the de-Witt equation?''

The reason why i questioned it, was because they are pretty much analogous - an equation which describes a universe in terms of static dimensions - with no change in time.

Astro · 2010-06-08 17:17:37

So now instead of treating them as being analogous space and time dimensions between the observer matrix and the observed eigenstate matrix, we have the identification of $\alpha_i(\beta_i) \in R^8$ . Any static dimension of space can be convertable into a time dimension, which is why this is an indication the reference principle must be stating a trivial truth: That the static objective space dimension is the one main vector connecting to the time vector dimension in the observers reference point... the reason why is because the observer experiences their own reality of time alongside the universe external to us, but the static dimension is what allows the psychological arrow of time to exist, as i have shown in the time theory essay http://www.scribd.com/doc/21379144/Time-Theory [1]. The next part, will be trying to understand more about the eighth dimension, and see how we can envision our place in this strange quantum world, if one can be so bold.

[1] - I was quite ill in hospital when i wrote this, so please, have mercy on some of the terrible spelling. lol

Astro · 2010-06-08 19:06:33

Now this is where eight dimensional spacetime comes into the equations. First we need to understand more clearly what the above indicated.

For a general solution in relativity to even permit a static dimension, it must a non-vanishing global system, which actually involved a very complicated form of math called ‘’Killing-Vectors’’ even though mathematically, they are generally considered the easiest, but it’s all relative I say . Anyway, from these solutions of general relativity we talk about so freely right now, is mathematically-said to be a standard static of spacetime which is part of the Lorentzian Group Warped Product (1). It is also said that the dimension cannot be rotated, but this is because the energy perturbations do not change (again to remind the reader), this is due to a diffeomorphism Invariance problem which leads to a solution of a timelessness in our universe.

(1) – This is symbolized as $R X S$ .

In Einsteins equation, the part of which describes the Killing Vector is wrapped in expression: $\frac{8 \pi G}{c^4}$ is part of what is called a Limitation of a family on orthogonal states which determine the evolution of that state, so determines curvature in many senses.

$R_{\mu \nu} - \frac{1}{2}(g_{\mu \nu})R = (\frac{8 \pi G}{c^4})T_{\mu \nu}$

This is part of his famous dynamical equations. And its importance for the static model in which I
have presented for some time now, will be explained as soon as possible.

Last edited by Astro (2010-06-08 19:18:17)

Astro · 2010-06-08 20:41:11

It is also stated that it should have a finite potential range form [1]. There equation given satisfies the notions I’ve presented through a range of different paths.

$V(x)=-\frac{V_0}{a} (a-|x|)$

The idea that a spinless particle has an potential which has a finite boundary state, clearly shows why that when a spinless particle appears in spacetime, it is itself one of the most fundamental of the family of the standard model, because it has all the properties one would expect from a unified approach: The assumption is bold, but there are only two spinless types of particles in nature and they inflated to produce all the particles in the universe.

Please, entertain these ideas so far. Imagine an innate change in a particle, a form of particle seen as the ‘’most elementary’’ or the ‘’mother of elementary’’ particles and their arrival in spacetime caused the energy and time uncertainty to act on them – the first injection of energy which trivially pressed throughout the universe in a sludge of possibilities.

[1] - EJTP 6, No. 20 (2009) 399–404 Electronic Journal of Theoretical Physics
Eigenfunctions of Spinless Particles in a
One-dimensional Linear Potential Well
Nagalakshmi A. Rao1∗ and B. A. Kagali2�
1Department of Physics, Government Science College, Bangalore-560001,
Karnataka, India
2Department of Physics, Bangalore University, Bangalore-560056, India
Received 27 October 2008, Accepted 10 January 2009, Published 20 February 2009

Astro · 2010-06-08 20:52:41

$i\hbar \frac{\partial}{\partial t}\frac{1}{c^2} \sum^{|\psi|^2}_{j=0} \Lambda^j |\psi^j> - (-\nabla \phi)|\psi>= \mathbf{L}\psi>+( \omega^2 \gamma M_0 c^2( e \frac{i}{\hbar} \hat{H}t))|\psi>$

Is the equation describing a second type of spinless particle, and the equivalance of states can be formulated to be precisely equal to the Klein-Gorden Equations as given trivially by substitution,

$\frac{1}{c^2} \frac{\partial^2}{\partial t^2} \psi - (-\nabla^2 \psi)+(\frac{Mc}{h})^2 = (\omega \gamma M_0c^2(e \frac{i}{\hbar} \hat{H}t)|\psi>$

Astro · 2010-06-10 15:56:12

The post axioms above is in fact a wild speculation: The worst of its kind, because in a temperamental-sense, the equation

$\frac{1}{c^2} \frac{\partial^2}{\partial t^2} \psi - (-\nabla^2 \psi)+(\frac{Mc}{h})^2 = (\omega \gamma M_0c^2(e \frac{i}{\hbar} \hat{H}t)|\psi>$

is simply accepting that the original equation i derived at which can describe spinless particles

$i\hbar \frac{\partial}{\partial t}\frac{1}{c^2} \sum^{|\psi|^2}_{j=0} \Lambda^j |\psi^j> - (-\nabla \phi)|\psi>= \mathbf{L}\psi>+( \omega^2 \gamma M_0 c^2( e \frac{i}{\hbar} \hat{H}t))|\psi>$

Is where i've made a huge assumption. What if someone else comes along and posts a wave equation of a first order time derivative to explain such a particle? There are two things to consider here.

First, many equations in physics actually can be replaced, and even in their entirity, be equivalent; this used to make Schrodinger wonder, as it did with many physicists what the true nature of mathematics is, and whether it could be used sensibly to make a final theory.

Though i assumed immediately that my solution would be equivalent to the Klein-Gorden equation, but why can't my equation simply be a different way of looking at the Klein-Gorden Equation? This would make more sense no...? Afterall, his equation is universal.... it applies to all spinless particle fields, or atleast so we seem to believe...

Astro · 2010-06-10 18:24:54

It's just occurred to me that gamma can not just equal zero, but for speeds less that c then it must effect an inertial description from another equation given as expressed $I_M$ , which itself determines inertial bodies, then the equation $i\hbar \frac{\partial}{\partial t}\frac{1}{c^2} \sum^{|\psi|^2}_{j=0} \Lambda^j |\psi^j> - (-\nabla \phi)|\psi>= \mathbf{L}|\psi>+( \omega^2 \gamma M_0 c^2( e \frac{i}{\hbar} \hat{H}t))|\psi>$ is indeed important when $\gamma=0$ for the structure of the dimensionless observables in $( \omega^2 \gamma M_0 c^2( e \frac{i}{\hbar} \hat{H}t))|\psi>$ , however, if $\gamma= \frac{1}{2}(\frac{v}{c})^2$ then there is no need to longer express the equation $i\hbar \frac{\partial}{\partial t}\frac{1}{c^2} \sum^{|\psi|^2}_{j=0} \Lambda^j |\psi^j> - (-\nabla \phi)|\psi>= \mathbf{L}|\psi>+( \omega^2 \gamma M_0 c^2( e \frac{i}{\hbar} \hat{H}t))|\psi>$ but rather, one could write:

$i\hbar \frac{\partial}{\partial t} \gamma \frac{1}{c^2} \sum^{|\psi|^2}_{j=0} \Lambda^j |\psi^j>= \mathbf{L}|\psi>$

I see this equation as a much simpler derivative of the original equation, but does not show how important these components are $(\omega^2 \gamma M_0 c^2( e \frac{i}{\hbar} \hat{H}t))|\psi>$ .

Last edited by Astro (2010-06-10 18:36:10)

Astro · 2010-06-18 21:40:29

In some standard wave equations i wrote, i've deleted them, because i recognized a way to represent to gravitational charge as something not associated to a quantum flux - but using gamma to represent speeds less than c.

$(-\hbar \frac{\partial}{\partial t})^2 \frac{1}{c^2} \omega^2 \psi=\frac{1}{c^2} \omega^2 \hbar^2 (\frac{Mc^2}{\hbar})^2 \psi$

If we multiply $r^3$ on the right hand side of this equation, we find the relation of $M \frac{G}{c^2}= \mu$ :

$(-\hbar \partial_t)^2 \frac{1}{c^2} \omega^2 r^3 \psi=\frac{1}{c^2} \mu \hbar^2 (\frac{Mc^2}{\hbar})^2 \psi$

To find the gravitational charge, we need to consider what $\frac{1}{c^2}$ tends to mean - in this wave equation, it would seem to dictate that the energies are of speeds which are equal to $c$ , the speed of light. Whilst it is dimensionless, we could remain ignorant and just remove, but a better approach is to give it a coefficient given as $\gamma = \frac{1}{2}(\frac{v^2}{c^2})$ to imply that these speeds must be less than that of light. To find out how the energy is related to the gravitational charge $\sqrt{M \frac{G}{c^2}}= \mu_g$ we take the square root of our wave equation and apply the coefficient:

$\sqrt{(-\hbar \partial_t)^2 \gamma \frac{1}{c^2} \omega^2 r^3} \psi= \gamma \frac{1}{c^2} \mu_g \hbar (\frac{Mc^2}{\hbar}) \psi = \gamma(- \hbar \partial_t) \mu_g \psi$

Using partial differential notation.

Last edited by Astro (2010-06-18 22:48:28)

Astro · 2010-06-18 22:02:21

$\gamma \frac{1}{c^2} \mu_g \hbar (\frac{Mc^2}{\hbar})d \psi = \gamma(- \hbar \partial_t) \mu_g \frac{1}{2}at^2\psi$ .1

By using a substitutional method involving the inertial mass equation i derived, which took the identity of:

$I_M d= \mu_g \frac{1}{2}at^2$ .2

means that if eq. 1 and 2. are sound enough, then they explicitely define inertial mass, so the substitution is given as:

$(\gamma \frac{1}{c^2} \mu_g \hbar (\frac{Mc^2}{\hbar})(\frac{1}{2}at^2)) \psi = \gamma(- \hbar \partial_t) I_Md \psi$

The left looks like a frightfull mess, but it has defined the right handside of the equation in a more simplistic approach. This is for an equation where its distance must be involved.

Last edited by Astro (2010-06-18 22:54:53)

Astro · 2010-06-24 15:01:58

Some misunderstandings.

After some critizing from a Ph.D in physics, i have decided for the sake of clarity to address his issues, even if i never make him aware of it - so that future readers will not fall into my terrible way to rush through explaining things, thus resulting in not explaining it all that well at all.

His incongruity resided in the formation of the logic which was presented, and yes, upon first glance it can materialize as something mistaken. His remarks where,

''If $\alpha$ where some operator on $\beta$ such that it satisfies $\alpha(\beta) = \beta$ then $\alpha$ belongs to the space of endomorphisms, defined on the space $\beta$ resides in...

he furthers

''But then you go on to state that $\beta \in R_3$ would be wrong,''

and this is because he see's the two as the same postulations as proposed from the $R^4$ model and the absence of $\alpha$ which is not but a mathematical operator, it is defined in this case specifically to some intelligent recording device.

There is however a unique difference between the two, and that is of course, the addaptation of the schrodinger equation to diffeomorphism invariants of the Wheeler - de Witt equation. If the Ph.D was so advantageous, he might have come to realize that when i stated that the observer/operator acted on $\beta$ , it was unique because we have a sense of time, whilst the geometry of General Relativity flies in the face of such an eigenstate problem concerning $\alpha(\beta) = \beta$ ~ simply because if time itself is governed by how systems change in a spacetime metric, then $\beta \in R_3$ state of space indicates how the diffeomorphism invariants cast problems on how one can even define time without the presence of the true given formula $\alpha(\beta)=\beta$ ... There are two paths. They may seem mathematically a paradox, but these mathematical assertions have been based on Fotini Markopoulou's remarks on the frozen time model, or the ''problem of time in relativity'' as being a statement about our roles in the universe.

This means sufficiently, that $\beta \in R_3$ for the Wheeler-de Witt equation, which is given as $\hat{H}|\psi>=0$ is true for a generalistic approach in relativity - but as stated, $\alpha(\beta)=\beta$ is itself stating a specialized relativistic solution to our roles inside the universe. It's not that my work was wrong, maybe badly rushed, but i have tried to explain this - more than once in the entirity of the OP.

I hope this clears up any incongruities.

Last edited by Astro (2010-07-05 19:29:44)

Astro · 2010-06-24 15:40:17

This is why the reference principle of the observed and the observer had to be implied for these two remarkable contradictions in special and general relativity. Using the operator/observer in a way as to represent our roles in defining change, would give us the reason why our roles are important when understanding the frozen model of time.

Taking the mathematical description of the reference principle in a relativistic approach (which is defined commutatively as $\alpha_i(\beta_i)$ , takes the dynamical changes for time in the albalian group, we find that $\alpha(\beta) = \beta$ is quite literally a statement which makes the variable $\beta$ as a time-dependant eigenstate problem.

The time chronological approach is a solution where the first observer defines the solutions and the entanglement between the other observer in The Mind of Wigners Friend, is one based on the ignorance that such a superpositioning could happen with a different result. Before either observer 1. $\alpha_1$ or observer 2. $\alpha_2$ interacts with the object $\beta$ , both observers do share a state of entanglement on the probability of finding a spin up or spin down, in this case; has a superpositioning equation described as:

$\zeta_{\delta_{ij}} (\alpha_j \beta_{j}_{(1,2)}) |\psi> = |\psi \alpha_1 \beta_j... ...\psi \alpha_2 \beta_j> + |\psi \alpha_j \beta_1 \psi... ... \alpha _j \beta_2 \psi>$

Which was what led me to the principle of reference.

Last edited by Astro (2010-07-05 19:34:03)

Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.

#1 2010-06-04 18:49:28

The Matrix of the Observer and the Observed (excerpt)

#2 2010-06-08 16:18:06

Re: The Matrix of the Observer and the Observed (excerpt)

#3 2010-06-08 17:17:37

Re: The Matrix of the Observer and the Observed (excerpt)

#4 2010-06-08 19:06:33

Re: The Matrix of the Observer and the Observed (excerpt)

#5 2010-06-08 20:41:11

Re: The Matrix of the Observer and the Observed (excerpt)

#6 2010-06-08 20:52:41

Re: The Matrix of the Observer and the Observed (excerpt)

#7 2010-06-10 15:56:12

Re: The Matrix of the Observer and the Observed (excerpt)

#8 2010-06-10 18:24:54

Re: The Matrix of the Observer and the Observed (excerpt)

#9 2010-06-18 21:40:29

Re: The Matrix of the Observer and the Observed (excerpt)

#10 2010-06-18 22:02:21

Re: The Matrix of the Observer and the Observed (excerpt)

#11 2010-06-24 15:01:58

Re: The Matrix of the Observer and the Observed (excerpt)

#12 2010-06-24 15:40:17

Re: The Matrix of the Observer and the Observed (excerpt)

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