Return to the First Question: Information and Quantum Future
Once we have a preliminary answer to the question “How come quantum?”, we can return now to the first question: “How come existence?”. It may have something to do with “information”. But what is information? How is it related to physics? And is information sufficient? What about organization?
If energy and momentum have their own separate dimensions, should not information and organization have dimensions of their own as well? And if so, what would they be, “real” or “imaginary”? Physical or “non-physical”? These are vague questions; therefore I need to be more precise.
Physics, in spite of quantum mechanics and its excursions into subjectivism, essentially deals with a material world exclusively. Not only does it not show any interest in a non-material reality, but it also often ridicules such ideas, except, perhaps, when physicists engage in a dialogue with theologians, but even then it is at most a “philosophy” and not physics. But must it be so?
The Non-Material World
Can physics be extended in such a way that it can deal with non-material world and still be objective rather than falling into the subjectivity sinkhole? It appears that some people are doing it and being allowed to get away with it.
Military Interest?
One example is the duo of Elizabeth Rauscher and Richard Amoroso. According to Wikipedia:
Elizabeth A. Rauscher (1937–2019) was an American physicist and parapsychologist. She was born in Berkeley, California on March 18, 1937.[1] She died on July 3, 2019 (aged 82). (...) She was a former researcher with the Lawrence Berkeley National Laboratory, Lawrence Livermore National Laboratory, the Stanford Research Institute, and NASA
Richard Amoroso seems to be a Director of Noetic Advanced Studies Institute, has an education in psychology, and PhD in “Cosmology and Philosophy of Mind” – rather unusual domain, I’m sure you will notice. You may want to check their website to see what I mean.
Looking at one of their typical papers, with the title “The Physical Implications of Multidimensional Geometries and Measurement” I see that they are trying to include non-material reality by adding “imaginary dimensions” to those that are real. Tugging on this thread a little leads us to the fact that the US military seems to be interested in this kindof research by Prof. Rauscher:
Naval Surface Weapons Center, White Oak, 70-126, Silver Spring, MD 20910, 1982, R34-2- 0176 contract monitor E. Byrd, E.A. Rauscher, P.I. and W.L. Van Bise, P.I. Solving Maxwell’s equations in complex space relating to the space relating to the solutions to solitons and nonlinear Schrödinger equation and other nonlinear phenomena and the study of field phenomena.
What a surprise! Elizabeth Rauscher is also busy with “Modulating Brain Signals”:
E.A. Rauscher and W. Van Bise, Non-invasive Method and Apparatus for Modulating Brain Signals Through an External Magnetic or Electric Field to Reduce Pain, U.S. Patent Number 4,889,526, issued December 26, 1989.
But is it real, or is it a joke? I am inclined to think that it is, to a large extent, a joke, a red herring, so to say. Why do I think so? I am thinking that is a joke, because one of the works cited by Amoroso and Rauscher is “A Theory of Physical Vacuum” by G. I. Shipov, and this published monograph of Shipov is full of nonsense. I know it, because I checked it and also published a couple of papers discussing Shipov's errors. I also had a private exchange with Shipov, and I did not change my opinion after this dialogue.
Nevertheless, with her military connections and funding, we may take it as a given that Rauscher and her colleagues are not doing what they do just for fun. They are searching, but, I guess, they have a really hard time finding the answer or they are not publishing the papers that do lead to the answer. The clue about what they are after is given in the last line of their paper. They say that just doubling the dimension, that is adding four imaginary dimensions to those four real dimensions of space and time, may not be enough. They want to go into twelve dimensions, not just eight.
Which leads us to Burkhard Heim.
Volume 3, Structures of the physical world and its non-material side, was published in 1996 .
To be continued....
P.S.1. I pictured a rainbow, You held it in your hands, I had flashes, But you saw the plan. To my wife - The Whole of the Moon
P.S.2. New version of my notes on the conformal group. Will add still more today
Only those thoughts are true the opposite of which is alsotrue in its own time and application; indisputable dogmas arethe most dangerous kind of falsehoods.Sri Aurobindo
P.S.4 I know it's vanity, but it always feels nice. My old paper A. Jadczyk and K. Pilch, “Superspaces and Supersymmetries", Commun. Math. Phys. 78 (1981), 373–390, has been cited in a recent paper by R.R. Suszek (Warsaw), "Equivariant Cartan-Eilenberg supergerbes, the Kostelecký-Rabin defect and descent to the Rabin-Crane superorbifold"
Abstract: A concrete geometrisation scheme is proposed for the Green-Schwarz cocycles in the supersymmetric de Rham cohomology of the super-minkowskian spacetime, which determine standard super-σ-model dynamics of super-p-branes. The scheme yields higher-supergeometric structures with supersymmetry akin to those known from the un-graded setting - distinguished (Murray-type) p-gerbe objects in the category of Lie supergroups. These are shown to carry a canonical equivariant structure for the action of the discrete Kostelecký-Rabin subgroup of the target supersymmetry group on the target, and thus to resolve the `topology' of the corresponding Rabin-Crane soul superorbifold of the super-minkowskian spacetime. The equivariant structure is seen to effectively define a novel σ-model of super-p-brane dynamics in the super-orbifold through the construction of the corresponding (gauge-)symmetry defect.
It has quite impressive graphics. Like for instance this one:
P.S.5. Recommended reading: Lies, damn lies and limited hangouts
“Precisely. But forget the machine for now. Thinking about the machine has led me to think about natural laws and the laws of thought. Here is the point: we must find, from the outside, a way of describing the Aedificium as it is inside. . . .”
“But how?”
“We will use the mathematical sciences. Only in the mathematical sciences, as Averroës says, are things known to us identified with those known absolutely.”
“Then you do admit universal notions, you see.”
“Mathematical notions are propositions constructed by our intellect in such a way that they function always as truths, either because they are innate or because mathematics was invented before the other sciences. And the library was built by a human mind that thought in a mathematical fashion, because without mathematics you cannot build labyrinths. And therefore we must compare our mathematical propositions with the propositions of the builder, and from this comparison science can be produced, because it is a science of terms upon terms. And, in any case, stop dragging me into discussions of metaphysics. What the Devil has got into you today?”
P.S.7. 30-06-23 12:44 Another Umberto's Eco quote suggested to me by Laura. This time from Foucault's Pendulum:
"Amid all the nonsense there are some unimpeachable truths... I invite you to go and measure [an arbitrarily selected] kiosk. You will see that the length of the counter is one hundred and forty-nine centimeters—in other words, one hundred-billionth of the distance between the earth and the sun. The height at the rear, one hundred and seventy-six centimeters, divided by the width of the window, fifty-six centimeters, is 3.14. The height at the front is nineteen decimeters, equal, in other words, to the number of years of the Greek lunar cycle. The sum of the heights of the two front corners is one hundred and ninety times two plus one hundred and seventy-six times two, which equals seven hundred and thirty-two, the date of the victory at Poitiers. The thickness of the counter is 3.10 centimeters, and the width of the cornice of the window is 8.8 centimeters. Replacing the numbers before the decimals by the corresponding letters of the alphabet, we obtain C for ten and H for eight, or C10H8, which is the formula for naphthalene.
...With numbers you can do anything you like. Suppose I have the sacred number 9 and I want to get the number 1314, date of the execution of Jacques de Molay—a date dear to anyone who professes devotion to the Templar tradition of knighthood.
...Multiply nine by one hundred and forty-six, the fateful day of the destruction of Carthage. How did I arrive at this? I divided thirteen hundred and fourteen by two, by three, et cetera, until I found a satisfying date. I could also have divided thirteen hundred and fourteen by 6.28, the double of 3.14, and I would have got two hundred and nine. That is the year Attalus I, king of Pergamon, ascended the throne
You see? ...The universe is a great symphony of numerical correspondences... numbers and their symbolisms provide a path to special knowledge. But if the world, below and above, is a system of correspondences where tout se tient, it’s natural for the [lottery] kiosk and the pyramid, both works of man, to reproduce in their structure, unconsciously, the harmonies of the cosmos."
Eco, Umberto, Foucault’s Pendulum, (San Diego, New York, London: Harcourt Brace Jovanovich 1988) pp. 288-289.
P.S.8.
Revised proof from the previous note:
ReplyDeleteOur goal is to demonstrate that the action of the orthogonal group $O(4,2)$ on the set $\hat{M}$ is faithful. In other words, we need to show that if a group element $A$ in $O(4,2)$ acts as the identity on all points of $\hat{M}$, then $A$ must be the identity element of $O(4,2)$.
Let's assume that a matrix $A \in O(4,2)$ implements the identity transformation on $\hat{M}$. This means that for every $X \in N$ there exists $\lambda_X > 0$ such that
\[ AX = \lambda_X X \]
Now suppose we have a basis $\{v_1, ..., v_6\}$ for $R^{4,2}$, consisting of vectors from $N$. Using this basis, we can write any vector $X$ in $N$ as
\[X = c_1 v_1 + \dots + c_6 v_6 \]
where $c_i$ are the coefficients.
Now we apply $A$ to $X$:
\[ AX = A(c_1 v_1 + \dots + c_6 v_6) = c_1 Av_1 + \dots + c_6 Av_6 = c_1 \lambda_1 v_1 + \dots + c_6 \lambda_6 v_6 \]
However, from the condition $AX = \lambda_X X$, we get
\[ AX = \lambda_X (c_1 v_1 + \dots + c_6 v_6) = c_1 \lambda_X v_1 + \dots + c_6 \lambda_X v_6 \]
By comparing the two expressions for $AX$, we deduce that $\lambda_1 = \dots = \lambda_6 = \lambda_X$.
Given that the above must hold for all $X$, we have $\lambda_X = \lambda_Y$ for all $X, Y \in N$. Hence, we can just denote this common scalar as $\lambda$. Thus,
\[ AX = \lambda X \]
for all $X \in \hat{M}$.
As $A$ is in $O(4, 2)$, it preserves the metric tensor $G$, i.e., if $X^T G Y$ is the metric for any two vectors $X, Y \in R^{4,2}$, we have $AX^T G AY = X^T G Y$ for all $X, Y$. Taking $X = Y$, we obtain
\[ AX^T G AX = X^T G X \]
\[ (\lambda X)^T G (\lambda X) = X^T G X \]
\[ \lambda^2 X^T G X = X^T G X \]
Since $X^T G X \neq 0$ for vectors in $\hat{M}$, we get $\lambda^2 = 1$, thus $\lambda = 1$ or $\lambda = -1$.
But we've already seen from the problem's statement that $\lambda > 0$, so we must have $\lambda = 1$.
In conclusion, we have shown that if a matrix $A \in O(4, 2)$ implements the identity transformation on $\hat{M}$, then $A$ must be the identity matrix itself. This shows that the action of $O(4, 2)$ on $\hat{M}$ is faithful.
"By comparing the two expressions for $AX$, we deduce that $\lambda_1 = \dots = \lambda_6 = \lambda_X$."
DeleteWe deduce? How? What happened to c1,...,c6?
"We deduce? How? What happened to c1,...,c6?".
DeleteComparing the two expressions for $AX$, we have
\[ c_1 \lambda_1 v_1 + \dots + c_6 \lambda_6 v_6 = c_1 \lambda_X v_1 + \dots + c_6 \lambda_X v_6 \]
This implies that if $c_i \neq 0$ for some $i$, then $\lambda_i = \lambda_X$. As this must hold true for all $X \in N$, it leads to the conclusion that if $c_i \neq 0$ for any $i$, then all such $\lambda_i$ are equal to a common scalar, $\lambda$. Therefore, for each $i$ such that $c_i \neq 0$, we have
\[ A v_i = \lambda v_i \]
Since the $v_i$ form a basis for $N$, this implies that $A$ acts as a scaling by $\lambda$ on each $v_i$ for which $c_i \neq 0$, and hence on $N$. Thus, for all $X \in N$, we have $AX = \lambda X$.
Then, by choosing $X$ to be each of the basis vectors $v_1, \dots, v_6$ in turn, it follows that $\lambda_1 = \dots = \lambda_6 = \lambda$, and thus that $A$ is a scalar multiple of the identity.
Now it looks good.
DeleteA Reader asked me what I think of this paper:
ReplyDeletehttps://arxiv.org/abs/0808.1023
I looked at it and found the following word salad:
"Quantum measurements can be defined relative to these structures, as self-adjoint Eilenberg-Moore coalgebras for the comonads induced by the above comonoids"
Consider a bubble in a bubble chamber detecting a high energy electron. It is one of the most primitive quantum measurements. Where is the "self-adjoint Eilenberg-Moore coalgebras for the comonads induced by the above comonoids"???
The same Reader asked me if I think the world is mathematical?
ReplyDeleteNo. Certainly not. The world consists of all kinds of "stuff". Mathematics is just a formalized tool that sometimes, but not always, helps us to deal with "stuff".
My next blog will be about "What is real?".
In the pdf file added a question concerning the attempted solution of Exercise 2.
ReplyDelete"In the pdf file added a question concerning the attempted solution of Exercise 2.".
DeleteThe vectors ${v_1, v_2, \ldots, v_6}$ form a basis of $\mathbb{R}^{4,2}$, therefore, they are linearly independent. This implies that we can find a vector $X \in N$ such that the representation of $X$ in this basis will have all nonzero coefficients $c(X)_i$. This is because, for any basis, we can find a vector that has a nonzero component along each basis vector.
Thus, for this particular vector $X$, none of the $c(X)_i$ will be equal to zero. Consequently, the condition $c(X)_i\lambda_i = c(X)_i\lambda_X$ holds for all $i$, forcing all the $\lambda_i$ to be equal to a common scalar $\lambda$, as desired.
"This is because, for any basis, we can find a vector that has a nonzero component along each basis vector."
DeleteIncorrect reasoning. Omitted the essential here condition that this vector must be in N.
"Incorrect reasoning. Omitted the essential here condition that this vector must be in N.".
DeleteGiven that the vectors ${v_1, v_2, \ldots, v_6}$ form a basis for $\mathbb{R}^{4,2}$, the existence of such a vector in $N$ isn't guaranteed because $N$ is a proper subset of $\mathbb{R}^{4,2}$ due to the light cone constraint. This is indeed a crucial subtlety.
To be honest, I don't know how to solve this. Since $N$ is the null cone, any vector $X \in N$ satisfies the null condition, $X^T G X = 0$, where $G$ is the metric tensor. The null cone $N$ is a quadric surface in $\mathbb{R}^{4,2}$, and is spanned by vectors satisfying this condition.
Given that $N$ has a non-empty interior (since it's not merely a boundary), one might be able to argue that, because we can vary $X$ continuously within $N$, we should be able to find some $X \in N$ for which all $c(X)_i$ are non-zero, but I do not know how to make a proof of this statement...
"To be honest, I don't know how to solve this. "
DeleteBjab?
Bjab -> Ark
ReplyDeletePróbuję zrozumieć o co chodzi w tym całym Twoim artykule. Jest ciężko.
Pod koniec punktu 3.0 jest takie zdanie:
"It follows that the points of M ̃ which are not in the image
τ (M) of M"
Czy tam nie powinna być falka nad tau?
Nieco poniżej jest zdanie:
These images define the “conformal
infinity”.
Co za images?
OK. Will fix it tomorrow.
Delete@Bjab
DeleteFixed. Thanks.
Modified-hinted the Solution part under Exercise 2. Using your famous "separating" v-s could be handy.
Added subsection 7.4. Now working on the next subsection - the graphic representation.
ReplyDeleteBjab -> Ark
ReplyDeleteNa diagramach używasz różnego rodzaju strzałek. Wyjaśnij mi proszę znaczenie ich kształtów.
(Czy na diagramie w Proposition 1 nad tau nie powinno być falki?)
Arrow - general morphism (map)
DeleteArrow with double arrowhead - epimorphism ("onto" map)
Arrow with bent end - monomorphism (injection)
As for diagram and Proposition 1 - I have to think to make everything clear and correct. Thank you.
Bjab -> Ark
DeleteThanks.
@Bjab
DeleteCorrected the diagram of Proposition 1. Thanks.
Bjab -> Ark
DeleteJeśli zaś chodzi o diagram na stronie 3 to czy strzałka przy u nie powinna być onto a nie injection?
(APO przestały się wyświetlac numery stron.)
Fixed, updated.
DeleteBjab -> Ark
DeleteW zdaniu poniżej (66) wystąpiło wielkie X.
Czy wzór (67) jest taki jaki chcesz aby był (chodzi o te minusy przy x) ?
Fixed (66),(67) added Remark after (68). Thanks.
DeleteAdded P.S.6 and P.S.7.
ReplyDeleteExercise 3
ReplyDeleteYes, this intersection is empty.
The proof is based on an analogon to Lemma 2 proof.
This time for the set Z2 for which there is a slightly different requirement - > (x5 - x6) = -1.
The sets Z and Z2 are parallel hyperplanes.
My kind request: when making a comment, please choose for yourself a unique nickname (to distinguish from a generic "Anonymous").
DeleteAnd yes, the answer is essentially correct. Thank you.
Bjab -> Ark
DeleteSorry, I sometimes forget to sign.
Added Exercise 7.
ReplyDeleteBjab -> Ark
ReplyDeleteCzy mnie oczy nie mylą, że zmieniło się Exercise 3 ?
@Ark
ReplyDeleteTo nie do końca tak. Tam były dwa ćwiczenia jedno o numerze 3 bez rzutowania pi^ (którego dowód podałem) i drugie ćwiczenie bez numeru z rzutowaniem pi^.
Skasowałeś to ćwiczenie bez rzutowania pi^ i nadałeś numer 3 ćwiczeniu z rzutowaniem pi^.
Dowód obecnego ćwiczenia nr 3 (z rzutowaniem pi^) jest nieco trudniejszy.
Workplace accident. You are right. Hopefully corrected. Thanks.
Delete@Ark
DeleteTeraz pokręciła się numeracja punktów (7.3 i 7.4 się zdublowało)
@Ark
DeleteDowód obecnego Exercise 4 (tego trudniejszego):
Trzeba dodatkowo zauważyć, że zarówno τ+(M) jak i π^°τ+(M) znajdują się po tej samej stronie hiperpłaszczyzny x5=x6, a τ-(M) jak i π^°τ-(M) są po przeciwnej stronie. Więc te zbiory są również rozłączne.
"jak i π^°τ+(M) znajdują się po tej samej stronie hiperpłaszczyzny x5=x6,"
DeleteI am not very happy with this statement. π^°τ+(M) is not in R^6, it is in the quotient set, so to say that it is on some side of some hyperplane is, so to say, "iffy".
Fixed the double numeration. Thanks.
@Ark:
Delete"I am not very happy with this statement ... "
You are right. My mistake.
Added P.S.9.
ReplyDelete@Ark
ReplyDeleteW dwóch pierwszych zdaniach punktu 7.3 daszki nie trafiły nad M.
Fixed. Thanks.
Delete@Ark
DeleteW zdaniu poniżej wzoru (78) zamiast x = 0 powinno być X = 0
@Ark
DeleteW pierwszym zdaniu strony 18 brakuje plusa i znak "mniejsze" powinien być "mniejsze lub równe".
@Ark
Deleteatart ->
start
@Ark
DeleteIn that case 11
, x2
satisfy the equation ->
In that case x1
, x2
satisfy the equation
@Ark
DeleteThe variable The (x
1
, x2
) circles cor-
responding ->
?
Added P.S.10.
ReplyDeleteYou can plot the doubled infinity all by yourself. Just go to
ReplyDeletehttps://www.wolframalpha.com
and paste in the input window this line:
ParametricPlot3D[{(2 Abs[Cos[u]] Sin[v])/(2-Abs[Cos[u]] Cos[v]),(2 Sin[u])/(2-Abs[Cos[u]] Cos[v]),(2 Cos[u])/(2-Abs[Cos[u]] Cos[v])},{u,0,2Pi},{v,0,2Pi}]
It is not very pretty, but it works.
Added graphic representation of the conformal infinity.
ReplyDelete