This is a continuation of the previous post: Problems in Objectizing.
Wheeler's Second Question: "How come the quantum?”
Let me consider the second question, which seems to me to be less abstract. Indeed, we do have “the quantum” It is embodied in the Planck constant
It has the physical dimension of a “Joule” multiplied by a “second”, energy multiplied by time – physicists call it “action”. This constant seems to be numerically rather small in our accepted human-size units, but it enters into almost every formula of quantum theory.
Planck’s constant, by itself, indeed defines a “quantum”, but not a quantum of energy, as it sometimes written in popular books, but a quantum of action. Energy is not always quantized, there is no such thing as “quantum of energy”, except in some particular cases1, but there is a “quantum of action”. Therefore we should pose another question: “How come action?” Evidently it is a very important quantity, but not directly observed or measured. Why?
Energy and Momentum: Action
The physical dimension of action is that of energy multiplied by time, which happens to be the same as momentum multiplied by length. We all have a good general feeling or impression about what the time distance between, any two events is, and what the length distance between two points is. Energy and momentum seem to be more complicated and not so evident, and yet they happen to be fundamental in mechanics. Do they not deserve to have their own dimensions associated with them? Indeed they do, and they have.
Phase-space and Planck’s Constant
In mechanics, for quite some time now, we have been using not just space and time, and not just space-time, but also what is called “phase-space”. If space-time has four dimensions, three for space and one for time, then phase-space has these dimensions doubled. There are four dimensions added: three for momentum, and one for energy. Thus in relativistic mechanics, phase-space has eight dimensions.2
Things are somewhat more complicated when we are dealing with a system of particles rather than just one particle – we need a separate phase-space for every particle, but that can be thought of as a purely formal mathematical trick – we do it, because it makes calculation simple, and because “it works”.
Now, once we have phase-space, eight or six dimensional, depending on whether we respect the special theory of relativity or not, we can measure the volume of a region in this space. Then the Planck constant tells us that there is something like a “fundamental volume” – this phase-space has a kind of granular structure. That is to say, there is a discreteness at the fundamental level, though it has nothing to do with space or with time alone, but rather with a more complicated space that we are not immediately aware of – the phase-space. Energy and momentum, the additional coordinates of the phase-space have, intuitively, to do with frequencies and wave-lengths, not just time and position.
Are we somehow sensitive to frequencies and wavelengths, not just to positioning of objects in space and time? I think that in order to answer these questions the theory should be developed a little bit further. So, let me continue with my speculations about a possible shape of future physics.
To be continued...
1 An ideal harmonic oscillation, with a fixed frequency. is one such example.
2 U.S. Army seems to be interested in extending Einstein’s general theory of relativity to phase space: Howard E. Brandt, U.S. Army Research Laboratory, “Finslerian Quantum Field Theory”,
P.S.1. To my wife. She likes it.
P.S.2.
"If mankind could but see though in a glimpse of fleeting experience what infinite enjoyments, what perfect forces, what luminous reaches of spontaneous knowledge, what wide calms of our being lie waiting for us in the tracts which our animal evolution has not yet conquered, they would leave all & never rest till they had gained these treasures. But the way is narrow, the doors are hard to force, and fear, distrust & scepticism are there, sentinels of Nature, to forbid the turning away of our feet from her ordinary pastures."
Sri Aurobindo
P.S.3. The video below was taken 10 days ago in our yard.
P.S.4. I wish I would have such muscles:
I present a revised proof of faithfulness. Please feel free to comment.
ReplyDeleteWe aim to show that the action of the orthogonal group $O(4,2)$ on the compactified Minkowski space $\hat{M}$ is faithful. In other words, we need to demonstrate 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 $O(4,2)$.
Firstly, let's consider a matrix $A \in O(4,2)$. By definition, $A$ is orthogonal (in the sense of preserving the quadratic form), which implies that $A^T G A = G$, where $G$ is the metric tensor of the pseudo-Euclidean space with signature $(4,2)$. Note that both $A$ and $G$ are 6x6 matrices.
Now, let's suppose that $A$ acts as the identity on all points of $\hat{M}$, i.e., for every equivalence class $[X] \in \hat{M}$, we have $A \cdot [X] = [X]$.
But the definition of the action of $O(4,2)$ on $\hat{M}$ is given by $A \cdot [X] = [AX]$, where $[AX]$ denotes the equivalence class of the point $AX$. So, if $A \cdot [X] = [X]$ for every equivalence class $[X]$, this implies that for each equivalence class $[X]$, $AX$ and $X$ belong to the same equivalence class.
In other words, for each $X \in N$, $AX = \lambda X$ for some non-zero scalar $\lambda$.
Since this holds true for all $X$, it must be especially true for the basis vectors. Hence, we have $A e_i = \lambda e_i$ for all $i = 1, ..., 6$, where ${e_i}$ is the standard basis of $\mathbb{R}^{4,2}$. This precisely means that $A$ is a dilation matrix in $O(4,2)$.
However, we note that $A$ is an orthogonal matrix, implying that $A^T G A = G$. For a dilation matrix, the left-hand side of this equation yields a diagonal matrix with the values $\lambda^2$ on the diagonal, while the right-hand side remains constant and equal to $G$. The only possible solution is therefore $\lambda = 1$, which leads us to conclude that $A$ must be the identity matrix.
Hence, the action of $O(4,2)$ on $\hat{M}$ is faithful.
Catch number 1: logical inconsistency in reasoning
Delete"for each $X \in N$"
in N!
"Since this holds true for all $X$,"
All X? All?
""for each $X \in N$"
Deletein N!".
Since this holds true for all $X$, it must be especially true for the basis vectors. -> Since this holds true for all $X$ in $N$, it must be especially true for the basis vectors.
Is that what you had in mind?
"Since this holds true for all $X$"
DeleteThis we don't know. It is a logically unjustified reasoning.
""Since this holds true for all $X$"
DeleteThis we don't know. It is a logically unjustified reasoning.".
Ok. It's not given that this holds true for all $X \in N$, but we have shown that it holds true for all $X \in N$ which belong to the same equivalence class in $\hat{M}$ as their images under the action of $A$.
We can't claim directly that for every $X \in N$, $AX = \lambda X$ for some non-zero scalar $\lambda$. This is only proven to be true for those $X$ that, through the action of $A$, remain in their initial equivalence class in $\hat{M}$.
So the correct rephrasing would be:
"In other words, for each $X \in N$ such that $A \cdot [X] = [X]$ in $\hat{M}$, we have $AX = \lambda X$ for some non-zero scalar $\lambda$."
This way, we correctly emphasize that the property only holds for those $X \in N$ that satisfy the specified condition in $\hat{M}$.
Are you satisfied now?
No.
DeleteThere is also catch number 2.
Correct logical reasoning would be:
for each $X \in N$ such that $A \cdot [X] = [X]$ in $\hat{M}$, we have $AX = \lambda_X X$ for some non-zero scalar $\lambda_X$.
That is: there is no logical reason, at this point of deductive thinking, to assume that lambda is the same for every X in N.
@M.S. A general direction of the reasoning is good. But there are these two catches that require additional attention and additional work.
DeleteTrue. I should say: in fundamental rules of quantization.
ReplyDeleteAll other formulas containing hbar are derived from these ruls.
@Bjab Also hbar Pauli appear as a multiplication factor in spin matrices.
ReplyDeleteI will fix that remark. Thank you!
I corrected the entire proof. Is there anything else wrong now?
ReplyDeleteWe aim to demonstrate that the action of the orthogonal group $O(4,2)$ on the compactified Minkowski space $\hat{M}$ 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 consider a matrix $A \in O(4,2)$. By definition, $A$ is orthogonal (in the sense of preserving the quadratic form), which implies that $A^T G A = G$, where $G$ is the metric tensor of the pseudo-Euclidean space with signature $(4,2)$. Note that both $A$ and $G$ are 6x6 matrices.
We suppose that $A$ acts as the identity on all points in $\hat{M}$, i.e., for every equivalence class $[X] \in \hat{M}$, we have $A \cdot [X] = [X]$.
The definition of the action of $O(4,2)$ on $\hat{M}$ is given by $A \cdot [X] = [AX]$, where $[AX]$ denotes the equivalence class of the point $AX$. So, if $A \cdot [X] = [X]$ for every equivalence class $[X]$, this implies that for each equivalence class $[X]$, $AX$ and $X$ belong to the same equivalence class.
In other words, for each $X \in N$ such that $A \cdot [X] = [X]$ in $\hat{M}$, we have $AX = \lambda_X X$ for some non-zero scalar $\lambda_X$.
Since this holds true for all $X \in N$ such that $A \cdot [X] = [X]$ in $\hat{M}$, it must be especially true for the basis vectors. Hence, we have $A e_i = \lambda_i e_i$ for all $i = 1, ..., 6$, where ${e_i}$ is the standard basis of $\mathbb{R}^{4,2}$, and $\lambda_i$ are scalars associated with these basis vectors.
However, we note that $A$ is an orthogonal matrix, implying that $A^T G A = G$. If $A$ were a scaling matrix, the left-hand side of this equation would yield a diagonal matrix with the values $\lambda_i^2$ on the diagonal. Yet, the right-hand side remains constant and equal to $G$. As the elements on the diagonal of $G$ are not all the same, it is clear that $\lambda_i = 1$ for all $i$, otherwise $A^T G A \neq G$.
Hence, we conclude that $A$ must be the identity matrix, and thus the action of $O(4,2)$ on $\hat{M}$ is faithful.
"it must be especially true for the basis vectors."
DeleteThis is still catch number 1 unresolved.
N is a SUBSET of the six- dimensional vector space. How do we know that it contains six linearly independent vectors?
"N is a SUBSET of the six- dimensional vector space. How do we know that it contains six linearly independent vectors?".
DeleteWe don't know that. I assumed we did. In that case, what can I do? I guess I am left with proof by contradiction.
"I guess I am left with proof by contradiction.".
DeleteSuppose $A$ is not the identity matrix, i.e., there exists a vector $v \in \mathbb{R}^{4,2}$ such that $Av \neq v$. But since $A$ acts as the identity on $\hat{M}$, this means that $v$ and $Av$ belong to the same equivalence class. Hence, $Av = \lambda_v v$ for some non-zero scalar $\lambda_v$.
However, since $A$ is orthogonal, we have $Av \cdot Av = v \cdot v$ (where $\cdot$ denotes the dot product). This implies that $\lambda_v^2 v \cdot v = v \cdot v$. Since $v \cdot v \neq 0$, it must be the case that $\lambda_v = \pm 1$.
Now, since $A$ is in $O(4,2)$, and by our assumption $Av \neq v$, the only possibility left is $\lambda_v = -1$, i.e., $Av = -v$. This would imply $A$ acts as the negative identity on $\mathbb{R}^{4,2}$.
But if $A$ acted as the negative identity on $\mathbb{R}^{4,2}$, it wouldn't act as the identity on $\hat{M}$, because it would send every equivalence class $[X]$ to $[-X]$, and $[X] \neq [-X]$ for $X \in N$. This is a contradiction, so our assumption that $A$ is not the identity is false. Therefore, $A$ must be the identity matrix in order for it to act as the identity on $\hat{M}$.
We know exactly what is N. Therefore we can find out whether we can find six linearly independent vectors in N or not.
Delete"Therefore we can find out whether we can find six linearly independent vectors in N or not.".
DeleteI can see it now. Formula (2)... Sorry, I didn't look at it earlier.
"Suppose $A$ is not the identity matrix, i.e., there exists a vector $v \in \mathbb{R}^{4,2}$ such that $Av \neq v$. But since $A$ acts as the identity on $\hat{M}$, this means that $v$ and $Av$ belong to the same equivalence class."
DeleteThis is an incorrect reasoning. W have no guarantee that our vector v is in N (which is used in the definition of hat(M)).
In the new proof I also replaced the term "dilation matrix" with "scalling matrix" because the term "scaling matrix" is broader, as it encompasses both uniform (dilation) and non-uniform transformations.
ReplyDeleteAnd I don't know for the moment whether there will be a system of six linearly independent vectors in N, and I don't know how to know that.
ReplyDeleteBjab - Mathilde S.
DeleteI'm not fully compiling your proofs because I'm only on page 10. What comes to my mind is that N is five-dimensional so it cannot have 6 independent vectors, and that since M is hatted there should be double square brackets and a lambda greater than zero.
Bjab -> Ark
ReplyDeleteCzytam punkt 6.4
" ... is singular on the light cone at the origin {x ∈ M : q(x) = 0}" ->
at the origin?
Wzór (45) - brak nawiasu w linijce: 1,3,4 tego wzoru.
Fixed, updated. Thank you.
DeleteCone at a is described by the equation q(x-a)=0.
Bjab -> Ark
DeleteJakoś skojarzyło mi się, że at the origin odnosi się do singular at the origin.
Czy "null cone" to "light cone" at 0 ?
Null cone is for a general quadratic form Q in any number of diemnsions. Light cone is only for q in 3+1 dimensions.
DeleteBjab -> Ark
ReplyDeleteDrugi wiersz od dołu strony 11 jest jakiś symbol dwukropka ze strzałką.
Fixed. Thank you.
DeleteBjab -> Ark
ReplyDeleteWzór (51)
Mnie wychodzi:
(x + q(x) a) / (1 + 2 a · x + q(x) a · a )
Indeed. Then, setting a=-b, it agrees with
Deletehttps://en.wikipedia.org/wiki/Special_conformal_transformation
Correcting. Thank you.
Gdzie będzie można obejrzeć tę korektę?
DeleteFixed. Uploaded. Thank you.
DeleteBjab -> Ark
DeleteOK, już widzę.
Zatem jeszcze trzeba by zmienić wzór w linijce pod wzorem (51)
I also realized that seconds after the first uploading. And I knew that you will notice it too. Uploaded again corrected.
DeleteBjab -> Ark
Delete"and let xi be a vector tangent" ->
xi nie zwinęło się.
"manifolds, such as ¶N" ->
manifolds, such as PN
We say that lift of γ through X. ->
We say that is lift of γ through X.
Fixed. Replaced. Thanks.
Deleteis another lift of gamma through p. ->
Deletegamma się nie zwinęła.
with X(0) = X.. ->
dwie kropki
Bjab -> Ark
DeleteTakeing ->
Taking
Let now ξ1, xi2 be any ->
xi2 się nie zwinęło
Bjab -> Ark
DeleteBrak przecinka rozdzilającego i dwa duże S we wzorze (57)
Both fixed, updated. Thanks.
DeleteBjab -> Ark
DeleteStrona 12, czwarty wiersz od dołu - gamma niezwinięta.
Strona 12, pierwszy wiersz od dołu - dwie kropki
Bjab -> Ark
DeleteDlaczego w (57) występują dwa wielkie S? Spodziewałbym się tam dwóch małych s.
Nie rozumiem wzorów w (58).
Corrected. Thanks.
DeleteCorrected correction.
DeleteBjab -> Ark
DeleteTeraz rozumiem (58)
(Brakuje tam teraz jeszcze jednego plusa).
W wierszu pod (58) brakuje zera.
Fixed. Thanks.
Delete
DeleteBjab -> Ark
Conjecture:
Every 6 non-zero vectors from N that are not pairwise scaled are linearly independent.
@Ark
ReplyDeleteBased on your paper (p. 1 - p. 3), the space N is defined as the set of points in the six-dimensional space R^4,2 (which is the direct sum of R^3,1 and R^1,1) that satisfy the condition Q(X) = 0, where X is not zero. The quadratic form Q is defined as the sum of two other quadratic forms q and q2, which operate on different four-dimensional and two-dimensional subspaces of R^4,2, respectively.
Regardless of what structure N has as a subset of R^4,2, it fundamentally contains points in a six-dimensional space, so six coordinates can be assigned to each point in N. However, this doesn't necessarily mean that N contains six linearly independent vectors.
The fact that points in N are solutions of the equation Q(X) = 0 suggests that N is some sort of hypersurface in R^4,2, suggesting that N is (potentially) five-dimensional, not six-dimensional. This is similar to how the surface of a sphere in three-dimensional Euclidean space is two-dimensional, not three-dimensional.
Specifically, despite the R^4,2 space having six dimensions, the subset N, which is defined by the condition Q(X) = 0, is restricted to points that satisfy this condition. For this reason, vectors that are in N must also lie in this "hypersurface", which means that they cannot be linearly independent in the full six-dimensional sense. Instead, they can only be linearly independent in the context of the constraints imposed by the condition Q(X) = 0.
That's how I see it based on what I read in your paper. If you have comments here they would be very valuable to me.
Revised Proof:
ReplyDeleteOur goal is to demonstrate that the action of the orthogonal group $O(4,2)$ on the compactified Minkowski space $\hat{M}$ 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 consider a matrix $A \in O(4,2)$. By definition, $A$ is orthogonal (in the sense of preserving the quadratic form), which implies that $A^T G A = G$, where $G$ is the metric tensor of the pseudo-Euclidean space with signature $(4,2)$. Note that both $A$ and $G$ are 6x6 matrices.
We suppose that $A$ acts as the identity on all points in $\hat{M}$, i.e., for every equivalence class $[X] \in \hat{M}$, we have $A \cdot [X] = [X]$.
The definition of the action of $O(4,2)$ on $\hat{M}$ is given by $A \cdot [X] = [AX]$, where $[AX]$ denotes the equivalence class of the point $AX$. So, if $A \cdot [X] = [X]$ for every equivalence class $[X]$, this implies that for each equivalence class $[X]$, $AX$ and $X$ belong to the same equivalence class.
In other words, for each $X \in N$ such that $A \cdot [X] = [X]$ in $\hat{M}$, we have $AX = \lambda_X X$ for some non-zero scalar $\lambda_X$.
However, due to the constraint Q(X) = 0 that defines N, this does not necessarily imply that $A$ is a scaling matrix in the full six-dimensional space R^4,2. Instead, the scalar multiplication may only be valid within the subspace defined by N.
If $A$ were a scaling matrix in R^4,2, then $A e_i = \lambda_i e_i$ for all $i = 1, ..., 6$, where ${e_i}$ is the standard basis of $\mathbb{R}^{4,2}$, and $\lambda_i$ are scalars associated with these basis vectors. However, due to the constraint Q(X) = 0 that defines N, this condition does not necessarily hold true in N, even though $A$ acts as the identity in $\hat{M}$.
Instead, we can conclude that $A$ must act as the identity matrix within the subspace defined by N, and thus the action of $O(4,2)$ on $\hat{M}$ is faithful.
What's been changed:
The main change in the revised proof is the recognition that, while $A$ acts as the identity on all points of $\hat{M}$, this does not necessarily mean that $A$ is a scaling matrix in the full six-dimensional space R^4,2. This change addresses the catch you raised about N being a subset of R^4,2, which does not necessarily contain six linearly independent vectors due to the constraint Q(X) = 0 that defines N. Instead, $A$ must act as the identity matrix within the subspace defined by N.
"may only be valid within the subspace defined by N."
DeleteThis concept "subspace defined by N" is unclear. What exactly it is? It would be good to have a clear definition of this concept. Logical reasoning, require precise definitions. Intuitive reasoning always requires a logical check - whenever possible. In our case it IS possible.
"This concept "subspace defined by N" is unclear. What exactly it is? It would be good to have a clear definition of this concept. Logical reasoning, require precise definitions. Intuitive reasoning always requires a logical check - whenever possible. In our case it IS possible.".
DeleteWe can define the "subspace defined by N" as follows:
N is defined as the null cone in the six-dimensional space $\mathbb{R}^{4,2}$, excluding the origin. In other words, N is the set of all points X in $\mathbb{R}^{4,2}$ that are not the origin (X $\neq$ 0) and which satisfy the equality Q(X) = 0, where Q is a quadratic form defined on $\mathbb{R}^{4,2}$.
This cone is determined by the equality relation Q(X) = 0. All points (vectors) in the cone N satisfy this equality. In a geometric context, a cone is a set of points (or vectors) that can be generated by scaling (multiplying by a scalar) and adding vectors that are already in the cone. In this way, N defines a subspace in $\mathbb{R}^{4,2}$, but it is a specific subspace - a cone - which has certain unique properties.
Therefore, when the proof states "may only be valid within the subspace defined by N", it means that the results may only hold for points (or vectors) that lie within the cone N, and not necessarily for all points in the entire space $\mathbb{R}^{4,2}$. This makes sense since the properties of the cone N (such as satisfying the equality Q(X) = 0) may not hold for points outside of this cone.
" In a geometric context, a cone is a set of points (or vectors) that can be generated by scaling (multiplying by a scalar) and adding vectors that are already in the cone."
DeleteThis is not what we can find here:
https://en.wikipedia.org/wiki/Cone_(disambiguation)
under "Cone (linear algebra), a subset of vector space"
N is a "cone", but not a "convex cone".
DeleteHint: the unit sphere in R^3 is two-dimensional, and yet three vectors (1,0,0), (0,1,0),(0,0,1) are linearly independent and all three belong to this sphere.
DeleteThus utmost attention and strict logic are needed
"Hint: the unit sphere in R^3 is two-dimensional, and yet three vectors (1,0,0), (0,1,0),(0,0,1) are linearly independent and all three belong to this sphere.".
DeleteSo I understand that while the cone N is defined in the six-dimensional space $\mathbb{R}^{4,2}$ and all points in N satisfy the quadratic equation Q(X) = 0, this does not imply that N has to be a lower-dimensional space. The cone N could contain six linearly independent vectors, although all of them must satisfy the equation Q(X) = 0.
So, when we talk about the "subspace defined by N", we are talking about a set of points satisfying certain conditions (in this case, Q(X) = 0) in $\mathbb{R}^{4,2}$, but we are not saying anything specific about the dimension of this subspace.
In other words, restricting to the "subspace defined by N" refers to restricting to points which satisfy the equation Q(X) = 0, not restricting to a subspace of a lower dimension. Whether vectors are linearly independent will depend on the specific vectors we consider in $\mathbb{R}^{4,2}$, not on the fact that they lie in N.
"Q(X) = 0, this does not imply that N has to be a lower-dimensional space"
DeleteThis is a very dangerous concept. N is not a vector space. N is a 5-dimensional surface (submanifold) in a 6-dimensional vector space.
But the "linear span" of N, which is a vector space, may well be 6-dimensional.
If we can find an explicit form of six linearly independent vectors, all in N, something like (1,0,0), (0,1,0),(0,0,1) in S^2, then the linear span of N will be the whole six-dimensional space R^{4,2}. We will be one step closer to completing proof.
In mathematics paying attention to details, to definitions and statements, is very important (even when we have many years of practice). In physics, if we have a good physical intuition, we can risk logical holes and errors in reasoning.
Are they in N?
DeleteBjab -> Ark
DeleteConjecture:
every 6 vectors that are not pairwise scaled are linearly independent.
Conjecture:
DeleteEvery 6 vectors from N that are not pairwise scaled are independent.
What about this example (other components assumed to be 0)
Deletex1=x4=1
x1=-x4=1
x2=x4=1
x2=-x4=1
x3=x4=1
x3=-x4=1
x5=x6=1
x5=-x5=1
x1=x6=1
x1=-x6=1
There are a maximum 6 linearly independent vectors in 6-dimension vector space.
DeleteAssuming your (wrong) eighth vector is x5=-x6=1:
v7 - v8 - v9 + v10 = 0
Bjab -> Ark
DeleteAnyway, since v7 - (v8) - v9 + v10 = 0 then the above conjecture has to be modified to be true.
My proposition:
Deletev1 = [1, 0, 0, 1, 0, 0]
v2 = [1, 0, 0, -1, 0, 0]
v3 = [0, 1, 0, 1, 0, 0]
v4 = [0, 1, 0, -1, 0, 0]
v5 = [0, 0, 1, 1, 0, 0]
v6 = [0, 0, 1, -1, 0, 0]
Q(v1) = (1^2 + 0 + 0 - 1^2) + (0 - 0) = 0
Q(v2) = (1^2 + 0 + 0 - (-1)^2) + (0 - 0) = 0
Q(v3) = (0 + 1^2 + 0 - 1^2) + (0 - 0) = 0
Q(v4) = (0 + 1^2 + 0 - (-1)^2) + (0 - 0) = 0
Q(v5) = (0 + 0 + 1^2 - 1^2) + (0 - 0) = 0
Q(v6) = (0 + 0 + 1^2 - (-1)^2) + (0 - 0) = 0
DeleteBjab -> Mathilde S.
The vectors you proposed are not linearly independent.
(v1 - v2 - v3 + v3 =0)
I have a different suggestion:
Deletev1 = [1, 0, 0, 1, 1, -1]
v2 = [-1, 0, 0, 1, 1, -1]
v3 = [0, 1, 0, 1, -1, 1]
v4 = [0, -1, 0, 1, -1, 1]
v5 = [0, 0, 1, 1, 1, -1]
v6 = [0, 0, -1, 1, 1, -1]
Here, too, there is Q(v1)=Q(v2)=...=Q(v6)=0.
But I think I got it wrong, because in both examples the determinant of the matrix composed of these vectors is 0.
@Bjab
Delete"Bjab -> Mathilde S.
The vectors you proposed are not linearly independent.
(v1 - v2 - v3 + v3 =0)".
I just know. I miscalculated the determinant the first time.
@Bjab
DeleteThe worst thing is that they have to meet two conditions. Both to be linearly independent and to satisfy Q(X)=0.
I don't want to substitute more vectors and check. We have to solve this task in a different way, backwards.
This is how I do it:
Deletev1 = {1, 0, 0, 1, 1, 1};
v2 = {0, 1, 0, 1, 1, 1};
v3 = {0, 0, 1, 1, 1, 1};
v4 = {1, 1, -Sqrt[2], 1, 1, 1};
v5 = {1, -Sqrt[2], 1, 1, 1, 1};
v6 = {-Sqrt[2], 1, 1, 1, 1, 1};
matrix = Transpose[{v1, v2, v3, v4, v5, v6}]
Det[matrix]
The determinant came out 0 again!
The condition only holds for v1 and for v3, if I have written it down correctly, and I have written it like this:
Deleteq[x_List] := x[[1]]^2 + x[[2]]^2 + x[[3]]^2 - x[[4]]^2;
q2[x5_, x6_] := x5^2 - x6^2;
Q[X_List] := q[Take[X, 4]] + q2[X[[5]], X[[6]]];
Q[v1] == 0
Q[v2] == 0
Q[v3] == 0
Q[v4] == 0
Q[v5] == 0
Q[v6] == 0
Of course, in this code of mine we can change v1 to v6 and see what happens.
DeleteExcuse me. This is all the code:
DeleteThis is all the code in Mathematica. Maybe it will be of use to someone:
v1 = {1, 0, 0, 1, 1, 0};
v2 = {-1, 0, 0, 1, 1, 0};
v3 = {0, 1, 0, -1, 0, 1};
v4 = {0, -1, 0, -1, 0, 1};
v5 = {0, 0, 1, 1, -1, 0};
v6 = {0, 0, -1, 1, -1, 0};
matrix = Transpose[{v1, v2, v3, v4, v5, v6}]
Det[matrix]
q[x_List] := x[[1]]^2 + x[[2]]^2 + x[[3]]^2 - x[[4]]^2;
q2[x5_, x6_] := x5^2 - x6^2;
Q[X_List] := q[Take[X, 4]] + q2[X[[5]], X[[6]]];
Q[v1] == 0
Q[v2] == 0
Q[v3] == 0
Q[v4] == 0
Q[v5] == 0
Q[v6] == 0
I have pasted the vector system twice before.
Bjab -> Mathilde S.:
Delete"Only then, what didn't Ark like about these vectors?"
I don't know if he didn't.
But if he didn't, it might be because there were squares.
Bjab -> Mathilde S.
Delete"Det = 1/27 (-54 + 54 Sqrt[3])"
That is 2√3 - 2
@Bjab
Delete"AnonymousJune 28, 2023 at 10:26 AM
Bjab -> Mathilde S.
"Det = 1/27 (-54 + 54 Sqrt[3])"
That is 2√3 - 2".
Yes, indeed. But I still prefer this Det=-2 result.
Bjab -> Mathilde S.
DeleteIs -2 nicer than 2√3 - 2 ?
Anyway, my first set of linearly independent 6 vectors separates R3,1 from R1,1. Nice?
"In mathematics paying attention to details, to definitions and statements, is very important (even when we have many years of practice). In physics, if we have a good physical intuition, we can risk logical holes and errors in reasoning.".
ReplyDeleteI see this as a weakness in physics. Note that this approach has led to a plethora of seemingly contradictory theories and gaps. For some time now, I have tended to agree that physics has oversimplified the world, reduced it to a description of selected observable phenomena and neglected this precision. I think this is not good, which is why I am in the process of retraining myself.
I believe that a significant new theory will not emerge from physics, but from abstract mathematics and metaphysics, and that what we know from physics will rather be some special case.
I admit that in my youth I made the decision to choose physics rather than mathematics. I do not regret that decision because it taught me a lot, but I am changing that decision now. I'm not saying I'll never do anything towards physics, but I see it differently now than I did a few years ago and I know that what I see now is better for me and I'm happier in this world. I also think that this new world is closer to the truth.
And I still have a question. What is physical intuition? Where does physical intuition end and abstract intuition begin?
"I see this as a weakness in physics. Note that this approach has led to a plethora of seemingly contradictory theories and gaps."
DeleteThe reason why we have a plethora of theories is the fact that theoretical physicists too often escape too far from reality. They go into abstractions and like to stay there - "The hell with reality, when the theory is so abstract and so beautiful!"
There is an uncountable number of structures that you can create. The more you learn abstract things, the more freedom you have in your creations. Paying attention to reality left and right - is absolutely necessary.
A Reader asked me a question: "What is Reality?
ReplyDeleteThere are different kinds of realities. In general paying attention to reality means staying close to earth. For instance finding if there are six linearly independent vectors in N, where N is precisely defined, is staying close to earth. It is still being somewhat abstract, but concrete.
Bjab -> Ark
ReplyDeletecane ->
can
equatios ->
equation
Bjab -> Ark
ReplyDeleteW (62) wystąpiła litera t.
W wierszu poniżej (62) wystąpiła litera t.
Bjab -> Ark
ReplyDeleteW (63) brak nawiasu zamykającego.
Bjab -> Ark
ReplyDeleteW wierszu poniżej (64):
terms terms ->
terms
Fixed. Updated. Thanks for all above.
DeleteThis comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeletehttps://we.tl/t-M42HpGOz5w
ReplyDeleteHere are the calculations for the latter vector system, where the determinant of the matrix is -2. I will try to complete the faithfulness proof this afternoon.
Now let's reconsider the proof, in light of the additional data (about six linear independent vectors in $N$) we have received:
ReplyDeleteOur goal is to demonstrate that the action of the orthogonal group $O(4,2)$ on the compactified Minkowski space $\hat{M}$ 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 consider a matrix $A \in O(4,2)$. By definition, $A$ is orthogonal (in the sense of preserving the quadratic form), which implies that $A^T G A = G$, where $G$ is the metric tensor of the pseudo-Euclidean space with signature $(4,2)$. Note that both $A$ and $G$ are 6x6 matrices.
We suppose that $A$ acts as the identity on all points in $\hat{M}$, i.e., for every equivalence class $[X] \in \hat{M}$, we have $A \cdot [X] = [X]$.
The definition of the action of $O(4,2)$ on $\hat{M}$ is given by $A \cdot [X] = [AX]$, where $[AX]$ denotes the equivalence class of the point $AX$. So, if $A \cdot [X] = [X]$ for every equivalence class $[X]$, this implies that for each equivalence class $[X]$, $AX$ and $X$ belong to the same equivalence class.
In other words, for each $X \in N$ such that $A \cdot [X] = [X]$ in $\hat{M}$, we have $AX = \lambda_X X$ for some non-zero scalar $\lambda_X$.
Now, referring to Ark's comment, let's precisely define the "subspace defined by N". N is the null cone in $R^{4,2}$ minus the origin, i.e. N = {X ∈ $R^{4,2}$ : X ≠ 0 and Q(X) = 0}. There exist six linearly independent vectors in N that nullify Q(X), so these are vectors in $R^{4,2}$ that satisfy the condition Q(X) = 0. These vectors form a subspace in $R^{4,2}$.
If $A$ were a scaling matrix in $R^{4,2}$, then we would have $A e_i = \lambda_i e_i$ for all $i = 1, ..., 6$, where ${e_i}$ is the standard basis of $\mathbb{R}^{4,2}$, and $\lambda_i$ are scalars associated with these basis vectors. However, due to the constraint Q(X) = 0 that defines N, this condition does not necessarily hold true in N, even though $A$ acts as the identity in $\hat{M}$.
Instead, we can conclude that $A$ must act as the identity matrix within the subspace defined by N, which means $A$ must preserve these six linearly independent vectors that are in N and nullify Q(X). In other words, for each vector $v_i$ from these six vectors, $Av_i = \lambda_i v_i$ must hold true, where $\lambda_i$ is a non-zero scalar. Given that these vectors are linearly independent, the only possible outcome is that for each $i$, $\lambda_i$ must be equal to 1, meaning that $A$ is the identity matrix.
Therefore, the action of $O(4,2)$ on $\hat{M}$ is faithful.
"$Av_i = \lambda_i v_i$ must hold true, where $\lambda_i$ is a non-zero scalar. Given that these vectors are linearly independent, the only possible outcome is that for each $i$, $\lambda_i$ must be equal to 1"
DeleteThis is called:
5. The Hasty Generalization Fallacy
This fallacy occurs when someone draws expansive conclusions based on inadequate or insufficient evidence. In other words, they jump to conclusions about the validity of a proposition with some — but not enough — evidence to back it up, and overlook potential counterarguments.
https://blog.hubspot.com/marketing/common-logical-fallacies
I started with the assumption that for every equivalence class $[X] \in \hat{M}$, we have $A \cdot [X] = [X]$. This translates to $AX = \lambda_X X$ for some non-zero scalar $\lambda_X$. Now, we are considering six specific vectors $v_i \in N$ that are linearly independent and nullify Q(X). For these vectors, $Av_i = \lambda_i v_i$ holds, with each $\lambda_i$ being non-zero.
DeleteIs there something wrong here?
Bjab -> Mathilde S.
DeleteArk marks elements of M^ by double brackets [[.]] not [.].
Lambda is not not equal 0 but grater than 0.
Is that important?
@Bjab
Delete"Is that important?"
Yes. The devil hides in such details.
Yes, you are right Bjab and Ark. Ok. These details will have to be worked out. On the other hand, I now make another request to Ark.
ReplyDeleteNamely: I am discussing with my co-author, who is a professor of mathematics. I think I would like to share your work on EEQT with her, nevertheless she is not a physicist, she is a mathematician and philosopher and specializes in category theory, but she is curious about physics. Could you provide me with links to your papers on EEQT that would be most accessible to a mathematician who deals mainly with very high abstraction? I am very interested in getting her to look at it and then for us to be able to refer to it in our paper.
Ch. 4.4. in my book "Quantum Fractals".
Delete"Ch. 4.4. in my book "Quantum Fractals.".
ReplyDeleteThank you very much, it's just that I have the book in paperback and she lives in Portugal and it's practically impossible for me to show it to her. Do you have a pdf version of Quantum Fractals? Sorry, I think I had a pdf at one time, but I can't find it at the moment.
https://www.amazon.es/-/pt/dp/B00P63YD5Y/ref=sr_1_fkmr0_1?crid=1T2C3EYX5X5S5&keywords=jadczyk+quantum+fractal&qid=1687967793&s=books&sprefix=jadczyk+quantum+fractals%2Cstripbooks%2C72&sr=1-1-fkmr0
Deletehttps://www.worldscientific.com/doi/suppl/10.1142/8992/suppl_file/8992_errata.pdf
DeleteAlso Sec. 4.7 - it has some kind of a "philosophy".
DeleteUpdated the pdf file. Added beginning of Sec. 7.
ReplyDeleteUpdated adding the beginning of the solution of Exercise 2, p. 15.
DeleteBjab -> Ark:
Delete"But X and −X are i the same equivalence class of ̃, so π(X) =
π(−X) - the define the same point of M ̃ ." ->
Myślę, że są trzy literówki w powyższym zdaniu.
Bjab -> Ark
DeleteBrak kropki kończącej zdanie nad Exercise 2
Thanks. Will correct tomorrow. Will also add another hint.
Delete@Bjab. Updated. Thanks!
Delete