The dimension of space is fundamentally related to the number of its independent variables. Consequently, spacetime is associated with existence in a three-dimensional space, with time as a fourth dimension. In my personal view, time is entirely independent of the presence of an object in a location; in fact, a change in time may affect its motion, as usually described using the Minkowski metric in special relativity. On the other hand, dimensions in mathematics are a topological concept that can be studied as a metric space.

Ismail Abbas

From a mathematical point of view, it is used to work in a N-dimensional space, defining PDEs on that. Thus, we can describe elliptic, parabolic, hyperbolic character.

But when we discriminate space and time independent variables, we must accept some limitation in time. For example, an elliptic equation in time (Div Grad 4D operators) would produce a solution violating the retro-casuality constraint. The physics of contunuum introduces constraints on the role of the time.

The mathematical description R4 is too poor to speak about geometry (or physics as an interpretation of geometry) of some space. It only states the dimension equal to 4 and linearity. There exist 27 linear 4-dimensional spaces, each can be described as R4.

To distinguish between all these spaces, we need to introduce metrics. If we speak about 4-dimensional Euclidean space E4, its metric is:

d2 = x12 + x22 + x32 + x42.

If we speak about physical space-time, its mathematical model is Minkowski space M1,3 with metric:

d2 = x12 - x22 - x32 - x42.

This is just a brief answer

to shed more light on the question and it's answer.

Believe it or not ! moving from classic 3D +t space to the unified 4D unitary space makes all solutions to physical and mathematical situations possible and simple.

You are moving to the unified field theory.

For example the Maxell equation :

Div D = Rho in the classic 3D+t space is incomplete and misleading.

On the other hand it's equivalence in unitary 4D space is;

dV/dt )partial = (D) . BC vector + (B)**N ,

(1)

Eq 1 has been successfully used since 2020 to solve Laplace and Poisson equations in most general form.

They are both four dimensional vector spaces. So algebraically they are the same. The geometry of the standard Euclidean space and the Minkowski space time are determined by the metrics. However, by the invariance of domain theorem which implies the invariance of dimension theorem - they are the same dimension. On one hand the dimension of a vector space is an algebraic property. One the other when a metric is assigned with it's topology, it is also a topological concept.

In relativity theory it is common to adjust units so that the speed of light is equal to one. In this formulation - the time has not special meaning. The basic model for relativity theory is a Lorentizan manifold which is a four dimensional manifold with the Lorentezian metric (pseudo Riemannian manifold with a meter with signature equal (3,1)).

I'm not convinced that as stated the original statement makes much sense.

Answer II continued

This brief answer aims to shed more light on the question amd it's answer and to thank our colleagues contributors for their helpful mathematical answers,

However, these answers are basically false showing that mathematics without physics understanding is merely nonsense,

In eq 1 in particular and the statistical chains in general real time t is completely lost and replaced by a dimension less integer N,

Time t is not continuous but discrete,

This implies that equation 1 (based on universal laws of physics plus modern definition of transition probability represents the unified field theory,)!

Eq 1 is eligible to solve classical physics,quantum physics problems as well as mathematical integrations and differentiation

Eq 1 and Cairo statistical techniques in general are not complicated but more difficult to understand than Schrodinger equation itself.

This is to be expected since SE is merely a subset of the theory of Cairo techniques

Ismail Abbas

I suppose we are within the continuum hypothesis to accept the concept of continuity and differentiation. If you introduce time as discrete variable the same must be assumed for space. And no classic partial derivative should be introduced.

General Relativity used a proper metric for time but within the hypothesis of differentiable time.

This is a new space never known before.

The classic vector space is something from the past while the Cairo techniques scalar space is the future.

Here, the vector quantity E is replaced by electric potential V , quantum wave function psi is replaced by its square and the sound intensity I is replaced by the sound energy density . . Etc.

The resulting pde in 4D unit save is always:

dU/dt)partial = C* Nabla ^2 U + Souce/Sink term.

It follows that the Stokes curl theorem and the divergence theorem are replaced by the so-called Laplacian theorem:

Integral U(X,y,X,t) over volume V = integral Nabla **2 over the surrounding surface.

This leads to ,

U(x , y , z ,t)=D * Boundary conditions vector b.

D is the Cairo techniques transfer matrix easy to find.

The a above rules are magical and help solving classical physics and quantum physics problems in addition to math integration and differentiation.

Answer III - continued

The question arises,

Did you arrive at the numerical statistical theory called the Cairo techniques by chance of try and error or via rigorous physical and mathematical axioms ?

The answer follows as,

1- The unified field theory must be the way where our mother nature is solving emerging physical situations in the fastest way satisfying the principle of least action.

2- Nature is stable, unique and symmetric.

3- Axioms 1,2 imply that there exist a well-defined symmetric transition matrix which controls the time dependent evolution of the energy density function U, namely,

U( x y ,z , t + dt ) = B * U ( X ,y , z , t ) , , (2)

Note that equation 2 lives and functions on unitary 4D space and not in the classic R^4 space.

It is obvious that the statistical transition matrix B must have a place for the boundary vector conditions b and a place for the source/sink term vector S.

The above statement is satisfied by the B-Matrix.

At the present time we know only two transition matrices , namely,

The Markov mathematical transition matrix M and the physical statistical B-transition matrix.

The physical matrix B is much more superior than the mathematical matrix M.

Answer IV - continued

A second question arises :

How do you show that the following PDE of energy density,

dU/dt )partial = a D(N)( X , y ,z ,t ) * ( b + S ) + B^N . IC . . . (2)

[ Note that the well known Heisenberg matrix in R^4 space is neither statistical nor complete ]

Is the unified field theory ?

A simple and straight forward answer is that it challenges and resolved all classical and quantum physical situations for any boundary conditions vector b, any source/sink term S and any initial conditions vector IC.

Hereafter are few usage examples among many:

1- The B matrix chains together with vacuum dynamics is able to describe the formation and explosion of the Big Bang.

The Big Bang i s a fact.

2- The B matrix chains with discrete dimension less time in the 4D unitary space space results in exponential rise/decay of the time component as,

F(t)= a Exp ( - Lambda . t )

More over the numerical value of Lambda is exactly equal to the experimental value obtained by W. Sabine for the reverberation time in audio rooms

Sabines formula did not have a rigorous proof for more than a century.

3- The 4D unit space B matrix chains predicts a solution for a quantum particle in 1D , 2D and 3D infinite potential well in agreement with that obtained by solving SE PDE.

4- The 4D unitary space B-Matrix chains is able to numerically evaluate 1D, 2D and 3D (Hyper cube) finte Simpson integrals without the need for any Lagrangian multipliers.

Moreover it is more accurate than Simpson rules.

etc ..etc..

Answer V - continued

A third question arises:

I am still unconvinced for two reasons,

1- The unified field theorem is hard to understand.

2- The old iron guards of Schrodinger equation are constantly denying any equation which is not contained in SE cause they erroneously beleve that the classic SE is the unified field theory.

Can you explain in details some illustrating examples?

Here are few examples out of many,

The classical Laplacian operator in the classical 3D + t space generates nothing, unlike the generalized Laplacian in the 4D unit space.

The reason is that in 4D unit space, the feald and its source are generalized.

Just as the loop and divergence differential operators have two forms differential and integral, the theorem of Laplacian operators in 4D unit space also does the same.

The differential form of Laplacian's theorem is provided in the first answer while the integral form of Laplacian's theorem follows:

∫∫ U dS = ∫∫∫ S(U) dV.

Note that,

d/dt)partial U = D Nabla^2 U +S(U),

The Laplacian theorem (integral form) above is equivalent to

U( X ,y z , t+ dt ) = B . U ( X,y ,z ,t )

Where the transition matrix B is subject to,

The transition matrix B is subject to,

D(N)=B + B^2 . . . . +B^N

And,

And D(N)=B+B^2 +.+..+B^N

Also,

D(N)= E as N tends to infinity.

Note also that,

E=1/(I-B)

And.

E=D+I

To be continued.

Finally, here is a simple understanding of spaces:

The Laplacian theorem lives and functions in the 4D-unit space but not in the classical 3D+t space.

The theory of relativity and the speed of light c lives and functions in 4D unit space but not in classical 3D+t space.

. . etc.

However, general relativity is a theory that uses the Lorentz metric - a pseudo Reimnnian metric of time (3,1) on a four dimensional manifold. The underlying local model of general relativity is nothing more than R^4 equipped with the Lorentz metric. There is nothing magic about the Laplacian. The Laplacian on a manifold is defined by the metric carried by the manifold. Sir Roger Penrose used this metric tracking the geodesics forward in time to prove black holes exist as gravity got stronger the geodesics converged to a singularity. Today we know black holes are abundant. Steven Hawkins reversed time in Penrose's argument to establish the existence of the "big bang." In reality time is not necessarily "unique" give the speed of light is constant - time and distance are the same. In fact that is encoded in the Lorentz metric = dx^2 + dy^2 + dz^2 - (c dt)^2.

The the form Laplacian on any manifold is a function of the underlying metric tensor. It is known as the Laplace-Beltrami operator. In physics one often sees the generalization, the Laplace- de Rham operator used which is the Laplace-Beltrami operator generalized to tensors and differential forms.

https://en.wikipedia.org/wiki/Laplace–Beltrami_operator

Normally I don't comment on other contributors' answers, but the excellent answer from our American professor motivates me to respond to his comments in detail and step by step.

1-The underlying local model of general relativity is nothing more than R^4 fitted with the Lorentz metric.

*The Lorentz transformation is a universal law of physics that cannot be derived and special relativity itself is one of its consequences.

2-**Well, there is nothing magical about the Laplacian described in R^4 but when the Laplacian is described in 4D unit space, magic appears everywhere in classical physics, quantum physics and pure mathematics such as numerical integration and the summation of integer geometric series.

3-Hawkins reversed time in Penrose's argument to establish the existence of the "big bang metric. Lorentz = dx^2 + dy^2 + dz^2 - (c dt)^2.

***The formation and explosion of the Big Bang can be described and proven via statistical matrix mechanics in 4D unit space without the need for Einstein's sphere of light:

dx^2 + dy^2 + dz^2 - (c dt)^2 being the same for inertial frames.

4-The Laplacian form on any manifold is a function of the underlying metric tensor. It is known as the Laplace-Beltrami operator. In physics, we often see the generalization, the Laplace-de Rham operator used which is the Laplace-Beltrami operator generalized to tensors and differential forms.

https://en.wikipedia.org/wiki/Laplace–Beltrami_operator

**** The unified 4d xt space considers the classical manifold of R^4 space to be inferior and incomplete and therefore there is no need to answer the following.