Breakdown of Navier-Stokes Equations

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Abstract:

Given limited energy and a small mass density or large kinematic viscosity, this work shows why blowups of velocity $u$ can occur within a short time period. Additionally, by examining the Eulerian and Lagrangian perspectives and exposing inequalities, this work demonstrates that, given an arbitrary existing vector field, there is no solution of $u$ via the Navier-Stokes momentum equation that is always physically reasonable and/or mathematically consistent and can successfully model the vector field. Because a valid calculation of pressure $p$ via the said equation depends on a valid solution of $u$, there is no solution of $p$.

1. Introduction:

The Navier Stokes equations (NSE) were created during the 19th century to model viscous fluids. It was a time before atoms were established, a time before special relativity placed a speed limit on how fast a fluid can move, so velocities approaching infinity did not raise eyebrows. Fluids were believed to be infinitely divisible, and could be incompressible. So why should we be shocked or surprised that the NSE sometimes yield non-physical results by today’s standards? There are many exact solutions to the NSE such as Couette or Poiseuille solutions which result from simple textbook problems consisting of steady, fully developed flows where many of the troublesome NSE terms can be cancelled or eliminated. The discussion below seeks to keep the "troublesome" terms in play for the purpose of demonstrating how they lead to a lack of a solution.

2. The Challenge

Within his paper (“Existence and Smoothness of the Navier-Stokes Equation”), posted at the Clay Mathematics Institute (CMI) website, Charles L. Fefferman states in pertinent part:

“The Navier-Stokes equations are to be solved for an unknown velocity vector $u(x,t)$ … and pressure $p(x,t)$… defined for position $x \in \mathbb{R}^n$ and time $t \geq0$ .”

Below are the Navier-Stokes (NS) equations, the conditions imposed by the CMI along with the statement to be proved: (C).

\[ \frac{\partial u_i}{\partial t} + \sum_{j=1}^{n} u_j \frac{\partial u_i}{\partial x_j} = \nu\Delta u_i - \frac{\partial p}{\partial x_i} + f_i(x,t) \hspace{1 cm} (1) \] \[ div\hspace{.1 cm} u = \sum_{i=1}^{n} \frac{\partial u_i}{\partial x_i} \hspace{1 cm} (2) \] We assume mass density $\rho$ is set to $1$. \[ u(x,0) = u^\circ(x) \hspace{1 cm} (3) \] \[ |\partial^\alpha_x u^\circ(x)| \leq C_{\alpha K}(1+|x|)^{-K} \hspace{.3 cm} on \hspace{.1 cm} \mathbb{R}^n \hspace{.1 cm} for \hspace{.1 cm} any \hspace{.1 cm} \alpha,K \hspace{1 cm} (4) \] \[ |\partial^\alpha_x \partial^m_t f(x,t)| \leq C_{\alpha m K}(1+|x|+t)^{-K} \hspace{.3 cm} on \hspace{.1 cm} \mathbb{R}^n \times [0,\infty) \hspace{.1 cm} for \hspace{.1 cm} any \hspace{.1 cm} \alpha,m,K \hspace{1 cm} (5) \] \[ p,u \in C^\infty \hspace{.1 cm}(\mathbb{R}^n \times [0,\infty)) \hspace{1 cm} (6) \] \[ \int_K |u(x,t)| dx \leq C \hspace{1 cm} (7) \]

(C) Take $\nu >0$ and $n = 3$ . Then there exist a smooth, divergence-free vector field $u^\circ(x)$ on $\mathbb{R}^3$ and a smooth $f(x,t)$ on $\mathbb{R}^3 \times [0,\infty)$, satisfying (4), (5), for which there exist no solutions of $(p, u)$ of (1), (2), (3), (6), (7) on $\mathbb{R}^3 \times [0,\infty)$.

3. Definitions and Authorities

We define functions, components, indices, and list authorities below: \[ x=\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} \hspace{1 cm} (8) \] \[ u=\begin{bmatrix} u_1\\ u_2\\ u_3 \end{bmatrix} \hspace{1 cm} (9) \]

Index $i = 1,2,$ or $3$; $j = 1,2$ or $3$.

Conservation of Energy: Energy can neither be created nor destroyed, but only transformed from one kind to another.

Newton's Second Law: If a net or resultant force acts upon an object of non-zero mass, the object accelerates in the direction of the force. The acceleration is proportional to the force and inversely proportional to the mass of the object.

Identity Principle: $u=u$.

Constant Rule: The derivative of a constant is zero.

Constant Multiple Rule: $\psi \delta u = \delta \psi u$, where $\delta$ is a constant and $\psi$ is an operator.

4. Preliminary Proofs

4.1. Divergence-free

Vector component $u_i$ is an instantaneous velocity: \[ u_i = \frac{dx_i}{dt}= \lim_{h\to0}\frac{x_i(t +h)-x_i(t)}{h} \hspace{1 cm} (10) \] Take the partial derivative with respect to $x_i$: \[ \frac{\partial u_i}{\partial x_i}= \frac{\partial}{\partial x_i}\frac{x_i(t +h)-x_i(t)}{h} = \frac{1-1}{h}= 0 \hspace{1 cm} (11) \] The result is zero. Since $i = 1,2,3$, then \[ \frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3} = 0 \hspace{1 cm} (12) \]

Equation (12) is also true if $x_i$ is constant, and $u_1=u_2=u_3 = 0$.

4.2. The Material Derivative

Take the total derivative of $u_i$: \[ \frac{d}{dt}u_i(x,t) = \frac{\partial u_i}{\partial t} + \sum_{j=1}^3\frac{\partial u_i}{\partial x_j}\frac{dx_j}{d t} \hspace{1 cm} (13) \] \[ u_j = \frac{dx_j}{dt} \hspace{1 cm} (14) \] Substitute, then take the derivative using the product rule: \[ \frac{\partial}{\partial x_j}u_i u_j = \frac{\partial u_i}{\partial x_j}u_j + u_i\frac{\partial u_j}{\partial x_j} \hspace{1 cm} (15) \] \[ \frac{\partial u_j}{\partial x_j} = 0 \hspace{.5 cm}(see \hspace{.1 cm}4.1) \hspace{1 cm} (16) \] The last term on the right of (15) vanishes. Then we have the symmetric property: \[ \frac{\partial u_i}{\partial x_j}u_j = u_j\frac{\partial u_i}{\partial x_j} \hspace{1 cm} (17) \] Therefore, we have the following: \[ \frac{d}{dt}u_i(x,t) = \frac{\partial u_i}{\partial t} + \sum_{j=1}^3 u_j\frac{\partial u_i}{\partial x_j} \hspace{1 cm} (18) \] The material derivative is as follows: \[ \frac{du_i}{dt} =\frac{Du_i}{Dt} = \frac{\partial u_i}{\partial t} + \sum_{j=1}^3 u_j\frac{\partial u_i}{\partial x_j} \hspace{1 cm} (19) \]

5. In Search of u(x,t)

We select an existing arbitrary, smooth, divergence-free vector field $u^\circ(x)$ on $\mathbb{R}^n$ and a smooth $f(x,t)$ on $\mathbb{R}^n \times [0,\infty)$, satisfying all CMI requirements including (4), (5), where $\nu >0$, $n = 3$, where if $t>0$, $u \neq 0$, and $u$ changes at each position $x$ in response to changes in $p$ and $f$.

5.1. The Bounded Energy Blowup

First, let us examine what can go terribly wrong and why. Assume the NS momentum equation is always an equation, that the material-derivative proof is valid, and assume the vector field at all times satisfies (7), i.e., always has limited energy $E$, where $E= Vx(\rho f-\nabla p)$ and $V$ is the fluid volume.

5.1.1. Infinitesmal Mass Density

Assume mass density $\rho$ is constant but also infinitesmal, that $\rho = 1 \hspace{.1 cm}\rho_{unit}$. Suppose that at $t=0$, energy $E = 0$: \[ V\rho_{unit}\frac{Du}{Dt}x= E + Vx\nu\Delta u = 0 \hspace{1 cm} (20) \] If $E=0$, assume that $u$ and all its derivatives $=0$, and (4) is satisfied. At an arbitrary finite time $t >0$ we add a limited amount of energy to the system: \[ V\rho_{unit}\frac{Du}{Dt}x= E +Vx\nu\Delta u \neq 0 \hspace{1 cm} (21) \] The material derivative is derived from and equal to the total derivative (see 4.2): \[ \frac{du}{dt} = \frac{Du}{Dt} \hspace{1 cm} (22) \] Make a substitution: \[ V\rho_{unit}\frac{du}{dt}x= E + Vx\nu\Delta u \hspace{1 cm} (23) \] Solve for acceleration: \[ \frac{du}{dt}= \lim_{\rho_{unit} \to 0}\frac{E + Vx\nu\Delta u }{x V\rho_{unit}} \sim \pm\infty \hspace{1 cm} (24) \] Integrate both sides, assume the integration constant = 0: \[u = \lim_{\rho_{unit} \to 0} \int_{dt}^t \frac{E + Vx\nu\Delta u}{xV\rho_{unit}}dt \sim \pm \infty \hspace{.3 cm}(0 < dt < t < \infty) \hspace{1 cm} (25)\] If the mass density $\rho$ is constant but infinitesimal (or set to 1 infinitesimal unit) and $x$ is constant, then to conserve energy, the acceleration at (24) and velocity at (25) must have a $\pm$ infinite upper limit at any arbitrary finite time $t > 0$. In Newton's universe, velocity is not limited to light speed: infinite speed in n-dimensional space is perfectly fine, where n = 1, 2, or 3. Ironically, any solution of $u$ that does not allow $u$ to blow up to $\pm$ infinity if the mass density is infinitesmal, violates the energy conservation principle. Granted, this is a physically unreasonable result by today's standards, but a plausible result when using Newtonian-based equations including the NSE.

5.1.2. Unlimited Viscosity

Here is what results if there is no upper bound on $\nu$ given the same finite conserved energy $E$ added to the system when $t>0$: \[ V\rho_{unit}\frac{Du}{Dt}x - \lim_{\nu\to\infty} Vx\nu\Delta u = E \hspace{1 cm} (26) \] To conserve energy $E$, $Du/Dt$ must blow up to $\pm$ infinity. It follows that a $\pm$ infinite acceleration will lead to a $\pm$ infinite velocity $u$ in a finite period of time $t$: \[ u= \lim_{\nu \to \infty} \int_{dt}^t\frac{E + Vx\nu\Delta u }{ xV\rho_{unit}}dt\sim \pm\infty \hspace{.3 cm}(0 < dt < t < \infty) \hspace{1 cm} (27) \] Equation (27) was derived from (20) through (26).

5.2. The Eulerian Perspective

If we focus on a single position in the vector field, i.e., if we focus on a fixed position $x$ and measure the velocity $u_j(x,t)$ located there, we find to our dismay that the math does not match the reality if $u_j \neq 0$. We restate (14): \[ \frac{d}{dt}x_j = u_j(x,t) \hspace{1 cm} (28) \] But coordinate $x_j$ of a fixed position $x$ is not a function of time $t$. In fact, it is constant for all time $t$. Thus, we are forced to conclude the following: \[ \frac{d}{dt}x_j = 0 \neq u_j(x,t) \hspace{.5 cm}(t>0) \hspace{1 cm} (29) \] The derivative of a constant is always zero, and satisfies (4) where $dx/dt = u^\circ(x) = 0$, but the velocities of our selected existing vector field are not always zero if $t>0$ (see section 5). From the Eulerian perspective, there is no solution of $u_j$ that models the vector field. Since index $j=1,2,3$, we deduce that there is no solution of $u$ as well.

5.3. The Lagrangian Perspective

If $x_j = x_j(t)$, then $dx_j(t)/dt \neq 0$. However, if $dx_j/dt = x_j/t$ where $t=0$, we have the following: \[ u_j(x,0) = \frac{x(t)}{0} \hspace{1 cm} (30) \] To avoid a singularity at $t=0$, we use the difference quotient, (10) and (14): \[ u_j(x,0) = \frac{dx_j}{dt} = \lim_{h\to0}\frac{x_j(0+h)-x_j(0)}{0 + h-0} \hspace{.3 cm}(h \neq 0) \hspace{1 cm} (31) \] The right side of equations (31) perfectly models a divergence-free velocity vector (see 4.1) and is perfectly consistent with the material derivative proof at 4.2 and equation (14). Since $j=1,2,3$, we deduce the following: \[ u(x,t) = \lim_{h\to0} \frac{x(t+h)-x(t)}{t+h-t} = \lim_{h\to0} \frac{x(t+h)-x(t)}{h} \hspace{1 cm} (32) \] The far right side of (32) shows that the input $t$ is inconsequential in the denominator. The $h$ is the valid time input there; however, its forever shrinking value cannot be pinned down exactly. We can replace it with the SI unit for time: 1 second or $s$. We can also replace $x(t+h)$ with $x_f$ which represents a future position of $x$ if the instantaneous velocity is sustained for a period of $s$. We then have the following: \[ \frac{x_f - x}{s} = \lim_{h\to0} \frac{x(t+h)-x(t)}{h} \hspace{1 cm} (33) \] Where $x = x(t)$ and \[ x_f = us + x \hspace{1 cm} (34) \] \[ u(x,t) = u(x,x_f) \hspace{1 cm} (35) \] Note that the function on the left side of (35) requires inputs $x,t$; whereas, the function on the right side requires inputs: $x,x_f$. If we plug in a value for $x$ on both sides of (35), we might have an equation, but if we proceed to plug in a value for $x_f$, the left side has no reason to change, but the right side could change, so we could have the following inequality: \[ u(x,t) \neq u(x,x_f) \hspace{1 cm} (36) \] An obvious solution to this inequality problem is to make $t$ and $x_f$ constants, then only an input of $x$ is required on both sides of (35). Unfortunately, each $x$ will have only one value of $u$ (see 33). This is inconsistent with the selected vector field where $u$ changes. Now, will the inequality problem be solved if we let $x_f = x_f(x)$? If so, once again, only the input of $x$ would be required for both sides of (35). The answer to the question posed is a resounding no. Equation (34) shows that $x_f$ is a function of $u$, not just $x$. Velocity $u$ is the unknown we are trying to solve. Thus, we need to know and input $x_f$ like we know and input $x$ to solve $u$. Equations (31) through (35) show that $dx/dt = u(x,x_f)$. Equation (36) shows the following can be true: \[ u(x,t) \neq \frac{dx}{dt} \hspace{1 cm} (37) \] At the material-derivative proof (4.2), it was assumed, not proven, that (14) is true. We have shown here that (14) may not always be true. Since $u(x,t)$ represents any solution of $u(x,t)$, the inequality at (36) can occur for any solution of $u(x,t)$ where the only input variables are $x$ and $t$. Thus, there is no solution of $u(x,t)$ that solves this inequality problem and successfully models the selected existing vector field.

5.4. The Navier-Stokes Inequality

Here is a compact version of the NS momentum equations: \[ \frac{\partial u}{\partial t} + (u \cdot \nabla) u - \nu\Delta u = f - \nabla p \hspace{1 cm} (38) \] Assume $u = u(x,t)$ and $F = F(x,t)=f(x,t)- \nabla p(x,t)$: \[ \frac{\partial u}{\partial t} + (u \cdot \nabla) u - \nu\Delta u = F \hspace{1 cm} (39) \] According to 4.2, the following is true: \[ \frac{D u}{Dt}=\frac{\partial u}{\partial t} + (u \cdot \nabla) u \hspace{1 cm} (40) \] We make a substitution: \[ \frac{D u}{D t} - \nu\Delta u = F \hspace{1 cm} (41) \] Multiply both sides of (41) by a constant $\delta$: \[ \delta\frac{Du}{D t} - \delta\nu\Delta u = \delta F \hspace{1 cm} (42) \] Apply the constant multiple rule in reverse: \[ \frac{D (\delta u)}{D t} - \nu\Delta (\delta u) = \delta F \hspace{1 cm} (43) \] Where $u(x',t')=\delta u(x,t)$, $F(x',t') = \delta F(x,t)$ and $t'>t$. We observe that $u \propto F$. Now, multiply both sides of (39) by $\delta$: \[ \delta\frac{\partial u}{\partial t} + \delta(u \cdot \nabla) u - \delta\nu\Delta u = \delta F \hspace{1 cm} (44) \] Once again, apply the constant multiple rule in reverse: \[ \frac{\partial (\delta u)}{\partial t} + ( u \cdot \nabla)(\delta u) - \nu\Delta (\delta u) = \delta F \hspace{1 cm} (45) \] At the (45) advection term, $u = u(x,t)$ and $u=\delta u(x,t) = u(x',t')$. This is a violation of the identity principle and we have ... \[ u \neq u \hspace{1 cm} (46) \] If we fix the identity problem and enforce the observation that $u \propto F$, we have ... \[ \frac{\partial (\delta u)}{\partial t} + (\delta u \cdot \nabla) \delta u - \nu\Delta (\delta u) \neq \delta F \hspace{1 cm} (47) \]

...the Navier-Stokes Inequality. This inequality occurs if $p,f,u$ change with time and space. Since $u$ represents any solution of $u$, there is no solution of $u$ that can fix this problem and model the selected existing vector field.

6. In Search of p

We calculate $p$ using (1): \[ \frac{\partial p}{\partial x_i}=\nu\Delta u_i -\frac{\partial u_i}{\partial t} - \sum_{j=1}^3 u_j\frac{\partial u_i}{\partial x_j} + f_i \hspace{1 cm} (48) \] \[ p= \int(\nu\Delta u_i -\frac{\partial u_i}{\partial t} - \sum_{j=1}^3 u_j\frac{\partial u_i}{\partial x_j} + f_i)dx_i- C_P \hspace{1 cm} (49) \]

Equation (49) reveals that a solution of $p$ depends on solutions of $u_i,u_j$. We have shown there is no solution of either $u_i$, $u_j$ or $u$ that is mathematically consistent and/or physically reasonable for all time and space. Thus, it follows that there is no solution of $p$ that is mathematically consistent and/or physically reasonable for all time and space.

Conclusion

Based on the foregoing, if $\nu >0$ and $n = 3$, then there exists a smooth, divergence-free vector field $u^\circ(x)$ on $\mathbb{R}^3$ and a smooth $f(x,t)$ on $\mathbb{R}^3 \times [0,\infty)$, satisfying (4), (5), for which there exist no solutions of $(p, u)$ of (1), (2), (3), (6), (7) on $\mathbb{R}^3 \times [0,\infty)$.

References

1. Fefferman, Charles L. Existence and Smoothness of the Navier-Stokes Equation. Clay Mathematics Institute.

2. Lo, Asane. October 28, 2018. Smooth Solutions of the Three Dimensional Navier-Stokes Problem. King Fahd University of Petroleum and Minerals.

3. Porton, Victor. Analyzing Navier-Stokes Equations Using A Linearly Extended Funcional. Orcid 0000-0001-7064-7975.

4. Tao, Terrence. Finite Blowup for an Averaged Three-Dimensional Navier-Stokes Equation. arxiv.org.

5. Ladyzhenskaya, O.A.. 1958. Solution 'in the Large' to the Bounary Value Problem for the Navier-Stokes Equations in Two Space Variables Doki Akad Nauk SSSR, Volume 123, No. 3, P. 427-429

6. Peng Shi. General Solution to 2D Steady Navier-Stokes Equation for Incompressible Flow without vorticity diffusion. Well Logging Key Laboratory, China National Logging Corporation,

© 2026 G.M. Jackson

Proof that Aleph Zero Equals Aleph One, Etc.

ABSTRACT:

According to the current dogma, Aleph-0 is less than Aleph-1, but is there evidence to the contrary? Is it really true that something with an unlimited upside is less than another thing with an unlimited upside? This paper offers proof to the contrary.

Equation 1 below is the formula used to build Cantor's cardinal infinities. The next higher-order of infinity is equal to two to the power of a lower-order infinity, assuming we are working with binary numbers. Take note that if we express the higher-order infinity as a binary number, we have a one followed by an infinite number of zeros. We might dare to ask, what kind of infinity is that infinity of zeros? The zeros are discrete and distinct and can be counted. Thus we can infer that the infinity of zeros is the same infinity associated with the set of natural numbers: Aleph-0. There is a one-to-one mapping between the zeros and natural numbers (see mapping below):

Therefore, when we express any order of infinity in binary form, the infinite digits (zeros) following the one must amount to Aleph-0. That means equation 1 is wrong. The correct equation is equation 2:

All cardinal infinites ranging from Aleph-1 to Aleph-infinity are equal to Aleph-1. What about Aleph-0? If we assume it is structured like the other cardinal infinities (2^n) and express it as a binary number, once again we have a one followed by infinite zeros. Since those digits can be counted, we are forced to conclude that their number also amounts to Aleph-0. So Aleph-0 also equals Aleph-1:

The conclusion at equation 5 shows all cardinal infinities are equal to Aleph-1. Of course we are told over and over that this can't be! So let's look at a complete list of whole numbers ranging from 000... to 111... and numbered or paired with natural numbers ranging from 1 to 1000... To have a complete list, we need 2^Aleph-0 rows and Aleph-0 digits or columns when we express the list in matrix form:

If we apply the diagonal method, the matrix needs to be square, i.e., the number of rows must equal the number of columns, so it is physically impossible to do the diagonal method with a complete list of whole numbers unless 2^Aleph-0 rows equals Aleph-0 columns. We could imagine the matrix being square if the number of rows is 2^Aleph-(-1), where Aleph-(-1) is a lower order of infinity than Aleph-0. That would be consistent with equation 1 above. Unfortunately, Aleph-0 is the lowest order of infinity possible--thus Aleph-0 must equal 2^Aleph-0 or Aleph-1. This implies, however, that the diagonal method should fail to produce a new number not on the list:

In the left diagram above, we flip each digit along the diagonal. The created number (in red) must have a leading digit of 1 and a last digit of 0. A complete list, however, has a range of 000...0 to 111...1. The created whole number is greater than 000...0 and less than 111...1, so it already exists on a complete list of whole numbers. This makes perfect sense if you stop and think about it. However, this implies that Cantor did not use a complete list for his diagonal-argument demonstration. He used a n X n matrix format when a complete list requires a 2^n X n format.

Notwithstanding the forging arguments and evidence in favor of all cardinal infinities being equal, it is counter-intuitive to claim that there is a bijective mapping between the set of natural numbers (N) and the set of real numbers (R). If we try to identify and count the next real number above or below zero, for instance, we get stuck. So how do we know there are Aleph-1 members in set R? Take a line of real numbers and divide it in two, then divide each new line segment in two and repeat this exercise an infinite number of times:

Notice that the final result is 2^Aleph-0 or Aleph-1 infinitesimal line segments. Also notice that line segments are discrete and can be counted:

Imagine each line segment with a length of dx shrinking towards zero. The total number of line segments grows to infinity as they continue to shrink, but because they are discrete, they are countable. They only stop being countable when they all have zero length, but by the time they all reach zero length, we have counted all the members of R.

Therefore there is a bijective mapping between sets N and R and Aleph-0 equals Aleph-1. Additionally, we have shown above the that following is true:

References:

1. Cantor, Georg. Grundlagen einer allgemeinen Mannigfaltigkeitslehre (Foundations of a General Theory of Sets). jamesrmeyer.com.

2. Cantor, Georg. Uber eine elemtare Frage de Mannigfaltigketslehre (On an Elementary Question of Set Theory). jamesmeyer.com. 3. Cantor's Theorem. Wikipedia

4. 2020. SP20:Lecture 9 Diagonalization. courses.cs.cornell.edu

5. Cantor's Diagonal Argument. Wikipedia

6. Cardinality of the Continuum. Wikipedia

7. Continuum Hypothesis. Wikipedia

Why Different Infinities Are Really Equal

ABSTRACT:

Assuming different infinities are unequal leads to strange and counter-intuitive mathematical results such as Ramanujan's finite solutions to infinite divergent series. On the flip side, assuming different infinities are equal leads to infinite solutions to divergent infinite series. This last assumption, however, flies in the face of the current dogma: that different infinities are unequal. This paper offers proofs that show that different infinities are in fact equal.

Cantor's diagonal argument is used to prove that bijection fails between the set of non-negative integers (Z) and the set of real numbers (R). Even though both sets have an infinite number of members, the number of members of Z are not considered equal to the number of members of R. However, the implicit assumption is that infinities behave like finite numbers. If we assume infinity has no upper limit, or fixed value, Cantor's diagonal argument can be turned on its head.

In the diagrams below, we have the diagonal argument:

The top diagram is a list of real numbers paired with non-negative integers to the left. At the bottom diagram, the numbers along the diagonal are changed (in red). This creates a real number that is not on the list. We place that number in red below the list.

As you can see, we started with a list of infinite real numbers, so allegedly there are no positive integers beyond infinity we can assign to this new real number. Thus this new real number is considered uncountable. In fact, real numbers are considered uncountable because they are a continuum, and not discrete, but Cantor's diagonal process shows a way to count real numbers discretely, because if it is repeated, it produces one real number at a time. We can just simply add each newly produced real number to the total. But does this imply that R's infinity is greater than N's? Let's assume, arguendo, this is true. We can then justify counting the new real number with a so-called larger infinity, so we pair it with infinity + 1.

Now, assuming infinity has no upper limit, is infinity + 1 really larger than infinity? Equations 1 through 5 below prove that different infinities are equal.

We can easily show that infinity = infinity + 1, but if that's true, then infinity + 1 = infinity + 2 ... and so on all the way up to infinity^infinity and beyond. Every time we create a new real number, we are adding 1 to infinity which gives us back infinity. Using the diagonal method, we can create a new real number as many times as we like, to a point where we imagine a value greater than infinity. Given that infinity has no upper bound, it should be sufficient to cover any set with infinite members. New members can be added to a an infinite-member set and infinity should still be large enough to cover all the members.

Now here's the rub: if different infinities are equal, why do two lines appear to be unequal? Below, line n and line k each have an infinite number of points; yet, line n is longer than line k. Surely there are more points along line n than line k; albeit, the proof beneath the lines tells a different story:

At 12 we see that an infinite number of zero-points (0 * infinity) can equal either line n or line k. However, at 16 we see that infinity equals n * infinity equals k * infinity. This means that each point along n can be mapped to a unique point along k. This seems incredulous, so let's further test our conclusion. Consider the divergent infinite geometric series below. We derive two different infinities: s and sx. Let's assume they are unequal.

At 23 we end up with an inequality. Infinity simply does not equal a finite number! Albeit, Ramanujan would disagree. As a side note, we can see how Ramanujan got finite solutions to divergent series. He treated different infinities as though they were unequal. Now, watch what happens when we assume different infinities are equal (s = sx):

At 29 we get infinity like we should. This result adds further support to the conclusion that different infinities are equal. Based on what we know so far, the following two equations seem valid:

References:

1. Cantor, Georg. Grundlagen einer allgemeinen Mannigfaltigkeitslehre (Foundations of a General Theory of Sets). jamesrmeyer.com.

2. Cantor, Georg. Uber eine elemtare Frage de Mannigfaltigketslehre (On an Elementary Question of Set Theory). jamesmeyer.com. 3. Cantor's Theorem. Wikipedia

4. 2020. SP20:Lecture 9 Diagonalization. courses.cs.cornell.edu

5. Cantor's Diagonal Argument. Wikipedia

6. Cardinality of the Continuum. Wikipedia

7. Continuum Hypothesis. Wikipedia

Debunking Ramanujan's Finite Solutions to Divergent Series

ABSTRACT:

This paper shows how an infinite series yields both an infinite solution and a finite solution and how to determine which one is correct and which one is just plain crazy!

Imagine the following geometric series:

Is s convergent or divergent? Does it blow up to infinity or have a finite value? So far, we can't say. However, according to Ramanujan and his apologists, even if s is divergent, it has a finite value. Let's see if this is really the case. With a little algebra we derive equation 5 below:

What solutions will work for equation 5? Here are two:

One solution is infinity. If k is finite, the other solution is finite. This means, if we want, we can always extract a finite solution from a divergent series. It also means we can always extract an infinite solution from a convergent series! Ramanujan claimed that divergent series have finite solutions. He could have (but didn't) claim that convergent series have infinite solutions. If such claims seem counter-intuitive and just plain crazy, it's because they are. Extracting finite solutions from divergent series (or infinity from convergent series) violates the geometric series rules. Here are the rules:

At 9 we have a system of equations that show that the value of x matters. We don't just get to choose willy-nilly which equation we like best. If we choose the appropriate equation for a convergent series, we get a finite solution as we should. If we choose the appropriate equation for a divergent series, we get infinity as we should when x is greater than or equal to 1. Albeit, our system of equations doesn't cover x when it is equal to or less than -1. Let's start with the case where x equals -1:

Equation 10 doesn't converge, nor does it blow up to infinity. The partial sums oscillate between 1 and 0 forever. Mathematical physicists like to take the average of .5. However, it is obvious there are two solutions: 1 and 0. In the case where x is less than -1, the partial sums oscillate toward infinity and minus infinity. Again, there are two solutions. For negative numbers less than or equal to -1 we need two equations:

The complete system of equations is as follows:

Where x is less than or equal to -1, we can take advantage of a peculiar property of infinity: Infinity minus infinity doesn't have to equal zero, since infinity equals any number plus infinity:

Thus when x is less than -1 we can take the average of the two solutions s1 and s2 and get a finite solution. In fact, since n can be any finite number, there can be an infinite number of finite solutions.

Where x equals -1, there is one finite solution: .5.

Now, given what we know about infinity, finite numbers and the geometric series rules, what kind of solution(s) can we extract from the following series:

This is a tricky one, since there is no x variable to tell us which equation to use. So let's go with the usual method. We first take the average of the "a" series:

Then we do some fancy Ramanujan algebra to get the average of the "b" series:

Now that we have the b-series average, we can find solutions to the c-series:

At 38 and 39 above, we have two solutions: infinity and -1/12. One solution seems correct and the other seems counter-intuitive. Are they both valid? There is a test: simply sum up the series. The fastest way to do that is find the average number in the series, then multiply it by the number of terms. The average number and the number of terms are both infinity:

As expected, c = infinity. For any series, a good rule of thumb is if the series is divergent, a finite solution is invalid and just plain crazy! We can tell if a series equals infinity by looking at the last term. If the last term is infinite, then the series sum is infinite because infinity plus a bunch of other numbers equals infinity. We get a finite solution because the series itself does not know if it is convergent or divergent. Both convergent and divergent series are structured the same way: a sum of numbers--so the algebra is the same for both and yields more than one solution. Thus it is up to us to take a further step and test each solution.

N-body Problem Solutions

ABSTRACT:

This paper reveals a paradox within our current understanding of gravity that makes problems involving more than two bodies seemingly unsolvable. By introducing the alpha factor, the paradox is resolved and the solutions to various n-body problems become clear.

Imagine a skydiver falling to the earth's surface. Normally we can treat this as a 2-body problem--the two bodies being the skydiver and the earth. But suppose we didn't have the benefit of Newton's formula? Suppose we see this as an n-body problem because the earth is made up of n-1 particles, each with an individual mass and distance from the skydiver. The position and momentum of each particle, of course, is uncertain, so let's settle for the expectation values of the particles' positions, i.e., where the particles are most likely to be:

The diagram below is a representation of the distance vectors for each particle to the skydiver (green dot):

To simplify the problem, we determine the average position of the expected value positions of the particles (see blue dots below). Our n-body problem becomes a 5-body problem.

At equation 1 below we sum the individual masses to get the total mass M. At equations 2 through 4 we calculate the resultant vector R which shows us which way the skydiver will move. As it turns out, R is the distance between the skydiver and the earth's center of mass. Via observations we determine the constant G. This leads us to equation 5 which is Newton's formula.

Thus, we have solved an n-body problem along an x-y plane. We could have included the z-axis, but it wasn't necessary. It is often possible to rotate or orient the coordinate system so the problem can be solved in the simpler 2D format.

Note that the direction the skydiver falls is down, or along the y-axis. This is because the x directions cancel each other. The skydiver's movement along x is always zero. This presents a paradox, however! The diagrams below demonstrate that the skydiver will always fall at the same rate no matter how far apart along x the red and blue masses are from each other. The distance to the mass center is always the same.

Here's a 3-body problem: Imagine the red and blue bodies above are an infinite distance apart. The center of mass between them is the same as when the two bodies are both located at the center of mass. In fact, all the earth's particles gravitationally behave as if they are all located at the center of mass, no matter how close or far away they are from the sky diver. Yet, the inverse square law tells us that distance matters. Paradoxically, if distance matters, then Newton's formula, as it is normally applied, wouldn't give us the right answer. It would make more sense to simply sum up each particle's gravitational impact on the skydiver.

Unfortunately, in earth's case, Newton's formula gives us the right answer if we pretend that all the earth's particles are located at the center of mass. It is also unfortunate that it won't enable us to solve the 3-body problem suggested in the above diagrams. Newton's method predicts gravitational acceleration is the same whether the red and blue bodies are close to the green body or far away. Albeit, perhaps there's a way to solve the paradox. Here is my proposed solution:

Equations 6 and 7 above show how the skydiver's position changes along x (delta-x) and along y (delta-y). At the right side of equation 6, dt is the increment of time; X/|X| is the direction along the x-axis; X^2/R^2 is the fraction of acceleration along x; then there's the summation of alpha factors, body masses, and r distances each body is from the skydiver. Ditto for the y-axis at equation 7.

Now, I've introduced something new: the alphas. So what exactly are the alphas and what is their purpose? Examine equations 9 and 10 below:

If each alpha has the value suggested at 10, then the inner product yields the familiar Newton's equation. If each alpha equals one, we get a different outcome:

At equation 11 above we have a solution that makes sense for our paradoxical 3-body problem. So when is it appropriate to use equation 11 instead of equation 10? Both the earth and the 3-body problem appear similar, but there is a difference. This difference becomes more obvious if we think of the earth as accelerating towards the skydiver rather than vice versa. Earth's particles, for the most part, move as one towards the skydiver, so they all move along the same vector instead of individual vectors. If, for example, one or more of the earth's particles are perturbed by an outside force, the remaining particles will also very likely accelerate along the same vector as the perturbed particles. By contrast, if, say, an outside force perturbed the red body in our 3-body problem, the blue body need not move in lockstep with the red body. The two bodies are more independent. The value we assign to each alpha is determined by how independently the bodies move.

Equations 6 and 7 seem fine for falling skydivers, but what about orbiting satellites? Equations 12 through 17 below provide the relevant math for orbiting bodies:

Now, that an n-body solution has been proposed, let's see if we can prove it mathematically. At 18 below we start with what we know: Newton's equaton, then go from there. The final result is equation 24.

Below we use our new formula to solve the seemingly unsolvable 3-body problem. All the alphas are set to 1. The distance between the skydiver and the center of mass is still the same, but the red and blue bodies don't have to move in lockstep, so their distance apart now matters.

Equations 29 and 30 above tell us how much the skydiver moves along x and y during time interval dt. To see some animated simulations of 3-body and 6-body solutions, visit https://garmichaels.pythonanywhere.com/n_body_index/.

References: ChatGPT4 and Wikipedia.com

Quantum Entanglement and Teleportation Do Not Work the Way You Think

ABSTRACT:

Quantum entanglement can be explained using decks of cards. This paper shows how these decks of cards violate Bell's inequality, and that "hidden operators" determine a quantum measurement outcome. If these hidden operators could be known prior to measurement, one could predict the outcomes of quantum events. The herein thought experiment demonstrates that no faster-than-light communication takes place between entangled pairs. Further, when teleportation is scrutinized, there is no evidence that information passes through a wormhole.

According to John Stewart Bell, the following thought experiment should not violate his famous inequality: Start with two decks of cards joined together along the red-dotted line:

Because they are joined together, when the decks are randomly shuffled, they correlate (or anti-correlate). Now, divide the two decks along the same red-dotted line. Send one to Alice and the other to Bob. On the face of each card is a |0> or |1>. Neither Alice nor Bob know what card they will each deal from the top of their respective decks. It is as if the decks are in a state of superposition. However, if Alice deals a |0>, she knows which card Bob has dealt. Note there is no communication going on between the decks. They can be any distance apart and Alice will instantaneously know which card Bob has dealt.

Now, if Alice could just take a peek at the cards hidden in her deck before she deals them, she would be able to predict which card she will deal next with 100% certainty.

So, does this thought experiment demonstrate, by analogy, quantum entanglement? There is a test. Let's have Alice and Bob send each of their dealt cards to three polarizers: A, B and C. There is a chance each card will pass through one or more of the polarizers, or none at all. Alice and Bob keep track of the results and each create a data table matching the one below:

Since their datasets are identical, let's focus on Alice's. If a card passed through polarizer A, Alice records a 1, otherwise she records a 0. Same goes for polarizers B and C. At the right of the table, Alice checks if a card passed through A but not B (A!B). If so, she records a 1; otherwise, she records a 0. Then she checks if a card passed through B but not C (B!C) and records the result, and finally, she checks if a card passed through A but not C (A!C). She then adds up the columns. The final result is 2 + 2 is greater than or equal to 2. This is consistent with Bell's inequality:

Thus, according to Bell, the above thought experiment fails to accurately describe quantum entanglement. But what if Bell is wrong? Let's explore this possibility. Equation 2 below is another way to express Bell's inequality. It is expressed in terms of probabilities, sines and cosines, and, most importantly, it is a function of the polarizers' angles. Equations 3, 4, and 5 show that the data at the data table's first row would not be possible if the polarization angles were anything but zero or ninety degrees. This implies that the cards were oriented either vertically or horizontally (90 degrees or 0 degrees) and the polarizer slits were also vertical or horizontal (90 degrees or 0 degrees).

What if the polarization angles were not just zero or ninety degrees? Would Bell's inequality still be valid or would it be violated? Let's modify the above thought experiment. Suppose the cards are made of elastic material that enables them to squeeze through any opening; although, this is not guaranteed--there is a probability a card will squeeze through a polarizer that has an arbitrary polarization angle. Let's see if the cards can violate Bell's inequality if the polarization angles are altered.

Let's assume Alice's deck and Bob's deck are anti-correlated. If Alice deals a |0>, Bob deals a |1>. Below are the probabilities that their cards will pass through polarizers A, B and C:

On the left side of each equation above, the first term is the probablity that a card will pass through a given polarizer; the second term is the probability that a card will fail to pass through. Note that the probabilities add up to one as they should. Also note that the polarization angles are represented by theta-A, theta-B and theta-C. Using the above information, we can create the following version of Bell's inequality:

The first term is the probability of Alice's card passing through A times the probability of Bob's card not passing through B. The second term is the probability of Alice's card passing through B times the probability of Bob's card not passing through C. Finally, the term on the right of the inequality is the probability of Alice's card passing through A times the probability of Bob's card not passing through C. Using equation 6 above, we can calculate what polarization angles are needed to violate Bell's inequality:

At 9 above, we see that if theta-B is less than the term to its right, Bell's inequality will in fact be violated. Of course, theta-B is an arbitrary label. Let's swap it with theta-A:

At 11 above, we assign new angles to polarizers A, B and C: 0 degrees, 22.5 degrees and 45 degrees, respectively. When we plug these angles into inequality 6 and inequality 10, we find that our two decks of cards do in fact violate Bell's inequality. This means that the above thought experiment is an accurate analogy of the workings of quantum entanglement. It also implies that Einstein, Podolsky and Rosen (EPR) had an idea that may be valid after all: "hidden variables." This idea is not so far-fetched when you consider the fact that quantum mechanics has hidden operators. Here are a couple of examples:

The hidden operators are labeled in red. They are the NOT (X) and Identity (I) operators. When given a state of superposition, they secretly conspire with the Hadamard operator (H) to determine the outcome of a measurement. The result seems random to us, since we can't examine these hidden operators ahead of any measurement. Even if we could, we would know the order of the cards in the deck, so to speak, but we still wouldn't know how they got that way. We would be forced to fall back on, "They were randomly shuffled." Nevertheless, Alice could predict what her next card is going to be. She could even know ahead of time if the card will make it through a polarizer, since she might also know the "hidden probability operators."

Even if Alice can never know any hidden operators, she knows which card Bob dealt even if he is light-years away, and, the cards violate Bell's inequality. It appears the two decks of cards are truly entangled--is there a wormhole connecting them? If not, how does quantum teleportation work? There is a five-step teleportation process that apparently shows that Alice can instantaneously send a qubit to Bob, allegedly through a wormhole. Let's examine this process.

Alice wants to send Bob the following qubit (greek letter chi):

Teleportation step 1: Alice and Bob share an entangled pair of particles:

Find the tensor product of chi and the entangled pair:

Note that chi, Alice, and Bob's zeros and ones are colored black, blue and red, respectively. This enables us to track them. Now, we can use the table below to represent the original state of the system. The table has n units of ones and zeros in each column. The top portion has (alpha^2)n units, the bottom portion has (Beta^2)n units.

Teleportation step 2: Alice applies a CNOT gate:

Let's update the table:

Notice that Alice and Bob are no longer entangled! There is no guarantee that their ones and zeros will match. Technically this implies the wormhole is now closed or non-existent, since wormholes allegedly depend on entanglement. Yet, somehow Alice still manages to send the qubit to Bob? Let's continue. Step 3: Alice applies a Hadamard gate:

Time to update the table:

Step 4: Alice measures her pair (which includes chi's one or zero, and her own). Looking at the table above, notice that no changes were ever made to Bob's ones and zeros. Bob's final state is identical to his original state; yet, for example, when chi and Alice have a |00>, there is an alpha^2 probability Bob will have a |0> and a beta^2 probability Bob will have a |1>. Equation 22 confirms this. It's as if Alice really teleported the chi qubit to Bob, but Bob has no clue Alice tried to teleport anything. That's why step 5 is necessary. Alice must send Bob a no-faster-than-light signal through normal channels to let him know about her new disentangled measurements and what gates he must apply. But wouldn't it be faster if Alice uses the wormhole for step 5? What wormhole?

It should be reasonably obvious at this point that entanglement exists without the aid of wormholes or faster-than-light communication. Teleportation seems to work if both Bob and Alice assume Alice's ones and zeros have not changed; therefore, when they compare notes, it seems like Bob's ones and zeros take on new probabilities. This reminds me of relativity, where the train passes the station, or, is it the station that passes the train?

Acknowledgements:

Special thanks to ChatGPT4.

References:

1. Einstein, A., Poldolsky, B., Rosen, N. 03/25/1935. Can Quantum-Mechanical Description of Reality Be Considered Complete. Physical Review, Volume 47.

2. Leonard Susskind | "ER = EPR" or "What's Behind the Horizons of Black Holes?"". Archived from the original on 2021-12-11 – via www.youtube.com.

3. Einstein, A. Rosen, N. 07/01/1935. The Particle Problem in the Theory of General Relativity. Physical Review. 48.

4. McMahon, David. 2008. Quantum Computing Explained. John Wiley and Sons Inc.

5. Fadelli, Ingrid. 09/21/2023. A model probing the connection between entangled particles and wormholes in general relativity. Phys.org

6. Bell's Theorem. Wikipedia, ChatGPT4, and various online sources.