KJW

Surely this statement cannot be right. Say there are G axioms and axiom G is found to be provable from the other axioms A to F. Unless the proof of G is independent of one or more axioms, say B, how does this lead to an inconsistent set ? The point is that if an axiom is proven or disproven, then it is no longer an axiom. If the "axiom" is proven, it is a theorem. If the "axiom" is disproven, the set of axioms is inconsistent. Sometimes this is difficult to avoid. For example, in group theory, the existence of inverse elements can be expressed as: For each element [math]g[/math], there exists an inverse element [math]g^{-1}[/math] such that [math]g^{-1} g = g g^{-1} = e[/math] (the identity element). However, only one of the statements [math]g^{-1} g = e[/math] or [math]g g^{-1} = e[/math] is actually an axiom as the other statement can be proven. Which one is the axiom, and which one is the theorem is an arbitrary choice, so it seems natural to include both in a single statement so as to not arbitrarily break the symmetry.

This understanding is wrong. Well, it certainly is true that "any finite set of axioms is limited in what can be proven (or disproven) from them", so why is this not what (in)completeness is about? However, this is what axioms are. The problem is that it doesn't actually differentiate between statements that are axioms and statements that are theorems.

I believe what I said went into greater depth than that covered by this. To me, axioms are definitions, and as such are not subject to proof. I do reject the commonly held notion of axioms being "self-evident truths". What is completeness or incompleteness of a theory in your understanding? My understanding of (in)completeness is that any finite set of axioms is limited in what can be proven (or disproven) from them, and that additional axioms must be included to extend what can be proven (or disproven), ultimately requiring an infinite number of axioms to cover everything. But if any axiom can actually be proven (or disproven) from the other axioms, then there are too many axioms, and the set of axioms is potentially inconsistent. I see Gödel's Incompleteness Theorem as a statement that the set of all axioms is an open set in the analogous sense that the set [math]\{x \in \mathbb{R}: 0 < x < 1\}[/math] is an open set for which there is no largest or smallest number, and that there is no final axiom that completes the set of axioms, with the closure of the set [math]\{0,1\}[/math] rendering the axioms inconsistent. I say Gödel's Incompleteness Theorem is trivial because it is obvious that if one has a set of axioms that define a class of objects, then any particular object from that class will have properties that are unique to that particular object and not derivable from the axioms that define the class as a whole. To derive those unique properties require axioms that define that particular object within the class. And because the number of objects in the class are infinite, the total number of axioms that define all the individual objects in the class are also infinite. Thus, completeness becomes impossible.

For me, this seems to raise the question of what precisely is an "axiom"? Consider, for example, the axioms of a group. Then the commutative axiom of an Abelian group acts as a constraint on the structure of a general group. More can be said about Abelian groups than about general groups because everything that can be said about general groups can also be said about Abelian groups, whereas there are things that can be said about Abelian groups that cannot be said about general groups. But the axioms of multiplication do not constrain the structure of arithmetic. Instead, they make explicit an operation that already implicitly exists. The mere fact that addition exists implies that multiplication exists. The notation associated with the operation still needs to be defined, but this doesn't change the intrinsic properties of numbers (for example, whether they are prime) in the Platonic sense. Admittedly, I subscribe to the philosophical view that everything in mathematics exists unless proven otherwise. In the case of group theory, the notion of automorphisms add to what can be said about groups, though they do not act as constraints on groups. The automorphisms of a given group existed even before the notion of automorphisms were discovered. And yes, they were discovered, not merely invented. Although I'm not sure I properly understand this theorem, I am inclined to think it is rather trivial.

It is so because sentences that cannot be decided (incompleteness) in a stronger theory with addition and multiplication, cannot even be formulated in the weaker one which has only addition. So, Gödel's Incompleteness Theorem is a matter of mathematical notation? For example, I provided a definition of multiplication based on recursion. Although it apparently doesn't conform to the [math]a\ \circ\ b = c \Leftrightarrow \varphi(a,b,c)[/math] rule, it is still logically valid and probably shouldn't be rejected on the basis of excessively limiting notation.

Well, maybe Ukraine should be given back the nuclear weapons it gave up at the end of the Cold War in return for guaranteed sovereignty.

How is it possible that an incomplete arithmetic becomes complete by removing axioms?

Hmmm. I wasn't aware of Gödel's β function, so I looked it up. Anyway, I can see why it can't be used for defining multiplication in terms of addition, but I also fail to see how it can be used in practice for anything else. So how about demonstrating the use of the formula above to evaluate [math]7^5[/math]. Ok, I do find it interesting that binary operations are to be defined in the form: [math]a\ \circ\ b = c \Leftrightarrow \varphi(a,b,c)[/math] although I'm not sure that I agree in principle. Recursive definitions seem relatively straightforward, and although Gödel's β function seem to be intended to replace recursive definitions, they don't seem to be in any way practical. There is one disadvantage of recursive definitions: when used in computerised mathematics, each recursion takes up memory on the call stack. Perhaps Gödel's β function is a way to avoid call stack usage.

So, in what way is exponentiation in the form [math]a^b =c \Leftrightarrow \varphi(a,b,c)[/math]?

But you said that "multiplication cannot be formally defined from addition", whereas I formally defined multiplication from addition. Although I added a symbol to the language, this doesn't mean that it isn't defined in terms of addition. Therefore, it is only the definition of the symbol itself that can be considered axiomatic. It should be noted that recursion is not circular. The initial case, which does not include the notion being defined, acts as a STOP to the repetition.

I don't accept this. For integers, multiplication can be defined in terms of addition as: [math]n \times a = \left\{\begin{array}{cl} 0 & : \ n = 0 \\(n - 1)\times a + a & : \ n > 0 \\(n + 1)\times a - a & : \ n < 0 \end{array} \right.[/math] One thing I have come to realise is that recursive definitions are necessary when something is to be repeated an arbitrary number of times. Ellipses are often used for such things, but this lacks rigor.

{{a}, b} ≠ {{b}, a}

Thanks for the clarification. However, the specific statement I said was incorrect was: In standard GR, you cannot calculate the absolute scale of the geometry ([math]R_s[/math]) without knowing the central mass [math]M[/math] and [math]G[/math]. The formula I provided was an expression for [math]R_s[/math] that did not require knowing the central mass [math]M[/math] and [math]G[/math]. Now you are saying that [math]a[/math], the semi-major axis of the orbit, is also unacceptable. That's a case of moving the goalposts. However, I did anticipate this. There are other parameters that one can choose to specify an orbit than [math]a[/math]. For example, on page 6 of this thread, I derived the time dilation for an object in a circular orbit: [math]\dfrac{\Delta t_R}{\Delta t_\infty} = \sqrt{1 - \dfrac{3GM}{c^2 R}}[/math] Substituting [math]r_s = \dfrac{2GM}{c^2}[/math] gives: [math]\dfrac{\Delta t_R}{\Delta t_\infty} = \sqrt{1 - \dfrac{3}{2} \dfrac{r_s}{R}}[/math] Rearranging leads to: [math]R = \dfrac{\dfrac{3}{2} r_s}{1 - \left(\dfrac{\Delta t_R}{\Delta t_\infty}\right)^2}[/math] Substituting into [math]r_s = \dfrac{8\pi^2}{c^2} \dfrac{R^3}{T^2}[/math] (where [math]R = a[/math]) gives after rearranging: [math]r_s = \dfrac{c\ T}{\sqrt{27} \pi} \left(1 - \left(\dfrac{\Delta t_R}{\Delta t_\infty}\right)^2\right)^{3/2}[/math] Thus, we have an expression for [math]r_s[/math] in terms of the orbit period and time dilation of a circular orbit. Obviously, it would be better to have an expression for [math]r_s[/math] in terms of a general orbit rather than just a circular orbit, but I don't have the appropriate formulae for general orbits. That does not mean that such formulae do not exist in principle. The point is that it is quite incorrect to suggest that GR is not sufficiently powerful as a theory. GR is more difficult than Newtonian theory because it needs to provide corrections to Newtonian theory due to spacetime curvature that make equations more difficult to solve. But not having solutions because the equations are difficult to solve is not the same as such solutions not existing in principle. There has been discussion on this forum about whether energy-momentum is spacetime curvature or energy-momentum causes spacetime curvature. But either way, spacetime curvature is physically real and associated with energy-momentum. You are going to have a very difficult time trying to convince me that the ontology of physical reality is different.

Is the helium-4 superfluid state a BEC?

Well, this does seem to throw a monkey wrench into the Heisenberg Compensators used to enable the transporters in Star Trek. I hope Swansont didn't have any input into the consultations for that explanation ... The fine print of the teleporter brochure states that passengers and cargo being teleported are destroyed at the origin before being recreated at the destination.

Unfortunately, your 'simple' involves a notation system that didn't make it into Chem Eng courses in the '70s. Think crayons and picture books. That's more my level. 😄 The mathematics is actually simpler than the notation suggests. But let's try this: Suppose we have a function called CLONE( ) defined such that for all X, CLONE(X) = XX So, CLONE(U) = UU and CLONE(V) = VV Quantum mechanics requires CLONE( ) to be linear. Therefore, by definition of linearity: CLONE(U + V) = CLONE(U) + CLONE(V) = UU + VV But, by definition of CLONE( ): CLONE(U + V) = (U + V)(U + V) = UU + UV + VU + VV ≠ UU + VV Therefore, the definition of CLONE( ) is not consistent with the required linearity of CLONE( )

... so a particular unitary operator cannot switch between Model Ts, As, Bs etc in the same production line? A simple mathematical way to illustrate the no-cloning theorem is as follows: Suppose one can clone state [math]|\psi\!\!>[/math] as well as state [math]|\varphi\!\!>[/math]. Cloning state [math]|\psi\!\!>[/math] produces tensor product state [math]|\psi\!\!>\!|\psi\!\!>[/math]: [math]|\psi\!\!>\ \longrightarrow \ |\psi\!\!>\!|\psi\!\!>[/math] Similarly, cloning state [math]|\varphi\!\!>[/math] produces tensor product state [math]|\varphi\!\!>\!|\varphi\!\!>[/math]: [math]|\varphi\!\!>\ \longrightarrow \ |\varphi\!\!>\!|\varphi\!\!>[/math] Linearity of quantum mechanics requires that cloning state [math]|\psi\!+\!\varphi\!\!>[/math] produces: [math]|\psi\!+\!\varphi\!\!>\ \longrightarrow\ |\psi\!\!>\!|\psi\!\!>\! +|\varphi\!\!>\!|\varphi\!\!>[/math] But, a clone of state [math]|\psi\!+\!\varphi\!\!>[/math] is actually the tensor product: [math]|\psi\!+\!\varphi\!\!>\ \longrightarrow\ |\psi\!+\!\varphi\!\!>\!|\psi\!+\!\varphi\!\!>[/math] [math]= |\psi\!\!>\!|\psi\!\!>\! +|\psi\!\!>\!|\varphi\!\!>\! +|\varphi\!\!>\!|\psi\!\!>\! +|\varphi\!\!>\!|\varphi\!\!>[/math] [math]\ne |\psi\!\!>\!|\psi\!\!>\! +|\varphi\!\!>\!|\varphi\!\!>[/math] The above is not saying that states [math]|\psi\!\!>[/math] and [math]|\varphi\!\!>[/math] can't be cloned. But if these two states can be cloned, then the sum state [math]|\psi\!+\!\varphi\!\!>[/math] can't be cloned, proving that arbitrary states cannot be cloned.

It should be noted that the no-cloning theorem says that an arbitrary quantum state can't be copied. It doesn't say that particular quantum states can't be copied. The no-cloning theorem is a consequence of the linearity of quantum mechanics.

I'm not sure, but the only black hole solution I'm aware of that is not in an otherwise flat background is the De Sitter-Schwarzschild metric. For the Schwarzschild metric: [math](ds)^2 = -(1 - a r^{-1}) (c\ dt)^2 + (1 - a r^{-1})^{-1} (dr)^2 + r^2 ((d\theta)^2 + (\sin\theta\ d\phi)^2)[/math] and the De Sitter metric: [math](ds)^2 = -(1 - b r^{2}) (c\ dt)^2 + (1 - b r^{2})^{-1} (dr)^2 + r^2 ((d\theta)^2 + (\sin\theta\ d\phi)^2)[/math] for [math]b > 0[/math] the De Sitter-Schwarzschild metric is: [math](ds)^2 = -(1 - a r^{-1} - b r^{2}) (c\ dt)^2 + (1 - a r^{-1} - b r^{2})^{-1} (dr)^2 + r^2 ((d\theta)^2 + (\sin\theta\ d\phi)^2)[/math] for [math]b > 0[/math] The thing I found to be surprising is that one can simply superimpose the Schwarzschild metric directly onto the De Sitter metric. However, given that the Schwarzschild and De Sitter metrics have the same form, I suppose it is to be expected that they can be superimposed to produce another metric of the same form. By contrast, the Schwarzschild metric and FLRW metric do not have the same form, and so are probably not amenable to being directly superimposed.

That's not correct. From the formula for escape velocity (or alternatively from the Schwarzschild metric): [math]r_s = \dfrac{2GM}{c^2}[/math] From Kepler's third law of planetary motion: [math]\dfrac{a^3}{T^2} = \dfrac{GM}{4\pi^2}[/math] Eliminating [math]GM[/math] from both formulae: [math]r_s = \dfrac{8\pi^2}{c^2} \dfrac{a^3}{T^2}[/math] Thus, we have an expression for [math]r_s[/math] in terms an orbiting object without [math]G[/math] or [math]M[/math]. I recently said that using [math]r_s[/math] as the arbitrary constant in the Schwarzschild solution is purely mathematical, not requiring any connection to physical mass. Even the original equation: [math]R_{\mu \nu} = 0[/math] does not contain any reference to physical energy-momentum. However, there is a natural connect to distance within the mathematics. Not only can GR work without [math]G[/math] and [math]M[/math], but it takes extra effort to work with [math]G[/math] and [math]M[/math].

Is this topic somehow about Gödel's incompleteness theorems?

Sulfur is IUPAC. Also, aluminium is IUPAC, so the Americans don't have it all their own way.

No, that will not do. That would be like trying to prove the Riemann hypothesis by brute force. Unfortunately, what you are asking me about is galaxy rotation curves, of which I have little knowledge or interest. I don't think I can help you. I should remark that I am firmly in the dark matter camp for two reasons: 1, the alternative is a modified law of gravitation, which I reject on the basis of my wholehearted acceptance of general relativity; and 2, the Bullet Cluster and other galaxy cluster collisions, exhibit gravitational lensing that strongly support dark matter over a modified law of gravitation. However, I don't have any particularly strong opinions about dark matter candidates. In the beginning, I rejected a cosmological constant on the basis that any such constant could be absorbed into the energy-momentum source term of the Einstein field equation. But sometime later, I had an epiphany. The cosmological constant has a property that is not possessed by energy-momentum in general: It is invariant to Lorentz transformations. This means it is not possible to establish one's velocity relative to the cosmological constant. The reason why one can determine the velocity of an ordinary object (relative to oneself) is because that object breaks Lorentz symmetry. That is, the object appears different at different velocity, and that difference is used to determine the velocity. But the inability to determine the velocity of the cosmological constant means that the cosmological constant looks like empty space. However, that empty space still appears curved though that curvature can't establish its velocity. It's worth noting that although the speed of light is invariant to Lorentz transformations, light itself is not. A Lorentz transformation changes both the frequency and wavelength of a given light. Empty space does not behave this way, so light does distinguish itself from empty space. Thus, the cosmological constant will naturally distinguish itself from the energy-momentum in the Einstein field equation.

I tend to think opposites sign unify them Why would opposite signs unify them? □ψ=0 is a second-order partial differential equation to be solved for ψ. The sign of the individual terms of the □ operator correspond to the signature of the metric. Solutions exist regardless of the signature, though the solutions themselves do depend on the signature. For the signature (+,+,+,+), □ψ=0 is called a "potential equation", though such an equation is best known in two and three dimensions. Just a question,if you make light keep orbiting the sun using whichever means for one year...what will be the length of helical trajectory? When I wrote this, it was my intention that it be visualised in an ordinary three-dimensional Euclidean space where the vertical dimension represents time that is scaled correctly but ignoring such things as time dilation or other relativistic effects. That is why I said "about one light-year" instead of "exactly one light-year" or even just "one light-year". In the case of light, relativity can't be ignored. The corresponding trajectory of light will have a completely different length in spacetime than the Euclidean space in which it is intended to be visualised. In spacetime, lightlike trajectories have zero length. And it is the opposite signs in the signature that enable such trajectories to exist.

https://en.wikipedia.org/wiki/Group_(periodic_table)