KJW

Am I sure that the metric describes flat spacetime? Yes, I am. Any metric of the form: (ds)² = T(t)² c² (dt)² – X(x)² (dx)² – Y(y)² (dy)² – Z(z)² (dz)² describes flat spacetime. The following coordinate transformation exists between this metric and the Minkowskian metric: t' = t'(t) ; x' = x'(x) ; y' = y'(y) ; z' = z'(z) where t'(t), x'(x), y'(y), and z'(z) are solutions to the differential equations: dt'(t)/dt = T(t) dx'(x)/dx = X(x) dy'(y)/dy = Y(y) dz'(z)/dz = Z(z) Ignoring your use of the (–,+,+,+) signature, your metric is of the above form: T(t) = X(x) = Y(y) = 1 Z(z) = H(z)(1 + z2)½ + H(–z)(1 + z4)½

You do realise that your high school teacher was teaching children? Well, maybe not children, but certainly not adults.

Checking the link... yes. The link opens to a webpage that asks if you are a robot. Verifying you are human takes you the webpage of the paper. From the webpage of the paper, you can view the PDF file, from which it can be saved.

Why would there be problems? So what if the second derivative has a jump discontinuity. If the source term of the EFE was describing a ball of matter, then there would be a jump discontinuity at the surface of the ball, and the solution metric would be a cookie-cut of the metric describing the ball of matter into the Schwarzschild metric, with matched zeroth and first derivatives at the surface of the ball. However, the metric: [math]\text{diag}(−1,1,1,1+H(z)z^2+H(−z)z^4)[/math] does describe a flat spacetime, so all the derivatives will cancel to zero in the Riemann tensor field.

Functions of this type do seem to crop up every now and then in various places. They seem to highlight the notion of how few the functions are that can be expressed in terms of elementary functions. Thanks. I think the authors invented the Leal-functions. I imagine being a contemporary of someone like Euler wondering if this newfangled notation will just be a flash in the pan or whether it will become a permanent part of mathematics.

It seems that the ads have come back.

This would also allow the neutrons to approach each other ~5 orders of magnitude closer than in atomic systems.

I have been trying to understand this for some time but still fail. A flat spacetime metric cannot be obtained by coordinate transformation from an non-flat expanding space-only metric? How does this refute an expanding time-only metric? A flat spacetime metric cannot be obtained from a non-flat spacetime metric by a coordinate transformation. A flat spacetime metric can only be obtained from a flat spacetime metric by a coordinate transformation. This is central to the mathematics of general relativity. In general, a tensor that is zero in one coordinate system is zero in every coordinate system, and curvature is a tensor, so a curvature of zero (ie flat) in one coordinate system is zero (flat) in every coordinate system. An FLRW metric of an expanding flat space is a non-flat spacetime (do not confuse space with spacetime). But I have shown that an expanding time only metric can be coordinate-transformed to a flat spacetime metric. Therefore, the expanding time only metric is a flat spacetime. And because it is a flat spacetime, it cannot be obtained from a non-flat spacetime such as an FLRW metric of an expanding flat space by a coordinate transformation.

I think I should clarify my question. I know that the strong force binds quarks within nucleons, and that the binding between nucleons is due to the residual strong force similar to the van der Waals force between neutral atoms or molecules. However, this was not what I meant by "the nuclear equivalent of electromagnetic van der Waals forces". Neutrons have both positive and negative electromagnetically charged quarks, and therefore it is not unreasonable to consider if there exists an induced electric dipole to induced electric dipole attractive force that is similar to the van der Waals force. That is, the attractive force between neutrons I was enquiring about was electromagnetic and the result of an induced charge distribution within the neutrons.

You appear to be talking about the strong force. But is there the nuclear equivalent of electromagnetic van der Waals forces that would also lead to attraction?

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Thanks.

One thing about the integral of [math]x^n[/math] that I find interesting is the case of [math]n = -1[/math]: [math]\displaystyle \int x^n\, dx = \begin{cases}\ \dfrac{x^{n+1}}{n+1} + C & \text{if } n \neq -1 \\ \\ \ \log(x) + C &\text{if } n = -1 \end{cases}[/math] Note that: [math]\displaystyle \int x^{-1 + \varepsilon} \ dx = \dfrac{x^{\varepsilon}}{\varepsilon} + C[/math] for all [math]\varepsilon \neq 0[/math] regardless of how small [math]\varepsilon[/math] is. Furthermore, note that [math]x^{-1 - \varepsilon}[/math] can be deformed to [math]x^{-1 + \varepsilon}[/math] without discontinuity at [math]x^{-1}[/math]. Therefore, one would expect that: [math]\displaystyle \int x^{-1 - \varepsilon} \, dx[/math] can be deformed to: [math]\displaystyle \int x^{-1 + \varepsilon} \ dx[/math] without discontinuity at: [math]\displaystyle \int x^{-1} \ dx[/math] even though the above formula seems to indicate that this is not the case. But let's consider the definite integral: [math]\displaystyle \lim_{\varepsilon \to 0} \displaystyle \int_{1}^{x} u^{-1 + \varepsilon} \ du[/math] [math]= \displaystyle \lim_{\varepsilon \to 0} \dfrac{x^{\varepsilon} - 1}{\varepsilon}[/math] [math]= \log(x)[/math] Thus, it can be seen that the definite integral of [math]x^{-1 + \varepsilon}[/math] is continuous with respect to [math]\varepsilon[/math] at [math]x^{-1}[/math]. Interestingly, this notion can be extended to the definite integral of [math]\log(x)[/math] as follows: [math]\displaystyle \int_{1}^{x} \log(v) \ dv[/math] [math]= x \log(x) - x + 1[/math] And: [math]\displaystyle \lim_{\varepsilon \to 0} \displaystyle \int_{1}^{x} \displaystyle \int_{1}^{v} u^{-1 + \varepsilon} \ du \ dv[/math] [math]= \displaystyle \lim_{\varepsilon \to 0} \displaystyle \int_{1}^{x} \dfrac{v^{\varepsilon} - 1}{\varepsilon} \ dv[/math] [math]= \displaystyle \lim_{\varepsilon \to 0} \dfrac{x^{\varepsilon + 1}}{\varepsilon (\varepsilon + 1)} - \dfrac{x}{\varepsilon} - \dfrac{1}{\varepsilon (\varepsilon + 1)} + \dfrac{1}{\varepsilon}[/math] [math]= \displaystyle \lim_{\varepsilon \to 0} \dfrac{x^{\varepsilon + 1}}{\varepsilon (\varepsilon + 1)} - \dfrac{x (\varepsilon + 1)}{\varepsilon (\varepsilon + 1)} - \dfrac{1}{\varepsilon (\varepsilon + 1)} + \dfrac{(\varepsilon + 1)}{\varepsilon (\varepsilon + 1)}[/math] [math]= \displaystyle \lim_{\varepsilon \to 0} \dfrac{x^{\varepsilon + 1}}{\varepsilon (\varepsilon + 1)} - \dfrac{x \varepsilon}{\varepsilon (\varepsilon + 1)} - \dfrac{x}{\varepsilon (\varepsilon + 1)} - \dfrac{1}{\varepsilon (\varepsilon + 1)} + \dfrac{\varepsilon}{\varepsilon (\varepsilon + 1)} + \dfrac{1}{\varepsilon (\varepsilon + 1)}[/math] [math]= \displaystyle \lim_{\varepsilon \to 0} \dfrac{x^{\varepsilon + 1}}{\varepsilon (\varepsilon + 1)} - \dfrac{x \varepsilon}{\varepsilon (\varepsilon + 1)} - \dfrac{x}{\varepsilon (\varepsilon + 1)} + \dfrac{\varepsilon}{\varepsilon (\varepsilon + 1)}[/math] [math]= \displaystyle \lim_{\varepsilon \to 0} \dfrac{x^{\varepsilon + 1}}{\varepsilon} - \dfrac{x \varepsilon}{\varepsilon} - \dfrac{x}{\varepsilon} + \dfrac{\varepsilon}{\varepsilon}[/math] [math]= \displaystyle \lim_{\varepsilon \to 0} \dfrac{x^{\varepsilon + 1}}{\varepsilon} - x - \dfrac{x}{\varepsilon} + 1[/math] [math]= x \Big(\displaystyle \lim_{\varepsilon \to 0} \dfrac{x^{\varepsilon} - 1}{\varepsilon}\Big) - x + 1[/math] [math]= x \log(x) - x + 1[/math] However, if one starts with [math]x^{\varepsilon}[/math] and form the derivative: [math]\displaystyle \lim_{\varepsilon \to 0} \dfrac{dx^{\varepsilon}}{dx}[/math] [math]= \displaystyle \lim_{\varepsilon \to 0} \varepsilon x^{\varepsilon - 1}[/math] [math]= 0[/math] If we consider [math]\varepsilon[/math] to be small but not infinitesimal, then for the integral, we start with [math]x^{\varepsilon - 1}[/math] and end with [math]\dfrac{x^{\varepsilon}}{\varepsilon}[/math], whereas for the derivative, we start with [math]x^{\varepsilon}[/math] and end with [math]\varepsilon x^{\varepsilon - 1}[/math]. That is, the derivative is smaller than the integral by factor of [math]\varepsilon[/math], becoming zero in the limit. Thus, although repeated integration starting from [math]x^{\varepsilon - 1}[/math] can use the power function integration formula, the resulting sequence of functions are distinct from power functions obtained by starting from, for example, [math]x^0[/math].

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I'm inclined to think that "dangerous chemical" means dangerous to those who work with the chemical as a chemical as well as to those in the vicinity of any accident from working with the chemical. Dangerous chemicals require more stringent safety protocols, which reduce the likelihood of deaths but not the danger. The danger from sugar does not come from it being a chemical, but rather from it being a food. Similarly, the danger from drowning in water does not come from water being a chemical. On the other hand, safety protocols demand that no one travel in an elevator with liquid nitrogen. That is, liquid nitrogen might not be especially dangerous, but it does have its hazards which can lead to death. Ethers are not especially dangerous... unless they're old, in which case, distilling them can lead to an explosion. Also, dangerous chemicals need not be just about death, but also serious injury. For example, osmium tetroxide is dangerous because it can lead to blindness if any gets on the eyeball.

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According to Wikipedia, nitrogen triiodide is more sensitive, being the only known chemical explosive that detonates when exposed to alpha particles and nuclear fission products. I doubt that. It is my understanding that the most toxic known substance is botulinum toxin, with an estimated human median lethal dose of 1.3–2.1 ng/kg Interestingly, what might actually be the strongest known acid, the only known acid to protonate carbon dioxide, carborane acid, is considered to be "gentle". I often walk past 1kg bags of sugar while shopping in a supermarket. I do so without any fear that my life is in danger. I can't exactly say the same about lithium-ion batteries in the home. And if I saw "chlorine trifluoride" written on a railway tanker somewhere, I think I would very much like to be somewhere else.

Sugar a most dangerous chemical??? You people have a weird notion of what a dangerous chemical is. I'm going with the stuff that burns through concrete.

@Max70, you appear to have the view that the expansion of the universe can be explained by the tidal effect external to a gravitational source. No, such a tidal effect has the property of being "volume preserving". In other words, a free-falling sphere distorts into the shape of a prolate spheroid of the same volume. On earth, this ideally gives rise to two antipodal high tides separated by a ring of low tide. By contrast, the universe is expanding in all directions. A free-falling sphere becomes a larger sphere... not volume preserving. It's worth noting that the flat-space FLRW spacetime that ideally describes our universe is entirely devoid of the type of curvature associated with a black hole.

You have the correct products, but you've drawn the loops incorrectly. The oxygen atom on each carboxylate anion comes from the hydroxide ion, not the glycerol molecule. The nucleophilic attack by hydroxide ion is on the carbonyl carbon atom, not the glyceryl carbon atoms. And we know this from isotopic labelling experiments.

Consider the differential equation: [math]\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - 2/x}} = \dfrac{\sqrt{x}}{\sqrt{x - 2}}[/math] Let: [math]x = u + 1[/math] ; [math]dx = du[/math] [math]\dfrac{dy}{du} = \dfrac{\sqrt{u + 1}}{\sqrt{u - 1}} = \dfrac{\sqrt{u + 1}}{\sqrt{u - 1}} \dfrac{\sqrt{u + 1}}{\sqrt{u + 1}}[/math] [math]= \dfrac{u + 1}{\sqrt{u^2 - 1}}[/math] [math]y - C = \sqrt{u^2 - 1} + \textrm{arccosh}(u)[/math] Let: [math]u = \cosh(v)[/math] [math]y - C = \sinh(v) + v[/math] [math]v = \textrm{Lsinh}_2(y - C)[/math] [math]u = \cosh(\textrm{Lsinh}_2(y - C))[/math] Therefore: [math]x = \cosh(\textrm{Lsinh}_2(y - C)) + 1[/math]

While attempting to solve the differential equation: [math]\dfrac{dr'}{dr} = \dfrac{1}{\sqrt{1 - \dfrac{2GM}{c^2 r}}}[/math] expressing [math]r[/math] in terms of [math]r'[/math], I encountered a novel family of transcendental functions called "Leal-functions". These functions are similar to the Lambert W function (the function [math]W(x)[/math] that solves [math]W(x)e^{W(x)} = x[/math]), but (apparently) can't be derived from it. The link to the full article about these functions: https://www.sciencedirect.com/science/article/pii/S2405844020322611 The link to the section that defines these functions: https://www.sciencedirect.com/science/article/pii/S2405844020322611#se0040 Below is a list of Leal functions and their definitions: [math]y(x) = \textrm{Lsinh}(x)[/math] [math]\iff[/math] [math]y(x) \sinh(y(x)) = x[/math] [math]y(x) = \textrm{Lcosh}(x)[/math] [math]\iff[/math] [math]y(x) \cosh(y(x)) = x[/math] [math]y(x) = \textrm{Ltanh}(x)[/math] [math]\iff[/math] [math]y(x) \tanh(y(x)) = x[/math] [math]y(x) = \textrm{Lcsch}(x)[/math] [math]\iff[/math] [math]y(x) \textrm{ csch}(y(x)) = x[/math] [math]y(x) = \textrm{Lsech}(x)[/math] [math]\iff[/math] [math]y(x) \textrm{ sech}(y(x)) = x[/math] [math]y(x) = \textrm{Lcoth}(x)[/math] [math]\iff[/math] [math]y(x) \coth(y(x)) = x[/math] [math]y(x) = \textrm{Lln}(x)[/math] [math]\iff[/math] [math]y(x) \ln(y(x) + 1) = x[/math] [math]y(x) = \textrm{Ltan}(x)[/math] [math]\iff[/math] [math]y(x) \tan(y(x)) = x[/math] [math]y(x) = \textrm{Lsinh}_2(x)[/math] [math]\iff[/math] [math]y(x) + \sinh(y(x)) = x[/math] [math]y(x) = \textrm{Lcosh}_2(x)[/math] [math]\iff[/math] [math]y(x) + \cosh(y(x)) = x[/math] The authors say that the Leal family of functions can be extended to solve other transcendental equations, and provide examples of other similar functions. They even say that users can propose their own functions, applying the methodology used in the article. It turns out that the solution to the above differential equation for the coordinate transformation of the [math]g_{rr}[/math] component of the Schwarzschild metric to [math]g_{r'r'} = -1[/math] involves the [math]\textrm{Lsinh}_2(x)[/math] Leal-function defined above.

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It's funny that you say this because I have also had the idea that the arrow of time is connected to spinors. If you disagree with the ontology, then in what way are you agreeing with special and general relativity? It seems to me that you think time dilation is a physical effect acting on clocks. This conflicts with the principle of relativity which says that the laws of physics are the same in all frames of reference. This means that an ideal clock ticks at the same intrinsic rate in all frames of reference, and therefore time dilation is the result of something other than a physical effect acting on the clock. You say you agree with the equations, but you seem to disagree with the principles upon which the equations are based. It's as if you think Einstein got lucky with a wrong theory that happens to make correct predictions.