KJW

The Proof of the 3X + 1 Conjecture

But you need sodium sulfate or magnesium sulfate to be present for the reaction to continue because these provide the ions that are migrating across the salt bridge to electrically neutralise the hydrogen ions and hydroxide ions that are formed at the electrodes. One other consequence of the low solubility of magnesium hydroxide is that the required voltage across the electrodes will be lower because the produced hydroxide ions will be removed from solution (Le Chatelier's principle).

One difference between using sodium sulfate and using magnesium sulfate is that the product magnesium hydroxide has very low solubility whereas the product sodium hydroxide is very soluble.

I don't know. I assume the quantum Zeno effect to be quite general. I have even simulated the effect on an Excel spreadsheet.

There is the quantum Zeno effect where the rate of decay is affected by the frequency of measurement.

In Australia at the moment, there is discussion of forcing social network platforms to age-verify their participants to prevent children from being exposed to inappropriate content. But this seems to me like a backdoor way to require everyone to provide ID to use the internet. We have a term for governments that wish to implement policies of this nature... we call them "nanny states".

You don't need to do anything. A thermos keeps hot things hot and cold things cold. Or do you need to make a thermos from scratch? If so, why? A thermos is basically a container inside another container with a vacuum in the space between the containers and the inner container having a mirrored surface.

Where did the idea come from that people in the time of Christopher Columbus thought the earth was flat?

The way I see it, you would have to do your measurements on water rather than on land because the effect you are trying to measure would be swamped by the topography of the land. On the water, you'd have to contend with waves, but an average over multiple measurements would tend to iron out that problem.

Well, I am assuming that we are not living on the edge of a disc, a case that would be invalidated by observing an eclipse that occurs around midnight. Anyway, I've observed enough lunar eclipses occurring at various times during the night to know that the earth can only be spherical.

One can observe the curvature of the earth's shadow on the moon during a lunar eclipse. And by observing lunar eclipses that occur shortly after sunset or shortly before sunrise, it is clear that the shadow is due to a spherical object.

Ok. This would appear to resolve the inconsistency I attempted to create by having a metric that can be coordinate-transformed to the Minkowskian metric but with a Riemann tensor that is not everywhere zero. I was thinking that a choice had to be made between the coordinate transformation and the Riemann tensor. However, because the metric: [math]\text{diag}(−1, 1, 1, 1 + H(z) z^2 + H(−z) z^4)[/math] is everywhere continuous and invertible, the coordinate transformation to the Minkowskian metric is a diffeomorphism.

I have a question for you: Suppose I have a metric that is pathological in some way. But there exists a coordinate transformation that is everywhere continuous and invertible from this metric to the Minkowskian metric. However, due to the pathological nature of the metric, the Riemann tensor is not everywhere zero. Is the spacetime described by the metric flat?

But, we are not dealing with mixed partial derivatives. So, even if [math]R_{ijkl} = -R_{jikl}[/math] doesn't hold in general due to discontinuity, [math]R_{zzzz} = -R_{zzzz} = 0[/math] must still hold.

I don't think you fully grasped the logic behind this sentence from my previous post: "And because the metric is flat regardless of the function [math]Z(z)[/math], in the Riemann tensor, the cancellation of derivatives must be exact regardless of any pathologies in the function [math]Z(z)[/math]." That is, the cancellation of derivatives in the Riemann tensor must occur for reasons that have nothing to do with the properties of [math]Z(z)[/math]. To be specific, of the 64 first-order partial derivatives of the above metric tensor components, the only one that is not exactly zero is [math]g_{zz,z}[/math], and of the 256 second-order partial derivatives of the above metric tensor components, the only one that is not exactly zero is [math]g_{zz,zz}[/math]. Therefore, the only component of the corresponding Riemann tensor that could possibly be non-zero is [math]R_{zzzz}[/math]. But the Riemann tensor is antisymmetric on the first two indices, [math]R_{ijkl} = -R_{jikl}[/math]. Therefore, [math]R_{zzzz} = -R_{zzzz} = 0[/math]. Thus, all components of the Riemann tensor for the above metric are exactly zero (everywhere).

Those are certainly involved in the calculation of the Riemann tensor, aren't they? I didn't calculate the Riemann tensor because I had an alternative method of proving the metric is flat. And because the metric is flat regardless of the function [math]Z(z)[/math], in the Riemann tensor, the cancellation of derivatives must be exact regardless of any pathologies in the function [math]Z(z)[/math]. However, I will attempt to answer your question: [math]H'(z) = \delta(z)[/math] [math]H''(z) = \dfrac{d}{dz} \delta(z) =\ ...[/math] I'll have to refresh my memory of distributions before answering this. Let [math]T(u)[/math] be an arbitrary test function of [math]u[/math]. Then: [math]\int_{-\infty}^{\infty} T(u)\ \delta(u-x)\ du = T(x)[/math] [math]\int_{-\infty}^{\infty} T(u)\ \dfrac{d}{du} \delta(u-x)\ du = -\int_{-\infty}^{\infty} \dfrac{d}{du} T(u)\ \delta(u-x)\ du = -T'(x)[/math]

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.