KJW

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.

I found this article titled "Calculus Before Newton and Leibniz - An in-depth article on the history of calculus". Here is the introductory section of the article: The Development of Calculus History has a way of focusing credit for any invention or discovery on one or two individuals in one time and place. The truth is not as neat. When we give the impression that Newton and Leibniz created calculus out of whole cloth, we do our students a disservice. Newton and Leibniz were brilliant, but even they weren’t capable of inventing or discovering calculus. The body of mathematics we know as calculus developed over many centuries in many different parts of the world, not just western Europe but also ancient Greece, the Middle East, India, China, and Japan. Newton and Leibniz drew on a vast body of knowledge about topics in both differential and integral calculus. The subject would continue to evolve and develop long after their deaths. What marks Newton and Leibniz is that they were the first to state, understand, and effectively use the Fundamental Theorem of Calculus. No two people have moved our understanding of calculus as far or as fast. But the problems that we study in calculus—areas and volumes, related rates, position/velocity/acceleration, infinite series, differential equations—had been solved before Newton or Leibniz was born. It took some 1,250 years to move from the integral of a quadratic to that of a fourth-degree polynomial. But awareness of this struggle can be a useful reminder for us. The grand sweeping results that solve so many problems so easily (integration of a polynomial being a prime example) hide a long conceptual struggle. When we jump too fast to the magical algorithm and fail to acknowledge the effort that went into its creation, we risk dragging our students past that conceptual understanding. This article explores the history of calculus before Newton and Leibniz: the people, problems, and places that are part of the rich story of calculus.

I don't know precisely how Newton or Leibniz obtained the product rule of differential calculus, but it seems rather easy to obtain to me: [math]\text{By definition:}[/math] [math]\dfrac{df(x)}{dx} \buildrel \rm def \over = \displaystyle \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}[/math] [math]\text{Therefore:}[/math] [math]\dfrac{df(x)g(x)}{dx} = \displaystyle \lim_{h \to 0} \dfrac{f(x + h) g(x + h) - f(x)g(x)}{h}[/math] [math]= \displaystyle \lim_{h \to 0} \dfrac{f(x + h) g(x + h) - f(x) g(x + h) + f(x) g(x + h) - f(x)g(x)}{h}[/math] [math]= \displaystyle \lim_{h \to 0} \dfrac{f(x + h) g(x + h) - f(x) g(x + h)}{h} + \displaystyle \lim_{h \to 0} \dfrac{f(x) g(x + h) - f(x)g(x)}{h}[/math] [math]= \displaystyle \lim_{h \to 0} \dfrac{f(x + h) g(x) - f(x) g(x)}{h} + \displaystyle \lim_{h \to 0} \dfrac{f(x) g(x + h) - f(x)g(x)}{h}[/math] [math]= \left(\displaystyle \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}\right) g(x) + f(x) \left(\displaystyle \lim_{h \to 0} \dfrac{g(x + h) - g(x)}{h}\right)[/math] [math]= \dfrac{df(x)}{dx} g(x) + f(x) \dfrac{dg(x)}{dx}[/math]

[math]\buildrel \rm def \over =[/math]

Yes. The expanding time only metric is flat, whereas the FLRW metric is not flat, so there is no coordinate transformation between them. In this thread, I demonstrated that an expanding space and time metric does not exist as distinct from an expanding space only metric. I did this by showing that an expanding space and time metric can be transformed to an expanding space only metric by a coordinate transformation, thus proving that they are physically the same. Mathematically however, the manifestly conformally flat form of the metric is useful for two reasons: 1, it indicates that the Weyl conformal tensor field is zero (the Weyl conformal tensor field represents pure gravitation, and therefore the only gravitation in the flat space FLRW spacetime is that which is intrinsic to the energy-momentum tensor field itself); and 2, it simplifies the equation of light-like geodesics (though this also assumes the homogeneity and isotropy of the FLRW spacetime¹). My most recent post in this thread demonstrated that there are limits to what coordinate transformations can do. By showing that an expanding time only metric can be transformed to a flat spacetime metric by a coordinate transformation, I established that it cannot be obtained from an expanding space only metric, which is not flat, by a coordinate transformation. It is worth noting that 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) ¹ Although the equations for the light-like trajectories in the manifestly conformally flat form of the metric given earlier in this thread are indeed light-like trajectories, I did not prove that these are geodesics. To prove that they are geodesics, note that in the two-dimensional tx-spacetime, the light-like trajectory passing through a given point from a given direction is unique, and therefore must be geodesic. But to prove that in the four-dimensional spacetime, the light-like geodesic doesn't deviate from the two-dimensional tx-spacetime, the symmetry arising from homogeneity and isotropy can be invoked, requiring that the unique light-like geodesic remain in the two-dimensional tx-spacetime (symmetry breaking can only occur if there are multiple solutions).

No, it is not possible to coordinate-transform a metric of the form: (ds)² = c² (dt)² – a(t)² ((dx)² + (dy)² + (dz)²) to a metric of the form: (ds)² = α(t)² c² (dt)² – ((dx)² + (dy)² + (dz)²) In general, a metric of the form: (ds)² = c² (dt)² – a(t)² ((dx)² + (dy)² + (dz)²) has non-zero Ricci curvature, whereas a metric of the form: (ds)² = α(t)² c² (dt)² – ((dx)² + (dy)² + (dz)²) describes flat spacetime. Note that this metric can be transformed to the Minkowskian metric by the coordinate transformation: t' = t'(t) ; x' = x ; y' = y ; z' = z where t'(t) is a solution to the differential equation: dt'(t)/dt = α(t)

Note that Markus Hanke said that "uniform acceleration is locally equivalent to a uniform gravitational field" [my bold]. The distant stars are not part of a local measurement.

All redshifts and blueshifts ultimately have the same cause: Consider some arbitrary spacetime in which there is an arbitrary trajectory of an emitter and an arbitrary trajectory of a receiver. From the emitter, consider two infinitesimally separated future-directed light-like geodesics that eventually intersect the trajectory of the receiver. Then the ratio of the proper time along the trajectory of the emitter between the two light-like geodesics, and the proper time along the trajectory of the receiver between the two light-like geodesics, is the Doppler ratio (redshift or blueshift) that the receiver observes at the instant of receiving the two light-like geodesics from the emitter. From the above, one may consider more special cases where the cause of the redshift and blueshifts can be differentiated between relative velocity, acceleration, gravitation, or cosmological, as well as combinations of these. For example, constant acceleration in the Minkowskian metric can be considered in terms of relative velocity, whereas the same constant acceleration in its own frame of reference can be considered in terms of pure acceleration. In other words, the change in metric, even though the physics has not changed, leads to a change in interpretation. But in all cases, the above applies. And note that the above has been expressed entirely in terms of invariants.

In the case of the twin paradox, the redshift and blueshift are due to the relative velocity between the source and receiver. In the case of gravitational redshift and blueshift, equivalent to accelerational redshift and blueshift, there is no relative velocity between the source and receiver. The redshift and blueshift is due to the acceleration itself, and depends on the displacement between source and receiver. If one examines the metric of an accelerated frame of reference, sources that are at rest and above the receiver (note that the acceleration is directed upward) will be blueshifted, while sources that are at rest and below the receiver will be redshifted. It is worth noting that the source that is at rest below the receiver has a greater acceleration than the receiver, and that the source that is at rest above the receiver has a lesser acceleration than the receiver. However, this does not determine the redshift or blueshift, and it might not apply to gravitation.

You also have to worry about whether particular codes are recognised on this site... some aren't. So even if you write something using a LaTeX editor and everything looks fine, some of it may not be recognised when transferred to this site. Thus, it is always a good idea to test something you've not used before in The Sandbox before posting it in a thread for real.

I find it quite bizarre that some people regard it as significant that the image on a retina is upside down. It's like wondering how the images on a DVD are turned upright on a TV screen.

One other way to view rapidity it to consider three observers, A, B, and C in relative collinear motion. The Doppler ratio between observers A and C is the product of the Doppler ratios between A and B, and B and C. Taking logarithms of the Doppler ratios, the resulting quantity combines additively.

The usual presentation of the twin paradox is that the travelling twin's journey is two inertial journeys, and I have not considered anything different. But although the journey is locally inertial at almost all times during the journey, the change in direction prevents the journey from being globally inertial. So, you have to take into account whether or not what you are considering is a local or a global notion. What does this actually mean? I'll repeat: There is no discontinuity in the time of the earth in the frame of reference of the travelling twin. You have made an error in your assessment of the frame of reference of the travelling twin. Minkowski spacetime doesn't exist, except locally or perhaps as part of a multiverse. Otherwise, it is a mathematical idealisation. That's why I dropped the "Minkowski" when I said, "I think spacetime is physical". But even in general relativity with curved spacetime, local physics is the physics of Minkowskian spacetime, so Minkowskian spacetime is more than just a mathematical idealisation or approximation. I should also state that one can calculate the result of the twin paradox without reference to the notion of simultaneity of the travelling twin by considering the Doppler effect that is observed by both twins of each other during the entire journey of the travelling twin. Each twin is literally observing the clock of the other twin.

"Physical or mathematical" is a false dichotomy. In the case of the notion of simultaneity, at a given instant of a given trajectory in spacetime, there is a set of physical points in spacetime that form a physical three-dimensional space that is simultaneous to the given instant of the given trajectory. But if you examine the points themselves, you'll find nothing that physically indicates that they are simultaneous to the given instant of the given trajectory. Different instants give rise to different three-dimensional spaces, and different spacetime trajectories have their own sets of three-dimensional spaces. But you are incorrect in what you consider to be the space of simultaneity for the travelling twin of the twin paradox. You assume that the simultaneity of the travelling twin is the same as that of an outbound inertial trajectory during the outbound part of the journey and the same as that of an inbound inertial trajectory during the inbound part of the journey. But the space of simultaneity for the travelling twin is distinct from that of any inertial trajectory, taking into account the change that occurred during the journey. It should be noted that the space of simultaneity of individual points of any trajectory depends on the entire trajectory, not just a part of it. I think spacetime is physical. If I look at myself in the mirror, I am seeing what I looked like in the past. And the further away the mirror is, the further back into my past I am looking. The principle of general relativity is that all coordinate systems are equally valid. But each observer has their own natural frame of reference. This is physical in the sense that it is based on the relationship of the physical observer to the spacetime. It is not however a physical property of the spacetime itself. However, the metric of a given coordinate system of a given physical spacetime is in principle the result of physical measurements of length and time using physical rulers and physical clocks.

I'll repeat: \not \! \partial [math]\not \! \partial[/math] Perfect!!! According to a reference PDF file I have, "\!" is a "negative thin space".

From what I can tell, "\displaystyle" does nothing. It was the "\not" placed in braces "{}" that you need, as well as the small space "\,". I actually tried to do something similar not long ago, so I have prior experience (which I didn't explicitly use). Let's try this: \not \! \partial [math]\not \! \partial[/math] Perfect!!! According to a reference PDF file I have, "\!" is a "negative thin space".

This is what I come up with: {\not}\tiny\,\normalsize\partial [math]{\not}\tiny\,\normalsize\partial[/math] This didn't render exactly as it did in Online LaTeX Equation Editor. Try this: {\not}\small\,\normalsize\partial [math]{\not}\small\,\normalsize\partial[/math] Try this: {\not}\,\partial [math]{\not}\,\partial[/math] Testing: X \tiny X \normalsize X [math]X \tiny X \normalsize X[/math] How about this: \displaystyle{\not}\tiny\,\normalsize\partial [math]\displaystyle{\not}\tiny\,\normalsize\partial[/math] Or this: \not \partial [math]\not \partial[/math] This: \not{\partial} [math]\not{\partial}[/math]

As a result of this thread, I did some exploration of factoring semiprimes. I even came up with a solution of sorts, but this didn't seem to be any more efficient than brute force.

Accelerated objects can be described perfectly well in special relativity. But accelerated frames of reference are outside the scope of standard special relativity. That's because accelerated frames of reference involve some of the mathematics of general relativity (though not the mathematics of spacetime curvature). Standard special relativity limits itself to the Minkowskian metric. The Minkowskian metric is invariant to Lorentz transformations, and inertial trajectories in spacetime transform to inertial trajectories under Lorentz transformations. Thus, all inertial trajectories in Minkowskian spacetime are on equal footing in that they all observe the same spacetime metric. The invariance of the Minkowskian metric to Lorentz transformations implies that it is not possible to measure one's velocity relative to Minkowskian spacetime, and that only velocities relative to other objects can be measured, which is made possible because symmetry to Lorentz transformations is broken. In the case of an accelerated frame of reference, the transformation from an inertial frame of reference to the accelerated frame of reference is not a Lorentz transformation, it is a transformation under which the Minkowskian metric is not invariant. That is, the metric of an accelerated frame of reference is not a Minkowskian metric. Thus, an observer in an accelerated frame of reference can distinguish between being in an accelerated frame of reference and being in an inertial frame of reference. Even though velocity is only relative, acceleration is absolute because one can measure one's current velocity relative to one's past velocity. Thus, absolute acceleration does not imply absolute velocity.

This is correct. I have used this to explain the asymmetry in the twin paradox. This asymmetry is often explained by the fact that only the travelling twin accelerates, whereas the above explains precisely how the fact that only the travelling twin accelerates manifests the asymmetry. Indeed, one can calculate the twin paradox result from the Doppler effect observed by each of the twins, which naturally agrees with the result obtained by calculating the proper time for each of the twins. Unlike velocity, which is only relative, acceleration is absolute. In the twin paradox, only the travelling twin accelerates. Not only that, the travelling twin knows that they accelerated, while the stay-at-home twin knows that they did not accelerate, and both the travelling twin and the stay-at-home twin knows that only the travelling twin accelerated. Thus, both the travelling twin and the stay-at-home twin observe the Doppler effect change from redshift to blueshift when they observe the travelling twin accelerate. .

According to Wikipedia, the peculiar velocity of the Sun relative to the comoving cosmic rest frame is 369.82 ± 0.11 km/s towards the constellation Leo. That doesn't seem "almost immobile" to me.