KJW

That doesn't mean they're not mathematical geniuses, although I have no opinion concerning the topic of this thread.

I decided to investigate this because of the mention of the fine-structure constant. I'm actually quite impressed with how close the derived Hubble constant matches the measured value. Unravelling the formula, what you appear to me to be saying is that: [math]\dfrac{\rho_{\text{electromagnetic}}}{\rho_{\text{critical}}} = \alpha^2 \approx \dfrac{1}{18779}[/math] However, I don't agree with your suggestion that you've figured out the fine-structure constant. The mystery of its particular numerical value remains, even though its relation to other physical notions is well known to physicists. In particular, it is known to be the value of [math]\beta[/math] of the electron in the lowest orbit of the Bohr model of the atom.

I think that's as much about class as anything else, although there is the "first in, best dressed" aspect as well. Bear in mind that the life of the elderly wasn't easy when they were young, either. That's not a reason to deny the elderly the vote for the short time they have left. That's why I said it was a fundamental problem. Although minorities having a vote doesn't prevent their interests from being trampled on, at least a vote gives them some sort of voice which would be silenced if they were disenfranchised. No, I'm looking to stop the disenfranchisement of the elderly.

Given that electoral terms are only a few years, and even old people expect to be still alive after the one they're voting in has completed, to suggest that old people no longer have a stake in the outcome of an election is wrong. A fundamental problem with democracy is that the majority don't always act in the interest of minorities and often act contrary to that interest, so to disenfranchise a minority only exacerbates that problem.

A little bit of background from https://en.wikipedia.org/wiki/Hadronization The top quark does not hadronize The top quark, however, decays via the weak force with a mean lifetime of 5×10−25 seconds. Unlike all other weak interactions, which typically are much slower than strong interactions, the top quark weak decay is uniquely shorter than the time scale at which the strong force of QCD acts, so a top quark decays before it can hadronize. The top quark is therefore almost a free particle.

You are overcomplicating my friend. If we strip the problem of any specific details and other additional anthropocentric components we will see the foundation of this phenomena. let's test the hypothesis that the age difference is manifested by the asymmetric observation of Doppler shifts (the optical delay of the turnaround) by removing the delay entirely. Consider a modified scenario: Observer A remains at rest. Observer B does not travel to a distant star, but instead orbits Observer A at a negligible, constant distance with a kinematic projection of [math]\beta = 0.8[/math]. No. The explanation I provided was specifically for the twin paradox scenario. If you want to modify the scenario, then the analysis of the scenario has to be modified as well. You can't say my explanation of the twin paradox scenario is incorrect or even overcomplicated because (although it is a thought experiment) my explanation is purely in terms of what is observed. What is observed by each twin is the redshift and blueshift of the other twin's clock, and the amount of time the redshift and blueshift are observed. As for your modified scenario, I'd like to first point out that you said nothing about how the observer in circular motion observes the observer in the centre. Your discussion of the scenario suggests that you are only concerned with the time dilation of the circularly moving observer that is observed by the central observer, without addressing the question of consistency between the perspectives of the two observers. And it is the question of consistency between the perspectives of the two observers that lies at the heart of the twin paradox, as well as precisely why the twins end up with different ages. Unfortunately, so far, I have been unable to provide an explanation of your modified scenario in terms I used for the twin paradox scenario. I'm not sure such an explanation even exists. However, I can say that the central observer observes the observer in circular motion as transverse Doppler redshifted, and that the observer in circular motion observes the central observer as accelerationally blueshifted. In the case of the twin paradox scenario, both legs of the travelling twin's journey were inertial, and although the turnaround is an acceleration, it wasn't treated as such and was merely a time and location where the redshift becomes a blueshift. Thus, the explanation I gave for the twin paradox scenario was the natural explanation. I believe (I have a possibly false recollection of doing the maths) that my explanation of the twin paradox scenario can be extended to arbitrary longitudinal motion of the travelling twin. Although such a case has the travelling twin in an accelerated frame of reference, ultimately all redshifts and blueshifts based on time dilation are Doppler effects. Thus, your modified scenario ought to be able to be treated like the generalised twin paradox scenario, although I am presently unable to connect the accelerated frame of reference associated with circular motion to the Doppler effect. That is, I am presently unable to explain the blueshift of the central observer that is observed by the circularly moving observer in terms of the Doppler effect. Nevertheless, the blueshift can be explained by invoking the metric. A problem I see with your explanation is that it seems to be unconnected to the actual physics of the situation. And by being unconnected to the actual physics, it is difficult to see how it can deal with subtleties present in the actual physics. For example, suppose I am in circular motion at constant speed around some object at the centre. I face the direction of my inward acceleration. Am I looking at the object at the centre?

In SI units: E is in joules m is in kilograms c is in metres per second There are a number of different systems of units, but as you appear to have discovered, for a formula to make sense, it is necessary for all the units to belong to the same system of units.

Consider two rooms and an open doorway between them. You are in one room, and a friend is in the other. There is no direct line of sight between the two of you. However, although you can't see each other, you can clearly hear each other. That is because sound can propagate around obstacles. This is called diffraction and is a property of waves in general. Because light is also a wave, light also exhibits diffraction. But you can't see your friend, so evidently the light isn't diffracting the way the sound is. That is because diffraction depends on the wavelength of the wave, and the wavelength of visible light is very small compared to the wavelength of audible sound.

Each particle will have a wavelength, and the pair of particles itself will also have a wavelength (corresponding to the sum of the momenta of the two particles). But note that momentum is a vector, and each component of the momentum vector has its own wavelength. Let Ψ1(x1,y1,z1,t1) be the wavefunction of particle 1 and Ψ2(x2,y2,z2,t2) be the wavefunction of particle 2. Then the two-particle wavefunction of particles 1 and 2 is the tensor product: Ψ(x1,y1,z1,t1,x2,y2,z2,t2) = Ψ1(x1,y1,z1,t1) Ψ2(x2,y2,z2,t2) Note that if the Ψ1(x1,y1,z1,t1) and Ψ2(x2,y2,z2,t2) exist in four dimensions, then Ψ(x1,y1,z1,t1,x2,y2,z2,t2) exists in eight dimensions. One significance of this is that macroscopic objects exist in very high dimensions as the tensor product of the wavefunctions of the many individual particles that make up the macroscopic objects. And the higher the number of dimensions, the more likely it is that two arbitrarily chosen vectors will be orthogonal. There is no interference between orthogonal vectors (wavefunctions), thus providing an explanation why there is no observed interference between classical (macroscopic) objects. I should remark that if particles 1 and 2 interact with each other, then the two-particle wavefunction can no longer be decomposed as the tensor product of two single-particle wavefunctions, and the two-particle wavefunction is said to be entangled.

If one has two particles of mass m1 and m2 and their corresponding single-particle wavefunctions, then the two-particle wavefunction is the tensor product of the two single-particle wavefunctions, and as such is a wavefunction corresponding to mass m1+m2, noting that exp(ζ)·exp(η)=exp(ζ+η).

In mathematics, "or" is "inclusive or", and "exclusive or" is "not equivalent".

Quantum superposition of molecules beyond 25 kDa (Published: 23 September 2019) Yaakov Y. Fein, Philipp Geyer, Patrick Zwick, Filip Kiałka, Sebastian Pedalino, Marcel Mayor, Stefan Gerlich & Markus Arndt Abstract Matter-wave interference experiments provide a direct confirmation of the quantum superposition principle, a hallmark of quantum theory, and thereby constrain possible modifications to quantum mechanics. By increasing the mass of the interfering particles and the macroscopicity of the superposition, more stringent bounds can be placed on modified quantum theories such as objective collapse models. Here, we report interference of a molecular library of functionalized oligoporphyrins with masses beyond 25,000 Da and consisting of up to 2,000 atoms, by far the heaviest objects shown to exhibit matter-wave interference to date. We demonstrate quantum superposition of these massive particles by measuring interference fringes in a new 2-m-long Talbot–Lau interferometer that permits access to a wide range of particle masses with a large variety of internal states. The molecules in our study have de Broglie wavelengths down to 53 fm, five orders of magnitude smaller than the diameter of the molecules themselves. Our results show excellent agreement with quantum theory and cannot be explained classically. The interference fringes reach more than 90% of the expected visibility and the resulting macroscopicity value of 14.1 represents an order of magnitude increase over previous experiments. (The remainder of the article is behind a paywall) https://www.nature.com/articles/s41567-019-0663-9

It's interesting that we are both saying the same thing in two different ways, suggesting a preference in how we view mathematics in general (me algebraic, you geometric).

I was not aware of it, either. But, [math]v_{2i} + v_{3i} = v_{1i}[/math] [math]\displaystyle \sum_{i} v_{2i}^2 + \sum_{i} v_{3i}^2 = \sum_{i} v_{1i}^2[/math] [math]\displaystyle = \sum_{i} (v_{2i} + v_{3i})^2 = \sum_{i} v_{2i}^2 + \sum_{i} v_{3i}^2 + \sum_{i} 2v_{2i} v_{3i}[/math] [math]\displaystyle \text{Therefore: }\sum_{i} v_{2i} v_{3i} = 0[/math]

Where have you been the past year?

@Anton Rize, you are correct in saying that only one of the twins actually turns around, but you are incorrect in how that turnaround manifests in that twin ending up younger than the stay-at-home twin. I actually provided a little bit of a hint when I mentioned the symmetry between observers who see each other either redshifted or blueshifted depending on whether they are moving away from or towards each other. Instead of a pair of twins, consider three observers A, B, and C. Let A be the same as the stay-at-home twin, B be the same as the travelling twin on the outbound leg of the journey, and C be the same as the travelling twin on the inbound leg of the journey. Whereas the travelling twin accelerates at the turnaround, observers B and C are two different observers, neither of whom accelerate when they meet at the turnaround location of the travelling twin. Observers A and B are moving away from each other and see each other redshifted by the same amount. Observers A and C are moving towards each other and see each other blueshifted by the same amount. Thus, there is symmetry between observers A and B as well as between observers A and C. Yet in spite of this symmetry, the total time of B and C is less than the time of A. It should be noted that A and B, as well as A and C, are observing each other's clocks (comparing with their own clocks), the comparison being according to the Doppler effect for light. That is, the observed Doppler effect can be used to establish that the travelling twin returns younger than the stay-at-home twin. Remarkably, it also explains precisely how the turnaround manifests this asymmetry. At the turnaround, the redshift of A's clock observed by B immediately becomes the blueshift of A's clock observed by C. By contrast, the redshift of B's clock observed by A doesn't become the blueshift of C's clock observed by A until light from the turnaround has reached A. For the travelling twin, half the time the stay-at-home twin's clock is observed as redshifted, and half the time the stay-at-home twin's clock is observed as blueshifted. For the stay-at-home twin, more than half the time the travelling twin's clock is observed as redshifted and less than half the time the travelling twin's clock is observed as blueshifted. And the faster the travelling twin travelled, the further away from home the turnaround location will be, and therefore the longer the time it takes light to travel the distance, and the greater the proportion of time the travelling twin's clock is observed to be redshifted. When one actually calculates the Doppler effect observed by each twin of the other twin's clock, in both cases, the time of the travelling twin is less than the time of the stay-at-home twin by the expected time dilation factor (consistent and not symmetric).

I'm not familiar with that. The basic problem will be who owns the machines and what will become of most of humanity when those in power no longer need us.

The way I see it, when machines become as generally capable as humans, capitalism will fail. This failure will either lead to a new golden age or an apocalypse. And without any specific plan to deal with such a future, I think an apocalypse would be inevitable.

Good point! But again - why waves in the first place? What precisely do you mean by "waves"? Do you know the general solution of the two-dimensional wave equation? [math]\dfrac{1}{c^2}\dfrac{\partial^2 \psi}{\partial t^2} - \dfrac{\partial^2 \psi}{\partial x^2} = 0[/math] [math]\psi = f(ct+x) + g(ct-x)\ \ \ \ \text{where }f()\text{ and }g()\text{ are arbitrary functions of a single variable.}[/math] Note that: [math]\dfrac{\partial \psi}{\partial t} = cf'(ct+x) + cg'(ct-x)\ \ \ \ \ ;\ \ \ \ \ \dfrac{\partial \psi}{\partial x} = f'(ct+x) - g'(ct-x)\\\dfrac{\partial^2 \psi}{\partial t^2} = c^2f''(ct+x) + c^2g''(ct-x)\ \ \ \ \ ;\ \ \ \ \ \dfrac{\partial^2 \psi}{\partial x^2} = f''(ct+x) + g''(ct-x)[/math] Thus, we see that "wave" refers to the motion of a function through spacetime rather than any sinusoidal character of the function. However, arbitrary functions can be decomposed into their sinusoidal Fourier component functions. In the case of the four-dimensional wave equation, a disturbance at each spacetime point propagates along the light cone emanating from that point. And because the wave equation is linear, superposition applies at the intersections of the various light cones, as well as any decomposition of a function into component functions. Also note that because the wave equation is a differential equation, the general solution is valid at each spacetime point, which is the basis of the Huygens-Fresnel principle.

How do you handle the twin paradox? If observers A and B are moving away from each other, then A sees B's clock tick slower while B also sees A's clock tick slower. And if observers A and B are moving towards each other, then A sees B's clock tick faster while B also sees A's clock tick faster. This is because of the symmetry of the relationship between A and B. But in the twin paradox, one of the twins is definitely younger than the other twin when they eventually meet. Thus, the symmetry of the relationship between A and B does not exist between the two twins. Why?

@Markus Hanke, I believe what @Anton Rize is saying is that Schwarzschild radius can act as a substitute for mass. In other words, mass and Schwarzschild radius are equivalent, and this equivalence can be applied even if the mass is not of a spherical object. [math]r_s = \dfrac{2GM}{c^2}\ \ \ \ \ ;\ \ \ \ \ M = \dfrac{c^2 r_s}{2G}\ \ \ \text{or}\ \ \ GM = \dfrac{c^2}{2}\ r_s[/math] Thus, any formula with [math]M[/math] or [math]GM[/math] can be replaced with the corresponding formula with [math]r_s[/math]. It is interesting to note that the radius of a Newtonian "black hole" (where the escape velocity at the surface is [math]c[/math]) is the Schwarzschild radius of its mass. I think this and other formulae for which the Newtonian version is the same as general relativity is due to the special nature of the [math]r[/math] coordinate of the Schwarzschild metric (the surface area at [math]r[/math] is equal to the Euclidean surface area). In this discussion, this appears to create an uncertainty as to whether @Anton Rize is in the realm of general relativity.

Huygens-Fresnel principle

How did you derive this (without invoking G or M)?

Actually, an exception to this was first year high school, when all of the classes were in the same building, and subjects not requiring specialised classrooms were in the same classroom. Different subjects had the same students but different teachers. First year high school students were grouped according to their primary school grades, somehow correlating the grades from different primary schools. First year subjects were somewhat different to the subjects of years two to four, and fifth and sixth year subjects were different again.

I had not come across "abnormal molar mass" in my studies, either. However, the excerpt from the book in the OP made it clear what "abnormal molar mass" meant, and this was the basis of my reply.