KJW

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.

In high school, I moved from classroom to classroom for each different subject. But the layout of all the buildings seems to be such that the windows were on the left when looking to the front. And the door was on the right, at the front of the classroom. However, I can't recall every classroom I was in, so I can't say there were no classrooms that were different. And there may have been classrooms that were different for particular reasons. In infants and primary school, I was in the same classroom with the same students and the same teacher for all subjects for a whole year. In different years, the classroom and teacher usually changed, but the students were usually the same. The students were grouped according to their overall grades. But in high school, each subject for each year had its own classroom and teacher, both of which were the same throughout the year, and the students were grouped according to their grades for that subject. And because particular subjects were taught by particular teachers in particular classrooms, the teachers and classrooms for a particular subject tended not to change from year to year.

At my high school, I seem to recall the windows being to my left while facing the blackboard in most if not all of my classrooms.

There is nothing wrong with the excerpt from the book. Do you know what "inversely proportional" means? Perhaps the following will help: [math]i = \dfrac{\text{Calculated molar mass}}{\text{Observed molar mass}}[/math] [math]i = \dfrac{\text{Observed number of moles}}{\text{Calculated number of moles}}[/math] [math]i = \dfrac{\text{Observed colligative property}}{\text{Calculated colligative property}}[/math] [If the above LaTeX doesn't render, please refresh browser.]

But that's the whole point. Trump needs to shift everything over which he doesn't have control to where he does have control, by any means possible, even if it means resorting to sophistry (such as likening drug trafficking to an armed attack on the US). But does that allow Trump to wrest Constitutional control of voting away from the states? Trump has already said he wants Republicans to take over the voting in 15 places. Presumably that means swing states. On the other hand, with the recent strong swings against the Republicans, the swing states may not be enough, and Trump probably knows that. What can the states do if the Supreme Court decides the judiciary doesn't have the authority to decide on matters of foreign relations? Surely a proven instance of foreign interference helps Trump make his case, regardless of the particular details. And if it doesn't, then foreign interference can be manufactured. What good is delaying the midterms? Trump needs a resounding Republican victory. And he needs that victory to have the appearance of legitimacy.

Even though I live nearly half a world away from the insanity that is the Trump administration, I have been taking a very keen interest in what is happening in the US. It seems clear to me that Trump is now shoehorning everything into "foreign relations" because that is where he claims to have presidential power beyond the reach of anyone, including the Supreme Court. It seems inevitable that he will claim "foreign interference" as a pretext to taking control of the midterm elections. And this may even become true if he enlists some of his mates at the "Board of Peace" to create "foreign interference". What do you think? And what can be done about it?

You asked a question and I answered it. Yes, I read what was written. My answer stands.

The number of moles is inversely proportional to the molar mass.