joigus

Very interesting topics. Looking forward to seeing them posted. I've got some other historical topics to suggest. I'm glad to have you back, @MSC, after the pachi-dermatological treatment.

Amen. I don't find you the least curious. What's the new theory? Mind you, a new brand of word salad is not a theory. Strawmen at work sign.

You're welcome.

Of course. I know Al-Bukhari is a Sunni source. That's why I mentioned him in the point I was making about Sunni and Shi'a source discrepancies. Second of owl is... the barn owl? What I definitely do not understand is how you have decided that the best place to discuss the finer points of the Qu'ran is an international scientific forum, as Sensei pointed out.

You assume too much. Shi'a and Sunni Muslims disagree about 'the roots': Who are the rightful heirs of Mohammed, and whether Al Bukhari was right about him and his doings, and probably many more things. I'm sure you know much more about it than most of us here do. It's a 'sources' problem (both about the authenticity of books and/or translations, and about the line of authority) very much like what was for Christians several hundred years ago in Europe between the many Protestant offshoots, and Catholics, and Jews. That led to unimaginable bloodshed between Christians and Jews. We know. Actually, we know much better than you guys do. We've killed each other, we've hated each other for so many more years. Most of us seem to have taken home the lesson. You, unfortunately, haven't. That's a very big part of the problem, guys. A part of your community seems unable to take home some lessons from your brethren religions that are much older than yours. Jewish and Christians being at each other's throats for centuries. You're still obsessed with a couple of lines in a several-centuries-old book. That's, allow me to say, pathetic. Both in the most ludicrous sense, and in the most tragic one. Take a look at Mandaeans, Yasidis, etc., and how they've become victims of unspeakable violence in recent years in the Middle East, just because they follow the rituals that their ancestors did. Probably with the same amount of doubt that you do yours. But also take a look at how some Muslims brothers die at the hands of each other because of a difference of opinion. And ask yourself: Is the interpretation of some lines on an ancient book worth the suffering that we see in the world? The suffering of a child is not worth ten thousand lines of a holy book.

The connection comes from the Robertson version of the uncertainty relations. The one you can find on Wikipedia is the Robertson-Schrödinger version. I will give you a proof of the Robertson version which does not involve the anti-commutator. Say you have any two operators \( A \) and \( B \), which in general do not commute. Say your system is in a pure quantum state \( \left|\psi\right\rangle \). The mean square deviation is defined as, \[\left(\triangle_{\left|\psi\right\rangle }A\right)^{2}=\left\langle A^{2}\right\rangle _{\left|\psi\right\rangle }-\left\langle A\right\rangle _{\left|\psi\right\rangle }^{2}=\left\langle \psi\left|A^{2}\right|\psi\right\rangle -\left\langle \psi\left|A\right|\psi\right\rangle ^{2}\] and similarly for \( B \). Now define operators \( A' \) and \( B' \) centred on their respective average values: \[A'=A-\left\langle A\right\rangle _{\left|\psi\right\rangle }\] \[B'=B-\left\langle B\right\rangle _{\left|\psi\right\rangle }\] It's easy to see that, \[\left[A',B'\right]=\left[A,B\right]\] Now you formally build the 1-parameter family of operators: \[C=A'+i\lambda B'\] This operator is not Hermitian, but it is always true that, \[\left\Vert C\left|\psi\right\rangle \right\Vert ^{2}=\left\langle \psi\left|C^{\dagger}C\right|\psi\right\rangle \geq0\] This gives you a polynomial condition in \( \lambda \): \[\left(\triangle_{\left|\psi\right\rangle }A\right)^{2}+\left(\triangle_{\left|\psi\right\rangle }B\right)^{2}\lambda^{2}+i\lambda\left\langle \left[A,B\right]\right\rangle _{\left|\psi\right\rangle }\geq0\] If this polynomial must always be above the real axis, the discriminant must be negative or zero: \[\left(\left\langle i\left[A,B\right]\right\rangle _{\left|\psi\right\rangle }\right)^{2}-4\left(\triangle_{\left|\psi\right\rangle }A\right)^{2}\left(\triangle_{\left|\psi\right\rangle }B\right)^{2}\leq0\] Keep in mind that if \( A \) and \( B \) are Hermitian, so is \( i\left[A,B\right] \). This immediately gives you the Robertson version of the uncertainty relations for arbitrary operators \( A \) and \( B \): \[\triangle_{\left|\psi\right\rangle }A\triangle_{\left|\psi\right\rangle }B\geq\left|\left\langle \frac{i}{2}\left[A,B\right]\right\rangle _{\left|\psi\right\rangle }\right|\] So when two operators do not commute, you cannot define "dispersions" or "precisions" (mean square deviations) better than those given by the above. I hope that helped. It's the simplest demonstration I know of the more general Robertson version for arbitrary operators.

But his is a denial on steroids. What if he doesn't realize he's dead? He would be the first political zombie in history. I picture him giving orders from his twitter account and letting Giuliani loose on passing 'trolls.'

Thank you. I wasn't aware of this. It seems that massive particles vanish into nonexistence in Penrose's model, rather than decaying or being ripped apart. I see a problem with particle-antiparticle asymmetry with it though.

Sweet deal. Plus they would be in charge of vaccination.

My feelings exactly. The average inter-species behaviour in this planet is trying to either outwit or outrun your predator or your prey. Other interesting possibilities are host-parasite, pet-master...

Try with, \[\det\left(\frac{\partial f_{i}}{\partial x_{j}}\right)=0\] Now that I think about it, the fact that the \( k \)'s don't appear linearly is not that bad. You would obtain critical values for such non-linear functions \( K\left( k \right) \). So it's a matter of re-defining the \( k \)'s. What would be more involved is if the \( k \)'s appeared mixed with the \( x \)'s.

My feelings exactly.

You are absolutely right. There is a deep connection. The Fourier transform of a function can be seen as the momentum representation of that function. It's based on the concept of considering functions as vectors in an infinite-dimensional space. Analogously to how the i-th component of a vector can be obtained by using the scalar product by the \( \boldsymbol{e}_{i} \) vector of the basis, \[v_{i}=\boldsymbol{v}\cdot\boldsymbol{e}_{i}\] you can obtain the p-component of a function by integrating (analogous to the scalar product): \[\hat{f}\left(p\right)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}dxf\left(x\right)e^{ipx/\hbar}=\left\langle f\left|p\right.\right\rangle\] where the matrix coefficients that give you the change of bases are the exponentials: \[\left\langle x\left|p\right.\right\rangle =\frac{e^{ipx/\hbar}}{2\pi\hbar}\] You must learn the concepts of operators, eigenvalues, and eigenvectors to complete the details. Do you know about those?

It's a natural thought and physicists did think in those terms in the past. But when the rules of quantum mechanics started to become clear, it was realised that the only states that make sense for quantum particles are some special average over all the identity tags. So ultimately there is no really meaningful sense in which you can say 'this electron' or 'that electron'.

One of the principles of quantum mechanics is that elementary particles have no labels. All electrons are the same electron in a sense that's very clear in the mathematics of QM. So no, we know it's not possible.

Easter Island is a microcosm of what could happen here. But still, back then you could make do with a catamaran before the last tree fell and start afresh. I don't see any chance of anything like that happening any time soon. What you see is what you get.

There was a time when the issue of who's the master of puppets was only of concern to those directly affected. Now it is a global issue. The situation has no historical precedent. Empires have come and gone, but now what's a stake is both management of the whole planet and a model of hierarchy and societal organization that's not just for restricted import-export, while keeping the balance of masters and underdogs; it's the way the world is going to be defined by, politically, economically, in terms of population dynamics... Never the stakes have been so high. And I can't deny that I'm worried. Paraphrasing @Alex_Krycek --if I remember correctly--, this is not empire business as usual. Communication bridges between similarly-minded people around the world will become essential. The concept of a nation in the usual sense is doomed. And we should be glad that it is. I hope it's for the best. That's what I think, anyway.

Right. He's an elephant-in-the-living-room kind of president we've never seen before. Literally and metaphorically.

I hope you're right. And your prediction seems very likely to come true. But another very worrying matter at issue should be the damage that the incumbent is doing by deciding not to brief the President elect on matters which are extremely important in a situation of national emergency. It seems to me that smooth transfer of power is not just a question about healthy institutions, important though it is. If national security weren't enough, there is an ongoing pandemic and it seems to be beyond his concern. I'd like to hear your --and everybody's-- thoughts about that. I have. I've just watched his reaction to Trump's tantrum and he was nothing if not reassuring.

No sorry. I made a mistake. \( \triangle y\geq D \) is not correct. For your example, the uncertainty (dispersion) in the \( y \) position is of order \( \triangle y\sim D \), so by HUP, \[ \triangle p_{y}\geq\frac{\hbar}{2D} \] which means that the dispersion in \( y \) is of order \( D \). That holds just after the particles go through the slit. What you wrote is, \[\triangle y\ll D\] That means that the dispersion in \( y \) is much less than \( D \), and that's not necessarily correct either. It depends on what regime of scattering you have after the electrons go through the slit, and it's certainly not true just after the electrons cross it, and before they scatter. The fact that you scatter particles after the passing of the electron through the slit of width \( D \) further changes the dispersion. So \( D \) no longer plays much of a role if you scatter the particles after they go through the slits. If you scatter the electrons with very energetic particles, you get better precision in \( y \) than \( D \), but you lose precision in \( p_y \). Dispersions are not added. But a detailed calculation requires scattering theory and is outside the scope of what we can deal with here. Whatever you do to the electrons to determine which way they go does destroy the interference pattern. Even if you set a detector in one of two alternative paths after they go through the slits and consider only the electrons that go through the other path --counterfactual measurement or so-called "interaction-free measurement." I do have a feeling that I'm not being very clear and I apologize for that. Maybe tomorrow.

Thanks a lot, Airbrush, for amazing pictures and captions with info.

I think you mean \( \triangle y\geq D \) or \( \triangle y\sim D \). They're gonna tell you about the observer effect vs HUP. One thing is 1st-kind measurements or preparations, and quite a different thing are 2nd-kind measurements, or "interventions" --I would call them. I'll be back.

Bouddi National Park (New South Wales, Australia.) This Australian National Park features the presence of sandstone, coloured by haloes of chemical change, brought up by differential oxidation of iron minerals in the presence of moisture (Liesegang rings*) as those in Maria Island presented in the previous series of photographs. ---------- (*) Liesengang rings are rings of chemical change that colour the interior of certain rocks. They are formed by oxidation of iron minerals due to moisture penetrating the rocks through their porous structure. Generally that moisture filters through fracture lines in the rocks, thus producing concentric rings towards the centre of the blocks that the fracture lines define. CREDITS: https://www.facebook.com/Geomorfologia.Para.Todos/posts/4021501111197661 (Translated from Spanish) (Geomorphology for everyone) https://forum.xcitefun.net/bouddi-national-park-australia-t83280.html

The word 'false' has only been used by you: Edit: my emphasis. Now I can say beyond any doubt that you're lying, unless you mean yourself. Nobody has used the word 'false' here, except you, or other users quoting you. It's a simple matter of using the search function in your browser.