joigus

No serious Republicans taking positions to save face when all this pissing in the wind blows over? I'm sure it must be embarrassing for somebody in their ranks. Just curious.

You're quite right. There is a caloric part in \( G \) that is usable for doing work. Thank you.

Gibbs free energy, roughly speaking, means the energy stored in all the chemical bonds. The \( \triangle \) quantity means difference between products and reactants, and 'molar' means 'per mole' or 'per molecule'. x-posted with Studiot.

You would have to breathe through a mask. There's no coronavirus in Titan, that's for sure.

Among other excellent points here, I think this puts the finger on the most likely misunderstanding of such creationist pseudo-arguments. Chimpanzees and orangutan, e.g., evolved from other apes as we did. But they did in a different direction. They are not an accurate picture of our ancestors by any means. The march-of-progress picture, as the one depicted in this creationist site: https://answersingenesis.org/human-evolution/lucy/a-look-at-lucys-legacy/ Is very naive and known to be wrong. They (creationists) keep obsessing with Lucy as the grand mother of all human kind. When, in fact, we know now there was no linear progression. They (creationists) are looking for Adam and Eve, as we all know. Apparently upright apes were very common 3 million years ago. Human evolution is more like a huge jigsaw puzzle, changing with time. https://www.nature.com/scitable/knowledge/library/overview-of-hominin-evolution-89010983/ It would be interesting to see how they explain Homo floresiensis, probably not a Homo at all, but an autralopithecine offshoot. Remember that about 6 million years ago savannas expanded into a huge range that got to cover an enormous uninterrupted area from northern Africa to the eastern coasts of Asia. It is believed that upright apes flourished back then like never before. And yes, we are apes.

Saturn seen from Titan amidst lakes and rivers of methane.

Clinton and JFK are not precisely the spitting image. Although both have the facial expression they're best remembered for. The only one that came out perfect is Bozo!

Can there be a set with fewer than zero elements? If so, then there can be negative-dimension spaces.

There are many good books that explain Christoffel symbols. It's not a notation; it's a real object that you need to correct for the fact that your reference directions change depending on the point when you're taking derivatives of vectors. A very good classic book is Lovelock & Rund: https://www.amazon.com/Tensors-Differential-Variational-Principles-Mathematics/dp/0486658406

You can live a long, healthy, happy life without going down that road. Are you sure you want to go where the buses don't stop? It depends on what you want QM for.

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.