joigus

Swansont's points well taken. Also: That's not an integral equation. There's extensive work in mathematics about how to solve integral equations. In what variable? Do you mean, \[ F\left(t\right)=\left[M+F\left(t\right)\int_{0}^{t}v\left(t'\right)dt'\right]\frac{v\left(t\right)}{t} \] as an equation for \( v \)? But \( F \) depends on position. What happens at \( t=0 \)?

Ferromagnetism,* superconductivity, and superfluidity are among quantum effects that can be seen with your own eyes. It is true that the context of QM par excellence is the very small, though. That's due to the smallness of the quantum of action when compared to ordinary experience. * https://en.wikipedia.org/wiki/Ferromagnetism#Explanation

Agreed. And it's the bizarre nature of your questions. If you keep going on like this, you'll never get around to mathematics. I cannot emphasize enough how much attention you must pay to these tips: I could hardly agree more. If you're trying to climb Mt. Everest, and you're looking at another summit in the distance, you're gonna fall through a crack. Does that make sense? And for Pete's sake, solve a simple linear equation, get pleasure from it, and step on towards a more difficult problem. And keep going. Get something under your belt, however modest, ASAP. No matter how simple. Edit: Oh, and another thing. You're getting excellent advice here. Don't pay heed to "Daft Science" or "CrackpoGenius" or similars who might tell you how much more intelligent than others you are. They're distracting you and you've gone astray by their compliments before. I've seen it happen. That's another crack in Mt. Everest. Compliments are very distracting. Reliable information is gold.

I did take a look at your channel, I don't obligate you to anything, and I do not have the power to ban you nor any will to do so. I agree with @studiot that it's a good example of a very bad simulation of science. If you care about these things, you should take time to learn them. Pressure does not require contact. We understand today that contact forces do not really exist, in the strict sense. It's all fields. You seem to ignore also that there are models of QM based on particles following the Hamilton-Jacobi equation with a quantum potential. They reproduce all the kinematical results. It's not that you don't understand pressure, or interactions monitored by fields, etc. It's, as very often happens in the realm of crackpottery, that you couldn't care less. You should open your ears and you could learn something in these forums from other users who know more than you. I have.

"Makus Henke" is Markus Hanke, "Goigus" is Joigus, and as to \( v_{12t} \) nobody knows what it is. It can be nothing other than a typo, you clearly meant either \( v_{1t} \) or \( v_{2t} \) (otherwise you're summoning something here which is neither \( a_1 \), \( a_2 \), \( b_1 \) or \( b_2 \), the four numbers of your previous (correct) mathematical lemma. Once you either correct this typo, or tell us what \( v_{12t} \) is, I'm afraid nothing's going to save the incorrect proof, for reasons very much already explained. Anamitra Palit should reconsider his writing before Joigus (at least) considers his views any further.

You obviously have problems processing what you read, or do not read at all. Go back to my words in and around "if that's the case". The rest of your blabbering is not worth my time.

No. There is a time component alpha and the corresponding Euclidean norm of the 3-vector alpha (that's why I wrote it in boldface). Same for beta. IOW, there are four numbers whose value I've chosen to be 1, producing a very clear counterexample of your assertion. It's very common notation in relativity. \[ \left( \alpha^{\mu} \right) = \left( \alpha, \boldsymbol{\alpha} \right) \] Do you understand the notation now? The question about constraints for physical 4-vectors, that Markus, Ghideon and I have told you about, still stands. As I said, Anamitra Palit must go back to elementary relativity books.

It's common courtesy, and how relevant it is remains to be seen. I also gave you the benefit of the doubt. So far you haven't used or mentioned any sensible physics, and you seem not to understand very basic principles (pressure, wavelength, action at a distance...) and deny evidence, so the doubt is vanishing fast.

In contradiction with experiments, then. Thanks for the heads up. Although I don't think living on welfare, if that's the case, automatically disqualifies you to do good science. Some folks may be able to keep a no-nonsense attitude and have a lot of time on their hands at the same time. I don't think that's impossible.

Duly noted. Is there a deadline to start my new funny self? Am I allowed to be funny in the sense?:

Well, yes. Japan is perhaps a better example of need for restraint. But the topic was China. That's why I mentioned China.

I'm in three minds about these news.

The turtle joke was hilarious. I'm under not obligation to be funny, may I remind you...

Or: funny 1. making you laugh 2. difficult to explain or understand; strange and not as you expect 3. (British English) humorous in a way that shows a lack of respect for somebody 4. (informal) slightly ill 5. (British English, informal) slightly crazy; not like other people 6. not working as it should (adv.) as in "My computer keeps going funny." https://www.oxfordlearnersdictionaries.com/definition/english/funny 😇

https://www.goodreads.com/quotes/582016-knowledge-exists-in-two-forms---lifeless-stored-in-books

Enough said.

Virtually everything. The most rigorous proofs are generally overlooked, like those based on epsilon and delta to prove existence of limits or continuity, etc. Multivariable calculus is used a lot. Also infinite series, limits, derivatives, integrals, improper integrals, complex analysis... The whole shebang!

That's what I said. Are you repeating what I said? Only the property is not unique. Any tensor product of 1-covariant 1-contravariant tensor also has that property. And same with epsilon tensors (in that case the components are 1, 0, and -1). The Kronecker delta \( \left. \delta^{\mu} \right._{\nu}\) is an isotropic tensor. The Kronecker deltas \( \delta^{\mu\nu}\), \(\delta_{\mu\nu} \) are not. On the other hand, from your document (ineq. 1), it does not follow, as you say, that, \[ \alpha\beta -\left| \boldsymbol{\alpha} \right| \left| \boldsymbol{\beta} \right| \geq 1 \] (expressed in a lighter notation). Your ineq. 1), e.g., would be, \[ \alpha\beta -\left| \boldsymbol{\alpha} \right| \left| \boldsymbol{\beta} \right| \geq \sqrt{ \alpha^2 - \left| \boldsymbol{\alpha} \right|^2 } \sqrt{\beta^2 - \left| \boldsymbol{\beta} \right|^2 } \] The above expression does not follow, as this counterexample shows: Pick \( \alpha = 1 = \left| \boldsymbol{\alpha} \right| \); \( \beta = 1 = \left| \boldsymbol{\beta} \right| \), but, \[ \alpha\beta -\left| \boldsymbol{\alpha} \right| \left| \boldsymbol{\beta} \right| =0 < 1\] So you're wrong here. Other mistakes have been pointed out to you repeatedly. Time to go back to a relativity book and do the exercises.

https://www.perimeterinstitute.ca/video-library/collection/2015/2016-complex-analysis-tibra-ali I generally recommend Perimeter Institute Lecture Series. I haven't followed this particular one, but quality is quite good at PI.

Indeed. Maybe we all are tools or someone who's someone else's tool. With no overriding handler of all tools. Paraphrasing @dimreepr, imagine that...

🎬 To waffling with waffling. peheh, pahah, poohooh. To me, to you, to us, is not significant. Unless disclaimers, caveats or qualifications are applied. 🎬

In fact, a Kronecker delta that is twice covariant or twice contravariant is not an isotropic tensor either. It must be 1-covariant 1-contravariant. Also arbitrary tensor products of 1-covariant 1-contravariant Kronecker deltas is an isotropic tensor. Arbitrary products are not. That's because contravariant indices transform with the inverse matrix with respect to covariant ones (that's why they're called "contra"). If you multiply twice by the same matrix you don't get back to Kronecker deltas. You must go carefully through all these checks in order not to make elementary mistakes. It's a natural rite of passage. The literature is full of mistakes of this nature. Not in the really prestigious books, of course. No. Mathematical physics is expected to cater to physics. Mathematics doesn't need any imput from physics. Mathematical physics is expected to be self-consistent, and further, it is expected to comply with what we measure in the laboratory.

invisible braces: \[ {\left. \delta^{\mu} \right. }_{\nu}\]

Funny that you think that, as nobody else does. Without going into details, it's quite clear for anybody who's worked with tensor calculus for some time that the most likely thing going on is that you're confusing invariant properties with coordinate-dependent properties, and mixing them all up in one big mess. Some tensor identities can be proved by appealing to some tensor being zero in one particular coordinate system. Then it must be zero in all coordinate systems. Conversely, non-zero in one system <=> Non-zero in all. On the contrary, the Christoffel symbols can always be chosen to be zero in one coordinate system, but non-zero in infinitely many other coordinate systems. Playing with these two properties facilitates some proofs, but you must know what you're doing. Handling index expressions without any care of what is a tensor and what only holds in one particular system is the wrong way to go. It is a real pain to go over every step of a calculation that somebody only too obviously did wrong, because you have proven the theorems forwards and backwards and gone through all the examples. A tensor being diagonal, e.g., is not an invariant property under SO(3) or O(3). On the other hand, tensors like the identity \( {\left. \delta^{\mu} \right. }_{\nu} \) or \( \epsilon_{\alpha\beta\cdots} \) are called isotropic tensors, because they look the same (have the same components) in all coordinates systems. So you cannot safely assume that a diagonal tensor takes part in any tensor equation. "Diagonal matrix" makes sense, meaning "a matrix that looks diagonal in a particular base". "Diagonal tensor" does not.