joigus

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.

I would have to be on top of the hill to look down on others here to evaluate them. I'm not in such position. How smug would I be if I did? I stand by my words: A superb explanation --especially considering the limited amount of time and manoeuvre we all have here-- by a person well versed in mathematics who has all my respect. MigL's explanation was also very helpful, although in a very different style and spirit. As to your kind offering of starting another thread, I'm not so interested in judging people as in examining ideas, and trying to understand some of the most difficult ones. But you're free to open that thread if you want. Here's smiling at you

Easy, because as @swansont told you, you're going around in circles. \[ \frac{\sqrt{FE_R} \left( \sqrt[4]{FE_R} \right)^2}{\frac{E_R}{c^2}} = \left( FE_R \right)^{\frac{1}{2}+\frac{2}{4}} \frac{c^2}{E_R} =\] \[ = Fc^2 = \frac{G}{c^2}c^2 = G \] You derive guess an equation from your definition. You substitute your definition, so you get to an identity. Doesn't matter that your definition dimensionally has no relevance. And your \( g_\text{photon} \) has the funny dimensions of (length)3/2(time).1.

\[ \frac{\sqrt{FE_R} \left( \sqrt[4]{FE_R} \right)^2}{\frac{E_R}{c^2}} = \left( FE_R \right)^{\frac{1}{2}+\frac{2}{4}} \frac{1}{c^2} =\] \[ =\frac{FE_R}{\frac{E_R}{c^2}} =Fc^2 = \frac{G}{c^2}c^2 \]

If I may say something... I was aware that @wtf was giving a superb mathematician's exposition of the topic, while @MigL who had had some previous experience with the OP, was quite deliberately trying to dumb it down. It was fun seeing you interact. But your effort is not in vain, wtf. Thank you. I appreciate it.

\( F= \frac{G}{c^2} \) has units of (mass)-1x(length) Energy has units of (mass)x(length)2x(time)-2 So \( \sqrt[4]{F\times E_\text{photon}} \) has dimensions of (length)3/4x(time)-1/2. So that's a non-starter from dimensional analysis alone. Sorry, I honestly thought you were joking in the Physics section. I immediately removed the neg reps.

Happy birthday and many happy returns!

You are serious, then? I couldn't believe you were serious.