joigus

Such a misnomer...

An alternative proof from direct Taylor expansion in the metric coefficients and counting how many parameters are left that I cannot set to zero by changing the coordinate system: https://www.youtube.com/watch?v=gf-G4QiAHLY&list=PLaNkJORnlhZnwjIXnOHrX50FEyoyiTh4o&index=5 Those must coincide with the number of independent components of the Riemann. \( \frac{1}{12}n^2\left(n^2-1\right) \) Uses Young tableaux, which allows you to count free parameters very easily.

Oh, but that's not because it's much worse than I pointed out. It's because it's bound to get worse if you make a notational blunder of that magnitude. If you want to discuss anything in terms of a 2-index tensor being diagonal in a certain point --o perhaps everywhere?, the OP didn't tell us--, you could arrange to distinguish this by using Latin capital letters, e.g., \[A^{BB}=\frac{\partial x^{B}}{\partial\bar{x}^{\mu}}\frac{\partial x^{B}}{\partial\bar{x}^{\nu}}\bar{A}^{\mu\nu}\] Meaning, \[A^{00}=\frac{\partial x^{0}}{\partial\bar{x}^{\mu}}\frac{\partial x^{0}}{\partial\bar{x}^{\nu}}\bar{A}^{\mu\nu}\] \[A^{11}=\frac{\partial x^{1}}{\partial\bar{x}^{\mu}}\frac{\partial x^{1}}{\partial\bar{x}^{\nu}}\bar{A}^{\mu\nu}\] etc. So it can be done, but not the way the OP is doing it. Not that it's very useful to consider tensors as objects that are or aren't diagonal in any invariant geometrical sense, as they are objects referred to two different bases. Absolutely. When I'm doing maths and I get to such surprising results as "the whole of tensor algebra/calculus is bonkers, because all tensors are null" --or something like that, I'm not completely sure if that's the point--, I try to retrace my steps and, sure enough, I can spot a silly mistake. The last thing that would cross my mind is to highlight the "result" and announce to the world, "hey, I've found an enigma".

Sorry. I made a mistake here. The delta tensor is an isotropic tensor only when it is a once-covariant, once-contravariant tensor --similarly for tensor products of them--. And the epsilon tensor is an isotropic tensor only when it's totally covariant or totally contravariant. I already pointed this out in a previous post. Same OP. Different thread.

Yes, this surfaced in the 1st post already. Yes, that's another problem. But also, the OP is not familiar with tensor algebra. They have a tendency to use repeated indices both for summation convention and for representing fixed diagonal elements, so no wonder the conclusions are wrong, already just at the mathematical level. Also, I've observed them being very cavalier in asserting other properties about tensors. Symmetric or antisymmetric only make sense for tensors twice covariant or twice contravariant, etc. The delta tensor, or tensor products of delta tensors; and the epsilon tensor, or tensor products of them--, are isotropic tensors only when they are pure-covariant or pure-contravariant. And so on. Seeing tensor algebra used like this brings tears to my eyes. That's why I'm taking certain distance from this particular OP's posts. I'm only too glad I have you and Kino to help with this.

The OP wants to fix the value of the alpha index and at the same time keep Einstein's summation convention. Very dangerous practice. No wonder they get inconsistent results.

Exactly. I can barely add anything significant. Contraction of pairs of symmetric indices with pairs of antisymmetric indices always gives zero, no matter what the value of the non-zero components of both. The OP clearly has problems with the tensor formalism. \[ A_{\alpha\beta}=-A_{\beta\alpha}\] \[ S_{\alpha\beta}=S_{\beta\alpha}\] \[ Q=S^{\alpha\beta}A_{\alpha\beta}=S^{\beta\alpha}A_{\alpha\beta}\] (swapping dummies) and, \[ S^{\beta\alpha}A_{\alpha\beta}=\left(S^{\alpha\beta}\right)\left(-A_{\alpha\beta}\right)=-Q\] (applying symmetry properties) As \( Q=-Q \), \(Q\) must be zero, even if \( A_{\alpha\beta} \) and \( S_{\alpha\beta} \) are not. The rest of the indices are along for the ride.

I share your concern, but I'm an optimist. I think --or wanna think-- many people will realise that either we learn to fork out some attention span on reliability of information, or else we become too vulnerable, and we're done for. Even fish will take the bait for so long. After a while they know it's bait. Either they wise up, or only the cleverer survive. Although you're the expert angler here, Moon.

Not that I know of. You do have a section to test your LaTeX beforehand, as you already know: https://www.scienceforums.net/forum/99-the-sandbox/

Yeah, nice insight. \( \gamma^{-1} \frac{d\gamma}{d\tau} \) doesn't have to be zero, even though \( \gamma^{-1} \frac{d\gamma}{d\tau} - \gamma^{-1} \frac{d\gamma}{d\tau} \) is identically zero. Welcome to the forums, @Kino.

You may be right. I'm not following this thread very closely, and I'm not sure if what you say is what the OP is trying to prove. But here's a more standard proof: This is obvious, but let's do a check. And gammas, of course, are in general time-dependent. SR can deal with accelerations, as Markus said. The 4-vectors are, and their 4-product is, It's necessary to keep in mind that, The derivative of the gamma is, So the 4-product is indeed identically zero: As Markus also said, the concept that in SR supersedes constant acceleration is that of hyperbolic motion. He also has been very careful to distinguish flat space-time from prescription to adopt inertial frames. Indeed, the Minkowski spacetime can be studied in terms of Rindler charts --hyperbolically accelerated frames--. It is not difficult to show that when the motion is completely collinear (spacial 3-vectors of velocity, acceleration, and force). It's in wikipedia, although the proof is not complete, and I can provide a completion, if anyone's interested. As I said, I'm not completely sure that what I'm saying is relevant to the discussion. It is the standard, reliable, mainstream formalism that we know and love.

I'll never forget your "beyond any reasonable doubt" correction. I stick it everywhere: "That was fun, beyond any reasonable doubt". I find Maher, if not hilarious, with one of the best ratios of being funny/making a case at the same time. Although you have to agree that lizard people, children-eating elites, and a saviour in the form of T**** is a gift from heaven as comedic material.

Observers at infinity are misinformed. Great summary, Hanke.

17 Republican Senators turning on Trump sounds like asking for a miracle* --although nobody would have thought 10 --10 was it?-- in the Congress was likely at the time of the 1st impeachment. Any cooling-off period will only cut down the chances. What with people already focused on other, more urgent matters, like Covid. How long does it take to get to the Senate, and how many Senators are positioning themselves? * Maybe that's the miracle that Trump will achieve: At the GOP's hands, his own sacrifice, Jesus-like. QAnon will have a martyr instead of an avenger --just managed to keep on-topic by the skin of my teeth.

This may be relevant to the topic at hand --especially 0:00 to 2:30--: https://en.wikipedia.org/wiki/Great_Disappointment Maybe QAnon believers will recycle the leftovers of their faith into a makeshift second-coming kind of faith. Adventists of 2024. That's why it's so important to impeach Trump --so that he can't run again. That and stopping him from getting a pension for life from American tax payers.

I don't understand this. How can a manifold "become" anything? And of all things, "threadbare"? Can you make this argument a little clearer? Metrics don't become space-like. Intervals do. I deeply mistrust Schwarzschild coordinates. I think they're good only to solve Einstein's field equations. OK. I'm looking forward to it. Thank you, @studiot.

Somehow a shockwave of disbelief is easier for me to swallow than a shockwave of understanding. I imagine going so deep into a rabbit hole implies some kind of inertia. It must take some time to get out. What ever happened to the denial-depression-anger-bargaining-acceptance chain?

Update QAnon believers are in disarray: https://www.bbc.com/news/blogs-trending-55746304 I don't know what it means when a group of numb nuts are "in disarray". Was there any array, to start with? They seem to imply that a shockwave of disbelief is going through their ranks.

Another possibility is that someone has been messing around with alternative histories. Parallel-universe plumbing if you will.

Bingo!

Is it? I wasn't aware of this.

There is no such thing as "absolute values of the metric coefficients" with an invariant meaning. Also, I don't think there's an invariant meaning to the concept of "relative sizes of the coordinate intervals". Another methodological comment on my part. If you're serious about relativity, try not to build up your arguments from coordinate patches. You're going to make mistakes. The literature is full of similar arguments, in the years before intrinsic formalisms were developed, which were proved to be wrong. The technique in GR today is: You use intrinsic formalism --vectors, forms, and tensors formed from them-- to establish the general results, and then go to a coordinate patch for a particular calculation with a particular distribution of energy-momentum. Allowed coordinate transformations in GR are diffeomorphisms, which means they're infinitely differenciable and never change the invariants of the metric --because their inverses are also differentiable--, so it's possible that you're going to a mathematical no-man's land. Nobody would bother to check due to your always working in coordinates.

Stands for either "original poster" or "original post".

Sorry, Markus. I hadn't seen this. But reading my comments you'll see I understood what you meant.

No "recalculation" requested this time, I suppose. He will probably be proud of it.