joigus

Hope this helps.

Hats off to Ghideon. Great summary.

I'm no expert either, so be my guest. And of course it would be nice that some of the local experts can give us a hand. Yes, DNA does get old. That's at the basis of cellular aging, and thereby the organism's aging itself, AFAIK. The replication mechanism is some kind of bi-directional zip assembly, so it's always imprecise at the ends. In one direction the replication process is very smooth, because the initial fragment (RNA primer) and the DNA polymerase work in the 5' to 3' direction, but in the opposite strand, primer and polymerase are forced to work against the uncoiling of the double strand, so it must interrupt and restart the copying work over and over again --the so-called Okazaki fragments. That's why there's always a mismatch at the end. Eukaryotes use a meaningless[?] chunk of DNA at the end --telomere-- which is partially replenished with every replication process, to kind of delay this ongoing degrading of the information. Also, as you point out, different cells down the line of cellular development, have different adjustments to their particular function. Red blood cells being the perfect examples of cells that will never go back to be able to produce anything in the way of stem-cells or higher-potent cells, because they've completely lost their DNA. Other extremes are neurons and cells from the digestive lining. The average life of the latter is, if I remember correctly, 48 to 72 hours. And neurons, because they never get replenished by sister cells mitotically splitting. Although new neurons do appear directly from stem cells, especially in the hippocampus*. Also, they retain some ability to reconnect, or change connections. That's about the summary of what I know. * Google search: "newborn neurons in hippocampus and olfactory bulb"

Sorry, wrong quote:

Suspicious of anyone? Eek!

That goes for neurons. But I meant it --more in general-- in the sense that the cell --every cell, including neurons-- is the basic unit that carries out a particular function within the organism. In order to do that, they specialize down the line of cellular development. Cells have a finite life though, so when they no longer work, they are replaced by releasing stress signals that activate their destruction and further mitosis in other sister cells. As long as the cell is performing its function, it's important that it does it well --cancer being an example of how bad it is that a cell stops working properly. Cancer cells get stuck in continual mitosis and just can't stop. It's their function that's essential. Gametes, on the contrary, are some kind of "inter-phase" between one organism and the next generation. They carry random arrangements of half the genetic material --haploid cells-- of the parent organism; and they're fundamentally like a throwing of the dice. Not a functional cell really. Not yet. So chromosomes are expendable. On the contrary, the organism cannot afford to have malfunctioning DNA in the nucleus of working cells. That's why eukaryotes have mechanisms to destroy tissue cells that are not working properly. It doesn't play around trying to fix it --replace it. During replication cells do have an impressive proofreading mechanism, very precise --transcription and translation don't have to be that accurate--. But when DNA that's being read for transcription is just too messed up, the cell must be destroyed. When the cell malfunctions, the DNA is replaced... by replacing the whole cell. Not taking any chances. But a gamete turns bad? No problem for this organism. That's more or less what I meant.

Blasphemy!

But, as I understand, in common slang "dark side of the Moon" means the side that we never see from Earth. Although it's not always dark. Ergo: misnomer. I think it was Eric the Red who decided to call it "Green land" so that his fellow Vikings would buy into the idea of going there looking for pastures new. May be an apocryphal story.

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?