joigus

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.

One of the girls in the last cover looks strangely familiar...

That would be the Lorenz gauge --different Loren*s --. https://en.wikipedia.org/wiki/Ludvig_Lorenz It's first order. It becomes second order when substituted in Maxwell's equations, which I think is probably what you meant, Markus. ----- Maybe a split is in order? I agree with @studiot and @MigL. Although I find the discussion with @Markus Hanke very interesting. It's OK. If somebody had to be in charge of the Inquisition I'd rather it be 21st-Century Canadians. Much more humane, I'm sure. Nobody expects... that joke. 😆

http://www.folksong.org.nz/soon_may_the_wellerman/index.html

Let me get back to you when I get more time to think about this and review some literature. It's correct that radiation is generally described as having "2 DOF" in the sense of having two polarisation states. But the term "degrees of freedom" is a bit ambiguous. Some people use it in the sense of counting spin states. But there are also space-time variables, according to, \[\left|\boldsymbol{k},s\right\rangle\] The \( k_x \), \( k_x \), \( k_x \), and \( s \) would be four variables, more in the sense that I was talking about (DOF as number of variables necessary to describe the states). As I said, let me get back to you. Also, I think you can do the counting on the E's and B's, which are gauge invariant. You can do it on the A's, but AFAIR there's just one gauge fixing condition, because you're in \( U\left(1\right) \)... Still thinking. Apparently I'm outside my comfort zone too. LOL

Ok. There's a couple of things that you're seeing there that are already taken care of in the mainstream formalism. And have been mentioned repeatedly. One of them is that for any physical 4-vector the pseudo-norm with signature (+,-,-,-) must be positive --causality. The other one that I at least forgot to mention --it's possible that either @Ghideon and/or @Markus Hanke have mentioned it and I've missed it--, is that the naught component must be positive. Those are called orthochronous 4-vectors. So you cannot add two arbitrary 4-vectors and hope that that has any physical meaning. When you add two orthochronous 4-vectors, you obtain another orthochronous 4-vector. Also, the orthochronous character ( \( v^0 > 0 \) ) is split in mutually disconnected subsets of the Minkowski space, so the idea that you suggest that there is a "perforated" structure is actually a misguided intuition. The reason is that both the sign of the Minkowski pseudometric and the zero component are continuous and differentiable function of their arguments: \[ f\left(\alpha^{0},\alpha^{1},\alpha^{2},\alpha^{3},\beta^{0},\beta^{1},\beta^{2},\beta^{3}\right)=\alpha\cdot\beta \] \[g\left(\alpha\right)=\alpha^0\] So you can't have either changes in sign of the product, nor changes in the sign of any component for arbitrarily close 4-velocities, as you posit.

(My emphasis.) Your tangents are always very interesting. Of course, you're right. I meant "in a sense" in the sense that it is an object that only has 6 non-zero components in general, which can be relevant when there is charged matter around, although in that case the new degrees of freedom can be attributed to matter densities --see below. Let me check the counting. Pure EM field must have identically null Lagrangian,* \[\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}=\frac{1}{2}\left(E^{2}-B^{2}\right)\] as in the vacuum \( \left|\boldsymbol{E}\right|=\left|\boldsymbol{B}\right| \), so the Lagrangian is identically null on solutions, as corresponds to massless particles. That leaves 5. If we now throw in the transversality, \[\boldsymbol{E}\cdot\boldsymbol{B}=0\] that leaves 4 DOF, which are the ones you're talking about. Correct me if I'm wrong. When light goes through matter, we no longer have the \( \boldsymbol{E} \) and the \( \boldsymbol{B} \), but also the \( \boldsymbol{D} \) and the \( \boldsymbol{H} \), so that, \[\mathcal{L}=\frac{1}{2}\left(\boldsymbol{E}\cdot\boldsymbol{D}-\boldsymbol{B}\cdot\boldsymbol{H}\right)\neq0\] with, \[\boldsymbol{D}=\varepsilon\left(x\right)\boldsymbol{E}\] \[\boldsymbol{B}=\mu\left(x\right)\boldsymbol{H}\] so we have two "additional" DOFs, the matter medium, characterised by \( \varepsilon\left(x\right) \) and \( \mu\left(x\right) \) that take us back to 6. But it's a fictional 6. *Edit: \( c=1 \) throughout.