joigus

Are you implying that asking whether science can prove the existence of 17 balls of jello, each with the mind of a baby, but whose mutual communication results into a common mega intelligence that rules our universe is a ridiculous question?

You're welcome.

It's not a metaphor; it's an analogy. Do you know the difference?

No. It's, \[\gamma^{2}\left(1-\beta^{2}\right)=1\] Gamma is a number always bigger than one. Beta is a number always less than one (in absolute value.) The absolute value of beta determines gamma.

Maybe I missed something all those years of measuring with ammeters and voltimeters, and oscilloscopes. Differentiating the equations to get the error formulas, actually doing the experiment, crunching the numbers for the analysis of errors, and reporting the results. I cannot guess the pointless next step of your "idea" because it's impossible to guess what's going to come next after a babbling nonsense like yours. Last time I measured it in a laboratory of electricity, Faraday's law was correct to within the error bars. Even though I was a terrible experimentalist, not even I was able to get it wrong. And then studying the theory and learning how to solve the equations forwards and backwards... Faraday's law is actually necessary for the Lorentz symmetry of the theory. Angular momentum in particular would not be well defined or conserved either. The symmetries are too tight. You need it for charge conservation too. The homogeneous terms of the Maxwell equations are actually forced upon you from a pure geometric identity (Bianchi identity) or from first principles of mechanics, after you arrange the E and B fields in their proper Lorentz-invariant form (Feynman's proof of Maxwell's equations.) But first you must understand also what a gauge theory is, something you cannot even begin to fathom. Moronic pseudo-science. That's what your sorry excuse for an idea is. Sleep well.

You don't make any sense, how can I guess what you're going to say next? You are clueless about electromagnetism. That's all I can say. No matter how many colours you use in your drawings. As to the mathematics (no wonder it's absent in your explanation), I can't even begin to tell you how inconsistent your idea is with everything we know. And finally, what you're saying is falsified by experiments every which way.

Utter nonsense.

No. Computers "solve" equation by getting approximate solutions. Humans who want to solve non-linear equations must be clever. That's what I mean.

No. It's neither delicate nor embarrassing. It's a little bit like asking: Can science prove the existence of 17 balls of jello, each with the mind of a baby, but whose mutual communication results into a common mega intelligence that rules our universe? Split of Theory of everything It's just as delicate and as embarrassing IMO.

Not really. Not in general. You might get lucky and pull it off in particular examples. For example, consider the (non-linear) system: \[x^{2}-y^{2}=a\] \[x+y=b\] And assume, \[b\neq0\] The first equation is non-linear, because it involves powers of x and y different from 1. But you could use the second one to substitute x+y in, \[x^{2}-y^{2}=\left(x+y\right)\left(x-y\right)\] and get to the linear system, \[x-y=\frac{a}{b}\] \[x+y=b\] which can be solved by Gauss (by adding and subtracting both eqs.) to get, \[2x=b+\frac{a}{b}\] \[2y=b-\frac{a}{b}\] So that, \[x=\frac{b^{2}+a}{2b}\] \[y=\frac{b^{2}-a}{2b}\] That's the problem with non-linear equations. Each one is different.

Absolutely not. Gauss elimination is for linear equations; GR is highly non-linear. "Linear" means that a combination of solutions S1, S2, like 2S1 + S2, or -S1 + 3S2, etc., is also a solution. That is, linear combinations of solutions are also solutions. Also, the Gauss method is only valid for numeric equations, not for differential equations. GR is set in terms of differential equations (equations that involve the derivatives of an unknown function.) I hope that helped.

How come identical situation (orientation, angular velocity,...) of the magnets and the rotating circuit gives rise to opposite effects in I-II and in III-IV?

Yes. I've noticed that OP was using the term in a different sense. Very inspired and inspiring comment, by the way.

Mating-brilliant. (@ Koti.)

Just to take up on @studiot's suggestion. An idea that you may find silly, but has helped me a lot during hard times is the thought that clouds keep rolling over my head no matter what I think. A similar mindset you can find here:

It's a super far-fetched idea that ignores that the universe is not a closed system and that there are sources of randomness at every level. Get some professional treatment for your depression. I join my voice to those who've told you before.

Extensive research? Really? It took you one hour to track down a paper where the authors sound like they're hedging the bets, or formulating disclaimers, which OTOH is common practice in the peer-reviewed world. So the viral RNA could not be efficiently assembled from certain techniques based on eukaryotic splicing. So what? What does that lead to, in the way of a conclusion? I can't build the Taj Mahal. Does that prove that there's no hard evidence that it exists? They're describing an assembly problem, not a signature problem. Even procaryotes have signatures of just 6 to 14-bases. It is well known that viruses have an even stronger adaptive pressure to be terse. I would expect re-assembly to be hard with eukaryotic techniques. Can you lead us to a paper that talks about signatures, instead of assembly?

Exactly, (either one of them, the reflection point or the reception point), and because the exact location of the reflection point is imprecise (we only know the Euclidean distance in that FOR), when the reception takes place is unknown. You can only take one of them as defining the x-direction with the emissor, if you will, but the other one could be at an angle. The three don't have to be collinear. I hope you agree.

But it does matter. If the mirror is on the same line where the reception point is, but in the same direction, the signal's delay will be minimal. On the contrary, if it's on the same line but in opposite direction, the delay will be maximal for the given distance, corresponding to the extreme values of the cosine in, \[\left(\boldsymbol{x}_{2}-\boldsymbol{x}_{1}\right)\cdot\left(\boldsymbol{x}_{3}-\boldsymbol{x}_{2}\right)\] The reception event is described only in terms of its Euclidean (spatial) distance to the emission in the given frame. The reflection point being at certain distance does not determine the time it takes for the photon to get to the reception point. It's not a "home-run," but an open trajectory in Minkowski space in which we only know the radial distance from the emission point. In what directions? The problem is under-determined. Maybe I misunderstood something...

Hawking worked in mysterious ways...

LOL. A noose, that's what it is. <giggle>

I was thrown away from that discussion with the escape velocity.

IOW, why are acids so important, so central to chemistry? They say the basic unit of chemical exchange is the electron. But that's only half the story. Protons are very powerful mediators of chemical reactions too. And the reason is that the size of a hydrogen atom compared to the size of just a proton (ionized hydrogen) is like the size of the Earth compared to the size of an orange. So when you have a substance that is capable of liberating protons, you're liberating myriads of little "positive versions of the electron," so to speak. That's why there is no central concept in chemistry of how easily a substance can liberate any other ion, like e.g. Na+. But liberating H+ is very powerful, very reactive. Protons are elementary particles, small as can be, and move about very freely, especially in aqueous solution. The mitochondria in your cells are powerful proton-pumps.

Mmmm. Interesting problem, but some technical difficulties I can see there. The first thing I see is that going from event 1 (emission) to event 2 (reflection), and from there to event 3 (reception) involves three relevant events, IMO. This implies two 4-vector translations, and their sum. Let's call them d12, d23, d13. They satisfy: d13=d12+d23. But unfortunately, \[s_{13}^{2}\neq s_{12}^{2}+s_{23}^{2}\] If you write down the correct expression, you get, \[s_{13}^{2}=s_{12}^{2}+s_{23}^{2}+\left(x_{2}-x_{1}\right)^{\mu}\left(x_{3}-x_{2}\right)_{\mu}=\] \[=s_{12}^{2}+s_{23}^{2}+c^2\left(t_{2}-t_{1}\right)\left(t_{3}-t_{2}\right)-\left(\boldsymbol{x}_{2}-\boldsymbol{x}_{1}\right)\cdot\left(\boldsymbol{x}_{3}-\boldsymbol{x}_{2}\right)\] You can see very clearly in this expression that how much it takes for the signal to reach the reception point depends on the relative positions (including orientation) of the whole "triangulation." It is not specified well enough by Euclidean distances only. This may be at the root of the problem that @md65536 sees. In your second post it looks like you're trying to get a differential equation that relates r and t. There's a problem there too. In the words of @md65536: I would re-word it as: your problem is under-determined, which I think is what md65536 is saying. The back and forth rays belong to different hyperbolas*. In fact, from the differential equation that you're trying to get at, which is correct: \[\frac{dr}{dt}=c^{2}\frac{t}{r}\] You get, not only the outgoing ray r=ct, but also the return trip r=-ct, plus the unwanted free gift of an infinite set of constant accelerated motions, \[r^{2}-c^{2}t^{2}=K\] with K being an arbitrary constant. So going back and forth implies jumping from one inertial system to another (in the case of the light rays, one relative sign to another in r, t). Sorry I took the whole thing to the language I understand better. I hope that helps, and I hope it has some bearing on the problem. *In the case of the light rays, they're "degenerate hyperbolas", meaning straight lines in Minkowski space. Corresponding to K=0.

You mean like magic?