On Wednesday, April 24, 2019 at 11:06:10 AM UTC-6, Bruno Marchal wrote: > > > On 23 Apr 2019, at 13:39, agrays...@gmail.com <javascript:> wrote: > > > > On Tuesday, April 23, 2019 at 4:00:26 AM UTC-6, Bruno Marchal wrote: >> >> >> On 20 Apr 2019, at 23:14, agrays...@gmail.com wrote: >> >> >> >> On Friday, April 19, 2019 at 2:53:00 AM UTC-6, Bruno Marchal wrote: >>> >>> >>> On 19 Apr 2019, at 04:08, agrays...@gmail.com wrote: >>> >>> >>> >>> On Thursday, April 18, 2019 at 6:53:33 PM UTC-6, Brent wrote: >>>> >>>> Sorry, I don't remember what, if anything, I intended to text. >>>> >>>> I'm not expert on how Einstein arrived at his famous field equations. >>>> I know that he insisted on them being tensor equations so that they would >>>> have the same form in all coordinate systems. That may sound like a >>>> mathematical technicality, but it is really to ensure that the things in >>>> the equation, the tensors, could have a physical interpretation. He also >>>> limited himself to second order differentials, probably as a matter of >>>> simplicity. And he excluded torsion, but I don't know why. And of course >>>> he knew it had to reproduce Newtonian gravity in the weak/slow limit. >>>> >>>> Brent >>>> >>> >>> Here's a link which might help; >>> >>> https://arxiv.org/pdf/1608.05752.pdf >>> >>> >>> >>> Yes. That is helpful. >>> >>> The following (long!) video can also help (well, it did help me) >>> >>> https://www.youtube.com/watch?v=foRPKAKZWx8 >>> >>> >>> Bruno >>> >> >> *I've been viewing this video. I don't see how he established that the >> metric tensor is a correction for curved spacetime. AG * >> >> >> ds^2 = dx^2 + dy^2 is Pythagorus theorem, in the plane. The “g_mu,nu” are >> the coefficients needed to ensure un non-planner (curved) metric, and they >> can be use to define the curvature. >> >> Bruno >> > > *Thanks for your time, but I don't think you have a clue what the issues > are here. And, as a alleged expert in logic, it puts your other claims in > jeopardy. Firstly, in the video you offered, the presenter has a Kronecker > delta as the leading multiplicative factor in his definition of the Metric > Tensor, which implies all off diagonal terms are zero. And even if that > term were omitted, your reference to Pythagorus leaves much to be desired. > In SR we're dealing with a 4 dim space with the Lorentz metric, not a > Euclidean space where the Pythagorean theorem applies. How does a diagonal > signature of -1,1,1,1 imply flat space? Why would non-zero off diagonal > elements have anything to do with a departure from flat space under > Lorentz's metric? AG * > > > > Oops sorry. Since long I do relativity only in its euclidian form, through > the transformation t' := it. (I being the square root of -1). This makes > Minkowski euclidean again. I should have mentioned this. > > Bruno >
*How does it make Minkowski euclidean if you're not dealing with spacetime. Euclidean and departures from flat require real coordinates. AG* -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
