On Wednesday, April 24, 2019 at 11:06:10 AM UTC-6, Bruno Marchal wrote: > > > On 23 Apr 2019, at 13:39, [email protected] <javascript:> wrote: > > > > On Tuesday, April 23, 2019 at 4:00:26 AM UTC-6, Bruno Marchal wrote: >> >> >> On 20 Apr 2019, at 23:14, [email protected] wrote: >> >> >> >> On Friday, April 19, 2019 at 2:53:00 AM UTC-6, Bruno Marchal wrote: >>> >>> >>> On 19 Apr 2019, at 04:08, [email protected] 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 [email protected]. To post to this group, send email to [email protected]. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
