Misleading Phrase: Collapse of wave function

Okay.

 

Log in or Sign up to hide all adverts.

All F is doing is pulling Einstein quotes out of my paper. But there's nothing wrong with the paper. Actually its basically an historical article. But the physics is solid. Especially since all the physics is either from Einstein or peer reviewed articles.

 

Log in or Sign up to hide all adverts.

Yes. Suppose a spacetime region where the source field is zero that is \(T_{\mu\nu} =0\). Then obviously the sum on your LHS must be identically zero in that region.

A necessary condition for this is that the curvature tensor field vanishes in this region. That is, \(R_{\mu\nu}=0\). Since this mandates the vanishing of the Ricci scalar, it is also a sufficient condition.

But as we discussed earlier, this does not mandate the vanishing of the metric field in this region - indeed, due to the relation between the metric tensor (at a point) and the Ricci tensor (at the same point) it means that the metric field \(g_{\mu\nu}\) need only be constant in our source-free region.

And strictly conversely - where the source field is no-zero, the Ricci tensor must be non-zero, and the metric field cannot be constant

Whether the term \(g_{\mu\nu}\Lambda\) represents curvature I am not sure - I know less cosmology than I do physics. My mathematical instinct says no, but I have taken sufficient mathematical hard knocks to distrust my instinct!

 

Log in or Sign up to hide all adverts.

Neither were Einstein's papers.

It isn't wrong.

It's a curvature in the inhomogeneity of space, and therefore the metrical properties of space as measured with rods and clocks, which you think of as curved spacetime. But it isn't curved space.

See this Baez article where you can read this: "Similarly, in general relativity gravity is not really a `force', but just a manifestation of the curvature of spacetime. Note: not the curvature of space, but of spacetime. The distinction is crucial".

I'm the one giving references to back up what I'm saying. You're giving up because you can't counter them.

 

FLRW cosmologies say it does; there are three possible topologies:

A bounded hypersphere (positive curvature)
An unbounded surface (zero curvature) Often referred to as "flat" in popular literature
An unbounded hyper-hyperboloid (negative curvature) Often called a "saddle shape" in popular literature

I think you will find this area of research of interest. Watch out for the Wikipedia article on FLRW; it's a bit messy and might be misleading. I expect at your level of expertise with GRT it's better to go to scholarly sources. For a good overview at the popular level I suggest Heinz Pagels' Perfect Symmetry, a bit dated but still highly relevant and correct in this area.

 

No, it is not at all necessary. The curvature is defined by the curvature tensor $R^{\kappa}_{\lambda\mu\nu}$ which has much more components. If they are all zero, the Ricci tensor is zero too. Thus, this would be sufficient. But it is not necessary at all. The curvature tensor for the Schwarzschild metric is not zero, but it is nonetheless a vacuum solution, thus,\(R_{\mu\nu}=0\). holds.

Correct, it has nothing to do with curvature.

 

In his 1920 Leyden Address. See the Einstein digital papers.

I know, I've read it. And I agree with your paper. A gravitational field is a place where spacetime is "tilted" as opposed to curved. The first derivative of potential relates to the "force" of gravity, the second derivate relates to the tidal force.

Take a look at this where Einstein said SR referred to a limiting case which is "nowhere precisely realized in the real world". It's an infinitesimal region, a region of no extent, hence your Synge quote that the midwife should be buried with full honours. Make the room as small as your like, and make your pencil as small as you like. Reduce it to a single electron if you like. It still falls down.

 

Never mind. Off topic; deleted.

 

Two out of three were always going to be wrong. And I'm afraid to say, they all are.

I cannot explain why Einstein didn't predict a flat finite expanding universe. For some reason his usual confidence seems to have deserted him when it came to cosmology.

 

Ha! You flatter me. I am a humble technician who pushes around more-or-less agreed symbols according to pretty-much agreed rules.

As it happens, I learned differential geometry in some depth, which included what were called "metric field theories", which of course nodded in Einstein's direction. I claim no "expertise" in the actual physics of GR, though I would be astonished to learn that the mathematics is incorrect - in fact I am convinced, on examination, that it is not (aside from some notational quirks I regard as misleading)

 

Please Register or Log in to view the hidden image!

I find it is an approach that is rarely rejected and since we'd clashed...

Please Register or Log in to view the hidden image!

I suspect I might learn a fair bit from your knowledge of differential geometry. I have a very wide amateur's knowledge of general physics and just enough math to get into serious trouble, most of which I got in an EE's education and the rest by bumbling around and poking a little deeper into things that interested me. That bumbling will now be assisted by knowing just exactly where I'm looking for those ten equations, and when I learn something new from someone it tends to impress me.