Reference Frames

[Qfwfq]

If you follow Einstein's paper, the one you linked to, and understand the differential geometry and then answer the questions I posed at the end of this post, you should be able to get the matter straight. A uniform gravity field is not a contradiction in terms, the intrinsic curvature is simply zero. Which tensor, then, represents the field? It follows from that paper and I think GR experts know the answer. The words of Thorne that your quote friend quotes suggest that he doesn't regard the field as the intrinsic curvature, because he specifies tidal. Perhaps it is your friend that is misinterpreting the meaning of what people say. :evil:

[spectrum disorder]

From using this forum to sitting in the room where your computer is, we are using several frames of reference. If we didnt have any of these how could we even hope to survive.

 

Reference frames can be seen as a kind of tool by which we measure things (speed, direction, up, down, etc) Or even a train going past, or being on a train itself, and seeing the outside pass by etc.

 

I'm not sure that we can manage without reference frames, because then how would we discuss things such as relativity?

[Farsight]

Yow, that book is $97.82.

 

If you follow Einstein's paper, the one you linked to, and understand the differential geometry and then answer the questions I posed at the end of this post, you should be able to get the matter straight. A uniform gravity field is not a contradiction in terms, the intrinsic curvature is simply zero. Which tensor, then, represents the field? It follows from that paper and I think GR experts know the answer. The words of Thorne that your quote friend quotes suggest that he doesn't regard the field as the intrinsic curvature, because he specifies tidal. Perhaps it is your friend that is misinterpreting the meaning of what people say.

 

Perhaps. Any chance you could point out something I could focus in on? I'm on page 176 looking at "the extension of the fundamental tensor" and I'm afraid I'm rather scratching my head. I'm a layman, my maths is weak. I don't understand it well enough. I'll have to come back to those questions, but to show willing:

 

How, then, would a uniform gravitational field be represented in GR?

I just don't know and that's the problem. In the real world, there are none.

 

Is the Riemann tensor the only one in GR?

No, but the question is most moot. I've been trying to compare a 3D mechanical stress tensor with GR tensors, and the picture I keep getting is "elasticity", where the stress and tension increases as we approach a central location. There's a local gradient which increases as you approach the central location, and you can plot this increase as a curve, but it's somewhat mathematical rather than actual. And I'm scratching my head trying to work out "What's actually there" and what best describes my mental model. It's a toy model I'm afraid, but one has to start somewhere.

 

What is the description from the rocket's point of view?

I'm not sure what you mean. Within an IRF, where the local curvature is deemed negligible?

 

Why do things "fall down" in it despite the field gradient being zero? I'll really have to study this properly to give you a worthy answer, but: because the local gradient isn't zero. There has to be a difference across the local frame of the falling body. If there was no difference and no gradient, there would be no "gravitational force" to cause the thing to fall down. If you take the derivative of this local gradient you're looking at the local change in gradient, the curvature, the tidal force, and you can plot another gradient in this "tidal force". Whilst this is slight, it is there. It has to be there. f there was no curvature you could not have gone from a "flat" location to one with a gradient. That's where my flat hill comes in.

 

How can a "hill" be "flat", without being a contradiction in terms?

I don't know. See above.

 

Sorry this isn't a great response. Thanks for talking.