Geometry for Intrinsic Curvature

[geordief]

If we set an observer at the centre  of a massive body** ,how will that observer create a geometry so as  model potential events?

 

Is this even a valid thought experiment?

 

Suppose the observer (call him O ?) wants to draw a circle  or the surface of a sphere from his stand point the  relationship of pi to the radius  will not be Euclidean ,will it and depending of  the ratio of the mass concentration to (what?)  that ratio will approach or depart from that Euclidean ratio.

Can a complete geometry be  created for O and will it be easy for him to understand or is Euclidean geometry the only easy geometry? 

Apologies in advance as per usual for any mistaken assumptions ,    foolish imaginings or poor expressivity (correct term?) 

 

**I am imagining an extremely massive body,perhaps as an example one where the size of the event horizon was  relatively close to some manageable multiple of the wavelength of a convenient em wave.

Edited December 24, 2017 by geordief

[Vmedvil]

This is an annoying question. The only way I can answer this in Standard physics is 

ds²  = dx² + dy² + dz²  - (Cdt')²

Which is from a Youtube video but is still the invariant version of 4-D Space-time, but I liked this picture, I want to give this Kid youtuber credit for knowing what he was talking about https://www.youtube.com/watch?v=zScn3tV9YPU , but I want to correct you on something if you notice why your youtube video got a bunch of hits that NO, that is not a new concept and has been a concept of SR for a long time even Quantum Gravity equations are bound to it at some point.

In any case, here is a link to Wikipedia with the same equation https://en.wikipedia.org/wiki/Special_relativity#4D_spacetime

ds  = S 

http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html

Where   dx² + dy² + dz²  = ∇ 

 

http://asonika.com/models/

then Laplace operator goes to Schrodinger's equation solved for it.

http://www.butterflyeffect.ca/Close/Pages/SchrodingersEquation.html

 

Where dt' 

https://physics.stackexchange.com/questions/110669/is-gravitational-time-dilation-different-from-other-forms-of-time-dilation

Where V² = C²  - V_Λ²  , C² in the Schwarzchild metric which is solved for C²

http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/blkhol.html

 

V_{Λ = H d(Kiloparsec)}

_{Hubble's constant, goes to Friedmann equation.} 

https://web2.ph.utexas.edu/~coker2/index.files/friedmann.htm

Edited December 25, 2017 by Vmedvil

[Vmedvil]

This form of the relationship is incorrect, I now realize  dx² + dy² + dz²  = ∇ 

Corrected form of Quantum Equivalence. 

dx² + dy² + dz² =  (d²/∇)

 

Edited December 26, 2017 by Vmedvil

[geordief]

My question was really ;If spacetime curvature is extremely curved (as in the region of something close to a mathematical singularity ) does a new type of geometry "hove into view" quite unlike Euclidean geometry but  very simple in  functionality?

 

If this geometry was understood would it be possible to derive Euclidean geometry from it in the  same way as the geometry of curved spacetime is  (painfully ,it seems) derived from Euclidean geometry?

Might this geometry be seen as a "special case" in a mirrior image way to the way that Euclidean geometry can be seen as a "special case" of "normal" curved geometry?

 

 

[Vmedvil]

7 minutes ago, geordief said:

My question was really ;If spacetime curvature is extremely curved (as in the region of something close to a mathematical singularity ) does a new type of geometry "hove into view" quite unlike Euclidean geometry but  very simple in  functionality?

 

If this geometry was understood would it be possible to derive Euclidean geometry from it in the  same way as the geometry of curved spacetime is  (painfully ,it seems) derived from Euclidean geometry?

Might this geometry be seen as a "special case" in a mirrior image way to the way that Euclidean geometry can be seen as a "special case" of "normal" curved geometry?

 

 

Say that again? I am in no way getting what you are saying.

Edited December 26, 2017 by Vmedvil

[geordief]

2 minutes ago, Vmedvil said:

Say that again? I am in no way getting what you are saying.

Did you read the OP carefully?

 

Imagine an observer  at the centre of an extremely curved BH. How would he "do geometry"?

Similar "impossible"questions  are asked regarding "flatlanders" So why cannot we ask this kind of a question too?

[Vmedvil]

3 minutes ago, geordief said:

Did you read the OP carefully?

 

Imagine an observer  at the centre of an extremely curved BH. How would he "do geometry"?

Similar "impossible"questions  are asked regarding "flatlanders" So why cannot we ask this kind of a question too?

He would use a Super-gravity equation "Quantum Gravity", due to the fact that he is really near the Schwarzchild radius or inside, that neither GR or QM can explain alone. 

Edited December 26, 2017 by Vmedvil

[MigL]

Don't have a clue what Vmedvil is talking about...
but anywhere inside an event horizon 'geometry' ( whether Euclidian or curved ) is impossible as there is only one direction.

In any strongly curved space-time ( but outside an event horizon ), the local approximation is always flat, so, no doubt, these 'curvedlanders' would also come up with Euclidian geometry.

[geordief]

2 hours ago, MigL said:

 

In any strongly curved space-time ( but outside an event horizon ), the local approximation is always flat, so, no doubt, these 'curvedlanders' would also come up with Euclidian geometry.

Yes I see that.(have just come to appreciate it actually). I have just realized that GR  is seemingly  put together  using Euclidean geometry  for all the  buzz about its  curved geometry(and arcane terminology).

 

It reminds me of my DIY work around the house and garden. Nothing fits at first  until I find a way of "making it fit" 

 

 

 

Edited December 28, 2017 by geordief