Good question. You don't actually need to have an 'extra' dimension to accommodate the distortion. In your 2D example, sure, we are accustomed to thinking of a 2D surface embedded in 3D space. However, the surface is still 2D, and we can geometrically encode the bending and warping of the surface by data on the surface, without the need for a higher dimensional ambient space within witch to place our 2D surface. For example, a torus (donut) has a well known form when its 2D surface is embedded in 3D space. However, I can equivalently represent a torus as a 2D sheet, with 'rules' for how to connect opposing edges of the sheet -- specifically, glue together opposite edges. The old Atari game 'Asteroids' is an example of toroidal topology even though the surface (screen) is only 2D!hey guys,
i jus got a question regarding space time fabric.....if we take an example of a 2d fabric...it is distorted in the third dimension...now if it is a 3d fabric ( which is what it is supposed to be ) then in which dimension will it get distorted??
Also possible - but bapowell's explanation is better because it shows that you don't need to introduce any new elements for its explanation (elements that might or might not get one into trouble later on). It is not necessary to invoke another dimension to have distortions.In my opinion, it will get distorted in the 4th dimension - time.
Right. This is key. The Ricci curvature is intrinsic to the surface, and completely independent of the (unnecessary) embedding. Silentbob -- the OP's 2D example helps visualize the problem, but really we have a 4D curved manifold in GR. The question the OP is asking then is...what is the 5th dimension? Time is already included in the 4D.It is not necessary to invoke another dimension to have distortions.
Right, sorry. I gave a topological example instead a geometric one! However, the point of the example was meant to convey that fact that you don't need to embed a manifold in order for it to have nontrivial properties. Of course, one could just say that the Ricci curvature (or the fundamental form, or whatever) is intrinsic to the surface and is independent of any embedding. But point taken, sorry.I don't think the example works quite well though? Although identifying opposite edges of a square gives a quotient topology of a torus, this torus is flat - it's gaussian curvature is zero.
You are too kind monsieur! Too kind! Nah... just right. Glad someone got the reference though... classic movie, absolutely classic. Lithgow is an AMAZING actor, and the rest... well.. rofl. just rofl!Nicely done John FrameDragger.
It is one of my top ten. And Lithgow was indeed a shining star.You are too kind monsieur! Too kind! Nah... just right. Glad someone got the reference though... classic movie, absolutely classic. Lithgow is an AMAZING actor, and the rest... well.. rofl. just rofl!
You are clearly a man of substance.It is one of my top ten. And Lithgow was indeed a shining star.
He is amazing, but it is VERY graphic. Frankly he's not the most violent in the show, or on tv/movies, but... well, the opening has him killing a woman and using a mirror to see her die. His acting makes it utterly convincing, and deeply DEEPLY frightning. For this, he won an emmy, I should add.Huh. Did not know that.