Recent content by nikolakis

I see... Because, all this time I have been thinking that all expansion-contraction was an issue of the curvature of space-time. But if the universe is truly flat, what is causing it to expand? More so, if the spatial universe boils down to an infinite stretching line, what could be the...

I was more thinking of a 3-torus, and trying to tile the infinite Minkowski plane with identical copies of this domain. The analogy to an infinite roll of paper folded to a cone for a constant-rate expanding universe is self-evident, but are cosmologists talking of the same thing here?

Oh, never mind... I am a mathematician and not familiar with the theory. I think, I was trying to say infinite, flat and expanding. I was trying to visualize the thing macroscopically, not from the differential point of view. Sorry, if i have offended you.:redface:

Hi, I can't visualize how the universe could be infinitely flat according to the big bang theory... The only way I can visualize this is like a cone surface, one dimension supressed and left with a circle line (the universe), the other dimension thus forming the cone, is time. Now, in a...

I am not happy with your definition that v = dr/dτ. I can't escape the idea that speed should be measured here, on my ds = rdτ worldline where I belong, not there where r=1. Is it a misconception of my part that velocity is a vector, whereas speed is a scalar? Has it anything to do, with our...

If OP stands for original poster, I think pervect makes perfect sense to me. In his example and for a uniform accelerating or de-accelerating frame, I consider the metric space (in geometric units) given by ds = \sqrt{dr^{2}-r^{2}dτ^{2}}, τ is proper time as measured by the accelerating...

Thank you, pervect!. This is exactly what I expected to hear. Another instance is the metric ds = \sqrt{dr^2 + r^2d\theta^2}, which describes what you've just said... Many thanks!

Har, har. Are you joking? How many coordinates span a 2-Sphere? Both examples are correct. I've made no mistake. It's a valid representation of my metric.

And there is certainly not co-ordinate independence in general, save for the metric in this particular example which I've chosen, deliberately. Look at my #25 post. Isn't it here, a more expert in this field to answer this question? Thank you.

There must be a big lapse in my understanding. In my previous example, let us use: X=x, T=t, Z=\sqrt{3} x thus giving ds = \sqrt{dt^{2}+4dx^{2}} You are saying that all ds's are frame invariant, aren't you? How far, how long, a displacement in dx should be on Z=0, if not ds=2dx...

Well, starting with the familiar ℝ^{3} manifold, let's examine what a natural representation of [SIZE="2"]a metric space equipped with ds = \sqrt{dx^{2}+4dy^{2}} would be. Let X=x, Y=y, Z=\sqrt{3} y ds = \sqrt{dX^{2}+dY^{2}+dZ^{2}} is inherited from ℝ^{3}, hence ds =...

Ookke, there is a way to measure the one-way speed in light and synchronize two distant clocks instantaneously, in theory! In cylindrical spacetime any light signal will return back to its origin, and any two extreme clocks on a line identify at the same point. Since their distance is now 0...

I know this! I have already said that c=1/2 for the dX/dT in the new metric space ! (Please use c=1 for the rest of this discussion.) What I am saying is since the new metric is ds = \sqrt{dX^{2}-dT^{2}/4}, by measuring a difference dT=1 on the T-axis, this same ds gives ds=dT/2. Since I, now...

Better still, if I make use of the transform: X=x T=2t, What do I take the new speed of light to be? Since I would be measuring ds on the T-axis in the new metric space defined by the transformation, wouldn't that imply that the speed of light is the same and not 1/2? When T=1, X=1/2. But...

I think any question without LATEX is being frowned upon. I'll ask properly this time. I will go for DrGreg's tranforms with c=1 and evaluate the new speed of light in the deformed frame. On setting x=t one gets: X = \frac{1}{2ε}T or upon evaluating the new...