Recent content by Zinger0

I've worked on this problem quite a while ago, but I can try to dig up the calculations. So, what values are you interested in - just those you've mentioned in your previous post? And could you specify the problem you've encountered; since many computer programs can calculate those values if you...

Just came across your post. If you are still interested, probably I may be of some assistance: just recently I've dealt with a problem of https://www.physicsforums.com/showthread.php?t=442930". I've calculated those values you look for: Christoffel symbols, Ricci tensors, etc, first by hand and...

Well, I've already made a disclaimer about its real-world value. Nevertheless, it might prove helpful as an analogy for a more complex 3d world. As for the Ricci tensor, you are right: in an ordinary (4-d spacetime) Schwarzschild solution R_{\mu \nu} = 0 too, but the metric is not flat.

You mean, to make x_3 invariant? But both setting it to zero and making it invariant would still keep us in 3d space (where we no longer want to be). So, I rather was calling 'omitting' either of those approaches. Actually, I've been trying to do the same at first. But I was told that such...

By omitting coordiante I've meant your suggestion . In my reply I was referring to that approach.

If I understand correctly, you suggest solving 4-dimensional spacetime equations omitting one of spherical coordinates. But that would not exactly yield a required (2+1) metric. Apparently, the "physics" in general is different when applied to space of different number of dimensions. Thus...

I think I'm starting to get the general idea. I'll have to do some research on interpretation, but at least the solution itself is correct. Thanks for everybody's help! I would never consider flat metric anything but a blunder.

That sounds a bit too advanced for me. You mean if I restrict the range of \phi to something less than the whole interval [0,2pi)? But the Schwarzshcild solution should apply not only to singularities, but to any symmetrical bodies in vacuum. Probably the flatness of metrics tells us there...

Let me first state that I don't claim this treatment to be anything other than pure math. That being said, however, given approach has a certain value. I give a citation from http://www.math.columbia.edu/~woit/wordpress/?p=555": But I can't really defend the model, because I've just...

Indeed we've proved it, but I can't quite get the hold of it. First, how can the metric be flat, when there is a massive body, forming the gravitational field and therefore curving the metric? Second, is the gravitational force in a plane really different from it in 3d space? I mean, if...

They are still R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R = {8 \pi G \over c^4} T_{\mu \nu} (if we ignore cosmological constant), only \mu, \nu are now restricted to 0..2 instead of 0..3. In Schwarzschild's case we consider vacuum solution, so T_{\mu \nu}=0 and, therefore, R_{\mu \nu}=0.

Thanks anyway! Well, it seems to be flat metric after all.

This is GREAT! Just what I was looking for! Did not manage to google it myself, not for the lack of trying. Just to be absolutely sure: do I understand correctly that Schwarzschild solution in 3-dimensional space-time without cosmological constant should be flat (just as I've obtained)?

Thanks for suggestion! But I have some questions. First, I'm rather convinced that Newton's law of gravity should be F=GMm/r^2 in a plane, same as in 3-dim. space. That is because: 1). what matters is the distance from the source, which is r in either case. 2). just the units analysis...

I'm trying to find Schwarzschild solution for 3-dimensional space-time (i.e. time\otimes space^2). The problem is, I can't take the 4-dimensional solution \[ds^2=\left(1-\frac{r_g}{r}\right) dt^2-\left(1-\frac{r_g}{r}\right)^{-1} dr^2-r^2\left(d\theta^2+sin^2\theta d\phi^2\right)\] and...