Schwarzschild metric in 3-dimensional space-time

  • Thread starter Zinger0
  • Start date
  • #26
15
0
No. When you solve for the Schwartzshild metric in 4-d spacetime, you actually omit 2 of the 3 spherical coordinates, namely the polar and azimuthal angles.
By omitting coordiante I've meant your suggestion
Another option is to place [itex]x_{3} = 0[/itex]
.
In my reply I was referring to that approach.
 
  • #27
2,967
5
In my reply I was referring to that approach.
Yes, but I gave another option as well.
 
  • #28
15
0
Yes, but I gave another option as well.
You mean, to make [itex]x_3[/itex] 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 derivation would be incorrect.
 
  • #29
2,967
5
Ok, I said what I had to say. Play with (1+2)-spacetime. However, it is a mathematical exercise with no connection to Physical Reality.

Just as a sidenote. If the Ricci tensor:

[tex]
R_{\mu \nu} = 0
[/tex]

is zero, it still does not necessarily mean space-time is flat. There is something called Riemann curvature tensor [itex]R^{\mu}{}_{\nu \rho \pi}[/itex], which is a 4-fold tensor. This one has to be equal to zero in order for space-time to be flat.

As you vary the dimensions, the number of independent components of the two change. In 2-d (total number of dimensions), they have the same number and are uniquely expressible between each other. There, it would mean that space-time is flat. However, in higher number of dimensions, I think there are more components to the Riemann tensor, then there are to the Ricci tensor and no such unique relationship exists.
 
  • #30
15
0
Ok, I said what I had to say. Play with (1+2)-spacetime. However, it is a mathematical exercise with no connection to Physical Reality.

Just as a sidenote. If the Ricci tensor: [tex] R_{\mu \nu} = 0 [/tex] is zero, it still does not necessarily mean space-time is flat.
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 [itex]R_{\mu \nu} = 0[/itex] too, but the metric is not flat.
 
  • #31
Stingray
Science Advisor
671
1
As Dickfore points out, the 3D Einstein equation has a completely different physical character than the 4D one. All degrees of freedom in the curvature are locally determined by the matter distribution. This means, for example, that there can be no gravitational radiation. And there can be no analog of the Schwarzschild solution.

What might be more interesting is to look at a 2+1 slice of a static, vacuum, cylindrically symmetric 3+1 solution. This describes the Levi-Civita metric:
[tex]
ds^2 = - \rho^{4 \alpha} d t^2 + \rho^{4 \alpha (2 \alpha-1)} ( d \rho^2 + d z^2 ) + \beta^{-2} \rho^{2-4\alpha} d \phi^2 .
[/tex]
[itex]\alpha[/itex] can be interpreted as a mass/length and [itex]\beta[/itex] is an angular defect parameter.
 

Related Threads on Schwarzschild metric in 3-dimensional space-time

Replies
58
Views
3K
  • Last Post
Replies
1
Views
3K
Replies
2
Views
2K
  • Last Post
Replies
10
Views
2K
Replies
3
Views
3K
Replies
5
Views
622
  • Last Post
Replies
3
Views
850
  • Last Post
Replies
10
Views
6K
  • Last Post
Replies
23
Views
1K
Top