Is this a curved or flat space?

  • #1
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Homework Statement:

All below ...

Relevant Equations:

\n
1605955820605.png

I would appreciate if someone check my work:
I tried to simplify the answer a lot: I imagined that, if we have this ds between two points different than the distance that should be if the space was flat, so it would be enough to generalize and say that space is not flat.

So, using this argument, i just analysed when ##\theta, \phi = \pi/2,0##. It would lead us to the ordinary "x" axis on the "xy plane".
So making r = 2, r =1, if the space is plane the distance between this points need to be 1.

##ds = \int_{1}^{2} \frac{r}{(r²+a²)^{1/2}}dr = (4+a²)^{1/2}-(1+a²)^{1/2} \neq 1## if ## a \neq 0##

Since a = 0 would be the ordinary spherical coordinates, the final answer is, yes, it is curved. Is it ok?
 

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  • #2
Office_Shredder
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I'm a bit confused about what the actual question you're trying to answer is. Whether the metric given describes a flat space?
 
  • #3
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I'm a bit confused about what the actual question you're trying to answer is. Whether the metric given describes a flat space?
exact
 
  • #4
PeroK
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Homework Statement:: All below ...
Relevant Equations:: \n

View attachment 272874
I would appreciate if someone check my work:
I tried to simplify the answer a lot: I imagined that, if we have this ds between two points different than the distance that should be if the space was flat, so it would be enough to generalize and say that space is not flat.

So, using this argument, i just analysed when ##\theta, \phi = \pi/2,0##. It would lead us to the ordinary "x" axis on the "xy plane".
So making r = 2, r =1, if the space is plane the distance between this points need to be 1.

##ds = \int_{1}^{2} \frac{r}{(r²+a²)^{1/2}}dr = (4+a²)^{1/2}-(1+a²)^{1/2} \neq 1## if ## a \neq 0##

Since a = 0 would be the ordinary spherical coordinates, the final answer is, yes, it is curved. Is it ok?
You need to do more than that. If you did a simple scaling of spherical coordinates, then the distance from ##r=1## to ##r =2## would not be ##1##.
 
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  • #5
PeroK
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PS You may want to check that the first term is not $$\frac{\rho^2}{r^2 + a^2}$$.
 
  • #6
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You need to do more than that. If you did a simple scaling of spherical coordinates, then the distance from ##r=1## to ##r =2## would not be ##1##.
At this stage the book don't talk about Gauss and Riemann way to determine if the space is curved or not (if there is, but probably there is, since they are names in this subject). The ways i imagine to see the curvature of the space is by check the area of know nfigures or the volume. I think we could calc the infinitesimal volume in this space and check if it agree with our intuition in flat space. Do you think this is a good approach?
 
  • #7
PeroK
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At this stage the book don't talk about Gauss and Riemann way to determine if the space is curved or not (if there is, but probably there is, since they are names in this subject). The ways i imagine to see the curvature of the space is by check the area of know nfigures or the volume. I think we could calc the infinitesimal volume in this space and check if it agree with our intuition in flat space. Do you think this is a good approach?
That's not going to work, because you don't know what the coordinates represent. I don't know any short cuts.

Is this a exercise from the book or something you decided to do?
 
  • #8
PeroK
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Since a = 0 would be the ordinary spherical coordinates, the final answer is, yes, it is curved. Is it ok?
If you don't know how to compute curvature, then it must be flat! If that is true, then you are looking for a coordinate transformation.
 
  • #9
WWGD
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Wouldn't it work if you changed coordinates to Euclidean see if you get the standard Id metric tensor ##ds^2=dx^2+dy^2##?
 
  • #10
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Wouldn't it work if you changed coordinates to Euclidean see if you get the standard Id metric tensor ##ds^2=dx^2+dy^2##?
I don't think so. This is the metric in cartesian, the metric in polar coordinates is different but yet is plane, no?

That's not going to work, because you don't know what the coordinates represent. I don't know any short cuts.

Is this a exercise from the book or something you decided to do?
From a book


If you don't know how to compute curvature, then it must be flat! If that is true, then you are looking for a coordinate transformation.
I think i don't understand what you mean here.
 
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  • #11
PeroK
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I think i don't understand what you mean here.
If you had the usual metric in spherical coordinates you could show the space is flat by using a coordinate transformation: $$x = r \sin \theta \cos \phi, \ y = r \sin \theta \sin \phi, \ z = r\cos \theta$$ Can you find a similar coordinate transformation here to show that this space is flat.

It's logical that if you haven't learned about curvature yet, then this space can't be curved - so it must be flat.

PS I'm still assuming it should be ##r^2 + a^2## in the denominator of the first term.
 
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  • #12
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1605955820605.png



Use pluginology with p2 = r2 +a2cos2Θ. Then, try to manipulate the equation to see if it gives a multiple of the metric tensor in the basis of the spherical coordinates ds2 = dr2+ r22 + r2sin2θdφ2
 
  • #13
PeroK
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View attachment 272934


Use pluginology with p2 = r2 +a2cos2Θ. Then, try to manipulate the equation to see if it gives a multiple of the metric tensor in the basis of the spherical coordinates ds2 = dr2+ r22 + r2sin2θdφ2
Do you have a solution with ##\rho^2 + a^2## in that first denominator? I think that should be ##r^2 + a^2##.
 
  • #14
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Do you have a solution with ##\rho^2 + a^2## in that first denominator? I think that should be ##r^2 + a^2##.
I haven't worked out the solution, but I think you are correct.

Swapping the denominator gives the Kerr metric, in the limit Mass → 0, which becomes a flat space.

Screen Shot 2020-11-25 at 3.54.31 PM.png
 

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