what is the “shape” of maximally extended schwarzschild












0












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I recall in GR class learning that Schwarschild solution was a radially symmetric solution to the field equations, independent of the time parameter $t$ with a coordinate singularity at $r=2m$ and a genuine curvature singularity at $r=0$. Thus we think of a family of 2-spheres parametrized by $rin (0, infty)$, $tin (-infty, infty)$, and the manifold is then topologically like $M= S^2 times mathbb{R}_{>0} times mathbb{R}$ or equivalently its like $mathbb{R}^4$ with a line removed, $M=(mathbb{R}^3setminus 0) times mathbb{R}$. We also had that timelike geodesics approach $r=0$ in finite time, so the metric is geodesically incomplete here.



It seems intuitive to visulize then a spacial cross section then like a punctured copy of $mathbb{R}^3$, with incompleteness and singularity etc. due to the puncture. A bit like in the popular stock images such as this



enter image description here



When people speak of the "maximally extended Schwarzschild", however, I understand it is usually represented by a penrose diagram in clever coordinates, which then references a spacetime with two asymptotically flat ends. But what is the exact shape? What does it look like topologically, and how do those two ends become glued together?










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$endgroup$












  • $begingroup$
    Maybe its better if you put penrose diagram instead of the picture above.
    $endgroup$
    – Sou
    Mar 29 '18 at 17:13










  • $begingroup$
    I realise that the above picture is not very scientific, just an "artists impression" but it does at least represent in some sense the 'shape' of a spacelike cross section of schwarzschild. Namely it is a punctured copy of $mathbb{R}^3$ which is asymptotically flat with some kind of singularity at the puncture. Not that I claim the picture is accurate! I included it instead of the penrose diagram precisely because I struggle to understand the shape of maximally extended schwarzschild from the penrose diagram. I was hoping someone could offer a similar "artists impression" heuristic.
    $endgroup$
    – Aerinmund Fagelson
    Mar 30 '18 at 12:28
















0












$begingroup$


I recall in GR class learning that Schwarschild solution was a radially symmetric solution to the field equations, independent of the time parameter $t$ with a coordinate singularity at $r=2m$ and a genuine curvature singularity at $r=0$. Thus we think of a family of 2-spheres parametrized by $rin (0, infty)$, $tin (-infty, infty)$, and the manifold is then topologically like $M= S^2 times mathbb{R}_{>0} times mathbb{R}$ or equivalently its like $mathbb{R}^4$ with a line removed, $M=(mathbb{R}^3setminus 0) times mathbb{R}$. We also had that timelike geodesics approach $r=0$ in finite time, so the metric is geodesically incomplete here.



It seems intuitive to visulize then a spacial cross section then like a punctured copy of $mathbb{R}^3$, with incompleteness and singularity etc. due to the puncture. A bit like in the popular stock images such as this



enter image description here



When people speak of the "maximally extended Schwarzschild", however, I understand it is usually represented by a penrose diagram in clever coordinates, which then references a spacetime with two asymptotically flat ends. But what is the exact shape? What does it look like topologically, and how do those two ends become glued together?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Maybe its better if you put penrose diagram instead of the picture above.
    $endgroup$
    – Sou
    Mar 29 '18 at 17:13










  • $begingroup$
    I realise that the above picture is not very scientific, just an "artists impression" but it does at least represent in some sense the 'shape' of a spacelike cross section of schwarzschild. Namely it is a punctured copy of $mathbb{R}^3$ which is asymptotically flat with some kind of singularity at the puncture. Not that I claim the picture is accurate! I included it instead of the penrose diagram precisely because I struggle to understand the shape of maximally extended schwarzschild from the penrose diagram. I was hoping someone could offer a similar "artists impression" heuristic.
    $endgroup$
    – Aerinmund Fagelson
    Mar 30 '18 at 12:28














0












0








0





$begingroup$


I recall in GR class learning that Schwarschild solution was a radially symmetric solution to the field equations, independent of the time parameter $t$ with a coordinate singularity at $r=2m$ and a genuine curvature singularity at $r=0$. Thus we think of a family of 2-spheres parametrized by $rin (0, infty)$, $tin (-infty, infty)$, and the manifold is then topologically like $M= S^2 times mathbb{R}_{>0} times mathbb{R}$ or equivalently its like $mathbb{R}^4$ with a line removed, $M=(mathbb{R}^3setminus 0) times mathbb{R}$. We also had that timelike geodesics approach $r=0$ in finite time, so the metric is geodesically incomplete here.



It seems intuitive to visulize then a spacial cross section then like a punctured copy of $mathbb{R}^3$, with incompleteness and singularity etc. due to the puncture. A bit like in the popular stock images such as this



enter image description here



When people speak of the "maximally extended Schwarzschild", however, I understand it is usually represented by a penrose diagram in clever coordinates, which then references a spacetime with two asymptotically flat ends. But what is the exact shape? What does it look like topologically, and how do those two ends become glued together?










share|cite|improve this question









$endgroup$




I recall in GR class learning that Schwarschild solution was a radially symmetric solution to the field equations, independent of the time parameter $t$ with a coordinate singularity at $r=2m$ and a genuine curvature singularity at $r=0$. Thus we think of a family of 2-spheres parametrized by $rin (0, infty)$, $tin (-infty, infty)$, and the manifold is then topologically like $M= S^2 times mathbb{R}_{>0} times mathbb{R}$ or equivalently its like $mathbb{R}^4$ with a line removed, $M=(mathbb{R}^3setminus 0) times mathbb{R}$. We also had that timelike geodesics approach $r=0$ in finite time, so the metric is geodesically incomplete here.



It seems intuitive to visulize then a spacial cross section then like a punctured copy of $mathbb{R}^3$, with incompleteness and singularity etc. due to the puncture. A bit like in the popular stock images such as this



enter image description here



When people speak of the "maximally extended Schwarzschild", however, I understand it is usually represented by a penrose diagram in clever coordinates, which then references a spacetime with two asymptotically flat ends. But what is the exact shape? What does it look like topologically, and how do those two ends become glued together?







general-relativity






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share|cite|improve this question











share|cite|improve this question




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asked Mar 29 '18 at 16:44









Aerinmund FagelsonAerinmund Fagelson

1,8041718




1,8041718












  • $begingroup$
    Maybe its better if you put penrose diagram instead of the picture above.
    $endgroup$
    – Sou
    Mar 29 '18 at 17:13










  • $begingroup$
    I realise that the above picture is not very scientific, just an "artists impression" but it does at least represent in some sense the 'shape' of a spacelike cross section of schwarzschild. Namely it is a punctured copy of $mathbb{R}^3$ which is asymptotically flat with some kind of singularity at the puncture. Not that I claim the picture is accurate! I included it instead of the penrose diagram precisely because I struggle to understand the shape of maximally extended schwarzschild from the penrose diagram. I was hoping someone could offer a similar "artists impression" heuristic.
    $endgroup$
    – Aerinmund Fagelson
    Mar 30 '18 at 12:28


















  • $begingroup$
    Maybe its better if you put penrose diagram instead of the picture above.
    $endgroup$
    – Sou
    Mar 29 '18 at 17:13










  • $begingroup$
    I realise that the above picture is not very scientific, just an "artists impression" but it does at least represent in some sense the 'shape' of a spacelike cross section of schwarzschild. Namely it is a punctured copy of $mathbb{R}^3$ which is asymptotically flat with some kind of singularity at the puncture. Not that I claim the picture is accurate! I included it instead of the penrose diagram precisely because I struggle to understand the shape of maximally extended schwarzschild from the penrose diagram. I was hoping someone could offer a similar "artists impression" heuristic.
    $endgroup$
    – Aerinmund Fagelson
    Mar 30 '18 at 12:28
















$begingroup$
Maybe its better if you put penrose diagram instead of the picture above.
$endgroup$
– Sou
Mar 29 '18 at 17:13




$begingroup$
Maybe its better if you put penrose diagram instead of the picture above.
$endgroup$
– Sou
Mar 29 '18 at 17:13












$begingroup$
I realise that the above picture is not very scientific, just an "artists impression" but it does at least represent in some sense the 'shape' of a spacelike cross section of schwarzschild. Namely it is a punctured copy of $mathbb{R}^3$ which is asymptotically flat with some kind of singularity at the puncture. Not that I claim the picture is accurate! I included it instead of the penrose diagram precisely because I struggle to understand the shape of maximally extended schwarzschild from the penrose diagram. I was hoping someone could offer a similar "artists impression" heuristic.
$endgroup$
– Aerinmund Fagelson
Mar 30 '18 at 12:28




$begingroup$
I realise that the above picture is not very scientific, just an "artists impression" but it does at least represent in some sense the 'shape' of a spacelike cross section of schwarzschild. Namely it is a punctured copy of $mathbb{R}^3$ which is asymptotically flat with some kind of singularity at the puncture. Not that I claim the picture is accurate! I included it instead of the penrose diagram precisely because I struggle to understand the shape of maximally extended schwarzschild from the penrose diagram. I was hoping someone could offer a similar "artists impression" heuristic.
$endgroup$
– Aerinmund Fagelson
Mar 30 '18 at 12:28










1 Answer
1






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oldest

votes


















1












$begingroup$

The Schwarzschild geometry has an apparent singularity at $r = 2 m$. This singularity is not a real one, since, when changing coordinates, the singularity can be removed.
When people speak of the "maximally extended Schwarzschild" they are referring to a coordinate system obtained by replacing the Schwarzschild coordinates $(t , r , theta , phi )$ by $(V , U , theta , phi )$ where for $r> 2m$:
begin{equation}
U=left(frac{r}{2 m}-1 right)^{1/2} e^{r/4m} cosh(frac{t}{4m})\
V=left(frac{r}{2 m}-1 right)^{1/2} e^{r/4m} sinh(frac{t}{4m})
end{equation}

while for $r< 2m$:
begin{equation}
U=left(1-frac{r}{2 m} right)^{1/2} e^{r/4m} sinh(frac{t}{4m})\
V=left(1-frac{r}{2 m} right)^{1/2} e^{r/4m} cosh(frac{t}{4m})
end{equation}

This are called the Kruskal–Szekeres coordinates.
The result of carrying this variable change out in either the region
$r> 2m$ or $r< 2m$, allows us to write the metric as:
begin{equation}
ds^2= frac{32 m^3}{r} e^{-r/2m} (dU^2-dV^2)+r^2(dphi^2+sin^2 theta dphi^2)
end{equation}

where $r$ is a function of U and V, since
begin{equation}
U^2-V^2=left(frac{r}{2 m}-1 right) e^{r/2m}
end{equation}

Null radial geodesics $ds^2 = 0$ have the equations dU = ± dV so they are straight lines through the origin of the UV plane. The maximally extended Schwarzschild geometry can be divided into 4 regions. Region (1) represents the exterior $r > 2m$ of the Schwarzschild metric. It is bounded by $U = pm V$, $U > 0$, i.e. by $r = 2m$, $t = pm infty$. The z axis is $t = 0$. The geodesic ends on $r = 0$, a genuine singularity. Kruskal–Szekeres coordinates are badly behaved at $r = 0$, which turns out to be a true singularity, since the curvature blows up there. We therefore require $r > 0$, which implies that
$U V < 1$. Thus, the U and V axes correspond to the horizons at r = 2m, and the singularities at $r = 0$ correspond to $U V = 1$.
Region (2) (defined between the line U=V and U=-V for V>0)is a black hole: everything can go in, nothing can come out. Once you’re in this
region, you can only move upward in the diagram (forward in time), at speeds less than $c$. There is also a time-reversed copy of this region in the bottom quadrant (region 4), from which everything must eventually escape; this region describes a white hole. In a white hole, time
goes ‘backwards’ and gravity is repulsive rather than attractive. It seems that white holes don't exist in nature, even if they are mathematically solution of the Einstein equations.
Since Kruskal–Szekeres coordinates U and V are well-defined at $r = 2m$,
we can use them to extend Schwarzschild geometry to $r < 2m$.
You can consider the spacelike hypersurface corresponding to V = 0 (then t = 0) and at we look at the equator of a 2-sphere obtained by setting $theta=pi / 2$. In this case the metric reduces to:
begin{equation}
ds^2= frac{32 m^3}{r} e^{-r/2m} dU^2+r^2 dphi^2= left( frac{r}{r-2m}right)+r^2 dphi^2
end{equation}

This is the metric on a 2D surface which is a paraboloid of revolution with equation
begin{equation}
r=frac{1}{8m}U^2+2 m
end{equation}

So
begin{equation}
dr=frac{1}{4m}U dU
end{equation}

And
begin{equation}
dr^2=frac{1}{2m}(r-2m) dU^2
end{equation}

hence on the paraboloid surface
begin{equation}
dU^2=frac{2m}{(r-2m)} dr^2
end{equation}

Taking a flat 3-dimensional space with cylindrical coordinates $r$,$phi$, and $U$ and metric
begin{equation}
ds^2= dr^2+r^2 dphi^2+dU^2
end{equation}

then on the paraboloid surface in this space is
begin{equation}
ds^2= dr^2 (1+frac{2m}{r-2m})+r^2 dphi^2= dr^2 frac{r}{(r-2m)}+r^2 dphi^2
end{equation}

It is obtained by taking the spacelike hypersurface $V = 0$, i.e. passing from
$U > 0$ to $U < 0$ , from region (1) to region (3) in the Kruskal diagram. It is also possible to take a different hypersurface $V=pm V_0$for passing from region (2) to (4) through the singularity at $r=0$. In the second case the two space-times are not connected. Then the Kruskal diagram shows us as two space-times separated spaces becomes a joined one as far as the time coordinate V moves from $-infty$ to 0 and then to $+infty$ when the spaces separate again.






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  • $begingroup$
    Thanks for the detailed answer :) Its funny I can spend all evening playing with these different coordinate systems, but I still find myself pretty clueless as to what the "shape" of schwarzschild is! I will have to keep practicing :p I get a bit lost calculating the paraboloid contained in ${V=0}$. Is there a term missing in your derivation of the 2d metric? How do you go from $ds^2 = frac{32m^3}{r} e^{-frac{r}{2m}} dU^2 + r^2 dphi^2$ to a metric of the form $ds^2=da^2+f(a)^2db^2$ i.e. the metric for a surface of revolution?
    $endgroup$
    – Aerinmund Fagelson
    Jan 13 at 21:15










  • $begingroup$
    I updated my answer. I hope this is helpful.
    $endgroup$
    – Upax
    Jan 30 at 15:59











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$begingroup$

The Schwarzschild geometry has an apparent singularity at $r = 2 m$. This singularity is not a real one, since, when changing coordinates, the singularity can be removed.
When people speak of the "maximally extended Schwarzschild" they are referring to a coordinate system obtained by replacing the Schwarzschild coordinates $(t , r , theta , phi )$ by $(V , U , theta , phi )$ where for $r> 2m$:
begin{equation}
U=left(frac{r}{2 m}-1 right)^{1/2} e^{r/4m} cosh(frac{t}{4m})\
V=left(frac{r}{2 m}-1 right)^{1/2} e^{r/4m} sinh(frac{t}{4m})
end{equation}

while for $r< 2m$:
begin{equation}
U=left(1-frac{r}{2 m} right)^{1/2} e^{r/4m} sinh(frac{t}{4m})\
V=left(1-frac{r}{2 m} right)^{1/2} e^{r/4m} cosh(frac{t}{4m})
end{equation}

This are called the Kruskal–Szekeres coordinates.
The result of carrying this variable change out in either the region
$r> 2m$ or $r< 2m$, allows us to write the metric as:
begin{equation}
ds^2= frac{32 m^3}{r} e^{-r/2m} (dU^2-dV^2)+r^2(dphi^2+sin^2 theta dphi^2)
end{equation}

where $r$ is a function of U and V, since
begin{equation}
U^2-V^2=left(frac{r}{2 m}-1 right) e^{r/2m}
end{equation}

Null radial geodesics $ds^2 = 0$ have the equations dU = ± dV so they are straight lines through the origin of the UV plane. The maximally extended Schwarzschild geometry can be divided into 4 regions. Region (1) represents the exterior $r > 2m$ of the Schwarzschild metric. It is bounded by $U = pm V$, $U > 0$, i.e. by $r = 2m$, $t = pm infty$. The z axis is $t = 0$. The geodesic ends on $r = 0$, a genuine singularity. Kruskal–Szekeres coordinates are badly behaved at $r = 0$, which turns out to be a true singularity, since the curvature blows up there. We therefore require $r > 0$, which implies that
$U V < 1$. Thus, the U and V axes correspond to the horizons at r = 2m, and the singularities at $r = 0$ correspond to $U V = 1$.
Region (2) (defined between the line U=V and U=-V for V>0)is a black hole: everything can go in, nothing can come out. Once you’re in this
region, you can only move upward in the diagram (forward in time), at speeds less than $c$. There is also a time-reversed copy of this region in the bottom quadrant (region 4), from which everything must eventually escape; this region describes a white hole. In a white hole, time
goes ‘backwards’ and gravity is repulsive rather than attractive. It seems that white holes don't exist in nature, even if they are mathematically solution of the Einstein equations.
Since Kruskal–Szekeres coordinates U and V are well-defined at $r = 2m$,
we can use them to extend Schwarzschild geometry to $r < 2m$.
You can consider the spacelike hypersurface corresponding to V = 0 (then t = 0) and at we look at the equator of a 2-sphere obtained by setting $theta=pi / 2$. In this case the metric reduces to:
begin{equation}
ds^2= frac{32 m^3}{r} e^{-r/2m} dU^2+r^2 dphi^2= left( frac{r}{r-2m}right)+r^2 dphi^2
end{equation}

This is the metric on a 2D surface which is a paraboloid of revolution with equation
begin{equation}
r=frac{1}{8m}U^2+2 m
end{equation}

So
begin{equation}
dr=frac{1}{4m}U dU
end{equation}

And
begin{equation}
dr^2=frac{1}{2m}(r-2m) dU^2
end{equation}

hence on the paraboloid surface
begin{equation}
dU^2=frac{2m}{(r-2m)} dr^2
end{equation}

Taking a flat 3-dimensional space with cylindrical coordinates $r$,$phi$, and $U$ and metric
begin{equation}
ds^2= dr^2+r^2 dphi^2+dU^2
end{equation}

then on the paraboloid surface in this space is
begin{equation}
ds^2= dr^2 (1+frac{2m}{r-2m})+r^2 dphi^2= dr^2 frac{r}{(r-2m)}+r^2 dphi^2
end{equation}

It is obtained by taking the spacelike hypersurface $V = 0$, i.e. passing from
$U > 0$ to $U < 0$ , from region (1) to region (3) in the Kruskal diagram. It is also possible to take a different hypersurface $V=pm V_0$for passing from region (2) to (4) through the singularity at $r=0$. In the second case the two space-times are not connected. Then the Kruskal diagram shows us as two space-times separated spaces becomes a joined one as far as the time coordinate V moves from $-infty$ to 0 and then to $+infty$ when the spaces separate again.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the detailed answer :) Its funny I can spend all evening playing with these different coordinate systems, but I still find myself pretty clueless as to what the "shape" of schwarzschild is! I will have to keep practicing :p I get a bit lost calculating the paraboloid contained in ${V=0}$. Is there a term missing in your derivation of the 2d metric? How do you go from $ds^2 = frac{32m^3}{r} e^{-frac{r}{2m}} dU^2 + r^2 dphi^2$ to a metric of the form $ds^2=da^2+f(a)^2db^2$ i.e. the metric for a surface of revolution?
    $endgroup$
    – Aerinmund Fagelson
    Jan 13 at 21:15










  • $begingroup$
    I updated my answer. I hope this is helpful.
    $endgroup$
    – Upax
    Jan 30 at 15:59
















1












$begingroup$

The Schwarzschild geometry has an apparent singularity at $r = 2 m$. This singularity is not a real one, since, when changing coordinates, the singularity can be removed.
When people speak of the "maximally extended Schwarzschild" they are referring to a coordinate system obtained by replacing the Schwarzschild coordinates $(t , r , theta , phi )$ by $(V , U , theta , phi )$ where for $r> 2m$:
begin{equation}
U=left(frac{r}{2 m}-1 right)^{1/2} e^{r/4m} cosh(frac{t}{4m})\
V=left(frac{r}{2 m}-1 right)^{1/2} e^{r/4m} sinh(frac{t}{4m})
end{equation}

while for $r< 2m$:
begin{equation}
U=left(1-frac{r}{2 m} right)^{1/2} e^{r/4m} sinh(frac{t}{4m})\
V=left(1-frac{r}{2 m} right)^{1/2} e^{r/4m} cosh(frac{t}{4m})
end{equation}

This are called the Kruskal–Szekeres coordinates.
The result of carrying this variable change out in either the region
$r> 2m$ or $r< 2m$, allows us to write the metric as:
begin{equation}
ds^2= frac{32 m^3}{r} e^{-r/2m} (dU^2-dV^2)+r^2(dphi^2+sin^2 theta dphi^2)
end{equation}

where $r$ is a function of U and V, since
begin{equation}
U^2-V^2=left(frac{r}{2 m}-1 right) e^{r/2m}
end{equation}

Null radial geodesics $ds^2 = 0$ have the equations dU = ± dV so they are straight lines through the origin of the UV plane. The maximally extended Schwarzschild geometry can be divided into 4 regions. Region (1) represents the exterior $r > 2m$ of the Schwarzschild metric. It is bounded by $U = pm V$, $U > 0$, i.e. by $r = 2m$, $t = pm infty$. The z axis is $t = 0$. The geodesic ends on $r = 0$, a genuine singularity. Kruskal–Szekeres coordinates are badly behaved at $r = 0$, which turns out to be a true singularity, since the curvature blows up there. We therefore require $r > 0$, which implies that
$U V < 1$. Thus, the U and V axes correspond to the horizons at r = 2m, and the singularities at $r = 0$ correspond to $U V = 1$.
Region (2) (defined between the line U=V and U=-V for V>0)is a black hole: everything can go in, nothing can come out. Once you’re in this
region, you can only move upward in the diagram (forward in time), at speeds less than $c$. There is also a time-reversed copy of this region in the bottom quadrant (region 4), from which everything must eventually escape; this region describes a white hole. In a white hole, time
goes ‘backwards’ and gravity is repulsive rather than attractive. It seems that white holes don't exist in nature, even if they are mathematically solution of the Einstein equations.
Since Kruskal–Szekeres coordinates U and V are well-defined at $r = 2m$,
we can use them to extend Schwarzschild geometry to $r < 2m$.
You can consider the spacelike hypersurface corresponding to V = 0 (then t = 0) and at we look at the equator of a 2-sphere obtained by setting $theta=pi / 2$. In this case the metric reduces to:
begin{equation}
ds^2= frac{32 m^3}{r} e^{-r/2m} dU^2+r^2 dphi^2= left( frac{r}{r-2m}right)+r^2 dphi^2
end{equation}

This is the metric on a 2D surface which is a paraboloid of revolution with equation
begin{equation}
r=frac{1}{8m}U^2+2 m
end{equation}

So
begin{equation}
dr=frac{1}{4m}U dU
end{equation}

And
begin{equation}
dr^2=frac{1}{2m}(r-2m) dU^2
end{equation}

hence on the paraboloid surface
begin{equation}
dU^2=frac{2m}{(r-2m)} dr^2
end{equation}

Taking a flat 3-dimensional space with cylindrical coordinates $r$,$phi$, and $U$ and metric
begin{equation}
ds^2= dr^2+r^2 dphi^2+dU^2
end{equation}

then on the paraboloid surface in this space is
begin{equation}
ds^2= dr^2 (1+frac{2m}{r-2m})+r^2 dphi^2= dr^2 frac{r}{(r-2m)}+r^2 dphi^2
end{equation}

It is obtained by taking the spacelike hypersurface $V = 0$, i.e. passing from
$U > 0$ to $U < 0$ , from region (1) to region (3) in the Kruskal diagram. It is also possible to take a different hypersurface $V=pm V_0$for passing from region (2) to (4) through the singularity at $r=0$. In the second case the two space-times are not connected. Then the Kruskal diagram shows us as two space-times separated spaces becomes a joined one as far as the time coordinate V moves from $-infty$ to 0 and then to $+infty$ when the spaces separate again.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the detailed answer :) Its funny I can spend all evening playing with these different coordinate systems, but I still find myself pretty clueless as to what the "shape" of schwarzschild is! I will have to keep practicing :p I get a bit lost calculating the paraboloid contained in ${V=0}$. Is there a term missing in your derivation of the 2d metric? How do you go from $ds^2 = frac{32m^3}{r} e^{-frac{r}{2m}} dU^2 + r^2 dphi^2$ to a metric of the form $ds^2=da^2+f(a)^2db^2$ i.e. the metric for a surface of revolution?
    $endgroup$
    – Aerinmund Fagelson
    Jan 13 at 21:15










  • $begingroup$
    I updated my answer. I hope this is helpful.
    $endgroup$
    – Upax
    Jan 30 at 15:59














1












1








1





$begingroup$

The Schwarzschild geometry has an apparent singularity at $r = 2 m$. This singularity is not a real one, since, when changing coordinates, the singularity can be removed.
When people speak of the "maximally extended Schwarzschild" they are referring to a coordinate system obtained by replacing the Schwarzschild coordinates $(t , r , theta , phi )$ by $(V , U , theta , phi )$ where for $r> 2m$:
begin{equation}
U=left(frac{r}{2 m}-1 right)^{1/2} e^{r/4m} cosh(frac{t}{4m})\
V=left(frac{r}{2 m}-1 right)^{1/2} e^{r/4m} sinh(frac{t}{4m})
end{equation}

while for $r< 2m$:
begin{equation}
U=left(1-frac{r}{2 m} right)^{1/2} e^{r/4m} sinh(frac{t}{4m})\
V=left(1-frac{r}{2 m} right)^{1/2} e^{r/4m} cosh(frac{t}{4m})
end{equation}

This are called the Kruskal–Szekeres coordinates.
The result of carrying this variable change out in either the region
$r> 2m$ or $r< 2m$, allows us to write the metric as:
begin{equation}
ds^2= frac{32 m^3}{r} e^{-r/2m} (dU^2-dV^2)+r^2(dphi^2+sin^2 theta dphi^2)
end{equation}

where $r$ is a function of U and V, since
begin{equation}
U^2-V^2=left(frac{r}{2 m}-1 right) e^{r/2m}
end{equation}

Null radial geodesics $ds^2 = 0$ have the equations dU = ± dV so they are straight lines through the origin of the UV plane. The maximally extended Schwarzschild geometry can be divided into 4 regions. Region (1) represents the exterior $r > 2m$ of the Schwarzschild metric. It is bounded by $U = pm V$, $U > 0$, i.e. by $r = 2m$, $t = pm infty$. The z axis is $t = 0$. The geodesic ends on $r = 0$, a genuine singularity. Kruskal–Szekeres coordinates are badly behaved at $r = 0$, which turns out to be a true singularity, since the curvature blows up there. We therefore require $r > 0$, which implies that
$U V < 1$. Thus, the U and V axes correspond to the horizons at r = 2m, and the singularities at $r = 0$ correspond to $U V = 1$.
Region (2) (defined between the line U=V and U=-V for V>0)is a black hole: everything can go in, nothing can come out. Once you’re in this
region, you can only move upward in the diagram (forward in time), at speeds less than $c$. There is also a time-reversed copy of this region in the bottom quadrant (region 4), from which everything must eventually escape; this region describes a white hole. In a white hole, time
goes ‘backwards’ and gravity is repulsive rather than attractive. It seems that white holes don't exist in nature, even if they are mathematically solution of the Einstein equations.
Since Kruskal–Szekeres coordinates U and V are well-defined at $r = 2m$,
we can use them to extend Schwarzschild geometry to $r < 2m$.
You can consider the spacelike hypersurface corresponding to V = 0 (then t = 0) and at we look at the equator of a 2-sphere obtained by setting $theta=pi / 2$. In this case the metric reduces to:
begin{equation}
ds^2= frac{32 m^3}{r} e^{-r/2m} dU^2+r^2 dphi^2= left( frac{r}{r-2m}right)+r^2 dphi^2
end{equation}

This is the metric on a 2D surface which is a paraboloid of revolution with equation
begin{equation}
r=frac{1}{8m}U^2+2 m
end{equation}

So
begin{equation}
dr=frac{1}{4m}U dU
end{equation}

And
begin{equation}
dr^2=frac{1}{2m}(r-2m) dU^2
end{equation}

hence on the paraboloid surface
begin{equation}
dU^2=frac{2m}{(r-2m)} dr^2
end{equation}

Taking a flat 3-dimensional space with cylindrical coordinates $r$,$phi$, and $U$ and metric
begin{equation}
ds^2= dr^2+r^2 dphi^2+dU^2
end{equation}

then on the paraboloid surface in this space is
begin{equation}
ds^2= dr^2 (1+frac{2m}{r-2m})+r^2 dphi^2= dr^2 frac{r}{(r-2m)}+r^2 dphi^2
end{equation}

It is obtained by taking the spacelike hypersurface $V = 0$, i.e. passing from
$U > 0$ to $U < 0$ , from region (1) to region (3) in the Kruskal diagram. It is also possible to take a different hypersurface $V=pm V_0$for passing from region (2) to (4) through the singularity at $r=0$. In the second case the two space-times are not connected. Then the Kruskal diagram shows us as two space-times separated spaces becomes a joined one as far as the time coordinate V moves from $-infty$ to 0 and then to $+infty$ when the spaces separate again.






share|cite|improve this answer











$endgroup$



The Schwarzschild geometry has an apparent singularity at $r = 2 m$. This singularity is not a real one, since, when changing coordinates, the singularity can be removed.
When people speak of the "maximally extended Schwarzschild" they are referring to a coordinate system obtained by replacing the Schwarzschild coordinates $(t , r , theta , phi )$ by $(V , U , theta , phi )$ where for $r> 2m$:
begin{equation}
U=left(frac{r}{2 m}-1 right)^{1/2} e^{r/4m} cosh(frac{t}{4m})\
V=left(frac{r}{2 m}-1 right)^{1/2} e^{r/4m} sinh(frac{t}{4m})
end{equation}

while for $r< 2m$:
begin{equation}
U=left(1-frac{r}{2 m} right)^{1/2} e^{r/4m} sinh(frac{t}{4m})\
V=left(1-frac{r}{2 m} right)^{1/2} e^{r/4m} cosh(frac{t}{4m})
end{equation}

This are called the Kruskal–Szekeres coordinates.
The result of carrying this variable change out in either the region
$r> 2m$ or $r< 2m$, allows us to write the metric as:
begin{equation}
ds^2= frac{32 m^3}{r} e^{-r/2m} (dU^2-dV^2)+r^2(dphi^2+sin^2 theta dphi^2)
end{equation}

where $r$ is a function of U and V, since
begin{equation}
U^2-V^2=left(frac{r}{2 m}-1 right) e^{r/2m}
end{equation}

Null radial geodesics $ds^2 = 0$ have the equations dU = ± dV so they are straight lines through the origin of the UV plane. The maximally extended Schwarzschild geometry can be divided into 4 regions. Region (1) represents the exterior $r > 2m$ of the Schwarzschild metric. It is bounded by $U = pm V$, $U > 0$, i.e. by $r = 2m$, $t = pm infty$. The z axis is $t = 0$. The geodesic ends on $r = 0$, a genuine singularity. Kruskal–Szekeres coordinates are badly behaved at $r = 0$, which turns out to be a true singularity, since the curvature blows up there. We therefore require $r > 0$, which implies that
$U V < 1$. Thus, the U and V axes correspond to the horizons at r = 2m, and the singularities at $r = 0$ correspond to $U V = 1$.
Region (2) (defined between the line U=V and U=-V for V>0)is a black hole: everything can go in, nothing can come out. Once you’re in this
region, you can only move upward in the diagram (forward in time), at speeds less than $c$. There is also a time-reversed copy of this region in the bottom quadrant (region 4), from which everything must eventually escape; this region describes a white hole. In a white hole, time
goes ‘backwards’ and gravity is repulsive rather than attractive. It seems that white holes don't exist in nature, even if they are mathematically solution of the Einstein equations.
Since Kruskal–Szekeres coordinates U and V are well-defined at $r = 2m$,
we can use them to extend Schwarzschild geometry to $r < 2m$.
You can consider the spacelike hypersurface corresponding to V = 0 (then t = 0) and at we look at the equator of a 2-sphere obtained by setting $theta=pi / 2$. In this case the metric reduces to:
begin{equation}
ds^2= frac{32 m^3}{r} e^{-r/2m} dU^2+r^2 dphi^2= left( frac{r}{r-2m}right)+r^2 dphi^2
end{equation}

This is the metric on a 2D surface which is a paraboloid of revolution with equation
begin{equation}
r=frac{1}{8m}U^2+2 m
end{equation}

So
begin{equation}
dr=frac{1}{4m}U dU
end{equation}

And
begin{equation}
dr^2=frac{1}{2m}(r-2m) dU^2
end{equation}

hence on the paraboloid surface
begin{equation}
dU^2=frac{2m}{(r-2m)} dr^2
end{equation}

Taking a flat 3-dimensional space with cylindrical coordinates $r$,$phi$, and $U$ and metric
begin{equation}
ds^2= dr^2+r^2 dphi^2+dU^2
end{equation}

then on the paraboloid surface in this space is
begin{equation}
ds^2= dr^2 (1+frac{2m}{r-2m})+r^2 dphi^2= dr^2 frac{r}{(r-2m)}+r^2 dphi^2
end{equation}

It is obtained by taking the spacelike hypersurface $V = 0$, i.e. passing from
$U > 0$ to $U < 0$ , from region (1) to region (3) in the Kruskal diagram. It is also possible to take a different hypersurface $V=pm V_0$for passing from region (2) to (4) through the singularity at $r=0$. In the second case the two space-times are not connected. Then the Kruskal diagram shows us as two space-times separated spaces becomes a joined one as far as the time coordinate V moves from $-infty$ to 0 and then to $+infty$ when the spaces separate again.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 20 at 7:18

























answered Dec 27 '18 at 9:23









UpaxUpax

1,522613




1,522613












  • $begingroup$
    Thanks for the detailed answer :) Its funny I can spend all evening playing with these different coordinate systems, but I still find myself pretty clueless as to what the "shape" of schwarzschild is! I will have to keep practicing :p I get a bit lost calculating the paraboloid contained in ${V=0}$. Is there a term missing in your derivation of the 2d metric? How do you go from $ds^2 = frac{32m^3}{r} e^{-frac{r}{2m}} dU^2 + r^2 dphi^2$ to a metric of the form $ds^2=da^2+f(a)^2db^2$ i.e. the metric for a surface of revolution?
    $endgroup$
    – Aerinmund Fagelson
    Jan 13 at 21:15










  • $begingroup$
    I updated my answer. I hope this is helpful.
    $endgroup$
    – Upax
    Jan 30 at 15:59


















  • $begingroup$
    Thanks for the detailed answer :) Its funny I can spend all evening playing with these different coordinate systems, but I still find myself pretty clueless as to what the "shape" of schwarzschild is! I will have to keep practicing :p I get a bit lost calculating the paraboloid contained in ${V=0}$. Is there a term missing in your derivation of the 2d metric? How do you go from $ds^2 = frac{32m^3}{r} e^{-frac{r}{2m}} dU^2 + r^2 dphi^2$ to a metric of the form $ds^2=da^2+f(a)^2db^2$ i.e. the metric for a surface of revolution?
    $endgroup$
    – Aerinmund Fagelson
    Jan 13 at 21:15










  • $begingroup$
    I updated my answer. I hope this is helpful.
    $endgroup$
    – Upax
    Jan 30 at 15:59
















$begingroup$
Thanks for the detailed answer :) Its funny I can spend all evening playing with these different coordinate systems, but I still find myself pretty clueless as to what the "shape" of schwarzschild is! I will have to keep practicing :p I get a bit lost calculating the paraboloid contained in ${V=0}$. Is there a term missing in your derivation of the 2d metric? How do you go from $ds^2 = frac{32m^3}{r} e^{-frac{r}{2m}} dU^2 + r^2 dphi^2$ to a metric of the form $ds^2=da^2+f(a)^2db^2$ i.e. the metric for a surface of revolution?
$endgroup$
– Aerinmund Fagelson
Jan 13 at 21:15




$begingroup$
Thanks for the detailed answer :) Its funny I can spend all evening playing with these different coordinate systems, but I still find myself pretty clueless as to what the "shape" of schwarzschild is! I will have to keep practicing :p I get a bit lost calculating the paraboloid contained in ${V=0}$. Is there a term missing in your derivation of the 2d metric? How do you go from $ds^2 = frac{32m^3}{r} e^{-frac{r}{2m}} dU^2 + r^2 dphi^2$ to a metric of the form $ds^2=da^2+f(a)^2db^2$ i.e. the metric for a surface of revolution?
$endgroup$
– Aerinmund Fagelson
Jan 13 at 21:15












$begingroup$
I updated my answer. I hope this is helpful.
$endgroup$
– Upax
Jan 30 at 15:59




$begingroup$
I updated my answer. I hope this is helpful.
$endgroup$
– Upax
Jan 30 at 15:59


















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