Schwarzschild radius and curvature

In summary: Basically, the curvature radius in a Schwarzschild spacetime is the radius at which the Gaussian curvature is maximum, and this radius is given by the ideal central mass parameter.
  • #1
TrickyDicky
3,507
27
In the Schwarzschild spacetime setting we have a vacuum solution of the Einstein field equations, that is an idealized universe without any matter at the geodesics that are solutions of the equations.
This spacetime has however a curvature in both the temporal and spatial component that comes determined by a constant of integration of the metric ([itex]\alpha[/itex]) and that in the Newtonian limit is a function of mass, so that we can introduce a central mass parameter in order to solve problems such as Mercury preccession,etc. This is also known as Schwarzschild radius ([itex]r_s= 2GM/c^2[/itex]), because it basically gives us a geometrized mass(x2 factor since we are at the Newtonian limit), a mass in terms of length.
If all this is basically correct, wouldn't follow from it that the [itex]r_s[/itex] is a precisely the curvature radius of the 3-space parabolic hypersurface in the Schwarzschild manifold?
 
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  • #2
Not quite sure what you are getting at but the coordinate r does relate to the Gaussian curvature.
 
  • #3
Passionflower said:
Not quite sure what you are getting at but the coordinate r does relate to the Gaussian curvature.

Sure, the coordinate r is a function of Gaussian curvature and radial distance, here I'm referring to the fact that in a spherically symmetric manifold like this one,the [itex]r_s[/itex] that appears in the metric sometimes a 2[itex]\mu[/itex] or 2m or [itex]2GM/c^2[/itex] related to the idealized central mass seems to be precisely the intrinsic Gaussian curvature of the spatial part of the Schwarzschild manifold, (I seem to recall the curvature of a 3-space that is spherically symmetric is completely determined by its Gaussian curvature, maybe someone confirm it).
 
  • #4
TrickyDicky said:
Sure, the coordinate r is a function of Gaussian curvature and radial distance, here I'm referring to the fact that in a spherically symmetric manifold like this one,the [itex]r_s[/itex] that appears in the metric sometimes a 2[itex]\mu[/itex] or 2m or [itex]2GM/c^2[/itex] related to the idealized central mass seems to be precisely the intrinsic Gaussian curvature of the spatial part of the Schwarzschild manifold, (I seem to recall the curvature of a 3-space that is spherically symmetric is completely determined by its Gaussian curvature, maybe someone confirm it).
What consists of the spatial and temporal part of any spacetime solely depends on the choice of coordinates. There is no such thing as an objective spatial and temporal part of spacetime.
 
  • #5
Passionflower said:
What consists of the spatial and temporal part of any spacetime solely depends on the choice of coordinates. There is no such thing as an objective spatial and temporal part of spacetime.

I'm centering on the usual line element for Schwarzschild vacuum solution. I'm not interested here in the real/unreal debate, it's just a simple question about differential geometry for a certain metric.
 
  • #6
TrickyDicky said:
If all this is basically correct, wouldn't follow from it that the [itex]r_s[/itex] is a precisely the curvature radius of the 3-space parabolic hypersurface in the Schwarzschild manifold?

More specifically I would say that the [itex]r_s[/itex] (2m) in the Schwarzschild line element
[tex]ds^2=(1-\frac{2m}{r})dt^2-\frac{dr^2}{1-\frac{2m}{r}}-d\Omega^2[/tex]
could be identified with the minimum Gaussian curvature (IOW the maximum curvature K) a specific central mass can produce in a Schwarzschild manifold. The total radius of curvature would of course depend on where we locate the test particle, that is, on the total distance from the central mass, so that at the limit at infinity of r the metric becomes Minkowski.
I'm not sure if I got this right, maybe someone knowledgeable in Riemannian geometry could confirm or correct?
 

Related to Schwarzschild radius and curvature

What is the Schwarzschild radius?

The Schwarzschild radius is a theoretical value that represents the radius at which an object must be compressed to become a black hole. It is named after the German physicist Karl Schwarzschild and is calculated using the object's mass and the gravitational constant.

How is the Schwarzschild radius related to the curvature of spacetime?

The Schwarzschild radius is directly related to the curvature of spacetime. As an object's mass increases, its Schwarzschild radius also increases, resulting in a more pronounced curvature of spacetime around the object. This curvature is what causes other objects to be pulled towards the black hole.

Why is the Schwarzschild radius important in understanding black holes?

The Schwarzschild radius is important because it is the defining characteristic of a black hole. Any object that is compressed beyond its Schwarzschild radius will have a gravitational pull so strong that not even light can escape, making it a black hole. It also helps us understand the extreme effects of gravity on spacetime.

Can the Schwarzschild radius be observed or measured?

The Schwarzschild radius itself cannot be directly observed or measured since it is a theoretical value. However, its effects on the surrounding spacetime and the objects within it can be observed and measured through various astronomical techniques and observations.

How does the Schwarzschild radius affect the behavior of matter and light near a black hole?

The Schwarzschild radius is directly responsible for the intense gravitational pull of a black hole. As matter or light approaches the Schwarzschild radius, it experiences extreme tidal forces and is stretched and distorted. Once it crosses the Schwarzschild radius, it is trapped within the black hole and cannot escape.

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