Ricci tensor for Schwarzschild metric

In summary: So I decided to look this up. It all fell into place pretty quickly. Saw your edit now - heh, we both immediately turned to the same place.This will be something great to look into, though - I'm only an undergrad, now, but I'm working with some faculty at my university and some of the first things I'm interested in really digging into have to do with constant density zero-pressure fluid solutions. Isn't this just a way of describing a uniform energy density in a region of space, though?Yes, this is exactly what it is - a way of describing a uniform energy density in a region of space. Yes, this is exactly what it is - a way of describing a
  • #1
AleksanderPhy
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Hello I am little bit confused about calculating Ricci tensor for schwarzschild metric:
So we have Ricci flow equation,∂tgμν=-2Rμν.
And we have metric tensor for schwarzschild metric:
Diag((1-rs/r),(1-rs]/r)-1,(r2),(sin2Θ) and ∂tgμν=0 so 0=-2Rμν and we get that Rμν=0.But Rμν should not equal to zero for schwarzschild metric. I may have some mistakes on symbols beacuse I use them first time.
 
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  • #2
The Ricci flow is not the same as the Einstein field equations. The Schwarzschild metric does not undergo Ricci flow.
 
  • #3
AleksanderPhy said:
Rμν should not equal to zero for schwarzschild metric.

Yes, it should. The Schwarzschild metric is a vacuum metric, so its Einstein tensor is zero; and it's simple to show that if the Einstein tensor is zero, the Ricci tensor must also be zero.
 
  • #4
PeterDonis said:
Yes, it should. The Schwarzschild metric is a vacuum metric, so its Einstein tensor is zero; and it's simple to show that if the Einstein tensor is zero, the Ricci tensor must also be zero.
The Schwarzschild solution is a vacuum solution?

I'm kind of surprised I haven't come across this yet. What does "m" represent in the metric, then?
 
  • #5
MattRob said:
The Schwarzschild solution is a vacuum solution?

I'm kind of surprised I haven't come across this yet. What does "m" represent in the metric, then?
If you compute the energy momentum tensor anywhere, you will get zero. So yes, it is a vacuum solution. The mass m is a parameter of a set of possible vacuum solutions.

Note that this is the exterior Schwarzschild solution - a solution outside a spherically symmetric object where there is no matter so you should not be surprised to find that it is a vacuum solition. The interior Schwarzschild solution is not a vacuum solution.
 
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  • #6
Orodruin said:
If you compute the energy momentum tensor anywhere, you will get zero. So yes, it is a vacuum solution. The mass m is a parameter of a set of possible vacuum solutions.

Note that this is the exterior Schwarzschild solution - a solution outside a spherically symmetric object where there is no matter so you should not be surprised to find that it is a vacuum solition. The interior Schwarzschild solution is not a vacuum solution.

First time I've heard "exterior" and "interior" Schwarzschild solution. Is this simply referring to regions where r > 2M or r < 2M (in geometric units)?
 
  • #7
MattRob said:
First time I've heard "exterior" and "interior" Schwarzschild solution. Is this simply referring to regions where r > 2M or r < 2M (in geometric units)?
No. The interior Schwarzschild solution is a solution inside a spherical mass distribution with zero pressure at the surface. It does not describe a black hole.

Edit: see https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric

Even in Newtonian gravity, the 1/r potential is a vacuum solution. It only describes the potential outside a spherically symmetric distribution.
 
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  • #8
Orodruin said:
No. The interior Schwarzschild solution is a solution inside a spherical mass distribution with zero pressure at the surface. It does not describe a black hole.

Edit: see https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric

Even in Newtonian gravity, the 1/r potential is a vacuum solution. It only describes the potential outside a spherically symmetric distribution.
So I decided to look this up. It all fell into place pretty quickly. Saw your edit now - heh, we both immediately turned to the same place.

But yeah, the difference in-between assuming you're outside a spherically symmetric mass or within a sphere of it - I think I get it, now.

This will be something great to look into, though - I'm only an undergrad, now, but I'm working with some faculty at my university and some of the first things I'm interested in really digging into have to do with constant density zero-pressure fluid solutions. Isn't this just a way of describing a uniform energy density in a region of space, though? I thought that's what "dust" was used to describe; a uniform energy density distribution in a region of space (in a static solution)?
 

1. What is the Ricci tensor for the Schwarzschild metric?

The Ricci tensor for the Schwarzschild metric is a mathematical object that describes the curvature of spacetime in the presence of a non-rotating, spherically symmetric massive object. It is a key component in Einstein's theory of general relativity and is used to calculate the gravitational field in the vicinity of the object.

2. How is the Ricci tensor calculated for the Schwarzschild metric?

The Ricci tensor for the Schwarzschild metric is calculated by using the Einstein field equations, which relate the curvature of spacetime to the distribution of matter and energy. The specific calculations involve solving a set of differential equations and can be quite complex.

3. What does the Ricci tensor tell us about the Schwarzschild metric?

The Ricci tensor provides information about the curvature of spacetime in the presence of a non-rotating, spherically symmetric massive object. It can tell us about the strength and direction of the gravitational field, as well as how spacetime is distorted by the object's mass.

4. How does the Ricci tensor relate to the Schwarzschild radius?

The Schwarzschild radius is a characteristic distance from the center of a non-rotating, spherically symmetric massive object at which the escape velocity becomes equal to the speed of light. The Ricci tensor is used to calculate the curvature of spacetime at this radius, which is a key factor in determining the object's gravitational pull.

5. What are the implications of the Ricci tensor for the Schwarzschild metric?

The Ricci tensor plays a crucial role in understanding the behavior of gravity in the presence of a massive object. It allows us to make predictions about the motion of objects in the vicinity of the object, as well as the bending of light and the overall structure of spacetime. It has also been used to test and confirm Einstein's theory of general relativity.

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