Ricci tensor of the orthogonal space

In summary, the Raychaudhuri equation is used to derive the density and the cosmological constant, but the result is different depending on whether the orthogonal space is 2-dimensional or not.
  • #1
PLuz
64
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While reading this article I got stuck with Eq.[itex](54)[/itex]. I've been trying to derive it but I can't get their result. I believe my problem is in understanding their hints. They say that they get the result from the Gauss embedding equation and the Ricci identities for the 4-velocity, [itex]u^a[/itex]. Is the Gauss equation they refer the one in the wiki article?

Looking at the terms that appear in their equation it looks like the Raychaudhuri equation is to be used in the derivation in order to get the density and the cosmological constant, but even though I realize this I can't really get their result.

Can anyone point me in the right direction?

Thank you very much

[itex]Note:[/itex]The reason why I'm trying so hard to prove their result is because I wanted to know if it would still be valid if the orthogonal space were 2 dimensional (aside some constants). It appears to be the case but to be sure I needed to be able to prove it.
 
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  • #2
Yes, the Gauss equation that they're referring to is the same Gauss equation mentioned in the Wikipedia article, relating the Riemann tensor of a surface to its second fundamental form. The second fundamental form, in turn, describes the embedding of the surface and can be expressed in terms of the kinematics of the normal congruence.

If you haven't already, I suggest you look up the cited articles, refs 5 and 6 by Ehlers and Ellis, where this relationship is proved.
 
  • #3
Bill_K said:
If you haven't already, I suggest you look up the cited articles, refs 5 and 6 by Ehlers and Ellis, where this relationship is proved.

I agree with Bill. This kind of "legwork" should be almost second nature. Another place to look is section 6.3 "The other Einstein field equations" in the new book "Relativistic Cosmology" by Ellis, Maartens, and MacCallum,

http://www.physicstoday.org/resource/1/phtoad/v66/i4/p54_s1
 
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  • #4
I just saw my error. I was forgetting some terms and in the end the expressions were obviously different. Thank you for the references, I had taken a quick look at them but, clearly, I had to read them with more attention...
 
  • #5

First of all, it's great that you are trying to understand and derive the result in the article. As a scientist, it is important to have a deep understanding of the concepts and equations we use in our work.

To answer your question, the Gauss embedding equation mentioned in the article is most likely the one from the Wikipedia article on Ricci curvature (https://en.wikipedia.org/wiki/Ricci_curvature). This equation relates the Ricci tensor of a manifold to its curvature tensor. It is a fundamental equation in differential geometry and plays a key role in the study of curved spaces.

In order to derive the result in Eq.(54), you will need to use the Gauss embedding equation and the Ricci identities for the 4-velocity, u^a. These identities relate the curvature tensors to the derivatives of the 4-velocity. The Raychaudhuri equation is also important in this derivation as it relates the expansion and shear of a congruence of geodesics to the Ricci tensor.

If you are having trouble understanding the hints in the article, it might be helpful to consult some textbooks on differential geometry or general relativity. They will provide a more detailed explanation of the concepts and equations involved in the derivation.

It is also worth mentioning that the result in Eq.(54) is valid for any dimension, not just in 4 dimensions. However, the specific values of the constants may differ depending on the dimension of the orthogonal space.

In conclusion, keep working on understanding and deriving the result in Eq.(54). It may take some time and effort, but once you have a solid understanding of the concepts and equations involved, you will be able to confidently use and apply them in your work. Good luck!
 

Related to Ricci tensor of the orthogonal space

1. What does the Ricci tensor of the orthogonal space represent?

The Ricci tensor of the orthogonal space is a mathematical tensor that represents the curvature of a given space. It describes how the shape of the space changes in different directions and is an important tool in the study of differential geometry.

2. How is the Ricci tensor of the orthogonal space calculated?

The Ricci tensor of the orthogonal space is calculated using the metric tensor, which describes the distance between points in a given space. It involves taking the derivative of the Christoffel symbols, which are calculated using the metric tensor and its derivatives.

3. What are some applications of the Ricci tensor of the orthogonal space?

The Ricci tensor of the orthogonal space has many applications in physics and geometry. It is used in general relativity to describe the curvature of spacetime and in cosmology to study the expansion of the universe. It is also used in differential geometry to study the geometry of manifolds.

4. How does the Ricci tensor of the orthogonal space relate to the curvature of a space?

The Ricci tensor of the orthogonal space is one of several tensors that can be used to describe the curvature of a space. It is related to the Riemann curvature tensor, which is a more general measure of curvature. The Ricci tensor is a contraction of the Riemann tensor and provides a more simplified description of the curvature.

5. Can the Ricci tensor of the orthogonal space be used to determine the topology of a space?

No, the Ricci tensor of the orthogonal space alone cannot determine the topology of a space. It is a local measure of curvature and does not provide information about the global structure of a space. Other mathematical tools, such as the Gauss-Bonnet theorem, are needed to determine the topology of a space.

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