Ricci Decomposition: Questions & Answers

In summary, the Ricci decomposition is a way of breaking down the Riemann tensor into three simpler pieces: the scalar part, semi-traceless part, and Weyl tensor. This allows for a better understanding of the physical meanings of each piece, particularly in relation to changes in volume and shape. The Ricci tensor is associated with non-gravitational stress-energy, while the Weyl tensor is associated with gravitational waves. The first two pieces do not have a special name and are equivalent to the curvature scalar and Ricci tensor, respectively. Semi-traceless means that it is built from the metric and the traceless Ricci tensor. Ramond's book on gravity and knowledge of group theory can provide further clarification.
  • #1
robousy
334
1
Hey folks,

I am trying to piece together what the point of the Ricci Decomposition is.

Wikipedia explains what it is but not really why.

I understand that we can break down the Ricci Tensor into:

1) Scalar part
2) Semi-traceless
3) Traceless (Weyl Tensor)


I have the following questions:

a) WHY break it down.
b) Do the first 2 have a special name in that the 3rd is called the Weyl Tensor.
c) What does semi traceless mean?


As always, thanks in advance.


:smile:
 
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  • #2
robousy,

The Ricci decompostion is actually a way of breaking up the Riemann tensor (not the Ricci tensor) into three simpler pieces. The first piece, the scalar part, is so called because it is built out of the curvature scalar and the metric. The second piece, the semi-traceless piece, is built out of the metric and the traceless Ricci tensor (hence the name semi-traceless). The third piece is what is left over and is called the Weyl tensor. This final piece is completely traceless.

What is the physical interpretation? Well, the curvature scalar and Ricci tensor are both contractions of the Riemann tensor and so they contain less information than the full Riemann tensor (low dimensions are exceptions: you need only the curvature scalar in 2d, and only the Ricci tensor in 3d). You know from the Einstein field equations that the stress energy tensor is coupled to the Einstein tensor which is made of only the metric, the curvature scalar, and the Ricci tensor. The Ricci tensor therefore measures the local non-gravitational stress-energy. In particular, the vacuum field equations are simply [tex] R_{a b} = 0 [/tex], but this doesn't mean the curvature is zero. Why? The freedom provided by the Weyl tensor! Gravitational wave solutions are vacuum solutions with vanishing Ricci tensor but non-vanishing Weyl tensor and therefore non-vanishing curvature.

But here I am droning on so let me answer your questions:
a) We break it down because the different pieces have different physical meanings. In particular, the Ricci tensor part is associated with changes in volume while the Weyl tensor is associated with changes in shape (loosely). Also, the Ricci tensor comes directly from the field equations, so it contains only the "here and now" if you will so perhaps it makes sense to treat it differently.

b) I don't think the first two have a special name. They are after all equivalent to the curvature scalar and Ricci tensor respectively.

c) Semi traceless means it is built from the metric (which is not traceless) and the traceless Ricci tensor which is.

Hope this helps! You can read a little more about here: http://www.polarhome.com:763/~rafimoor/english/GRE2.htm I'll see what else I can dig up for you.
 
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  • #3
Hi Physics Money,

Thanks for your explantion...and please feel free to drone! Its all useful to me - a GR newb.

Oh, I found your link really useful to, especially the parallel transport explanation.

I have another question actually but I'll make a new thread as the subject is different. Hope to hear from you there too!

Richard
 
  • #4
I hope the chapter on gravity in Ramond's book [1] and some thorough knowledge of group theory will clear things up for you.

Daniel.

P.S.If you've done your share of tensor operators in QM, then it's a useful analogy.

----------------------------------------------------------
[1]P.Ramond "Field Theory: A Modern Primer", 2.ed.
 

What is Ricci Decomposition?

Ricci Decomposition is a mathematical technique used in differential geometry to decompose a Riemannian manifold into its traceless Ricci part and its scalar curvature part.

What is the purpose of Ricci Decomposition?

The purpose of Ricci Decomposition is to simplify the study of Riemannian manifolds by breaking them down into more manageable components. This allows for easier analysis and computation of geometric properties.

How is Ricci Decomposition performed?

Ricci Decomposition involves using the Levi-Civita connection to split the Riemann curvature tensor into its traceless Ricci part and its scalar curvature part. This can be expressed mathematically as R = Ric + (n-1)S, where R is the Riemann curvature tensor, Ric is the Ricci tensor, S is the scalar curvature, and n is the dimension of the manifold.

What are some applications of Ricci Decomposition?

Ricci Decomposition has applications in various fields such as general relativity, differential topology, and mathematical physics. It is also used in the study of Einstein's field equations and the geometry of space-time.

Can Ricci Decomposition be extended to other types of manifolds?

Yes, Ricci Decomposition can be extended to other types of manifolds such as pseudo-Riemannian manifolds and Kähler manifolds. However, the decomposition may not have the same form as in the Riemannian case and may require different techniques.

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