Raychaudhuri equation for shear

  • Context: Graduate 
  • Thread starter Thread starter julian
  • Start date Start date
  • Tags Tags
    Shear
Click For Summary

Discussion Overview

The discussion revolves around the Raychaudhuri equation in the context of shear, specifically focusing on the time derivative of shear and the decomposition of the Riemann tensor. Participants explore the mathematical relationships and symmetries involved in these concepts.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant discusses deriving the time derivative for shear and expresses a need to show a specific relationship involving the Riemann tensor and its decomposition.
  • Another participant questions the symmetry of the Riemann tensor with respect to its indices.
  • A response clarifies that the Riemann tensor can be considered symmetric when contracted by specific vectors, noting the implications for its symmetry over other indices.
  • Another participant suggests that the Riemann tensor should be replaced by its decomposition into the Weyl tensor, referencing a specific equation from Wald's book.

Areas of Agreement / Disagreement

Participants appear to agree on certain properties of the Riemann tensor, particularly its symmetry when contracted. However, the discussion includes multiple viewpoints regarding the decomposition of the Riemann tensor and the approach to proving the proposed relationship, indicating that no consensus has been reached on the best method to proceed.

Contextual Notes

Some assumptions regarding the properties of the Riemann tensor and the specific forms of the tensors involved are not fully detailed, which may affect the clarity of the proposed relationships.

julian
Science Advisor
Gold Member
Messages
861
Reaction score
366
Following Wald I have nearly got the right answer out for time derivative for shear...what I am left with is showing that [itex]R_{cbad} V^c V^d + h_{ab} R_{cd} V^c V^d / 3[/itex] (which is obviously symmetric and trace-free) can be written as [itex]C_{cbad} V^c V^d + \tilde{R}_{ab} / 2[/itex] where [itex]\tilde{R}_{ab}[/itex] is the spatial, trace-free part of [itex]R_{ab}[/itex], i.e. [itex]h_{ac} h_{bd} R^{cd} - h_{ab} h_{cd} R^{cd} / 3[/itex].

Is there an easy way of proving this?
 
Last edited:
Physics news on Phys.org
Is the Riemann tensor symmetric in ba?
 
It is when contrcted by [itex]V^c V^d[/itex] cus that means you can take it to be symmetric over c and d, this plus the usual symmetries of [itex]R_{cbad}[/itex] makes [itex]R_{cbda} V^c V^d[/itex] symmetric over a and b.
 
Last edited:
  • Like
Likes   Reactions: Takeaki
Right
 
You have to replace the Riemann by it's decomposition into Weyl tensor ... which is given by the eq. 3.2.28 in Wald's book.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K