A problem from Sean Carroll's about Killing vectors

  • Thread starter shichao116
  • Start date
  • Tags
    Vectors
In summary, the problem is to prove that the left hand side of the equation for the commutator of the covariant derivative for rank 2 tensors is zero.
  • #1
shichao116
13
0
I'm now stuck in the second part of problem 12 in Chapter 3. The problem is " Show that any Killing vector [itex]K^\mu[/itex] satisfies the following relations:
[tex]\nabla_\mu\nabla_\sigma K^\rho = R^\rho_{\sigma\mu\nu}K^\nu[/tex]
[tex]K^\lambda\nabla_\lambda R = 0[/tex]

Where R is Riemann tensor.

I can prove the first one by using the definition of Riemann tensor, i.e. the commutator of two covariant derivatives, Killing equations associated with Killing vector, and Bianchi identity.

But for the second one, in the book it is said that we can prove it by contracting the first equation, i.e.
[tex]\nabla_\mu\nabla_\sigma K^\mu = R_{\sigma\nu}K^\nu[/tex]
and the contracted Bianchi identity
[tex]\nabla_\mu(R^{\mu\nu}-1/2g^{\mu\nu}R)=0[/tex]

What I do is multiplying Killing vector to the contracted Bianchi identity and then I get to where I stuck:
[tex]1/2K^\mu\nabla_\mu R = K_\nu\nabla_\mu R^{\mu\nu} [/tex]

obviously the left hand side is what we need to prove to be zero. But I failed to show the right hand side to be zero after tried many ways.

Can anyone give me some clue how to do that ?

Thanks a lot
 
Last edited:
Physics news on Phys.org
  • #2
Hi there mate! This is what I used to do the problem. First note that you can use the first and second equations under the second paragraph of your post (the Bianchi identity and the one directly above it) to very easily write [itex]\triangledown _{\nu }\triangledown _{\mu }\triangledown ^{\nu }\xi ^{\mu } = \frac{1}{2}\xi ^{\nu }\triangledown _{\nu }R[/itex] (can you indeed show this? It is rather trivial). Now utilize the formula [itex](\triangledown _{a}\triangledown _{b} - \triangledown _{b}\triangledown _{a})T^{c_1c_2} = -R_{abe}^{c_1}T^{ec_2} - R_{abe}^{c_2}T^{c_1e}[/itex] (which you can derive yourself - it may be a teensy bit tedious but it is straightforward)

EDIT: IMO it is worth noting the equation for the commutator of the covariant derivative for rank 2 tensors because I have needed it quite a few times (another notable time I had to use it was to show that the inhomogeneous curved space - time Maxwell equations, the first one in my signature, implied [itex]\triangledown ^{a}j_{a} = 0[/itex])
 
Last edited:
  • #3
Hey bro, thanks a lot. That clear things up. I used to get where you showed in the first equation in your reply but did not proceed because I never tried to use a Riemann tensor on a tensor of rank 2 or higher. Now I get some new experience. :)
 
  • #4
No problem! Post again if you get stuck or something.
 
  • #5


I would first commend you for your efforts in attempting to solve this problem. It is clear that you have a good understanding of the concepts involved and have made progress in solving the first part of the problem.

In order to help you with the second part, I would suggest breaking down the expression K_\nu\nabla_\mu R^{\mu\nu} into smaller parts. One way to approach this is to use the fact that the Riemann tensor is antisymmetric in its last two indices, i.e. R_{\sigma\nu} = -R_{\nu\sigma}. This can help simplify the expression and make it easier to manipulate.

Another approach would be to use the contracted Bianchi identity in a different way. Instead of multiplying it by the Killing vector, try using it to substitute for the Riemann tensor in the first equation. This may lead to a simpler expression that can then be manipulated to prove the desired result.

Overall, my advice would be to keep trying different approaches and breaking down the expression into smaller parts. It may also be helpful to consult other resources or seek guidance from a colleague or mentor. Persistence and perseverance are key qualities in scientific problem-solving. Good luck!
 

1. What is a Killing vector?

A Killing vector is a vector field on a manifold that preserves the metric of the manifold. In other words, it is a vector field that leaves the geometry of the manifold unchanged.

2. How is a Killing vector related to symmetries?

A Killing vector is related to symmetries because it represents a transformation that leaves the geometry of the manifold unchanged. This means that if we apply this transformation to a physical system described by the manifold, the laws of physics will remain the same.

3. What is the significance of Killing vectors in physics?

Killing vectors have many applications in physics, particularly in the study of general relativity. They are used to find solutions to Einstein's field equations, which describe the curvature of spacetime in the presence of matter and energy.

4. How do Killing vectors relate to conserved quantities?

Killing vectors are closely related to conserved quantities, as they are associated with symmetries that lead to conserved quantities in physical systems. For example, the time translation symmetry of a system is associated with a Killing vector that leads to the conservation of energy.

5. Can Killing vectors be used to solve physical problems?

Yes, Killing vectors can be used to solve physical problems, particularly in the field of general relativity. By finding the Killing vectors of a given spacetime, we can determine the symmetries and conserved quantities of the system, which can greatly simplify the problem and lead to solutions.

Similar threads

  • Special and General Relativity
Replies
19
Views
1K
  • Special and General Relativity
Replies
12
Views
1K
  • Special and General Relativity
Replies
2
Views
1K
  • Special and General Relativity
Replies
6
Views
1K
  • Special and General Relativity
Replies
1
Views
2K
  • Special and General Relativity
Replies
3
Views
949
  • Special and General Relativity
Replies
6
Views
2K
  • Special and General Relativity
Replies
10
Views
717
  • Special and General Relativity
Replies
9
Views
4K
  • Special and General Relativity
Replies
9
Views
1K
Back
Top