Killing Vector and Ricci curvature scalar

In summary, the task at hand is to prove the identity K^\lambda \nabla _\lambda R = 0, where K is a Killing vector and R is the Ricci Scalar. This can be done by using the Bianchi identity, the Killing equation, and the symmetry of the Ricci tensor. After some calculations, it is shown that \nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma = 0, which implies the desired result. However, to obtain this result, metric compatibility must be used.
  • #1
Psi-String
79
0

Homework Statement



I'm currently self-studying Carroll's GR book and get stuck by proving
the following identity:

[tex] K^\lambda \nabla _\lambda R = 0 [/tex]

where K is Killing vector and R is the Ricci Scalar

Homework Equations



Mr.Carroll said that it is suffice to show this by knowing:

[tex] \nabla _\mu \nabla _\sigma K^\mu = R_{\sigma \nu}K^\nu [/tex]

Bianchi identity [tex] \nabla ^ \mu R_{\rho \mu} = \frac{1}{2} \nabla _\rho R [/tex]

and Killing equation [tex] \nabla _\mu K_\nu + \nabla _\nu K_\mu = 0 [/tex]

The Attempt at a Solution



The work I done so far :

[tex] K^\lambda \nabla _\lambda R = 2 K^\lambda \nabla ^\mu R_{\mu \lambda} = 2 \left( \nabla ^\mu R_{\mu \lambda} K^\lambda -R_{\mu \lambda} \nabla ^\mu K^\lambda \right) = 2 \nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma [/tex]

Note that [tex]R_{\mu \lambda} \nabla ^\mu K^\lambda =0 [/tex], since

[tex]R_{\mu \lambda} \nabla ^\mu K^\lambda = - R_{\mu \lambda} \nabla^\lambda K^\mu = -R_{\lambda\mu} \nabla^\lambda K^\mu = -R_{\mu \lambda} \nabla ^\mu K^\lambda[/tex]

And I can't get any further :cry:

Could someone help?? Thanks in advace
 
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  • #2
[itex]\nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma = 0[/itex] in the same way that [itex]R_{\mu \lambda} \nabla ^\mu K^\lambda =0 [/itex].
 
  • #3
George Jones said:
[itex]\nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma = 0[/itex] in the same way that [itex]R_{\mu \lambda} \nabla ^\mu K^\lambda =0 [/itex].

Hi~

[itex]R_{\mu \lambda} \nabla ^\mu K^\lambda =0 [/itex]
is due to the Killing Equation and the symmetry of Ricci tensor [tex] R_{\mu \lambda} = R_{\lambda \mu}[/tex]

But it seems like
[tex]\nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma \neq \nabla _\sigma \nabla ^\mu \nabla _\mu K^\sigma [/tex] ??

Thanks for Reply!
 
Last edited:
  • #4
Psi-String said:
Hi~

[itex]R_{\mu \lambda} \nabla ^\mu K^\lambda =0 [/itex]
is due to the Killing Equation and the symmetry of Ricci tensor [tex] R_{\mu \lambda} = R_{\lambda \mu}[/tex]

But it seems like
[tex]\nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma \neq \nabla _\sigma \nabla ^\mu \nabla _\mu K^\sigma [/tex] ??

Thanks for Reply!

I think I figure it out! :redface:

[tex] \nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma [/tex]

[tex] = (g^{\mu \sigma} \nabla_\sigma) (g_{\sigma \mu}\nabla^\mu)(\nabla_\mu K^\sigma)[/tex]

[tex] = g^{\mu \sigma}g_{\sigma \mu} (\nabla _\sigma \nabla^\mu \nabla_\mu K^\sigma) + g^{\mu \sigma}(\nabla^\mu \nabla_\mu K^\sigma)(\nabla_\sigma g_{\sigma \mu})[/tex]

[tex] = \nabla_\sigma \nabla^\mu \nabla_\mu K^\sigma [/tex]

the second term of the third line vanishes due to metric compatibility.

Is the above correct??:uhh:
 
Last edited:
  • #5
Psi-String said:
Is the above correct??:uhh:

Yes, metric compatibility is used, but be careful with the indices. You have used [itex]\mu[/itex] and [itex]\sigma[/itex] four times each in the product [itex](g^{\mu \sigma} \nabla_\sigma) (g_{\sigma \mu}\nabla^\mu)(\nabla_\mu K^\sigma)[/itex].
 
  • #6
I noticed that. But I can't derive it out if I use different indices...

I don't know how to turn

[tex]g^{\mu \sigma}g_{\lambda \rho}(\nabla_\sigma \nabla^\rho \nabla _\mu K^\lambda)[/tex]

into

[tex] \nabla _\lambda \nabla^\mu \nabla_\mu K^\lambda [/tex]
 

1. What is a Killing vector?

A Killing vector is a vector field on a Riemannian manifold that preserves the metric tensor (and therefore the length and angle measurements) of the manifold. It represents a symmetry of the manifold and corresponds to a conserved quantity in the physical system described by the manifold.

2. How is a Killing vector related to Killing curvature?

Killing curvature refers to the components of the Riemann curvature tensor that are parallel to the Killing vector. These components are conserved along the flow of the Killing vector, making them important in studying the geometry and symmetries of the manifold.

3. What is the significance of the Ricci curvature scalar?

The Ricci curvature scalar is a measure of the intrinsic curvature of a manifold at a given point. It is related to the Einstein field equations in general relativity, which describe the relationship between the curvature of spacetime and the distribution of matter and energy. In general, a higher Ricci curvature scalar indicates a more curved and potentially more dynamic manifold.

4. What is the Ricci flow and how does it relate to the Ricci curvature scalar?

The Ricci flow is a mathematical tool used to study the evolution of a Riemannian manifold over time. It involves deforming the metric of the manifold according to the Ricci curvature scalar, with the goal of reaching a manifold with constant or scalar curvature. This flow is important in understanding the long-term behavior of a manifold and its curvature properties.

5. How does the presence of Killing vectors affect the Ricci curvature scalar?

The presence of Killing vectors can have a significant impact on the Ricci curvature scalar and the overall geometry of a manifold. In some cases, the existence of certain Killing vectors can lead to a reduction in the number of independent components of the Ricci curvature tensor, simplifying the analysis of the manifold. Additionally, the symmetries represented by Killing vectors can provide insight into the behavior of the manifold and its curvature properties.

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