# Problem: perturbation of Ricci tensor

• I
• dpopchev
In summary: I see what is happening... will try to solve it and post it here!In summary, the conversation discusses the calculation of the Ricci tensor in terms of small perturbations over an arbitrary background metric. The conversation follows the expressions given in Michele Maggiore's Gravitational Waves vol 1 and Yvonne Choquet-Bruhat's Introduction to General Relativity Black Holes Cosmology. It is found that there is an extra contraction between the perturbation and background Riemann tensor in Maggiore's expression, which is not present in Choquet-Bruhat's expression. This leads to further calculations and discussions on the difference between the two expressions. Further work is needed to resolve this discrepancy.
dpopchev
I am trying to calculate the Ricci tensor in terms of small perturbation hμν over arbitrary background metric gμν whit the restriction
$$\left| \dfrac{h_{\mu\nu}}{g_{\mu\nu}} \right| << 1$$

Following Michele Maggiore Gravitational Waves vol 1 I correctly expressed the Chirstoffel symbol in terms of the perturbation (equation 1.205)
$$\tilde\Gamma^\alpha_{\mu\nu} = \Gamma^\alpha_{\mu\nu} + \dfrac{1}{2}(\nabla_\mu h_\nu^\alpha + \nabla_\nu h_\mu^\alpha - \nabla^\alpha h_{\mu\nu})$$

After that I correctly obtain the perturbed Rieman tensor
$$\tilde R^\alpha_{\beta\mu\nu} = \partial_\mu \tilde \Gamma^\alpha_{\beta\nu} - \partial_\nu \tilde \Gamma^\alpha_{\beta\mu} + \tilde \Gamma^\alpha_{\tau\mu} \tilde \Gamma^\tau_{\beta\nu} - \tilde \Gamma^\alpha_{\tau\nu} \tilde \Gamma^\tau_{\beta\mu} = R^\alpha_{\beta\mu\nu} + \underbrace{R^{(1)} + R^{(2)}}_{\delta R}$$

Where the first part of the remaining terms is in agreement with equation 1.206
$$2R^{(1)} = \nabla_\mu \nabla_\beta h^\alpha_\nu + \nabla_\nu \nabla^\alpha h_{\beta\mu} - \nabla_\nu \nabla_\beta h^\alpha_\mu - \nabla_\mu \nabla^\alpha h_{\beta\nu}$$

But the second is problematic, my expression:
$$2R^{(2)} =( \nabla_\mu \nabla_\nu - \nabla_\nu \nabla_\mu ) h^\alpha_\beta = R^\alpha_{\tau\mu\nu} h^\tau_\beta - R^\tau_{\beta\mu\nu} h^\alpha_\tau$$

Maggiore expression, which I suppose has a typo
$$2R^{(2)} = h^{\alpha\tau} R_{\tau\beta\mu\nu} + h_{\beta}^\tau R^\alpha_{\tau\mu\nu}$$

After getting the contracted Rieman tensor to obtain Ricci tensor
$$\tilde R_{\beta\nu} = \tilde g^{\alpha\mu} \tilde R_{\alpha\beta\mu\nu} = ( g^{\alpha\mu} - h^{\alpha\mu} )( R_{\alpha\beta\mu\nu} + \delta R) = g^{\alpha\mu}\delta R - h^{\alpha\mu}R_{\alpha\beta\mu\nu}$$

Then here are is the first part of the remaining terms
$$2g^{\alpha\mu}R^{(1)} = \nabla_\mu \nabla_\beta h^{\mu}_\nu + \underbrace{\nabla_\nu \nabla_\mu h^{\mu}_{\beta}}_{-R_{\tau\nu}h^\tau_\beta + R_{\tau\beta\mu\nu}h^{\mu\tau} + \nabla_\mu\nabla_\nu h^\mu_\beta} - \nabla_\nu \nabla_\beta h^\mu_\mu - \nabla^\mu \nabla_\mu h_{\beta\nu}$$

and the second part of the remaining terms, my expression
$$2g^{\alpha\mu}R^{(2)} = R_{\tau\nu}h^{\tau}_{\beta} - R_{\tau\beta\mu\nu}h^{\mu\tau}$$

Maggiore expression
$$2g^{\alpha\mu}R^{(2)} = R_{\tau\beta\mu\nu}h^{\tau\mu} + R_{\tau\nu}h^{\tau}_\beta$$

Now taking into account the underbrace rewriting of ∇νμ hμβ in R(1) and focus only on the terms containing hR in δR it follows

using my expressions
$$2g^{\alpha\mu}R^{(2)} = R_{\tau\nu}h^{\tau}_{\beta} - R_{\tau\beta\mu\nu}h^{\mu\tau} -R_{\tau\nu}h^\tau_\beta + R_{\tau\beta\mu\nu}h^{\mu\tau} = 0$$
thus I am left with one contraction between the perturbation and background Rieman tensor
$$\tilde R_{\beta\nu} = R_{\beta\nu} + R^{(1)} - h^{\alpha\mu}R_{\alpha\beta\mu\nu}$$

but using Maggiore expression
$$2g^{\alpha\mu}R^{(2)} = R_{\tau\beta\mu\nu}h^{\tau\mu} + R_{\tau\nu}h^{\tau}_\beta -R_{\tau\nu}h^\tau_\beta + R_{\tau\beta\mu\nu}h^{\mu\tau} = 2R_{\tau\beta\mu\nu}h^{\tau\mu}$$
thus
$$\tilde R_{\beta\nu} = R_{\beta\nu} + R^{(1)}$$

Which suggests that there is no typo... or some other way to rewrite my expression... I tried the first Bianchi identity but with no success...

I am totally stumped on this for the past week and any help will be appreciated.

Here I will post detailed calculations, also will reorganize the post as a whole since as I see it now, it is very chaotic.

For the rest of the post I am working with torsion free connection, and thus my Christoffel symbols will be defined as
$$\Gamma^\sigma_{\mu\nu} = \dfrac{1}{2} g^{\sigma\tau}\left( \partial_\mu g_{\nu\tau} + \partial_\nu g_{\mu\tau} - \partial_\tau g_{\mu\nu} \right)$$

I want to expand the Riman tensor, Ricci tensor and Ricci scalar up to linear terms of small perturbation hμν around background metric gμν
$$\underbrace{\tilde g_{\mu\nu}}_{\tilde a \text{ above marks physical spacetime}} = g_{\mu\nu} + h_{\mu\nu}, \left| \dfrac{h_{\mu\nu}}{g_{\mu\nu}} \right| << 1$$

In the calculations below, because of the symmetry of the Christoffel, it is easy to check that all uninteresting terms involving them are vanishing. Thus for simplicity I am presuming a frame where Christoffel is vanishing, but keeping in mind that its derivatives are not! Thus
$$\Gamma^\sigma_{\mu\nu} = 0, \text{ but } \partial_\tau \Gamma^\sigma_{\mu\nu} \neq 0$$

Now using the definition of the Christoffel symmbol we can write it down up to linear terms of the small perturbation
$$\tilde\Gamma^\alpha_{\mu\nu} = \Gamma^\alpha_{\mu\nu} + \dfrac{1}{2}\underbrace{(\nabla_\mu h_\nu^\alpha + \nabla_\nu h_\mu^\alpha - \nabla^\alpha h_{\mu\nu})}_{T_{\mu\nu}^\alpha}$$

Now let's calculate the Rimena tensor using its definition
$$\tilde R^\alpha_{\beta\mu\nu} = \partial_\mu \Gamma^\alpha_{\beta\nu} - \partial_\nu \Gamma^\alpha_{\beta\mu} = R^\alpha_{\beta\mu\nu} + \partial_\mu T_{\beta\nu}^\alpha - \partial_\nu T_{\beta\mu}^\alpha$$

Since we are working in a frame of vanishing Christoffel symbols
$$\partial_\mu \rightarrow \nabla_\mu \Longrightarrow \partial_\mu T_{\beta\nu}^\alpha - \partial_\nu T_{\beta\mu}^\alpha = \\ = \nabla_\mu\nabla_\beta h^\alpha_\nu + \nabla_\mu\nabla_\nu h^\alpha_\beta - \nabla_\mu\nabla^\alpha h_{\beta\nu} - ( \nabla_\nu\nabla_\beta h^\alpha_\mu + \nabla_\nu\nabla_\mu h^\alpha_\beta - \nabla_\nu\nabla^\alpha h_{\beta\mu} ) = \\ = \nabla_\mu\nabla_\beta h^\alpha_\nu + \nabla_\nu\nabla^\alpha h_{\beta\mu} - \nabla_\mu\nabla^\alpha h_{\beta\nu} - \nabla_\nu\nabla_\beta h^\alpha_\mu + \nabla_\mu\nabla_\nu h^\alpha_\beta - \nabla_\nu\nabla_\mu h^\alpha_\beta$$

Lets focus on the last two terms, by keeping in mind that the Γ vanish, their derivitives not!:
$$\nabla_\mu\nabla_\nu h^\alpha_\beta - \nabla_\nu\nabla_\mu h^\alpha_\beta = \\ = ( \partial_\mu\Gamma^\alpha_{\nu\tau} - \partial_\nu\Gamma^\alpha_{\mu\tau} )h^\tau_\beta - (\partial_\mu \Gamma^\tau_{\nu\beta} - \partial_{\nu} \Gamma^\tau_{\mu\beta} )h^\alpha_\tau = R^\alpha_{\tau\mu\nu} h^\tau_\beta - R^\tau_{\beta\mu\nu} h^\alpha_\tau$$

When we contract to obtain the Ricci tensor
$$\tilde R_{\beta\nu} = \tilde g^{\alpha\mu} \tilde R_{\alpha\beta\mu\nu} = ( g^{\alpha\mu} - h^{\alpha\mu} )( R_{\alpha\beta\mu\nu} + \delta R) = \\ = R_{\beta\nu} + \\ + \dfrac{1}{2}\left( \nabla_\mu\nabla_\beta h^\mu_\nu + \underbrace{\nabla_\nu\nabla^\mu h_{\beta\mu}}_{A} - \nabla_\mu\nabla^\mu h_{\beta\nu} - \nabla_\nu\nabla_\beta h^\mu_\mu + R^\mu_{\tau\mu\nu} h^\tau_\beta - R^\tau_{\beta\mu\nu} h^\mu_\tau \right) - h^{\alpha\mu}R_{\alpha\beta\mu\nu}$$

term A expressed in Riman tensor terms
$$\nabla_\nu \nabla^\mu h_{\mu\beta} = \nabla_\nu \nabla_\mu h^\mu_\beta = R^\mu_{\tau\nu\mu}h^\tau_\beta - R^\tau_{\beta\nu\mu}h^\mu_\tau + \nabla_\mu\nabla_\nu h^\mu_\beta = - R^\mu_{\tau\mu\nu}h^\tau_\beta + R^\tau_{\beta\mu\nu}h^\mu_\tau + \nabla_\mu\nabla_\nu h^\mu_\beta$$

Thus pluging it back in we end up with
$$\tilde R_{\beta\nu} = R_{\beta\nu} + \dfrac{1}{2}\left( \nabla_\mu\nabla_\beta h^\mu_\nu + \nabla_\mu\nabla_\nu h^\mu_\beta - \nabla_\mu\nabla^\mu h_{\beta\nu} - \nabla_\nu\nabla_\beta h^\mu_\mu \right) - h^{\alpha\mu}R_{\alpha\beta\mu\nu}$$

Thus ending with extra contraction between the perturbation and background Riman tensor...

I found expression for linearized Ricci tensor in Michele Maggiore Gravitational Waves vol 1 (eq 1.207) and Yvonne Choquet-Bruhat Introduction to General Relativity Black Holes Cosmology (eq. I.11.5)
The expression in both books are the same and are lacking the contracted with perturbation term(-hR).

From Maggiore, whos expression for the Rieman tensor is given in eq 1.206, his and my expression differ in the covariant commutator of hαβ, his expression is
$$\nabla_\mu\nabla_\nu - \nabla_\nu\nabla_\mu h^\alpha_\beta = h^\alpha_\tau R^{\tau}_{\beta\mu\nu} + h^\tau_\beta R^\alpha_{\tau\mu\nu}$$

which when contracted with the background metric and summed with term A it yields:
$$2h^{\alpha\mu} R_{\alpha\beta\mu\nu}$$
Which, taking into account the negative one half in front of the brackets is my extra term with positive sign...

EDIT:
Posted it before finishing the whole post.
Spellcheck.

Last edited:
SOLUTION:

I should have been more careful with the indices juggling, keeping track in which spacetime I am working and definitions.

The definition of Rieman tensor comes from the commutator of two covariant derivatives acting on vector field
$$\left[ \nabla_{\mu}, \nabla_{\nu} \right] V^\alpha = R^{\alpha}_{\beta\mu\nu} V^\beta$$

R_SOLUTION:

I should have been more careful with the indices juggling, keeping track in which spacetime I am working and definitions.

The definition of Rieman tensor comes from the commutator of two covariant derivatives acting on vector field
$$\left[ \nabla_{\mu}, \nabla_{\nu} \right] V^\alpha = R^{\alpha}_{\beta\mu\nu} V^\beta$$

And thus is a shorthand of writing the derivatives and contractions between the resulting Γ

This means that to lower the leading α index
$$R_{\alpha^\prime\beta\mu\nu} = g_{\alpha^\prime \alpha} R^{\alpha}_{\beta\mu\nu}$$

This lead me to think that I can just rise/lower or replace an index in the Rieman tensor, without doing the extra work of writing it down with the metric, and this was my mistake, since

$$\tilde R_{\sigma\beta\mu\nu} = \tilde g_{\sigma\alpha} \tilde R^{\alpha}_{\beta\mu\nu} = (g_{\sigma\alpha} + h_{\sigma\alpha})( R^\alpha_{\beta\mu\nu} + \partial_\mu T_{\beta\nu}^\alpha - \partial_\nu T_{\beta\mu}^\alpha ) = R_{\sigma\beta\mu\nu} + g_{\sigma\alpha}\left( \partial_\mu T_{\beta\nu}^\alpha - \partial_\nu T_{\beta\mu}^\alpha \right) + \underbrace{h_{\sigma\alpha}R^\alpha_{\beta\mu\nu}}_{\text{which is the term I have been missing}}$$

## 1. What is the Ricci tensor?

The Ricci tensor is a mathematical object in the field of differential geometry that is used to describe the curvature of a space. It is a symmetric tensor that is derived from the Riemann curvature tensor and is named after the Italian mathematician Gregorio Ricci-Curbastro.

## 2. How does perturbation affect the Ricci tensor?

Perturbation refers to a small change or disturbance in a system. In the context of the Ricci tensor, perturbation can refer to changes in the curvature of a space. These changes can affect the values of the Ricci tensor, causing it to deviate from its original values.

## 3. What are some real-world applications of studying the perturbation of Ricci tensor?

Studying the perturbation of the Ricci tensor has various applications in physics and cosmology. For example, it can be used to understand the behavior of gravitational waves and black holes, as well as to study the evolution of the universe.

## 4. How is the perturbation of Ricci tensor calculated?

The perturbation of the Ricci tensor is calculated using mathematical equations and methods from differential geometry and tensor calculus. It involves analyzing how small changes in the curvature of a space affect the values of the Ricci tensor.

## 5. What are some current research developments related to the perturbation of Ricci tensor?

Recent research in this field has focused on using the perturbation of the Ricci tensor to study the properties of dark matter and dark energy, as well as to explore the concept of cosmic inflation in the early universe. There is also ongoing research into the effects of perturbation on the formation and behavior of gravitational waves.

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