# Problem: perturbation of Ricci tensor

• I
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.

## Answers and Replies

dpopchev
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:
dpopchev
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
$$\left[ \nabla_{\mu}, \nabla_{\nu} \right] V^\alpha = R^{\alpha}_{\beta\mu\nu} V^\beta$$

dpopchev
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

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