Pertubation of Riemann tensor in a general curved space-time

Question

It is a direct and simple question. I am fully developing the perturbation of Einstein Field Equations, and I need to calculate the perturbation of the Riemann tensor. However the background metric is not Minkowski, but a general curved space-time.

As usual, we have:

$g^t_{\mu \nu} = g_{\mu \nu} + h_{\mu \nu}$,

then

$\delta \Gamma^\gamma_{\alpha \beta} = 1/2 (g^{\gamma \lambda}(\partial_\alpha h_{\lambda \rho)} + \partial_\rho h_{\lambda \alpha} - \partial_\lambda h_{\alpha \beta} ) - 2 g_{\lambda \rho} h^{\gamma \lambda} \Gamma^{(0)\rho}_{\alpha \beta})$.

I am in the middle of the calculation and it is really tricky and huge, some direction on how it should look like, or an easier way to do it, would also be valuable.

My question is: what is the expression of the first-order perturbation of the Riemann tensor?

Answer 1

We have $$ \Gamma^c_{ab} = \frac{1}{2} g^{cd} \left[ \partial_a g_{bd} + \partial_b g_{ad} -\partial_d g_{ab} \right] . $$ Varying this, we find \begin{align} \delta \Gamma^c_{ab} = \frac{1}{2} g^{cd} \left[ \partial_a h_{bd} + \partial_b h_{ad} -\partial_d h_{ab} \right] - \frac{1}{2} h^{cd} \left[ \partial_a g_{bd} + \partial_b g_{ad} -\partial_d g_{ab} \right] \qquad (1) \end{align} where $$ h_{ab} = \delta g_{ab} \quad \implies \quad h^{ab} = - \delta g^{ab} . $$ I'll now give you a pro tip which is often extremely useful in GR calculations. You should however, confirm what I say by explicit calculations as well.

You have to remember that even though the Christoffel symbol itself is not a tensor, its variation is! Thus, it must be that all the partial derivatives in equation (1) must combine with the derivatives on the metric to become covariant derivatives (You should check this!). We must therefore have $$ \delta \Gamma^c_{ab} = \frac{1}{2} \left[ \nabla_a h_b{}^c + \nabla_b h_a{}^c - \nabla^c h_{ab} \right] $$

We can use the same pro tip to determine the variations of any tensor. The Riemann tensor is $$ R^a{}_{bcd} = \partial_c \Gamma^a_{bd} - \partial_d \Gamma^a_{bc} + \Gamma^a_{ce} \Gamma^e_{bd} - \Gamma^a_{de} \Gamma^e_{bc} $$ so that $$ \delta R^a{}_{bcd} = \partial_c \delta \Gamma^a_{bd} - \partial_d\delta \Gamma^a_{bc} + \delta \Gamma^a_{ce} \Gamma^e_{bd} + \Gamma^a_{ce} \delta \Gamma^e_{bd} - \delta \Gamma^a_{de} \Gamma^e_{bc} - \Gamma^a_{de} \delta \Gamma^e_{bc} . $$ Again, the above quantity is a tensor so all terms with explicit Christoffel symbols should combine with the partial derivative $\partial_a$ to reduce to $\nabla_a$. Thus, it must be (again, check this yourself!) \begin{align} \delta R^a{}_{bcd} &= \nabla_c \delta \Gamma^a_{bd} - \nabla_d\delta \Gamma^a_{bc} \\ &= \frac{1}{2} \left[ \nabla_c\nabla_b h_d{}^a + \nabla_c \nabla_d h_b{}^a - \nabla_c \nabla^a h_{bd} - \nabla_d \nabla_c h_b{}^a - \nabla_d \nabla_b h_c{}^a + \nabla_d \nabla^a h_{cb} \right]. \end{align} The variation of the Ricci tensor is then \begin{align} \delta R_{bd} &= \delta R^a{}_{bad}= \frac{1}{2} \left[ \nabla_a \nabla_b h_d{}^a + \nabla_a \nabla_d h_b{}^a - \nabla^2 h_{bd} - \nabla_d \nabla_b h \right]. \end{align} Finally, the variation of the Ricci scalar is \begin{align} \delta R &= \delta ( g^{bd} R_{bd} ) \\ &= - h^{bd} R_{bd} + g^{bd} \delta R_{bd} \\ &= - h^{bd} R_{bd} + \nabla_a \nabla_b h^{ab} - \nabla^2 h . \\ \end{align}

Answer 2

You don't need the perturbation of the Riemann Tensor: you just need the perturbation of the Ricci Scalar $R= R_{\mu\nu} g^{\mu\nu}$ and the Ricci tensor $R_{\mu\nu}$ for the left hand side of the Einstein Equation $G_{\mu\nu} =R_{\mu\nu} -\frac{1}{2} R g_{\mu\nu}$.

I wrote it as an answer cause I can't comment, apparently. If the source linked below isn't enough I'll expand the answer with the explicit computations tomorrow.

Take a look here: A. Riotto - Inflation and the Theory of cosmological perturbations, Section 7.1, 7.2 and 7.3 should have what you're looking for. There are some typos but computations are very explicit. With these tools you can probably derive the expression for the Riemann Tensor very easily, even if it's not useful for the Einstein Equations it can still be a useful exercise. If you still don't know how to do it tell me and I'll write the computation here.