# Pertubation of Riemann tensor in a general curved space-time

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?

• When you are done, you can check by comparing it with the linearized EFE around Minkowski space here. – G. Smith Sep 26 at 19:03
• I am well aware how it works in Minkowski (I have done it before), but I need the full expression for curved space-time, since I need to write the Einstein tensor with a general back-ground. – Edison Santos Sep 26 at 19:16
• As a general rule, whenever you need to find a formula and your field has a book by Weinberg, look there. In this case you have not one but two (Gravitation and Cosmology and Cosmology), so I'm sure one of them has to have it. – Javier Sep 26 at 23:45
• You can find your answer to a related question here - physics.stackexchange.com/a/487808/133418 – Avantgarde Oct 1 at 23:01

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_a h_b{}^c - \nabla_d \nabla_b h_a{}^c + \nabla_d \nabla^c h_{ab} \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}

• Thanks a lot for the answer, it works (I am checking each relation, as you have suggested). However I do have some questions about this: why is the variation of the Christoffel symbol a tensor? Is the variation of any non-tensor a tensor? Would this whole scheme work also for higher order perturbations? – Edison Santos Sep 27 at 15:41
• Think about how a connection transforms under coordinate transformations. It should then be easy to see that difference between any two connections is always a tensor (to all higher orders perturbations). The variation of random non-tensors is not a tensor. – Prahar Sep 28 at 4:23

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.