# Massive primordial tensor perturbations?

1. Jun 4, 2015

### 302021895

I am studying the generation of tensor perturbations during inflation, and I am trying to check every statement as carefully as possible. Starting from the metric

$ds^2 = dt^2 - a^2(\delta_{ij}+h_{ij})dx^idx^j$

I make use of Einstein's equations to find the equation of motion for the perturbation $h_{ij}$. For the Ricci tensor I obtain

$R_{00}=-3(\dot{H}+H^2)$
$R_{0i}=0$
$R_{ij} = \frac{a^2}{2}\left[2(3H^2+\dot{H})(\delta_{ij}+h_{ij})+3H\dot{h}_{ij}+\ddot{h}_{ij}-\frac{1}{a^2}\nabla^2h_{ij}\right]$
and
$R=-(12H^2+6\dot{H})$

This coincides with the results shown in Dodelson's textbook, equations (5.47) and (5.57). Dodelson then claims that since the Ricci scalar contains no perturbation (true), the Einstein tensor can be calculated as

$\delta G^i_j = \delta R^i_j$

(where $\delta$ means the perturbed part), which supposedly leads to the standard result for the massless tensor perturbation. However, if I just blindly substitute,

$G_{ij} = R_{ij}-\frac{1}{2}g_{ij}R = \frac{a^2}{2}\left[\ddot{h}_{ij}+3H\dot{h}_{ij} - \frac{1}{a^2}\nabla^2h_{ij} - 2(2\dot{H}+3H^2)(\delta_{ij}+h_{ij})\right]$

The energy-momentum tensor for the inflaton is diagonal $T_{ij}\propto\delta_{ij}$, which means that the previous equation leads to

$\ddot{h}_{ij}+3H\dot{h}_{ij} - \frac{1}{a^2}\nabla^2h_{ij} - 2(2\dot{H}+3H^2)h_{ij} = 0$

i.e., an equation for a 'massive' perturbation. Why is this incorrect? I agree with Dodelson in that $\delta R=0$, but $\delta g_{ij}\neq0$, and it seems to me that this would introduce the extra mass term. Moreover, by contracting with the metric, $G^a_b=g^{ac}G_{cb}$, we do 'get rid' of the extra term, since $g_{ab}\rightarrow\delta^a_b$, but it seems to me that this would turn the enegy-momentum tensor in the right-hand side of Einstein's equation to a non-diagonal form, which would then introduce an extra term proportional to $h^i_j$.

I'm probably making a stupid mistake, but I am now very frustrated and I would appreciate any help. Thanks.

2. Jun 5, 2015

### Chronos

That is a good reason to include a cosmological constant.

3. Jun 5, 2015

### 302021895

I'm not sure I follow you...