I'm trying to follow the calculation done by Will in his book Theory and experiment in gravitational physics, and I was hoping for some help in calculating the Ricci tensor components in Section 5.2 (for general relativity). To give some context, this is for the PPN formalism, where the metric is given by $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$, where $h_{\mu \nu}$ is the perturbation of the metric.

I'm hoping for some help in calculating, say, the Ricci tensor component $R_{00}$. In the book, the result is written as:

$$ R_{00} = - \frac{1}{2} \nabla^2 h_{00} - \frac{1}{2} \left( h_{jj,00} - 2 h_{j0,j0} \right) + \frac{1}{2} h_{00,j} \left( h_{jk,k} - \frac{1}{2} h_{kk,j} \right) - \frac{1}{4} \vert \nabla h_{00} \vert^2 + \frac{1}{2} h_{jk} h_{00,jk}.$$

Here, the Latin indices are spatial ones. I'm trying to reproduce this answer, but I can't seem to get it. I'm guessing that I'm making some basic mistakes with the tensor algebra, so if anyone could point them out, I'd appreciate it. Note that Will writes that this expression is what one gets when calculating the component to the required perturbation in $h_{\mu \nu}$. The order that Will states for the post-Newtonian formalism, the order of expansion for $h_{\mu \nu}$ is $$h_{00} \sim \mathcal{O}(2) + \mathcal{O}(4), \qquad h_{0j} \sim \mathcal{O}(3), \qquad\text{and} \qquad h_{ij} \sim \mathcal{O}(2).$$

To calculate this result, I began with the usual definition for the Ricci tensor:

$$ R_{\mu \nu} = \partial_{\rho} \Gamma^\rho_{\nu \mu} - \partial_{\nu} \Gamma^\rho_{\rho \mu} + \Gamma^\rho_{\rho \lambda} \Gamma^\lambda_{\nu \mu} - \Gamma^\rho_{\nu \lambda} \Gamma^\lambda_{\rho \mu}.$$

For the component I wrote above, I need to compute:

$$ R_{00} = \partial_{\rho} \Gamma^\rho_{00} - \partial_{0} \Gamma^\rho_{\rho 0} + \Gamma^\rho_{\rho \lambda} \Gamma^\lambda_{00} - \Gamma^\rho_{0 \lambda} \Gamma^\lambda_{\rho 0}.$$

I have some questions regarding the computation. For example, let's look at the first term. For the $\partial_{\rho} \Gamma^\rho_{00}$ term, I'll start with the Christoffel symbol.

$$ \Gamma^\rho_{00} = \frac{1}{2} g^{\rho \gamma} \left( \partial_0 g_{\gamma 0} + \partial_0 g_{\gamma 0} - \partial_\gamma g_{00} \right) \\ = \frac{1}{2} g^{\rho \gamma} \left( 2 \partial_0 \left[ h_{\gamma 0} + \eta_{\gamma 0} \right] - \partial_\gamma \left[ h_{00} + \eta_{00} \right] \right) \\ = \frac{1}{2} g^{\rho \gamma} \left( 2 \partial_0 h_{\gamma 0} - \partial_\gamma h_{00} \right).$$

How do I incorporate the metric $g^{\rho \gamma}$ with the terms in the parentheses? I want to expand this in terms of the Minkowski metric and the perturbation, but I'm not sure what it is when expanding to the correct order. When I looked online, most references state that $g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu}$, but is this appropriate here? I will also have to take partial derivatives, which means a bunch of product rules with $g^{\mu \nu}$. This will lead to lot more terms than needed, which suggests they won't cancel out at the end. I think I might be going down a rabbit hole here, so if anyone has any experience with these sorts of calculations, I would love the help.


As a related question, I'm trying to interpret the order of the various terms in the expression for $R_{00}$. Since $\partial_t / \partial_x \sim \mathcal{O}(1)$, which order is the term $- \frac{1}{2} \nabla^2 h_{00}$? What about the other terms in $R_{00}$? I don't find this is super clear from the text, so any help would be illuminating.

  • $\begingroup$ Yes, $g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu}$ to first order in $h$. Do you understand why? $\endgroup$
    – G. Smith
    Commented May 30, 2019 at 3:06
  • $\begingroup$ @G.Smith Yeah, it has to do with the fact that $g_{\mu \gamma} g^{\gamma \nu} = \delta_\mu^\nu$, and you can work it out from there. However, I'm wondering if this can be applied for the PPN formalism, because don't we want to expand $h_{\mu \nu}$ to different orders? $\endgroup$
    – Germ
    Commented May 30, 2019 at 13:09
  • $\begingroup$ I think you can use this to get the stated result for $R_{00}$, without going to higher orders, although I haven't worked it out. All the terms in that result are either (1) second derivatives of $h$, (2) the product of two first derivatives of $h$, or (3) $h$ times second derivatives of $h$. I think you can get (1) and (2) from just $g^{\mu\nu}=\eta^{\mu\nu}$, while (3) requires the first-order correction to the inverse metric. Give it a try. $\endgroup$
    – G. Smith
    Commented May 30, 2019 at 16:25
  • $\begingroup$ @G.Smith Thanks for the tip. I'm going to try what you suggested, but how am I supposed to do the calculation with only $g^{\mu \nu} = \eta^{\mu \nu}$ for parts (1) and (2), and then add in the first-order correction for (3)? In other words, should I just calculate everything with $g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu}$ and then drop terms? I'll report back here with my results. $\endgroup$
    – Germ
    Commented May 30, 2019 at 16:59
  • $\begingroup$ It's been a long time since I studied PPN, but I think order $\mathcal{O}(n)$ refers to terms of order $n$ when expanding in the parameter $\epsilon \sim v \sim~\sqrt{GM/r}$. This means that $\partial_i \sim \mathcal{O}(2)$ and $\partial_t \sim \mathcal{O}(3)$. Thus $\nabla^2h_{00}$ would be $\mathcal{O}(6)+\mathcal{O}(8)$. It appears that Will's $R_{00}$ is keeping all terms that contribute to orders 2, 4, 6, and 8. Other terms he leaves out, like $h_{0i}h_{00,0i}$ are higher than order 8. $\endgroup$
    – G. Smith
    Commented May 31, 2019 at 19:18

1 Answer 1


In the PPN formalism, one considers a collection of non-relativistic masses $m_i$, with velocities $\mathbf{v}_i$ and separations $r_i$, and does expansions to various orders in an expansion parameter $\epsilon$ where

$$\epsilon \sim v \sim \sqrt{Gm/r}.$$

Will refers to terms of order $\epsilon^n$ as being of "order $n$", written $\mathcal{O}(n)$. (I believe that there is another convention to say that terms of order $(\epsilon^2)^n$ are of order $n$, but we'll stick with Will's terminology.)

If $1/r$ is order 2, then so is $\partial/\partial r$. Thus

$$\partial_i \sim \mathcal{O}(2).$$

Since $\partial_t=\frac{\partial}{\partial t}=\frac{\partial x^i}{\partial t}\frac{\partial}{\partial x^i}=v^i \partial_i$,

$$\partial_t \sim \mathcal{O}(3).$$

Will states that $h_{00}$ is order 2, $h_{0j}$ is order 3, and $h_{ij}$ is order 2, from which we can derive the following table of orders.

$$\begin{array}{|cc|cc|cc|} \hline h_{00} & 2 & h_{0j} & 3 & h_{ij} & 2 \\ h_{00,0} & 5 & h_{0j,0} & 6 & h_{ij,0} & 5 \\ h_{00,k} & 4 & h_{0j,k} & 5 & h_{ij,k} & 4 \\ h_{00,00} & 8 & h_{0j,00} & 9 & h_{ij,00} & 8 \\ h_{00,0l} & 7 & h_{0j,0l} & 8 & h_{ij,0l} & 7 \\ h_{00,kl} & 6 & h_{0j,kl} & 7 & h_{ij,kl} & 6 \\ \hline \end{array}$$

From this we see that the first term in Will's $R_{00}$ is order 6 and the remaining terms are order 8. We can thus assume that he has calculated $R_{00}$ through order 8.

To do this calculation, start with the definition of $h_{\mu\nu}$, which is


Notice that in Will's expression for $R_{00}$, there are no Greek indices, only Latin ones. That is because he is separating the temporal components from the spatial components. So we start by writing

$$\begin{align} g_{00}&=-1+h_{00}\\ g_{0j}&=h_{0j}\\ g_{ij}&=\delta_{ij}+h_{ij}\\ \end{align}$$

where we have taken $\eta$ to have a (-+++) signature for the metric so that we can ignore whether a spatial index is raised or lowered.

Before computing Christoffel symbols and Ricci components, we must compute the inverse metric tensor $g^{\mu\nu}$. To first order in $h$, it is given by


Now, notice that Will's expression involves only the covariant perturbation $h_{\mu\nu}$, not the contravariant perturbation $h^{\mu\nu}$. Again to first order in $h$, we can raise and lower indices on $h$ using $\eta$ rather than $g$. The result is

$$\begin{align} g^{00}&=-1-h^{00}=-1-h_{00}\\ g^{0j}&=-h^{0j}=h_{0j}\\ g^{ij}&=\delta^{ij}-h^{ij}=\delta_{ij}-h_{ij}\\ \end{align}.$$

Now we can compute the Christoffel symbols using


Splitting the Greek indexes into temporal and spatial ones, there are only six of them:

$$\Gamma^0_{00}, \Gamma^k_{00}, \Gamma^0_{0j}, \Gamma^k_{0j}, \Gamma^0_{ij}, \Gamma^k_{ij}$$

I'll do just the first one in detail.

$$\begin{align} \Gamma^0_{00}&=\frac{1}{2}g^{0\mu}(2g_{\mu0,0}-g_{00,\mu})\\ &=\frac{1}{2}g^{00}g_{00,0}+\frac{1}{2}g^{0m}(2g_{0m,0}-g_{00,m})\\ &=\frac{1}{2}(-1-h_{00})h_{00,0}+\frac{1}{2}h_{0m}(2h_{0m,0}-h_{00,m})\\ &=-\frac{1}{2}h_{00,0}-\frac{1}{2}h_{00}h_{00,0}-\frac{1}{2}h_{0m}h_{00,m}+h_{0m}h_{0m,0} \end{align}.$$

The first term is order 5, the second and third are order 7, and the fourth is order 9. To compute the Riici tensor to order 8, there is no need to keep any term in the Christoffel symbols greater than order 6. (This is because they are either going to be multiplied together or differentiated, and differentiation increases the order by either 2 or 3.) So we conclude that


Five similar calculations yield






Proceeding on to calculate

$$ R_{00} = \partial_{\rho} \Gamma^\rho_{00} - \partial_{0} \Gamma^\rho_{\rho 0} + \Gamma^\rho_{\rho \lambda} \Gamma^\lambda_{00} - \Gamma^\rho_{0 \lambda} \Gamma^\lambda_{\rho 0},$$

we once again split Greek indices into temporal and spatial ones, producing

$$\begin{align}R_{00}&= (\partial_0\Gamma^0_{00}+ \partial_r\Gamma^r_{00})\\ &-(\partial_0\Gamma^0_{00}+ \partial_0\Gamma^r_{r0})\\ &+(\Gamma^0_{00}\Gamma^0_{00}+ \Gamma^0_{0l}\Gamma^l_{00}+ \Gamma^r_{r0}\Gamma^0_{00}+ \Gamma^r_{rl}\Gamma^l_{00})\\ &-(\Gamma^0_{00}\Gamma^0_{00}+ \Gamma^0_{0l}\Gamma^l_{00}+ \Gamma^r_{00}\Gamma^0_{r0}+ \Gamma^r_{0l}\Gamma^l_{r0}). \end{align}.$$

Note that here the index $r$ is just a generic spatial index, chosen to correspond to $\rho$, and not a radial component.

The first and third terms cancel, as do the fifth and ninth, and the sixth and tenth, leaving

$$R_{00} = \partial_r\Gamma^r_{00} -\partial_0\Gamma^r_{r0} +\Gamma^r_{r0}\Gamma^0_{00} +\Gamma^r_{rl}\Gamma^l_{00} -\Gamma^r_{00}\Gamma^0_{r0} -\Gamma^r_{0l}\Gamma^l_{r0}. $$

To calculate $R_{00}$ to order 8, use only the appropriate-order terms in each Christoffel symbol:

$$\begin{align} R_{00}&= \partial_r\left(-\frac{1}{2}h_{00,r}+h_{0r,0}+\frac{1}{2}h_{rm}h_{00,m}\right)\\ &-\partial_0\left(\frac{1}{2}h_{rr,0}\right)\\ &+0\\ &+\left(\frac{1}{2}h_{rr,l}\right)\left(-\frac{1}{2}h_{00,l}\right)\\ &-\left(-\frac{1}{2}h_{00,r}\right)\left(-\frac{1}{2}h_{00,r}\right)\\ &-0\end{align}.$$

The result,

$$\begin{align} R_{00}=&-\frac{1}{2}h_{00,rr} +h_{0r,0r} +\frac{1}{2}h_{rm,r}h_{00,m} +\frac{1}{2}h_{rm}h_{00,rm}\\ &-\frac{1}{2}h_{rr,00} -\frac{1}{4}h_{rr,l}h_{00,l} -\frac{1}{4}h_{00,r}h_{00,r}, \end{align}$$

is equivalent to Will's.

P.S. I don't have a copy of the book by Will that you are using, so I relied on my recollections of PPN. He has co-authored a more recent book, but I don’t have that one either.

  • $\begingroup$ Thanks for the detailed write up. I ended up getting the same kind of calculation as you did, though you have everything explained in detail. It makes a lot of sense now. I wish I had something like this in the margins of the book! $\endgroup$
    – Germ
    Commented Jun 3, 2019 at 12:59

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