The below quote is from Gravitation and Cosmology by Weinberg. I don't understand the calculations leading to equation $(9.1.2)$. Any help or alternative resources will be helpful.

1 The Post-Newtonian Approximation

Consider a system of particles that, like the sun and planets, are bound together by their mutual gravitational attraction. Let $\overline M$, $\overline r$, and $\overline v$ be typical values of the masses, separations, and velocities of these particles. It is a familiar result of Newtonian mechanics that the typical kinetic energy $\frac12\overline M\overline v^2$ will be roughly of the same order of magnitude as the typical potential energy $G\overline M^2/\overline r$, so

$$\overline v^2\sim\frac{G\overline M}{\overline r}\tag{9.1.1}$$

(For instance, a test particle in a circular orbit of radius $r$ about a central mass $M$ will have velocity $v$ given in Newtonian mechanics by the exact formula $v^2=GM/r$.) The post-Newtonian approximation may be described as a method for obtaining the motions of the system to one higher power of the small parameters $G\overline M/r$ and $\overline v^2$ than given by Newtonian mechanics. It is sometimes referred to as an expansion in inverse powers of the speed of light, but since in our units this speed is unity we prefer to say that our expansion parameter is $\overline v^2$, or equivalently, $G\overline M/\overline r$.

We must begin by asking what we need. The equations of motion of the particles are


From this we may compute the accelerations as


This may be written in more detail as


  • $\begingroup$ Hello! I have converted your screenshot to text using MathJax (LaTeX) math typesetting (for future questions, you can refer to MathJax basic tutorial and quick reference). Based on your question, I didn't add the parts after eq. 9.1.2 since they appeared irrelevant to your question to me. If you think they should be there, feel free to add them back in. Thanks! $\endgroup$
    – jng224
    Aug 2, 2021 at 10:46
  • $\begingroup$ Yes, they are indeed irrelevant. Thanks. $\endgroup$
    – Ben Stark
    Aug 2, 2021 at 11:17

1 Answer 1


Step by step: $$ \frac{d^2x^i}{dt^2}=\left(\frac{dt}{d\tau}\right)^{-1}\frac{d}{d\tau}\left[\left(\frac{dt}{d\tau}\right)^{-1}\frac{dx^i}{d\tau}\right] \tag{1} $$ This is just the chain rule of introductory calculus, along with the fact that $(dt/d\tau)^{-1} = (d\tau/dt)$. $$ =\left(\frac{dt}{d\tau}\right)^{-2}\frac{d^2x^i}{d\tau^2}-\left(\frac{dt}{d\tau}\right)^{-3}\frac{d^2t}{d\tau^2}\frac{dx^i}{d\tau} \tag{2} $$This is the product rule applied to the quantity in square brackets in (1). $$ =-{\Gamma^i}_{v\lambda}\frac{dx^v}{dt}\frac{dx^\lambda}{dt}+{\Gamma^0}_{v\lambda}\frac{dx^v}{dt}\frac{dx^\lambda}{dt}\frac{dx^i}{dt}\tag{3} $$ Here, we have applied the geodesic equation to eliminate the second derivatives of the coordinates in (2) and replace them by first derivatives. We have then used the chain rule again to eliminate the derivatives with respect to $\tau$. For example, we have $$ \frac{d^2x^i}{d\tau^2} = -{\Gamma^i}_{v\lambda}\frac{dx^v}{dt}\frac{dx^\lambda}{dt} = -{\Gamma^i}_{v\lambda} \left( \frac{dt} {d \tau}\frac{dx^v}{dt} \right) \left(\frac{dt}{d \tau} \frac{dx^\lambda}{dt} \right). $$ A similar manipulation is applied to the $d^2t/d\tau^2$ term in (2).

Finally, Eq. (9.1.2) is obtained from (3) by pulling the "time" terms out all of the "Greek-index" summations over all four indices. This leaves the spatial ("Roman-index") summations behind.


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