I want to study the movement of a particle along geodesics in an expanding universe with metric (FRW metric) $$ ds^2 = -dt^2 + a^2(t) \left( \dfrac{1}{1-kr}dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2 \right)\ . $$

My usual approach for getting the equations of the geodesics is the one usually taken in A. Zee's book, where instead of calculating the symbols $\Gamma^\mu_{\nu\sigma}$, we define the action $$ S = \int d\tau\, L = \int d\tau\, \sqrt{ -g_{\mu\nu} \dfrac{dx^\mu}{d\tau} \dfrac{dx^\nu}{d\tau} } $$ and minimize it using the Euler-Lagrange equations $$ \dfrac{d}{d\tau} \left( \dfrac{\partial L}{\partial \frac{dx^\mu}{d\tau}} \right) - \dfrac{\partial L}{\partial x^\mu} = 0\ . $$

Therefore, I can get the differential equations for the geodesics just by calculating this equation for each coordinate $t$, $r$, $\theta$ and $\phi$.

However, I have also seen this metric written as $$ ds^2 = -dt^2 + a^2(t) d\Sigma^2\ , $$ where $d\Sigma^2$ accounts for the 3-dimensional spatial part. Minimizing the action with respect to $t$ and $\Sigma$ is much simpler than doing it for the individual four coordinates, so I came up with the following questions:

  • Is it possible to derive the equations of the geodesics that come from the first metric but using the second one?
  • If the answer turns out to be yes, what is the correct way of doing it?
  • Finally, in which cases is it possible to reduce the metric in the way the second metric does and make calculations with it? For example, it is common to see the simplification $d\Omega^2 = d\theta^2 + \sin^2\theta d\phi^2$. In which situations is it possible to simplify things in this way?
  • $\begingroup$ That is possible if you have spherical symmetry so you can rotate your coordinate system in a way that for any given geodesic there is no motion along one angle. You can do it for Schwarzschild or the FLRW, but you can't do it for Kerr since the latter doesn't have spherical symmetry. $\endgroup$
    – Yukterez
    Commented May 15, 2022 at 14:41

1 Answer 1


I guess you may be making a reasonable point here, when speaking about this particular metric.

Assume the space-time is split into $(t , \, x) \, \in \, \mathbb{R} \times M_3$, where $M_3$ is one of the three possible 3D geometric manifolds: Euclidean, spherical or hyperbolic. Then I will write the 4D space-time metric as

$$d\tau^2 \,=\, dt^2 \, - \,a(t)^2\big(\, dx^T g(x) \, dx\,\big)$$

where $g(x)$ is the $3 \times 3$ matrix of the geometric metric of the 3D manifold $M_3$.

When parametrized with respect to proper time (this is a crucial assumption so I will go with it!), the geodesic equations can be written as

\begin{align} &\frac{d}{d\tau} \left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) \, =\, \frac{\partial \mathcal{L}}{\partial {x}}\\ &\\ &\frac{d}{d\tau} \left(\frac{\partial \mathcal{L}}{\partial \dot{t}}\right) \, =\, \frac{\partial \mathcal{L}}{\partial {t}} \end{align}

where the function $\mathcal{L} $ is defined as $$\mathcal{L} \,=\, \frac{1}{2}\left(\frac{dt}{d\tau}\right)^2 - \,\frac{1}{2} \,a(t)^2\left(\,\frac{dx}{d\tau}^T g(x) \, \frac{dx}{d\tau}\,\right)$$

Consequently, the first set of equations $$\frac{d}{d\tau} \left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) \, =\, \frac{\partial \mathcal{L}}{\partial {x}}$$ can be written as

$$\frac{d}{d\tau} \left( -\, a(t)^2 g(x) \, \frac{dx}{d\tau}\, \right) \, =\, -\, \frac{1}{2} \, a(t)^2\left(\,\frac{dx}{d\tau}^T \frac{\partial g}{\partial x} (x) \, \frac{dx}{d\tau}\,\right)$$ or after cancelling the minus sign $$\frac{d}{d\tau} \left(\, a(t)^2 g(x) \, \frac{dx}{d\tau}\, \right) \, =\, \frac{1}{2} \, a(t)^2\left(\,\frac{dx}{d\tau}^T \frac{\partial g}{\partial x} (x) \, \frac{dx}{d\tau}\,\right)$$ By introduce the new parametrization $$\frac{d}{d\lambda} \,=\, a(t)^2\,\frac{d}{d\tau} \,\,\,\,\,\,\,\,\,\,\,\, {d\lambda} \,=\, \frac{1}{a(t)^2}{d\tau}$$ and thus $$\frac{d}{d\tau} \, =\,\frac{1}{a(t)^2}\,\frac{d}{d\lambda}$$ the equations simplify to $$\frac{1}{a(t)^2}\, \frac{d}{d\lambda}\left(\, g(x) \, \frac{dx}{d\lambda}\, \right) \, =\, \frac{1}{2} \, a(t)^2\left(\,\frac{1}{a(t)^2}\,\frac{dx}{d\lambda}^T \frac{\partial g}{\partial x} (x) \, \frac{1}{a(t)^2}\, \frac{dx}{d\tau}\,\right)$$ and by factoring out the common factor $1/a(t)^2$ in the righthand side $$\frac{1}{a(t)^2}\, \frac{d}{d\lambda}\left(\, g(x) \, \frac{dx}{d\lambda}\, \right) \, =\, \frac{1}{2} \, \frac{a(t)^2}{a(t)^4}\, \left(\,\frac{dx}{d\lambda}^T \frac{\partial g}{\partial x} (x) \,\frac{dx}{d\lambda}\,\right)$$ we arrive at the geodesic equations for the 3D geodesics of the geometric manifold $M_3$ $$\frac{d}{d\lambda}\left(\, g(x) \, \frac{dx}{d\lambda}\, \right) \, =\, \frac{1}{2} \,\left(\,\frac{dx}{d\lambda}^T \frac{\partial g}{\partial x} (x) \,\frac{dx}{d\lambda}\,\right)$$ which in all three cases are easy and explicit to write, so let's simply write them as $x = x(\lambda)$. Moreover, the parameter $\lambda$ is actually the arclength parameter of the geometric 3D metric $$d\lambda^2 \,= \, dx^T g(x) \, dx$$ This yields the conservation law $$\frac{dx}{d\lambda}^T g(x) \,\frac{dx}{d\lambda} \, =\, 1$$ The final equation for the coordinate time variable $t$ can be derived by cutting corners and use the fact that when parametrized by proper time $\tau$, the metric is a conserved quantity. This is equivalent to simply going back to the original metric
$$d\tau^2 \,=\, dt^2 \, - \,a(t)^2\big(\, dx^T g(x) \, dx\,\big)$$ and reparametrizing $$d\tau \,=\, a(t)^2 d\lambda$$ which leads to $$a(t)^4 \, d\lambda^2 \,=\, dt^2 \, - \,a(t)^2\big(\, dx^T g(x) \, dx\,\big)$$ so from here we get the differential equation $$1 \,=\, \frac{1}{\,a(t)^4} \left(\frac{dt}{d\lambda}\right)^2 \, - \, \frac{1}{\,a(t)^2}\left(\, \frac{dx}{d\lambda}^T g (x) \,\frac{dx}{d\lambda}\,\right)$$ and since $\frac{dx}{d\lambda}^T g (x) \,\frac{dx}{d\lambda} \, = \, 1$ the equations is now $$1 \,=\, \frac{1}{\,a(t)^4} \left(\frac{dt}{d\lambda}\right)^2 \, - \, \frac{1}{\,a(t)^2}$$ After rearranging it, it becomes $$ \left(\frac{dt}{d\lambda}\right)^2 \,=\, a(t)^4 \, + \, {\,a(t)^2}$$ or if you prefer $$ \frac{dt}{d\lambda}\,=\, \sqrt{\, a(t)^4 \, + \, {\,a(t)^2} \, }$$ To put things together, the space-time geodesics, parametrized by the geometric 3D arclength $\lambda$ can be written as \begin{align} & x \,=\, x(\lambda)\\ & \frac{dt}{d\lambda}\,=\, \sqrt{\, a(t)^4 \, + \, {\,a(t)^2} \, } \end{align} Recall that the link between the 3D geometric arclength $\lambda$ and the proper time $\tau$ is described by the differential equation $$\frac{d\tau}{d\lambda} \, =\, a(t)^2$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.