How to simplify the process of calculating spacetime geodesics?

Question

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?

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 $$\frac{d\tau}{d\lambda} \, =\, a(t)^2$$