How to simplify the process of calculating spacetime geodesics? 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?

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