The correct statement is that we can always construct a geodesic such that $x(s)=y(s)=0$ for every value of the affine parameter $s$. All that independently from our initial choice of the origin and orientation of orthogonal Cartesian coordinates $x,y,z$ in the $3$-manifolds normal to $\partial_t$ (the natural rest space of the considered spacetime).
The geodesics are solutions of the Euler-Lagrange equations of the Lagrangian
$${\cal L} = \sqrt{|-\dot{t}^2 + a(t(\xi))^2(\dot{x}^2+\dot{y}^2+\dot{z}^2)|}\:,\tag{1}$$
where the used parameter is a generic one $\xi$ ad the dot denotes the $\xi$-derivative.
As ${\cal L}$ does not explicitly depend on $x,y,z$, from E-L equations, we have the three constants of motion:
$$\frac{\partial {\cal L}}{\partial \dot{x}}\:, \quad \frac{\partial {\cal L}}{\partial \dot{y}}\:, \quad \frac{\partial {\cal L}}{\partial \dot{z}}\:.$$
Passing to describe the curves with the geodesical length $s$, with
$$ds = \sqrt{|-\dot{t}^2 + a(t(\xi))^2(\dot{x}^2+\dot{y}^2+\dot{z}^2)|} d\xi$$
these constants read, in fact,
$$a(t(s))^2 \dot{x}(s)\:,\quad a(t(s))^2 \dot{y}(s)\:, \quad a(t(s))^2 \dot{z}(s)\:,$$
where now the dot denotes the $s$-derivative.
In other words, there is a constant vector $\vec{c}\in\mathbb R^3$, such that, for every $s$:
$$a(t(s))^2 \frac{d\vec{x}}{ds} = \vec{c}\tag{2}$$
where $\vec{x}(s) = (x(s),y(s),z(s))$. The geodesics are described here by curves
$$\mathbb R \ni s \mapsto (t(s), \vec{x}(s)) \tag{3}\:.$$
Looking at the Lagrangian (1), one sees that it is invariant under spatial rotations. That symmetry extends to solutions of E-L equations. In other words we have that, if (3) is a geodesics, for $R\in SO(3)$,
$$\mathbb R \ni s \mapsto (t(s), \vec{x}'(s)) := (t(s), R\vec{x}(s)) \tag{4}$$
is a geodesic as well.
Correspondingly, due to (2) we have the new constant of motion
$$a(t(s))^2 \frac{d\vec{x}'}{ds}= a(t(s))^2 \frac{dR\vec{x}}{ds} = a(t(s))^2 R\frac{d\vec{x}}{ds} = R\vec{c}\tag{2'}$$
Unless $\vec{c}=0$(*), we can rotate this constant vector in order to obtain, for instance, $R\vec{c} = c \vec{e}_z$. This means that the new geodesic verifies
$$a(t(s))^2 \frac{d\vec{x}'}{ds}\:\: ||\:\: \vec{e}_z$$
the spatial part is parallel to $\vec{e}_z$. I will omit the prime $'$ in the following and I assume to deal with a geodesic with spatial part parallel to $\vec{e}_z$ and thus, as $a\neq 0$, it holds $x(s)= x_0$, $y(s)=y_0$ constantly.
Let us finally suppose that the initial point of the geodesic is $\vec{x}(0) = \vec{x}_0$. As the Lagrangian is also invariant under spatial translations, we also have that if (3) is a geodesic, for $R\in SO(3)$,
$$\mathbb R \ni s \mapsto (t(s), \vec{x}'(s)) := (t(s), \vec{x}(s)+ \vec{r}_0) \tag{5}$$
is a geodesic as well. Choosing $\vec{r}_0 := - \vec{x}_0$, we have a geodesic with $x(s)=y(s)=0$ as requested.
(*) We can always choose $\vec{c}\neq 0$ assuming that the initial tangent vector of the geodesic verifies this requirement (notice that $a^2 \neq 0$). And we know that there is a geodesics for every choice of the initial conditions.
This post imported from StackExchange Physics at 2014-06-27 11:29 (UCT), posted by SE-user V. Moretti
Q&A (4871)
