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Killing vectors in flat FLRW metric

+ 6 like - 0 dislike
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I have the flat FLRW metric,

$$ ds^2=-dt^2+a(t)^2(dx^2+dy^2+dz^2) $$ and a geodesic $\gamma(s)=(t(s),x(s),y(s),z(s))$ with parameter $s$. Two of the Killing vectors of the metric are $ \partial_x$ and $\partial_y$ which give rise to the constants of motion $ a^2\dot x$ and $ a^2 \dot y $ respectively. This is how far I am. Now I need to show that from this follows that I can assume that the geodesic has $y(s)=0=x(s)$. This is not obvious for me from the above constants of motions, because they also permit $\dot x=\frac{const.}{a^2}$

If I can rule this out, I get $\dot x =0$ and therefore $x=const.$ and because I can set $x$ to zero locally, it is zero on the whole spacetime.

So my question is: Why is $\dot x =0$

Edit: Or am I completely on the wrong track?

This post imported from StackExchange Physics at 2014-06-27 11:29 (UCT), posted by SE-user SimonS
asked Jun 24, 2014 in Theoretical Physics by SimonS (30 points) [ no revision ]
This is a geodesic, but not the one unique geodesic. As shown below, you can translate any of the purely spatial geodesics into one of this form through translations and rotations, though.

This post imported from StackExchange Physics at 2014-06-27 11:29 (UCT), posted by SE-user Jerry Schirmer
You're allowed to set $\dot{x} = 0$, so why don't you do so? Everything else looks correct to me.

This post imported from StackExchange Physics at 2014-06-27 11:29 (UCT), posted by SE-user auxsvr

2 Answers

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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
answered Jun 24, 2014 by Valter Moretti (2,025 points) [ no revision ]
Do you happen to know under which conditions we are allowed to use the lagrangian without the square root to derive the geodesics? I faintly recall an inequality that is related, but I haven't seen anywhere a rigorous proof about this.

This post imported from StackExchange Physics at 2014-06-27 11:29 (UCT), posted by SE-user auxsvr
If you use the Lagrangian without the square root you obtain the right equations provided the parameter is assumed to be an affine one. With the square root instead, the parameter is generic.

This post imported from StackExchange Physics at 2014-06-27 11:29 (UCT), posted by SE-user V. Moretti
I believe I've found it: the Hölder inequality connects the minimum of the energy functional with that of the length functional, according to this, but this argument applies only for a riemannian manifold, not a lorentzian one.

This post imported from StackExchange Physics at 2014-06-27 11:29 (UCT), posted by SE-user auxsvr
Indeed, these reasonings hold for positive metrics only. Conversely, my argument is very direct: if you compute Euler-Lagrange equations for the Lagrangian without square root you find the parallel transport version of geodesic equation.

This post imported from StackExchange Physics at 2014-06-27 11:29 (UCT), posted by SE-user V. Moretti
@V.Moretti : Unless I misunderstood something, it seems to me that there is a typo, it is $\dot t^2$, and not $\dot x^2$, in the first term of the expressions of $\mathcal L$ and $ds$ ?

This post imported from StackExchange Physics at 2014-06-27 11:29 (UCT), posted by SE-user Trimok
@Trimok : You are right! Thanks, I am correcting.

This post imported from StackExchange Physics at 2014-06-27 11:29 (UCT), posted by SE-user V. Moretti
+ 0 like - 0 dislike

My only issue is that, while the work above is obviously correct, there are actually 6 Killing vectors for the FLRW metrics, for spatial homogeneity (the 3 spatial ones above), and the 3 for rotations, which is not immediately obvious from the form of the metric above. Indeed, it may be more helpful to perform a coordinate transformation, and write the FLRW metric in the more standard form: $ds^2 = -dt^2 + a^2(t) \left[dr^2 + r^2 \left(d\theta^2 + \sin^2 \theta d \phi^2\right)\right]$, from this form, the $G_{3}$ isotropy group around every point is quite clear which implies by definition, the existence of a $G_{3}$ simply transitive subgroup, but any comments on this would be helpful.

This post imported from StackExchange Physics at 2014-06-27 11:29 (UCT), posted by SE-user Ikjyot Singh Kohli
answered Jun 24, 2014 by Ikjyot Singh Kohli (0 points) [ no revision ]

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