A Motivation for Theta = Pi/2 on Wald GR p138

[strangerep]

On p138 of Wald's General Relativity, 4th para, he says:

[...] First, we note that because of the parity reflection symmetry, $\theta \to \pi - \theta$, of the Schwarzschild metric, if the initial position and tangent vector of a geodesic lie in the "equatorial plane" $\theta = \pi/2$, then the entire geodesic must lie in this "plane." Since every geodesic can be brought to an initially (and hence everywhere) equatorial geodesic by a rotational isometry, this means that without loss of generality we may restrict attention to study of the equatorial geodesics, and we shall do so.

Firstly, I already understand (of course) that this sort of simplification is easy and immediate in Newtonian mechanics, since conservation of angular momentum always occurs in central force problems (the proof is just a few lines) and one can re-orient the coordinates so that motion occurs in the plane perpendicular to the angular momentum axis.

Secondly, in GR, if one actually computes the connection coefficients for Schwarzschild, to write down the geodesic equations explicitly, one then sees immediately that $\theta=\pi/2$ (say) is a convenient choice that one can make without loss of generality.

But I don't follow Wald's particular argument, i.e., that the symmetry $\theta\to \pi-\theta$ allows one to conclude that "if the initial position and tangent vector of a geodesic lie in the equatorial plane $\theta=\pi/2$, then the entire geodesic must lie in this plane.''

What am I missing?

 

[Ibix]

Isn't he just using the "which way would it go" rhetorical question? If your particular initial conditions don't pick out a direction of increase/decrease for $\theta$ the symmetry of the metric implies any argument for "$\theta$ will increase" must apply equally to "$\pi-\theta$ will increase", so such an argument must be self-contradictory?

 

[PeroK]

I think Hartle argued this for initially $\phi =0$ to show the motion is planar. And then took the plane to be $\theta = \frac \pi 2$.

 

[vanhees71]

I think the most simple argument is to use "isotropic coordinates" first. There the Schwarzschild line element reads
$$\mathrm{d} s^2 =\frac{[1-m/(2 R)]^2}{[1+m/(2R)]^2} c^2 \mathrm{d} t^2 - [1+m/(2R)]^4 \mathrm{d} \vec{x}^2,$$
where $R=|\vec{x}|^2$ and $\vec{x}=(x^1,x^2,x^3)$.

The square-form Lagrangian for the geodesic reads
$$L=\frac{1}{2} \dot{s}^2,$$
where the dot denotes the derivative wrt. an arbitrary world-line parameter, $\lambda$, which is automatically affine, because since $L$ doesn't depend explicitly on $\lambda$, i.e., $H=p_{\mu} q^{\mu}-L=L=\text{const}$, where $p_{\mu} = \partial_{\dot{q}^{\mu}} L$, and for a massive particle we can set $\lambda=\tau$, such that $(\mathrm{d}_{\tau} s)^2=c^2$.

It's now also manifest that the solution is rotationally symmetric, i.e., the Lagrangian is invariant under rotations $\vec{x}'=\hat{R} \vec{x}$ with $\hat{R} \in \mathrm{SO}(3)$. This implies that the angular momentum $\vec{x} \times \vec{p}$ is conserved as in Newtonian physics, and thus the trajectory is in a plane.

Since you know that the trajectory is in a plane, you can make the angular momentum pointing in $x^3$ direction, which means that $\vartheta=\pi/2=\text{const}$ for the usual spherical coordinates,
$$\vec{x}=R \begin{pmatrix} \cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \cos \vartheta \end{pmatrix},$$
Since you get from the isotropic coordinates to the Schwarzschild coordinates by only transforming $R$ to $r$, this also holds in Schwarzschild coordinates.

 

[PeterDonis]

I don't follow Wald's particular argument, i.e., that the symmetry $\theta\to \pi-\theta$ allows one to conclude that "if the initial position and tangent vector of a geodesic lie in the equatorial plane $\theta=\pi/2$, then the entire geodesic must lie in this plane.''

What am I missing?

The parity reflection symmetry in general means that we can take any solution to the geodesic equation and substitute $\theta \to \pi - \theta$ (and make all other changes that go along with this, such as changing derivatives with respect to $\theta$) to get another solution. In general this will yield a pair of distinct solutions.

However, the particular value $\theta = \pi / 2$ is a fixed point of the parity reflection symmetry, which means that the symmetry takes any solution to the geodesic equation whose initial conditions are entirely in the plane $\theta = \pi / 2$ into itself. That can only be the case if the solution lies entirely in that plane.

If your particular initial conditions don't pick out a direction of increase/decrease for $\theta$ the symmetry of the metric implies any argument for "$\theta$ will increase" must apply equally to "$\pi-\theta$ will increase", so such an argument must be self-contradictory?

If you add the qualifier that this argument only works for $\theta = \pi / 2$, then I think it amounts to the same argument I gave above.

 

[strangerep]

The parity reflection symmetry in general means that we can take any solution to the geodesic equation and substitute $\theta \to \pi - \theta$ (and make all other changes that go along with this, such as changing derivatives with respect to $\theta$) to get another solution. In general this will yield a pair of distinct solutions.

However, the particular value $\theta = \pi / 2$ is a fixed point of the parity reflection symmetry, which means that the symmetry takes any solution to the geodesic equation whose initial conditions are entirely in the plane $\theta = \pi / 2$ into itself. That can only be the case if the solution lies entirely in that plane.

Thank you! (And thanks also to @Ibix , @vanhees71 and @PeroK.)

I must be getting older than I realize.