A geometric argument for geodesics in manifolds also being geodesics in hypersurafces?

Question

I've read that when a hypersurface within a manifold contains a curve, if the curve is a geodesic in the manifold, it is also a geodesic in the hypersurface.

This is quite abstract for me, I've only recently started learning GR, could someone provide a geometric (or if necessary, algebraic) argument for why this is true? Is the converse ever/always/never true?

(Based on my limited intuition, I'd guess sometimes but not always, but I can't think of anything a professor wouldn't call 'handwavey').

Cheers

Answer 1

Ehlers (1961) in the classic paper "Contributions to the relativistic mechanics of continuous media" states:

A vortex-free flow is rigid precisely when its orthogonal hypersurfaces are totally geodesic.

(1993 translation). As for "flow", this need not be an actual fluid, but simply a 4-velocity field. Background fact: hypersurfaces orthogonal to the 4-velocities exist if and only if the motion is vorticity-free (this follows from Frobenius' theorem). Totally geodesic means every geodesic in the hypersurface is also a geodesic on the full manifold.

For example, think of static observers in Schwarzschild spacetime (the ones with fixed $r>2M$, $\theta$, and $\phi$ coordinates; with 4-velocities proportional to the Killing vector field which is timelike at infinity). This velocity field is rigid, and the orthogonal hypersurfaces have $t=\textrm{const}$, and these must be totally geodesic. As another example, consider observers which "fall from rest at infinity". This is not a rigid velocity field, because tidal forces lead to shear and expansion, and it turns out the radial direction within the hypersurfaces is not a geodesic within the manifold. A final example, consider a homogeneous and isotropic universe, with the velocity field "comoving" with the average local matter flow. This is not rigid in an expanding universe, hence the hypersurfaces of constant cosmic time are not totally geodesic! This surprised me when I first realised it.

This response probably does not address everything in your question, but I hope it helps.

Answer 2

Easy to prove when you find the right theorem.

In the GR background, let's consider $g_{ab}$ in orthonormal basis being (-+++) signature.

Theorem 1. Consider a hypersurface $Σ$ in manifold $(M,g_{ab})$, with unit normal vector field on it denoted as $n^a$, then it's induced with a natural metric $h_{ab}$ on $Σ$:

$$ h_{ab}=g_{ab}+n_an_b $$

Theorem 2. Relation between the covariant derivative $∇_c$ of $M$ and that of the hypersurface $\mathrm D_c$: Consider acting on any arbitrary tensor field $T^{a_1\cdots a_k}{}_{b_1\cdots b_l }$ on $Σ$.

$$ \mathrm D_cT^{a_1\cdots a_k}{}_{b_1\cdots b_l }=h^{a_1}{}_{d_1}\cdots h^{a_k}{}_{d_k} h_{b_1}{}^{e_1}\cdots h_{b_l}{}^{e_l} h_c{}^f ∇_f T^{d_1\cdots d_k}{}_{e_1\cdots e_l }. $$

Now let's prove the geodesic w.r.t $M$ must be also geodesic w.r.t $Σ$. Set tangent vector as $w^a$. Then we have $w^b∇_bw^a=0$. Let's check it $\Rightarrow$ $w^bD_bw^a=0$:

$$ w^b\mathrm D_bw^a= {h^a}_α\underline{{h^β}_bw^b}∇_βw^α={h^a}_α\underline{w^β∇_βw^α}=0 $$

in which the second step using $w^b$ stay unchanged after $h^β{}_{b}$ projection. Q.E.D.