B What path does matter take after entering a black hole?

[timmdeeg]

I equated the inner product of the spatial part of $U^\mu$ (the infaller's four velocity) and $R^\nu$ (a radial unit vector) with $\cos\psi$, forgetting that the spatial part of $U^\mu$ isn't normalised. The correct normalisation is $\sqrt{|1-g_{tt}U^tU^t|}$, and $\cos \psi$ is therefore $g_{rr}U^r/\sqrt{|1-g_{tt}U^tU^t|}$, or $$\cos\psi={{\sqrt{\left(E^2-1\right)r^3+R_Sr^2-L^2r+L^2R_S} }\over{r\sqrt{\left(E^2-1\right)r+R_S}}}$$

Setting $r=R_S$ I obtain $\cos\psi=R_S/R_S=1$ if I calculated correctly.
Hmm does this mean that your formula can be applied at the horizon? That shouldn't be possible in Schwarzschild coordinates. And "purely radial" ($\cos\psi=1$) would be in contrast to what @PeterDonis explained in #82.

 

[timmdeeg]

So the objects hit the floor almost vertically.

And how about the extrapolation to infinite rocket thrust, still "almost" and not exactly vertical?

 

[Ibix]

Could you please elaborate on the meaning of E and L and how ψ\psi is defined

$E$ and $L$ are the energy per unit mass at infinity and angular momentum per unit mass. There's more discussion in chapter 7 of Carroll's notes. $\psi$ is simply the angle between the path of a free-falling body and what a co-located hovering observer would call "straight down".

I fixed the LaTeX. Not sure what typo I introduced there.

Hmm does this mean that your formula can be applied at the horizon?

No. The underlying reasoning doesn't make sense. The maths just happens to be well-behaved. Formally you could substitute something like $r=R_S(1+rho)$ and let $rho$ get arbitrarily small, corresponding to approaching arbitrarily close to the horizon.

And "purely radial" (cosψ=1\cos\psi=1) would be in contrast to what @PeterDonis explained in #82.

Well, the point is that when the maths says $\cos\psi=1$ is the point at which the logic underlying it breaks down. So we don't ever expect the path to be radial. It's measured arbitrarily close to radial by hovering observers near the horizon, but that's in part a consequence of choosing hovering observers as our experimenters. Other coordinate choices would see other angles. For example, two free-falling observers with different $L$ who happen to pass each other at the horizon would move apart inside the event horizon, which would bot be the case if they were falling perfectly radially.

And how about the extrapolation to infinite rocket thrust, still "almost" and not exactly vertical?

Infinite thrust is nonsense, so the conclusions drawn would be meaningless.

 

[timmdeeg]

Well, the point is that when the maths says $\cos\psi=1$ is the point at which the logic underlying it breaks down. So we don't ever expect the path to be radial. It's measured arbitrarily close to radial by hovering observers near the horizon, but that's in part a consequence of choosing hovering observers as our experimenters.

Got it.

Other coordinate choices would see other angles. For example, two free-falling observers with different $L$ who happen to pass each other at the horizon would move apart inside the event horizon, which would bot be the case if they were falling perfectly radially.

Yes, understand.

Infinite thrust is nonsense, so the conclusions drawn would be meaningless.

Ok, same like horizon, hovering isn't possible.

Thanks for your explanations.

 

[jartsa]

And how about the extrapolation to infinite rocket thrust, still "almost" and not exactly vertical?

If you feel an infinite acceleration while hovering above a black hole, that means that your energy, parallel transported to a higher place, is zero.

If you now drop something to the black hole, the mass increase of the black hole, parallel transported to a higher place, is zero.

This can not work with a black hole with a finite size, its entropy would stay constant while entropic stuff is dropped into it.Hey how about if we say that as some dropped stuff approaches the event horizon, the event horizon becomes non-smooth, and what "vertical" means becomes complicated.

What happens if a massive spaceship tries to land on a smallish black hole like an airplane lands on a runway? The spaceship 'lands' on a moving bulge on the event horizon?

 

[PeterDonis]

If you feel an infinite acceleration

You can't. That's physically impossible. Everywhere above the horizon, the acceleration needed to hover is finite; and at the horizon, hovering is impossible.

 

[jartsa]

You can't. That's physically impossible. Everywhere above the horizon, the acceleration needed to hover is finite; and at the horizon, hovering is impossible.

Yes. I was trying to say that for various reasons it is impossible for the path of an infalling object to approach a vertical path as the object is approaching the event horizon. Like forexample, how can the black hole get the angular momentum that the original system had, if the rotational energy of the system approaches zero, as the system's rotational energy is being converted to system's non-rotational kinetic energy.

By non-rotational kinetic energy I mean the kinetic energy of two objects moving straight towards each other.

Of course 99.99999999999% of the horizontal kinetic energy of a potato thrown into a 10 solar mass black
hole being converted to vertical kinetic energy is not any problem. But 100% would be.

Of course this may be as unclear as my previous post. But saying something is infinite was avoided this time.

 

[PeterDonis]

the system's rotational energy is being converted to system's non-rotational kinetic energy

This doesn't happen. What gets converted to kinetic energy as the object falls is gravitational potential energy, not "rotational energy".

 

[timmdeeg]

$$\cos\psi={{\sqrt{\left(E^2-1\right)r^3+R_Sr^2-L^2r+L^2R_S} }\over{r\sqrt{\left(E^2-1\right)r+R_S}}}$$Differentiating with respect to $r$ gives us a mess: $${{2\left(E^2-1\right)L^2r^2+\left(4-3E^2\right)L^2R_Sr-2L^2R_S^2}\over{\sqrt{\left(E^2-1\right)R_S}\sqrt{\left(E^2-1\right)r^3+R_Sr^2-L^2r+L^2R_S}\left(2\left(E^2-1\right)r^3+2R_Sr^2\right) }}$$However it's definitely a negative mess for $r$ close to ##R_S# so infalling massive objects do have trajectories that get closer to radial in their final stages of infall.

Interestingly there is yet another possibility to define an angle between a radial and a non-radial path. From the metric of the rain frame one obtains
$\cos\psi_{shell}=-[1-(1-\frac{2M}{r})\frac{b^2}{r^2}]^{1/2}$ (*) where $\psi$ is the angle between a radial and a non-radial path as seen by a shell observer (hovering at $r$). The impact parameter $b$ is the "perpendicular distance between their initially parallel paths (at this great distance)".(*)

Approaching the singularity yields $\cos\psi_{rain}\longrightarrow(\frac{r}{2M})^{1/2}$ as $r\longrightarrow 0$ (*), meaning that all stars except outward radially are seen at 90 degrees.

(*) Taylor&Wheeler [Exploring Black Holes]