What path does matter take after entering a black hole?

[PeterDonis]

I'd have to think a bit about exactly what "radial" means at the horizon.

The simplest way is to use coordinates for which $r$ is always spacelike, such as Painleve coordinates. Then $\partial / \partial r$ always defines the outward radial direction.

 

[kent davidge]

The simplest way is to use coordinates for which $r$ is always spacelike, such as Painleve coordinates. Then $\partial / \partial r$ always defines the outward radial direction.

Only if you first define outwards to be positive, correct?

 

[PeterDonis]

Only if you first define outwards to be positive, correct?

The definition of the $r$ coordinate already takes care of that: larger values of $r$ are further outward.

 

[Ibix]

This is the effective potential formulation, and is only valid outside the horizon. It doesn't help for the question under discussion here, which is to describe what happens to an object that falls through the horizon.

Of course - potential isn't defined inside the horizon. But it works fine for something approaching the horizon, which seemed to be the topic of discussion, at least in part of this thread:

Will a non-radial path be deflected towards radial (though not reaching radial at the horizon) with decreasing distance to the horizon?

 

[jartsa]

This certainly isn't generally correct, as discussed in #57. And I'm not sure what you mean by "horizontal momentum". The angular momentum, $L$, is a constant of the motion. The component of linear momentum in the tangential direction ($\propto L/r$) is not.

Well, I saw a question about infalling objects' paths approaching vertical. Then I thought about a guy shining a flashlight towards the wall of his spaceship, the motors of which are blasting like crazy. The light hits the floor instead of the wall.

As the power of the motors approaches infinity the path of the light approaches vertical. Seems like that answers the question that was asked, if we remember that the closer to the black hole an observer is hovering, the more proper thrust the engines of his spaceship must generate.

 

[Ibix]

The simplest way is to use coordinates for which $r$ is always spacelike, such as Painleve coordinates. Then $\partial / \partial r$ always defines the outward radial direction.

The problem I was thinking of was that outside the horizon you can define "radial" as perpendicular to the three Killing fields - moving in that direction puts you on a smaller/larger nested sphere. But the "$\partial_t$" KVF is null on the horizon and spacelike inside, and we don't have nested spheres any more - the nesting happens into the future inside and the horizon is a single surface anyway.

What you are doing using Painleve coordinates is just having the infaller point radially, then keep pointing in the same direction as he falls through the horizon, I think. Which is fine, but the direction he's pointing isn't really a radial one anymore, right?

 

[Ibix]

Well, I saw a question about infalling objects' paths approaching vertical. Then I thought about a guy shining a flashlight towards the wall of his spaceship, the motors of which are blasting like crazy. The light hits the floor instead of the wall.

As the power of the motors approaches infinity the path of the light approaches vertical. Seems like that answers the question that was asked, if we remember that the closer to the black hole an observer is hovering, the more proper thrust the engines of his spaceship must generate.

Fair enough - but there are quite a lot of caveats to that. You are specifically launching light almost tangentially from very close to the black hole (in fact, from within the $3R_S/2$ limit) and it's not at all clear that your claim is generally true if you launch at another angle, although it seems likely (note that my calculation above was done for a massive test object, not light). It's certainly not true in general if you launch from above $3R_S/2$. And I think your statement about "horizontal momentum" makes sense in the context that your spaceship floor is flat - implying a large enough black hole that this can be true (which leads to $L$ is conserved, $L/r$ is not, but $r\simeq\text{const}$ so $L/r$ is approximately conserved too).

 

[timmdeeg]

And why were you assuming that? Where would this magical influence that somehow changes the trajectory of the object come from?

Aren't things in free fall towards a mass attracted by the center of mass and hence are changing their trajectory accordingly (in case their initial trajectory isn't radial)?

 

[timmdeeg]

You can't from the equations you are looking at. Those equations are only valid outside the horizon. See my response to @Ibix just now.

Therefor I said "approaches". I understand @Ibix' formula such that $\psi$ approaches zero with decreasing $r$, whereby $r>R_S$.

 

[PeterDonis]

it works fine for something approaching the horizon, which seemed to be the topic of discussion, at least in part of this thread

For the part outside the horizon, yes, your math will be applicable.

 

[PeterDonis]

The problem I was thinking of was that outside the horizon you can define "radial" as perpendicular to the three Killing fields

More precisely, four Killing fields (3 for SO(3), the symmetry group of the 2-sphere, plus the extra one that must be there by Birkhoff's Theorem), one of which is timelike. What seems to be giving you pause is the fact that the fourth Killing field is no longer timelike at or below the horizon, so the "radial" vector orthogonal to it is no longer spacelike. That's true.

What you are doing using Painleve coordinates is just having the infaller point radially, then keep pointing in the same direction as he falls through the horizon, I think. Which is fine, but the direction he's pointing isn't really a radial one anymore, right?

Sure it is; it's just not a "radial" vector that's orthogonal to the fourth Killing field any more. If you want a "radial" vector to be spacelike, you have to drop that orthogonality condition anyway at or inside the horizon. But it will still be radial in the sense of being orthogonal to the 2-spheres; at any point there is a range of spacelike vectors that are orthogonal to the 2-spheres and point outward. To see why, pick any local inertial frame and decompose it into two orthogonal 2-surfaces: one tangent to the 2-sphere at the frame's origin and the other orthogonal to the 2-sphere. The second 2-surface can be described, within the local inertial frame, using a standard Minkowski 2-d spacetime diagram, and any spacelike vector pointing in the positive $x$ direction will be pointing radially outward.

 

[PeterDonis]

Aren't things in free fall towards a mass attracted by the center of mass and hence are changing their trajectory accordingly (in case their initial trajectory isn't radial)?

"Changing their trajectory" is a very poor choice of terminology. The geodesic determined by the initial position and velocity of the object is its trajectory. It doesn't "change"; it's already a fully determined curve in 4-d spacetime as soon as you pick the initial conditions. That's true above the horizon, at the horizon, and below the horizon.

There are particular properties of the curve that can change as you move along the curve, such as the angle @Ibix described. But none of that "changes" the curve itself.

Nor is it true that multiple different trajectories--curves in 4-d spacetime--somehow "merge" into one at the horizon. Different initial conditions lead to different geodesics; the mapping is one to one. So if you pick two different test objects and launch them from the same altitude above the horizon but with different tangential velocities, assuming neither one has enough angular momentum to prevent it from falling through the horizon, the trajectories of those two objects will be different, even though both of them will have @Ibix's angle "changing" as they fall. They will not become the same at the horizon; they will not become "purely radial" at the horizon.

 

[timmdeeg]

"Changing their trajectory" is a very poor choice of terminology. The geodesic determined by the initial position and velocity of the object is its trajectory. It doesn't "change"; it's already a fully determined curve in 4-d spacetime as soon as you pick the initial conditions. That's true above the horizon, at the horizon, and below the horizon.

Got it and thanks for being so accurate. A trajectory of a free fall object is predetermined and what eventually changes is the angle (not the trajectory itself as I meant wrongly).

Nor is it true that multiple different trajectories--curves in 4-d spacetime--somehow "merge" into one at the horizon. Different initial conditions lead to different geodesics; the mapping is one to one. So if you pick two different test objects and launch them from the same altitude above the horizon but with different tangential velocities, assuming neither one has enough angular momentum to prevent it from falling through the horizon, the trajectories of those two objects will be different, even though both of them will have @Ibix's angle "changing" as they fall. They will not become the same at the horizon; they will not become "purely radial" at the horizon.

Thanks for this excellent explanation.

A question regarding the limiting case a formula may yield. An example is the radial velocity at the horizon which is c "as a limiting case" [Exploring Black Holes, Taylor&Wheeler]. Does the formula $\cos\psi=\sqrt{{{E^2}\over{1-{{R_S}\over{r}}}}-{{L^2}\over{r^2}}-1}$ @Ibix has shown predict a limiting case at the horizon? Is "not purely radial" the limiting case? Can one say it predicts arbitrary close to the horizon an angle which is close to zero but its value depends on $E$ and $L$ (the properties of the trajectory)? And is it correct that the value of $\cos\psi$ diverges at the horizon?

How does this formula look like for a radial path (L = 0 ? but this is just guessing).

 

[Orodruin]

This is the effective potential formulation, and is only valid outside the horizon. It doesn't help for the question under discussion here, which is to describe what happens to an object that falls through the horizon.

This is not entirely true. While Carroll does work in Schwarzschild coordinates, the exact same result (equation 7.48) can be derived from working in Eddington-Finklestein coordinates with $R = r$, where the metric is not singular at $R = R_S$.

 

[PeterDonis]

Does the formula $\cos\psi=\sqrt{{{E^2}\over{1-{{R_S}\over{r}}}}-{{L^2}\over{r^2}}-1}$ @Ibix has shown predict a limiting case at the horizon?

No, because the angle that formula is describing is an angle with respect to stationary observers (observers hovering at a constant altitude above the horizon), and there are no stationary observers at or below the horizon.

Is "not purely radial" the limiting case?

No. An object with a tangential 4-velocity component above the horizon still has it at or below the horizon. It never disappears.

 

[timmdeeg]

No, because the angle that formula is describing is an angle with respect to stationary observers (observers hovering at a constant altitude above the horizon), and there are no stationary observers at or below the horizon.

No. An object with a tangential 4-velocity component above the horizon still has it at or below the horizon. It never disappears.

Ok, thanks for clarifying.

 

[Ibix]

And is it correct that the value of $\cos\psi$ diverges at the horizon?

No. I normalised the four velocity, not the three velocity. The expression is something like $\gamma\cos\psi$, where $\gamma$ is the Lorentz gamma factor of the infaller as measured by a hovering observer.



Dinner time - will post the correct calculation later.

 

[Ibix]

Ok - correcting #63:

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 $$\cos\psi={{\sqrt{\left(E^2-1\right)r^3+R_Sr^2-L^2r+L^2R_S} }\over{r\sqrt{\left(E^2-1\right)r+R_S}}}$$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.

 

[jartsa]

This looks like an analysis of Newtonian gravity with some SR concepts thrown in.

Yeah. If we think about how strong gravity field affects falling things, then it seems clear what happens: Same thing as in an spaceship with really strong rocket motors. Thrown things hit the floor with high downwards speed. So the objects hit the floor almost vertically.

Hovering observers near a black hole say that there's a strong gravity field here, and things fall accordingly. A distant observer can't disagree much.

 

[timmdeeg]

@Ibix Could you please elaborate on the meaning of E and L and how $\psi$ is defined? Also please make
$${{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+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) }}
readable.

Thanks

 

[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]