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

[timmdeeg]

Yes sure. There's a downwards force, causing an increase of downwards speed, and a decrease of horizontal speed. It's the time dilation effect. Or the the effect that sometimes in relativity acceleration points to other direction than force.

Ok, thanks.

 

[PeroK]

Yes sure. There's a downwards force, causing an increase of downwards speed, and a decrease of horizontal speed. It's the time dilation effect. Or the the effect that sometimes in relativity acceleration points to other direction than force.

If we are interested about momentum, downwards momentum increases, horizontal momentum stays constant.

(I'm not saying anything about the "within the black hole" -part)

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

 

[Ibix]

I seem to be wrong what "orbit" means, I thought orbit means circling around.

I certainly use the word to describe anything like planets' and comets' paths, which can be open or closed. Newtonian physics doesn't have the concept of a "must crash" trajectory except for a purely radial one, so I think "must crash" trajectories in GR are called orbits as a natural generalisation. I could be wrong.

I am talking about an object on a non-radial trajectory falling through the event horizon.

Edit: there's an error in the following - see #89 for the correction.
For a particle with four-velocity $U^\mu$, the cosine of the angle its path makes with a radial inward four vector $R^\nu$ is $g_{\mu\nu}U^\mu R^\nu$ (strictly I should restrict myself to summing over the spatial components, but that turns out not to matter here). In Schwarzschild coordinates, the metric is diagonal and the only non-zero component of $R^\nu$ is $R^r=-1/\sqrt{g_{rr}}$, which means that $\cos\psi=-\sqrt{g_{rr}}U^r$, where $\psi$ is the angle between the path and the radial direction.

For a free-falling particle on an inward path, Carroll's GR lecture notes 7.47 and 7.48 tell us that $$U^r=-\sqrt{E^2-\left(1-\frac {R_S}r\right)\left(1+\frac{L^2}{r^2}\right)}$$which means that the angle $\psi$ the path makes with the radial-inwards direction is given by$$\cos\psi=\sqrt{{{E^2}\over{1-{{R_S}\over{r}}}}-{{L^2}\over{r^2}}-1}$$The derivative of this with respect to $r$ is$${{\sqrt{R_S-r}\left(2L^2R_S^2+\left(-4rL^2-r^3E^2\right)R_S+2r^2L^2\right)}\over{\sqrt{rL^2-\left(L^2+r^2\right)R_S-r^3E^2+r^3}\left(2r^2R_S^2-4r^3R_S+2r^4\right)}}$$Evaluating this at $r=3R_S/2$, the limit below which you cannot dip and recover, gives us$${{2\left(4L^2-27E^2R_S^2\right)}\over{9R_S^2\sqrt{27E^2R_S^2-4L^2-9R_S^2}}}$$If this is positive, falling $r$ means falling $\cos\psi$ and hence increasing $\psi$ - so the path is getting further away from radial. If it is negative then the path is getting closer to radial. But note the denominator - it is only real if $27E^2R_S^2-4L^2>9R_S^2$, which implies that the numerator is negative.

So, yes, if my maths is correct, paths that strike the black hole are getting closer to radial, at least as they cross $3R_S/2$. As noted earlier, this is not true of paths that do not strike the black hole.

 

[Ibix]

And is the intuition correct that an increasing non-radial worldline within the black hole will take increasing proper time to reach the singularity?

I missed this earlier - no, it's not true. The maximal survival time from striking the black hole comes from being dropped from rest just above the horizon. That is, accelerating to get as close to the path of a radially free-falling object coming from just above the horizon maximises your survival time.

 

[Ibix]

Yes sure. There's a downwards force, causing an increase of downwards speed, and a decrease of horizontal speed. It's the time dilation effect. Or the the effect that sometimes in relativity acceleration points to other direction than force.

If we are interested about momentum, downwards momentum increases, horizontal momentum stays constant.

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.

 

[timmdeeg]

Thanks for investing so much time.

For a free-falling particle on an inward path, Carroll's GR lecture notes 7.47 and 7.48 tell us that $$U^r=-\sqrt{E^2-\left(1-\frac {R_S}r\right)\left(1+\frac{L^2}{r^2}\right)}$$which means that the angle $\psi$ the path makes with the radial-inwards direction is given by$$\cos\psi=\sqrt{{{E^2}\over{1-{{R_S}\over{r}}}}-{{L^2}\over{r^2}}-1}$$
...
So, yes, if my maths is correct, paths that strike the black hole are getting closer to radial, at least as they cross $3R_S/2$. As noted earlier, this is not true of paths that do not strike the black hole.

I am trying to see what happens at $r=R_S$. Here $1-R_S/r$ approaches zero. So it seems that $\cos\psi$ approaches a maximum and thus $\psi$ approaches zero. (however $\psi=0$ requires $\cos\psi=1$). Does that mean a radial path at the event horizon regardless any arbitrary initial non-radial path?

 

[PeterDonis]

I was assuming that the trajectory of a non-radially infalling object is bent towards the center of mass

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

 

[Ibix]

Does that mean a radial path at the event horizon regardless any arbitrary initial non-radial path?

That means those coordinates don't work at the horizon. I'd have to think a bit about exactly what "radial" means at the horizon.

 

[PeterDonis]

For a free-falling particle on an inward path, Carroll's GR lecture notes 7.47 and 7.48 tell us that

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.

 

[PeterDonis]

I am trying to see what happens at $r=R_S$.

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.

 

[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