# Entering a black hole, does the incident angle matter at all?

Inside the black hole (as you enter the EH), all objects (massive or massless) must move towards the singularity. The singularity becomes a moment in future.

In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it. In general relativity, an object in free fall is subject to no force and is an inertial body moving along a geodesic.

https://en.wikipedia.org/wiki/Free_fall

Inside the EH, if no other force acts on the objects, as per GR they should be in free fall.

Now if they are in free fall, they must follow geodesics.

I have not found any visual description of any objects entering the EH, and their trajectory afterwards, except this one:

And this one with the map:

Now this is where it needs clarification, what will be the trajectory of an object (massive or massless) moving towards the singularity, does it have to follow the shortest path (geodesic)?

So basically when an object enters the EH it could either follow the geodesic path or the non-geodesic path.

There are two cases:

1. As per GR, it should follow the geodesic path, but that looks weird, in a way because it includes an effect that seems something like changing the angle (at the EH) towards the singularity.

2. It follows the more curved (non-geodesic) path, but that is not the shortest way towards the singularity.

Question:

1. Which way will the object (massive or massless) follow entering the EH?

Your picture is wrong, there is no abrupt change in the angle. A particle approaching the black hole with an angle will have a trajectory like this (the simulation was done in Kerr Schild coordinates, but it looks the same in Raindrop and Droste coordinates):

r can only decrease, but that does not mean that the motion has to be purely radial. Angular momentum is conserved.

• thank you. Are all objects inside the EH in freefall? Do all objects have to follow geodesics inside the EH or not? Jan 26 '20 at 3:01
• A rocket ship with its engines on, inside or outside the horizon, is not in freefall and does not follow a geodesic. Freefall means zero nongravitational force. You can have radial freefall or nonradial freefall. Jan 26 '20 at 6:25
• A geodesic is not the shortest path in space but the path with the longest proper time between two events Jan 26 '20 at 16:46
• I didn't talk about Kruskal coordinates, I was talking about Kerr Schild, Raindrop and Droste coordinates. If you think that my simulation is wrong feel free to create your own, and if you want to show us the trajectory in other coordinates just do so. Jan 26 '20 at 18:12
• @BenCrowell Respectfully, I think you're being too pedantic here. I don't see how you think that the claim "it looks the same in Raindrop and Droste coordinates" "is meant to imply that the trajectory will look the same regardless of coordinates". Even if there's no preferred time slicing to project down to, I think that the (coordinate-dependent) projection shown here still gives a useful answer to the OP's question, even if not a unique one. (BTW, I upvoted your answer as well.) Jan 27 '20 at 14:18

Your picture is wrong because it depicts the black hole as a circle with the singularity as a point at the center. The singularity is a spacelike surface that lies in the future of every observer. See Is a black hole singularity a single point? .

The naive concept of a trajectory doesn't make much sense on the interior of a black hole, because the interior is not static. For this reason, we can't define things like proper distances on the interior. See What's the proper distance from the event horizon to the singularity? . If you can't define distances, you clearly can't define a trajectory.

You can certainly pick a coordinate chart that is well behaved (meaning not Schwarzschild coordinates) and plot graphs on those, but that's not showing you a trajectory in the sense of a geometrical shape in the Euclidean plane. For example, if you have one timelike coordinate and three spacelike coordinates, then, depending on the motion and the choice of coordinates, the graph of an infalling particle's motion could be simply a dot when you project out the time coordinate. That is, the spacelike coordinates could all be constant. Yes, the graph will depend on the angle of the infalling object.

• Comments are not for extended discussion; this conversation has been moved to chat. Jan 28 '20 at 10:28

which way will the object (massive or massless) follow entering the EH?

A really good question. Intuition may be misleading here.

Any object is crossing the event horizon on an almost radial path.

Imagine you are sitting in a rocket and feel very high acceleration. If you throw something horizontally, the thing will fall down and hit the floor almost vertically.

For the same reason, an observer very close to the horizon (who needs tremendous rocket thrust to keep a stationary position) will see an infalling object crossing the horizon almost vertically (i.e. almost radially). So the pictures aren't correct.

How does that look like in Schwarzschild coordinates? Based on Carroll's Notes (Chapter 7) one obtains: $$\cos\psi={{\sqrt{\left(E^2-1\right)r^3+R_Sr^2-L^2r+L^2R_S} }\over{r\sqrt{\left(E^2-1\right)r+R_S}}}$$ This shows that $$\cos\psi$$ approaches $$1$$ if $$r$$ approaches $$R_S$$.
Here $$E$$ and $$L$$ are energy at infinity and angular momentum, each per unit mass. $$\psi$$ is the angle between the path of an infalling object and "vertical" from the viewpoint of an observer close to the horizon.

Setting $$r$$=$$R_S$$ yields $$\cos\psi=1$$. However, this can't be understood as a prediction of the above formula because of the coordinate singularity at the horizon. So the trajectory at the horizon is almost, but not exactly, radial.

The Schwarzschild metric isn't applicable regarding motion inside the horizon. Here one can use the Rain Frame metric without infinities at $$r = R_S$$, which is given by

$$d\tau^2=(1-2M/r)dt^2-2(2M/r)dtdr-dr^2-dr^2-r^2d\psi^2$$ and which yields the motion of a rain drop inside the horizon (a bit complicated, search the web).

Infalling light-like geodesics will also be bent towards radial but presumably to a lesser extent.

• I don't think this is right. You're talking about a trajectory as seen by a static observer, but the OP is asking about trajectories inside the horizon, where there are no static observers. An object anywhere in the Schwarzschild spacetime can be moving in any direction whatsoever. By the equivalence principle, spacetime is locally Minkowski, so it has the full symmetry of Minkowski space, including the existence of geodesics having tangents in all directions inside the future light cone.
– user4552
Jan 26 '20 at 17:15
• My answer concerns "entering the EH". I fully agree with what you said regarding the trajectory inside the black hole.
– timm
Jan 26 '20 at 21:53
• @Ben Crowell:I have improved my answer and +1 for your description of the interior.
– timm
Jan 27 '20 at 9:19

what will be the trajectory of an object (massive or massless) moving towards the singularity, does it have to follow the shortest path (geodesic)?

Seems there is confusion about what "geodesic" means. In general, a geodesic is a path that minimizes or maximizes a specific quantity$$^1$$. For example, Fermat's principle states that light moves from one point to another in such a way that the needed time is minimal.

Similarly, General Relativity can be expressed in terms of geodesics, but you have to be careful what geodesics to use!

Take for example a point $$A$$ on the surface of earth and a point $$B$$ one meter beneath it. Then the shortest path connecting them is a straight line, i.e. a line in Euclidean$$^2$$ space.

Now consider a free-falling body that starts at $$t=0$$ at $$A$$ and ends up 1 second later at $$B$$, i.e. we search for a path that connects $$(A,t=0)$$ with $$(B,t=1)$$ which is the trajectory taken by the free-falling object. This path is no more in familiar 3-space, it is in some mathematical space, spacetime for short, and it's characterized by proper time along the path being maximized: For all observers moving from $$(A,0)$$ to $$(B,1)$$, the observer that ends up with the maximal clock reading on his clock is the one that is free-falling. And this motion is not along a straight line in 3-space: Time passes slower due to gravity (closer to earth in this case), hence the object can speed up its clock by moving up. However, when moving (up) time dilation also plays a role, and the best compromise to maximize proper time is the known parabola of free fall from classical mechanics.

$$^1$$ The mathematical definition is more sophisticated.

$$^2$$ In General Relativity, 3-space is no more Euclidean in general, but for all practical purposes space can be considered to be Euclidean (flat) around the earth or inside of earth.

• Thank you so much. Are you saying that the non radial freefall is a geodesic too? Jan 31 '20 at 16:07
• Non-radial fall is geodesic, too, provided there are no non-gravitational forces (all forces in General Relativity are non-gravitational because gravity is modeled as spacetime curvature). It's just like free fall on earth's surface is geodesic (ignoring air resistance) and free-falling around the earth is geodesic. From a specific point, there are geodesics in all directions because an object can have speed pointing in any direction an still be free-falling. Jan 31 '20 at 16:28