# Slingshot around a black hole, dipping into event horizon

Suppose we have an object swinging by a black hole. My understanding is that if it remains outside the black hole, it just swings by and doesn't lose any energy (except maybe gravitational waves?). But if it enters the event horizon, it's gone forever. i.e. the blue trajectory in the picture is possible, but the red one isn't.

How does this work? How does the object on the red trajectory lose its kinetic energy? Is it an "all or nothing" thing where the object will retain all its energy unless it dips into the event horizon and then lose it all? Or does the object gradually lose more energy as it gets closer to the event horizon?

## Edit:

Thank you to the respondents. Summary (if I've understood correctly - please correct me if not) is:

• In Newtonian physics, a small object orbiting a big object without losing energy will follow either an ellipse or hyperbola, depending on the speed. What goes down, must come up. In Relativity, this is not True - objects follow Schwarzchild geodesics - and the difference from Newtonian orbits becomes larger as we get very close to very massive objects. Neither the "red" nor the "blue" object in the below image loses its energy - in relativity, objects can spiral in without losing energy.

• Re: The all-or-nothing thing. There exists a sphere (the Photon Sphere - dotted line on below image) outside the event-horizon, where any photon entering it will eventually spiral into the black hole. So at least for massless objects, if the object dips into the photon sphere, it's in for good.

Image of Schwartzchild geodesics near a black hole from General-Relativistic Visualization.

• What happens to the red object after falling past the event horizon depends on who's watching. For an outside observer, the red object will just park at the event horizon and hang out there forever. For the observer aboard the red object, it will just cross the event horizon and hit the singularity in a finite (and small) time. After that, the rules don't say what happens.

• There appears to be some controversy on that last point (at least in the answers below). Some seem to argue that an observer on the red object ceases to exist once it hits the event horizon, rather than the singularity (i.e. the left image above is also true for the observer on the red trajectory). It may be a moot point since there's no way to verify this.

• Veritasium explains what happens as light passes near a black hole in How to Understand the Black Hole Image. youtube.com/watch?v=zUyH3XhpLTo&feature=youtu.be – mmesser314 Dec 23 '19 at 3:12
• How is it different from a particle around the Earth? – safesphere Dec 23 '19 at 4:11
• BTW, the Schwarzschild coordinates do not directly correspond to the perceptions of a single observer. From en.wikipedia.org/wiki/… A Schwarzschild observer is a far observer or a bookkeeper. He does not directly make measurements of events that occur in different places. Instead, he is far away from the black hole and the events. Observers local to the events are enlisted to make measurements and send the results to him. The bookkeeper gathers and combines the reports from various places. – PM 2Ring Dec 23 '19 at 23:16
• Schwarzschild coordinates have a coordinate singularity at the event horizon, but mathematically (& physically) that's quite different to the singularity that standard GR says occurs at the core of a Schwarzschild black hole. A coord singularity is removeable, in the sense that you can eliminate it simply by using different coords, eg Gullstrand-Painlevé. It's like how things get a bit weird in standard latitude & longitude coords on Earth near the North & South Poles. It just seems weirder with black holes because time is involved. – PM 2Ring Mar 27 at 17:01
• You can use analytical continuation to extend Schwarzschild coords to the interior of the event horizon, but they become kind of meaningless to our distant bookkeeper. Time at the horizon is in the bookkeeper's infinitely distant future, so beyond the horizon is even further into the future than infinitely far. ;) John Rennie briefly touches on this topic in this answer. If you want to discuss this further, you can do that in The h Bar, the main Physics.SE chat room. – PM 2Ring Mar 27 at 17:09

In a plain old Newtonian context, there are some things in your question that show some incorrect understanding. Kinetic energy isn't conserved, it's total energy that's conserved. It's not true in general, for motion under the influence of a central force, that conservation of energy forbids a collision with the origin. That depends on how the force varies with distance.

Re general relativity, Wikipedia has a nice article on the orbits of test particles in the Schwarzschild spacetime: https://en.wikipedia.org/wiki/Schwarzschild_geodesics . Another good reference is the book Exploring Black Holes by Taylor and Wheeler.

A test particle in the Schwarzschild spacetime has a conserved energy and a conserved angular momentum. These are conserved both inside and outside the horizon. The energy doesn't have a nice interpretation in terms of a split into kinetic and potential terms.

As a particle infalls past the event horizon and approaches the singularity, its energy and angular momentum remain constant. The singularity represents the end of time in this spacetime, so the particle's energy and angular momentum are literally conserved until the end of time. If these remarks about the end of time don't make sense to you, I would suggest you learn how to interpret Penrose diagrams. I have a simple nonmathematical presentation of Penrose diagrams in my book Relativity for Poets, which is free online: http://www.lightandmatter.com/poets/

• "As a particle infalls past the event horizon" - This never happens in the coordinate system of the external observer implied in this question. And this system does exist in GR and extends arbitrarily close to the horizon. – safesphere Dec 23 '19 at 4:05
• @safesphere: Coordinate systems don't correspond to observers. See physics.stackexchange.com/questions/458854/… . And in any case I don't see the relevance of your comment to the topic of discussion. – user4552 Dec 23 '19 at 14:35
• @Peter See this post by two accomplished mathematicians (Indiana University): physicalprinciples.wordpress.com/2016/07/29/… - And here is the actual publication (p. 43): newton.ac.uk/files/preprints/ni14098.pdf – safesphere Dec 23 '19 at 16:52
• @Peter: Everything you said in your comment sounds right to me except the following: An object falling towards the singularity continues to do so until the end of time. Stated this way, it makes it sound as if the object takes infinite time to get there. It takes finite proper time to get there. For a solar-mass black hole, the time is on the order of milliseconds. – user4552 Dec 23 '19 at 23:11
• @safesphere: This is a point of contraversy. No, it is not controversial. – user4552 Dec 23 '19 at 23:12

As long as the object is outside of the event horizon it will not lose kinetic energy (disregarding any gravitational wave phenomena). You can think of this case as an elastic scattering of two bodies.

If the object falls inside the event horizon, it will become a part of black hole and will not go out. Let's assume that the object has no charge and falls straight into the black hole so that it doesn't add any angular momentum to the black hole. The falling object will increase the size of the black hole according to the first law of black hole thermodynamics:

$$d E = \frac{1}{32 \pi M} d A$$

where $$d E$$ is the change in energy and $$dA$$ is the change in the horizon area of the black hole. Hence, in both cases energy will be conserved. The extra kinetic energy of the object will be compensated by an increase in the gravitational potential of the black hole.

• "If the object falls inside the event horizon" - As observed by whom? Or more precisely, in the coordinate system of which physical observer is this statement made and what are the spacetime coordinates of this event in this system? – safesphere Dec 23 '19 at 3:41