Tidal forces inside BH in horizontal direction

1. Feb 8, 2010

Dmitry67

Spaghettification outside of a BH rips apart objects in vertical direction, but compresses them in horizontal. Inside the BH, tidal forces rips objects apart in ALL directions - at least I was thinking so. But recently I read:

http://arxiv1.library.cornell.edu/abs/0903.4717 [Broken]
The edge of locality: visualizing a black hole from the inside

Figure 9 shows that as the observer approaches the singularity, the view above and below the observer (small affine distances) becomes highly redshifted, while the view in the horizontal direction (large affine distances) becomes highly blueshifted. The effect is caused by the enormous tidal force near the singularity. Looking up, the observer feels an upward gravitational force pulling their head off. Looking down, the observer feels a downward gravitational force pulling their feet off. The up and down gravitational forces away from the observer cause the view above and below to appear gravitationally redshifted. The same tidal force that pulls the observer apart vertically also compresses them horizontally. The horizontal tidal compression concentrates and blueshifts the view in the horizontal direction.

Is bold correct or not inside the black hole?

Last edited by a moderator: May 4, 2017
2. Feb 8, 2010

bcrowell

Staff Emeritus
It's correct both inside and outside the horizon. If you translate the vacuum field equations into the physics they represent, they essentially say that the volume V of a swarm of test particles obeys $d^2V/dt^2=0$. If the swarm is initially at rest in some frame, then this implies that the volume remains constant. This is the defining characteristic of vacuum solutions: they are volume-conserving. The vacuum field equations hold both inside and outside the horizon, so the horizontal compression and vertical elongation holds both inside and outside. An easy way to see this is that an observer free-falling through the event horizon does not detect any sudden change in the *local* properties of spacetime.

3. Feb 8, 2010

Dmitry67

When you are talking about the volume, what coordinates do you use?

Then I am confused. On diagrams I see that in Finkelstein's coordinates lightcones become narrower and narrower when they approach singularity:

http://www.valdostamuseum.org/hamsmith/DFblackIn.gif [Broken]
http://www.etsu.edu/physics/plntrm/relat/eventho2.gif

Lets follow the trajectories of radially falling particles inside the BH. As lightcones become narrower, sooner or later any particles lose vision of each other. This is an indication of very strong horizontal tidal forces, ripping matter apart and creating horizons.

I can obtain the same in Kruskal–Szekeres coordinates:

If you rotete this diagram over Kruskal R=0 axis forming 3D body, and plot trajectories of infalling particles, you will see. Kruskal coordinates preserves the angle of the lightcone, but there is more and more "space", so sooner or later any horizontally separated particles dissapear behind the apparent horizons.

Last edited by a moderator: May 4, 2017
4. Feb 8, 2010

bcrowell

Staff Emeritus
Locally Minkowski coordinates. (If you want to integrate volume in coordinates that aren't locally Minkowski, you have to throw in a factor of something like $\sqrt{-det(g)}$.)

Anyway, without worrying about all the technical details, I think there's a very easy way to see that the answer to your question is that spaghettification works the same way on both sides of the horizon. Spaghettification is basically expressing a locally observable property of spacetime. If a locally observable property of spacetime were to change discontinuously from one side of the horizon to the other, then the horizon would have to be a real singularity. But it's not a real singularity, it's a coordinate singularity.

In slightly more technical language, the tidal forces that lead to spaghettification are expressed by a certain part of the Riemann tensor. They can be expressed as a tensor built out of the Riemann tensor and its derivatives. (I think what you'd want is actually the Weyl tensor, or some part of it.) If it had discontinuous behavior across the horizon, then that would require that either the Riemann tensor or its derivatives be discontinuous across the horizon, and that the discontinuity be one that could not be removed by a change of coordinates. But of course we know that that isn't the case.

5. Feb 9, 2010

Dmitry67

No, I dont think they are discontinious: horizontal tidal forces are 0 at the horizon.
Also, whats wrong with the pictures?
Finally, do Kruskal–Szekeres coordinates respect lighcones angles in the horizontal direction too? (if we rotate them in the vertical axis)?

6. Feb 9, 2010

Naty1

bcrowell is correct...
to restate what he has explained with slightly different language: a free falling observer feels increasing extension forces in the vertical direction and compression forces orthogonally as he moves towards the singularity. This is true above the horizon, at the horizon, and inside the horizon...the free falling observer is never even aware of the event horizon and passes that calculated/theoretical spherical shell completely unaware.

7. Feb 9, 2010

Thank you.