When 2 black holes approach each other, they both bend space in an opposite direction. There must always be a flat space between 2 colliding black holes.

However, I heard that they actually merge, becoming one...

I wondered, how can this happen?

Also, is it a good way to get very-very close to the center of a black hole by pushing another one close to it, while our probing spacecraft occupies the flat space in between?

EDIT: Obviously, the flat space between the black holes only exists at a single point (L1). But around that point there is a relatively flat space, where a small enough spacecraft can have enough structural strength to survive. Also, around L1, time dilatation is small, so a probe can get out within reasonable time, or can send signals.

EDIT2: Ok, maybe space at L1 is not flat... Let me try to explain what I was thinking:


B1,2: black hole1,2
L1: Lagrange 1
X,Y: 2 extreme points on the spacecraft

Forces pulling the spacecraft apart:


Forces crushing the spacecraft:

X-->F_B2 ~L1~ F_B2<--Y

Since the forces are equal at L1, pulling apart and crushing should cancel out each other, the closer to L1, the more they cancel.

I wonder if this thinking is wrong because gravity is a pseudo-force, and they cannot simply be added together.

If this is the case, do the tidal forces strengthen or weaken between 2 massive objects?

  • 5
    $\begingroup$ Space-time curvature is associated with tidal forces. Any object between the black holes is under tidal forces of both. There is no flat space there. $\endgroup$ Jan 20 at 21:39
  • 4
    $\begingroup$ And thus the premise of both questions is false. $\endgroup$
    – Ghoster
    Jan 20 at 23:23
  • $\begingroup$ Either the tidal forces cancel out each other at Lagrange 1, or the gravitational field is not continuously differentiable. According to wiki, it is c.d. en.wikipedia.org/wiki/Gauss%27s_law_for_gravity $\endgroup$
    – Zoltan K.
    Jan 20 at 23:38
  • $\begingroup$ Physical tidal forces are not an approximation. They are real forces on finite objects. Only the leading order disappears in certain places. All higher orders do not. What you are referring to is merely an effect of the tangent space approximation. Nature doesn't care about that. $\endgroup$ Jan 21 at 5:37
  • $\begingroup$ "they both bend space in an opposite direction." -> What do you mean by this, exactly? $\endgroup$
    – Avantgarde
    Jan 23 at 11:31

1 Answer 1


You are mixing up the closely-related ideas of gravitational force versus spacetime curvature.

Spacetime curvature is associated with a gravitational field which changes as you move through space. For example, the gravitational field near the surface of the Earth is stronger than the gravitational field at the altitude of geostationary satellites. This variation in the gravitational field strength as you move nearer and farther from a massive object means that extended objects are subject to a gravitational stretching force. If you had a rock near the Earth whose diameter were roughly the width of the United States (which we actually do have), the side of that rock nearer to the Earth would be attracted to Earth more strongly than the far side. The different accelerations of the different pieces of the effect of elongating the object. This stretching is called a tidal force.

You are correct that a point exists on the centerline between two black holes where their gravitational attractions cancel each other out. However, the tidal interaction does not cancel out. An extended object at this zero-field point would have its two ends attracted in opposite directions, towards the two black holes. This tidal force demonstrates that the spacetime at this location is still curved.

  • $\begingroup$ At exactly L1 the space is flat. Near L1 there are tidal forces, it is true but in a small vicinity around L1 they are very small. This means, for a spacecraft, there is a tradeoff between size and structural strength. So a small spacecraft can safely be put in L1, tidal forces won't pull it apart, it won't experience significant time dilatation, etc. $\endgroup$
    – Zoltan K.
    Jan 21 at 12:37
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    $\begingroup$ @ZoltanK. - Why do you think space is flat at L1? The Riemann tensor is not going to vanish there. $\endgroup$ Jan 21 at 13:54
  • $\begingroup$ @ZoltanK. If you like, I can update the answer to show that the tidal force does not vanish anywhere on the line connecting the two centers of mass. But I have to use the Newtonian approximation in order to do it, because my general relativity is a little wobbly. $\endgroup$
    – rob
    Jan 21 at 14:41
  • $\begingroup$ I tried to clarify in EDIT2 $\endgroup$
    – Zoltan K.
    Jan 23 at 13:57
  • $\begingroup$ @ZoltanK. The tidal force is related to the derivative of the gravitational force. This means the tidal force will always be small over some "small vicinity," unless the gravitational force is discontinuous. You might take this argument for tides and apply it to a situation where there is more than one large mass. $\endgroup$
    – rob
    Jan 23 at 16:52

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