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I'm not a physicist and I do not understand maths. But I watch documentaries about "how it all began", "the big bang", "What is time", etc etc just really fascinating.

I was wondering if a blackhole would get in contact with another blackhole, which one would consume time and space? Because they have both infinite mass and gravity.

Will there be a point in space, right in the center between two blackholes, where time and space will move away from you in multiple direction?

What would happen if two blackholes suck on eachother?

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Possible duplicate: physics.stackexchange.com/q/26478/2451 –  Qmechanic Jun 17 '13 at 8:30

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up vote 3 down vote accepted

When two black holes collide, they coalesce to form a larger black hole. The event will produce gravitational waves that will have a particular signature that depends on the details of the collision. Black hole merger events are one of the main gravitational wave signals that are being sought at LIGO.

There are a few misconceptions in your question:

1) black holes do not have infinite mass. They in fact have a finite mass. The orbit of the Earth around a black hole with the same mass as the sun's would in fact be identical to the Earth's current orbit. So, the idea of black holes sucking up the space around them is inaccurate.

2) The gravity at the event horizon of a black hole is finite and (generally) inversely proprortional to the mass of the black hole--a sufficiently large black hole would not exert any detectable local strains on objects at all!

3) The geometry of spacetime will be well-defined everywhere except at the central singularity of the black hole. Every observer will report only one direction of time. This direction, in an absolute sense, will be observer dependent, just as it is in special relativity (the rate of aging will depend on speed and position, but will always be well-defined and knowable).

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Inversely proportional to its mass? So the less mass a black hole has, the greater the gravity at the event horizon? –  Mike Nov 22 '12 at 0:00
Event horizon sizes are proportional to mass, so a lighter black hole has a smaller event horizon, hence stronger gravity at the horizon. –  Muphrid Nov 22 '12 at 0:07
@jmi4: what Murphrid said, noting that the Newtonian force (or the Einstein geodesic deviation) is $\propto \frac{M}{r^{2}}$, but the radius of a black hole is $\propto 2M$. –  Jerry Schirmer Nov 22 '12 at 4:31

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