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Crossing the event horizon of a black hole 
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#1
Aug208, 01:49 PM

P: 13

I am struggling with an understanding on what the longest proper time an observer can spend before he will be destroyed into the singularity. How should I approach this problem?



#2
Aug208, 02:33 PM

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P: 16,996

Try this paper.



#3
Aug208, 02:41 PM

Sci Advisor
P: 1,893

Hey, I just looked for exactly this paper with keywords "proper time singularity" in the abstract. Nothing.



#4
Aug208, 02:48 PM

P: 58

Crossing the event horizon of a black hole
The word singularity means division by zero. This is not allowed. Therefore, the solution of the GR equations for the observer crossing the event horizon is not valid. Instead, we observe someone falling down the black hole, and note that it takes forever to reach the horizon. The horizon is like the end of the universe. GR tells us that mass changes geometry.



#5
Aug208, 03:54 PM

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#6
Aug208, 10:54 PM

P: 58

Yes, the singularity is at the center of the black hole. However, the center does not exist. There is no such thing as the inside of the black hole, as it takes forever to reach the surface, i.e., the event horizon. When we talk about the "inside", we are referring to the solution of the General Relativity (GR) equations from the point of view of the observer falling down. However, this solution is not valid, due to the existence of the singularity. This is like boundary conditions restricting which solutions can be allowed; this is the basis of the physics of music. In other words, since it is impossible to observe an object crossing the horizon, then nothing can cross the horizon. Can you travel past the end of the universe? Remember, geometry is not Euclidean near the horizon. The horizon is an end of the universe, as it takes forever to get there.
Ah, but great physicists have discussed this singularity! So what! They are wrong! Very simple! Division by zero is not allowed! 


#7
Aug208, 11:03 PM

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#8
Aug2008, 10:55 AM

P: 58

tan(x) has a singularity at x = 90°. The tan function is not defined here. It is not infinity, but not defined. A singularity is simply a point where the function or equation is not defined.



#9
Aug2108, 12:09 AM

P: 608

[tex]\tau_{max}\text{[seconds]}=\frac{\pi M}{c}=\frac{\pi Gm}{c^3}\ \equiv\ 1.548\times10^{5}\ \times\ \text{sol mass}[/tex] where τ is the wristwatch time (proper time) in seconds, M is the gravitational radius (M=Gm/c^2), G is the gravitational radius, m is mass and c is the speed of light. For a 10 sol mass black hole, the maximum freefloat horizon to crunch time is 1.548x10^4 seconds or 0.155 milliseconds, for a 3 million sol mass black hole, the time is ~46 seconds. The maximum freefloat horizon to crunch distance is [tex]\tau_{max}\text{[metres]}=\pi M[/tex] 


#10
Aug2108, 02:42 PM

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#11
Aug2208, 08:01 AM

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#12
Aug2208, 10:06 AM

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#13
Aug2208, 04:40 PM

P: 58

The singularity is at the center. This means that the equation of motion, the solution of GR, is not valid at the center. This means the equation is not valid anywhere inside the black hole. The singularity acts like a boundary condition, restricting the validity of equations. A valid solution of the wave equation must satisfy the boundary conditions.
But wait! How can a black hole, which in the simplest case, can be imagined as a sphere, not have a center? Answer: the geometry near the event horizon is not Euclidean. 


#14
Aug2308, 03:40 AM

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#15
Aug2308, 05:40 AM

P: 52

Well... where is located the mass of the black hole? If it is inside the event horizon, how its gravitational atraction acts upon anything outside the horizon? Wouldnt it involve travel faster than light?



#16
Aug2308, 07:02 AM

P: 608

There seems to be plenty of maths that supports the fact that there is an 'inside' to the event horizon, basically put [itex]s^2<0[/itex] exists inside the event horizon where [itex]s^2=c^2\Delta t^2\Delta r^2[/itex] (or [itex]c^2\Delta t^2<\Delta r^2[/itex]) while [itex]s^2>0[/itex] exists in space outside the EH (or outside the ergosphere as in the case of a rotating black hole). There seems to be plenty of metric out there that support this with SR taking care of infinities that crop up at the EH. I cannot remember who said the following but 'the event horizon is not where GR ends but where GR begins to end as it starts to unravel towards the singularity'. Of course, the idea of GR unravelling might change as a theory of quantum gravity is established and the 'singularity' is better understood.
Steve 


#17
Aug2308, 07:45 AM

P: 58

I suggest you do some research on BC. The concept of BC is very sophisticated in mathematics. 


#18
Aug2308, 11:34 AM

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