
#1
Jun210, 07:40 AM

P: 1,555

In the Schwarzschild solution consider two test point particles, p1 and p2, both at the same distance from the black hole (e.g. both particles have the same r value) but separated a distance d from each other (so they make an angle with respect to the black hole)
So now the question: when will their paths cross? At the event horizon or at the singularity? If you say at the singularity then what is their distance in terms of d when they both cross the event horizon? 



#2
Jun210, 07:51 AM

Mentor
P: 6,044

You haven't been specific enough. Are the particles falling radially? Falling initially from rest (with respect to Schwarzschild coordinates)? Falling form rest at infinity?
Distance is tricky and ambiguous in general relativity. What does "separated a distance d from each other" mean? 



#3
Jun210, 08:00 AM

P: 1,555

However if it is not possible to express any kind of distance then the question becomes: how can you prove they do not meet at the event horizon. 



#4
Jun210, 08:20 AM

Sci Advisor
P: 1,883

When will they meet?In case you're referring to Antoci: r=2m at the horizon, not 0. 



#5
Jun210, 08:29 AM

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PF Gold
P: 1,809

One way of looking at this is not to try to draw a "straight line" between the two points, but instead to consider the arc length of the circle that they both lie on (centred on the black hole). If the two points are close together (i.e. d is tiny compared with the radius), that's a good approximation anyway.
Choose your coordinate system so that both points are on the "equator" [itex]\theta = \pi / 2[/itex]. The circumferential distance between the points is just [itex]r (\phi_2  \phi_1)[/itex] and both particles move along worldlines of constant [itex]\phi[/itex] and [itex]\theta[/itex]. So the circumferential distance in the limit as they approach the event horizon is just [itex]r_s (\phi_2  \phi_1)[/itex] where [itex]r_s[/itex] is the Schwarzschild radius. Oops, Ich beat me to it! 



#6
Jun210, 08:33 AM

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P: 6,044

Schwarzschild coordinates are really two disjoint coordinate patches, and neither patch covers the event horizon. Another coordinate patch that does cover the event horizon should be used to determine if particles meet on the event horizon. 



#7
Jun210, 08:38 AM

Sci Advisor
PF Gold
P: 1,809

Just had a thought. The picture of Flamm's paraboloid seems to indicate that as you approach the event horizon (where the paraboloid becomes vertical), the circumference really is the geodesic distance (using the Schwarzschild coordinates to decompose spacetime into space+time).




#8
Jun210, 08:47 AM

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#9
Jun210, 08:51 AM

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#10
Jun210, 08:54 AM

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#11
Jun210, 10:55 AM

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#12
Jun210, 01:05 PM

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#13
Jun210, 04:51 PM

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#14
Jun210, 07:00 PM

Sci Advisor
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#15
Jun210, 10:07 PM

P: 101





#16
Jun310, 11:43 AM

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