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#73
Jan1509, 04:49 PM

P: 32

I don't know why I was being so dense on this... it would take longer than the 5.8 days for the lightpulse roundtrip. So, at what point in time will the distant observer be unable to receive a reflected returnpulse? If the reflective ball ("object") is tossed at the black hole at .5 c, from a distance of 2.9 lightdays, then you're saying the object will have crossed the event horizon in less than 5.8 days (using an unaccelerated clock, stationary WRT the black hole). (correct?) You're saying that, if the distant observer sends a light pulse 6 days after tossing the object, he will never receive a return pulse? The light pulse will pass through the event horizon, because there will be nothing for it to reflect from? 


#74
Jan1509, 05:01 PM

P: 32

You and I are in the same place, on this. I cannot (yet) see how two logically incompatible solutions can both be correct within any logically consistent system. This requires that one problem has two correct solutions: BOTH (Yes = Falls through event horizion) AND (No = Does not fall through EH). Doesn't this require that there is some statement, within the physical laws of our universe, for which "BOTH A AND NOT A" is true? I'm not at all comfortable with that. 


#75
Jan1509, 05:09 PM

P: 32

Does this not yield a Relativistic mass increase? Since the net mass of the system will be unchanged, wouldn't a considerable amount of energy (mass) be transferred from the black hole to every infalling object? In fact, it seems (to me) entirely possible that all of the mass of the black hole would be transferred to the infalling matter surrounding the black hole. I expect that xantox might have something to say about this. 


#76
Jan1509, 05:23 PM

P: 32

Since radius and mass are directly proportional, the mass density of a Schwarzschild black hole drops with radius. Density is proportional to R^{2}. Interestingly, (according to the WMAP 5year results) the mass density of a Schwarzschild black hole with a radius equal to that of the observable universe would equal the density of the observable universe. In fact, our observable universe satisfies all requirements for a Schwarzschild black hole:
Wilkinson Microwave Anisotropy Probe: http://map.gsfc.nasa.gov/universe/WMAP_Universe.pdf 


#77
Jan1509, 10:32 PM

Sci Advisor
P: 4,837

Furthermore, the idea of the relativistic mass is no longer used, as it leads to too many mistakes. This would be one of them, because the energy of the infalling object certainly does not diverge at it crosses the event horizon. 


#78
Jan1509, 10:41 PM

Sci Advisor
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#79
Jan1609, 04:49 AM

P: 4,512

Just in case anyone missed it, The Schwazschild metric applies to space without any matter in it. The Schwarzchild metric is equally applicable to the space around the Earth, for instance. It's just not applicable within the Earth itself. All that is required of it, is that the mass be mass be spherically symmetric and unchanging in qauantity over time.
There is nothing that requires the central mass to occupy a single point. 


#80
Jan1609, 04:57 AM

P: 4,512

A test particle, where by convention the mass of the test particle has no perturbative effect, is very small. It won't cross a static horizon. 


#81
Jan1609, 05:03 AM

P: 4,512




#82
Jan1609, 07:44 AM

Sci Advisor
P: 4,837

But as I said, this may not be the case for a realistic black hole if the horizon fails to form before it evaporates. 


#83
Jan1609, 11:51 AM

P: 32

Regarding "the idea of the relativistic mass is no longer used", perhaps I should refer to "momentum" or "energy"? 


#84
Jan1609, 11:55 AM

P: 32

The model I use is no less valid than the one Stephen Hawking used in lowering a box filled with thermal energy, via a rope, to the event horizon. 


#85
Jan1609, 12:06 PM

Sci Advisor
P: 4,837

1. What is the total energy of the particle very far away from the black hole? 2. What is the potential energy of a particle that falls from very far away to the event horizon? Add the two and you get the massenergy that is added to the black hole (assuming it's not spinning...things get a bit more interesting for spinning black holes). 


#86
Jan1609, 01:13 PM

P: 32

Consider a massive, hollow spherically symmetrical shell of matter. Suppose that the shell initially formed as a point, but mass has been added and the shell has grown. The external perspective is the first one to consider because it is, initially, the only valid perspective. All valid models will be consistent with models from external perspectives. The gravitational characteristics of all spherically symmetrical objects of equal mass are identical:
The astronomical characteristics would be the same. 


#87
Jan1609, 01:28 PM

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P: 4,837




#88
Jan1609, 04:51 PM

P: 1,822

When I picture the formation of a BH, I see a star collapsing through the white dwarf stage, through the neutron star stage and continuing to collapse until the matter, most likely near the center, reaches a density sufficient to form an EH. Matter at that radius cannot collapse further because of the contraction of space and dilation of time. The matter above that radius may continue to collapse until it reaches its own radius for an event horizon. The star essentially freezes from the inside out. This is a little different from MuggsMcGinnis' example because it doesn't involve a shell but still is similar. This permits the formation of an EH without any matter falling through it. Perhaps there is no singularity and no inside to the black hole.
There is another issue that MuggsMcGinnis just touched on. If the escape velocity is the velocity that would propel a projectile to come to rest at infinity, then matter infalling from infinity to a BH should just reach c at the EH. Another reason to consider why matter cannot fall into a BH is the same as why matter cannot escape a BH. To do so means exceeding c. 


#89
Jan1609, 11:14 PM

PF Gold
P: 247

The socalled "gravitational redshift" of signals sent from an observer hovering near the horizon to an observer hovering farther away is due simply to spacetime curvature. One of the fundamental interpretations of spacetime curvature is that initially parallel geodesics lying in a negatively (positively) curved twosurface will diverge (converge). So if we consider the ideal Schwarzschild vacuum outside some isolated massive object and suppress the angular coordinates, the "t, r plane" has negative curvature, so initially parallel null geodesics diverge as they head radially outward. That means that if we draw two radially outgoing null geodesics (ie light rays, in the geometric optics approximation), intersecting the world line of a static observer at r1, where horizon < r1, and see how they intersect the world line of a static observer at r2, where r1 < r2, we see that the "proper time interval" between two wave crests of a radio transmission (as measured by the ideal clock of the r1 observer) will be separated by a larger proper time interval when received by the r2 observer, because the two initially parallel null geodesics diverge. That's why the r2 observer measures a lower frequency at reception than the r1 observer measures at emission. Notice that in GR, discussions of a "frequency shift" always refer to specific signals sent by an observer with one world line and received by an observer with another world line. That's two distinct timelike world lines plus at least two null geodesics intersecting both of these. Frequency shifts are never welldefined in GR unless you specify all this information. As we imagine asking the lower observer to hover closer and closer to the horizon, the redshift becomes more and more extreme, and we see that distant observers cannot watch anything fall through the horizon because as an object crosses the horizon signals from it to the outside are infinitely redshifted. 


#90
Jan1709, 02:52 AM

P: 4,512




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