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Black Holes - the two points of view.

 
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Dec14-12, 05:28 AM   #375
 

Black Holes - the two points of view.


Austin0, I gather that I did not reply to this post of yours.

Quote by Austin0 View Post
you say the falling observers clock is never stopped in either frame because the distant observers clock never reaches infinity.
I agree. but you seem to ignore the fact that this is only true in the region where the faller has NOT reached the singularity.
you then want to magically have the faller PASS the horizon without ever having reached it.

It appears you interpret time dilation in a way that creates alternate contradictory realities.
If your premise that reaching the horizon requires infinite coordinate time for the distant observer is correct, that means that at all points in that interval the times at the two locations will be related by the SC metric. Both observers will agree on these relative elapsed times and both observers will agree that the faller has not reached the horizon.
The answer here is that all points on the two time scales ARE related by the SC metric, all the points from 0 to infinity on the distant observer's clock are rerelated to the points from 0 to T on the faller's clock, where T is his local time when he gets to the horizon. Obviously it is not a linear relationship, more like a tangent graph where tangent goes to infinity as angle goes to 90 degrees, and so they don't agree on relative elapsed times. Each sees the other's clock ticking at a different rate to his own, an ever increasing difference.
Dec14-12, 08:46 AM   #376
 
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Quote by Mike Holland View Post
Each sees the other's clock ticking at a different rate to his own, an ever increasing difference
This isn't quite correct; SC coordinates can be thought of as the "natural" ones for the distant observer, but they are not the "natural" ones for the infalling observer. So it's not really correct to equate SC coordinate values to anything the infalling observer "sees".
Dec26-12, 08:30 PM   #377
 
i have a question, if protons are massless why in the event Horison are they unable to escape. what is grabbing hold of um, they have no mass?????????? why can they no escape
Dec26-12, 08:41 PM   #378
 
it has mass or it or it don't pick one
Dec26-12, 09:14 PM   #379
 
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Quote by ianpaul12345 View Post
i have a question, if protons are massless why in the event Horizon are they unable to escape. what is grabbing hold of um, they have no mass?????????? why can they no escape
You said "proton", but I think that might be a typo and you meant to say "photon", as protons are most certainly not massless.

Assuming that you meant to say "photon"... In general relativity gravity is not treated as a force that grabs onto massive objects. Instead space and time are curved so that anything moving through them is affected by the curvature; whether massive or massless makes no distance. Below the event horizon, the curvature is such that no paths, even that of a massless photon moving at the speed of light, cross the event horizon into the space outside.
Dec26-12, 10:31 PM   #380
 
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Quote by ianpaul12345 View Post
i have a question, if protons are massless why in the event Horison are they unable to escape. what is grabbing hold of um, they have no mass?????????? why can they no escape
Nugatory gave a good response for GR, but even classical Newtonian gravity could in principle affect photons. The acceleration of gravity in Newtonian physics is independent of the mass, so even a photon would be accelerated the same.
Dec27-12, 02:33 AM   #381
Saw
 
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Quote by DaleSpam View Post
... even classical Newtonian gravity could in principle affect photons. The acceleration of gravity in Newtonian physics is independent of the mass, so even a photon would be accelerated the same.
An acceleration, however, which would happen without a force, since the product M x m would be zero... (?)
Dec27-12, 06:11 AM   #382
 
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Quote by Saw View Post
An acceleration, however, which would happen without a force, since the product M x m would be zero... (?)
For small test bodies, you normally divide out the the test body m anyway to get an acceleration formula rather than a force formula. For astronomy purposes, a many body approximation is often written in terms of acceleration rather than force. In any case, to take the limit as m/M->0, you just divide out m.
Dec27-12, 06:35 AM   #383
 
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Quote by Saw View Post
An acceleration, however, which would happen without a force, since the product M x m would be zero... (?)
Yes, of course. Consider Newtons 2nd law; how much force does it take to produce a finite acceleration on a massless object?
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