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What's inside the event horizon 
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#73
Sep2510, 12:07 PM

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#74
Sep2510, 12:14 PM

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#75
Sep2510, 12:16 PM

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From the point of view of those on board the ship, no time passed, but for those outside 300 years went by. But yes, the gravity would pancake everyone on board and you would need some wicked engine power to get away from the event horizon. In the TV show they were 'towed' out with what I can only describe as the worlds greatest tow rope! Ironically, I just finished watching all five seasons of it. So pretty sharp on it's content at the moment. 


#76
Sep2510, 12:31 PM

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by Bob. If Alice tried to hover longer by turning on her retrorockets it would be in vain; better to conserve fuel for finding an earth like planet after the BH has evaporated. And she wouldn't retro very long, anyway. Wait, Bob would see Alice's rockets fire for a long time; he might even think she is going to run out of fuel. No way, Alice would only use a few minutes of fuel. Bob is way off track about the fuel issue. 


#77
Sep2510, 12:50 PM

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As for the tow rope, it doesn't necessarily have to have been absurdly strong, because the ship was supposedly keeping itself from falling into the black hole under its own power. They only need to give it a little extra pull to get it out. However, what should have happened then is the ship rocketing off under its own power away from the black hole, after that initial bit of outward pull was provided. 


#78
Sep2510, 01:01 PM

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There are no earthlike planets, after bh finishes evaporating. Universe is almost empty, with few galaxies full of dying stars. 


#79
Sep2510, 01:08 PM

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You might be able to get stars to last a bit longer with exotic physics, but I doubt you can get them to last [itex]10^{26}[/itex] times as long... 


#80
Sep2510, 01:50 PM

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#81
Sep2510, 01:53 PM

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#82
Sep2510, 10:35 PM

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largescale features of the blackhole that could be symmetrical. Do you(plural) know of any possible candidates that might be symmetrical on the event horizon? 


#83
Sep2610, 12:23 AM

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#84
Sep2610, 12:26 AM

P: 221

that a *metric* is just another name for what I call a *coordinate system*? say a Cartesian coordinate system, (x, y, z) or a Spherical coordinate system, (R, theta, phi)? And, if so, is the coordinate system on our side of the event horizon in any way symmetrical with the coordinate system chosen for the other side of the event horizon? I am looking for symmetries that imply conservation of entropy, as always. 


#85
Sep2610, 12:53 AM

P: 221

(left or righthand rule) of electrons in a charge field due to the magnetic field created by the motion of the charge around the nucleus. As per other posts on this thread V_escape = root(2)*V_orbital and V_orbital = root(GM/R) for Newton and Einstein (but derived via different assumptions). "Only coincidental" seems to imply that they came together randomly, but I suspect they arise due to identities in each of the derivations(not between the derivations). Can you enlighten me a bit? 


#86
Sep2610, 01:13 AM

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One way to think about it is that you can perform a wide variety of coordinate transformations, and the metric changes as you change coordinates so that any lengths you calculate stay the same. 


#87
Sep2610, 01:39 AM

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#88
Sep2610, 04:33 AM

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http://en.wikipedia.org/wiki/Schwarz..._photon_sphere ...which aids understanding  a bit more anyway. 


#89
Sep2610, 06:41 AM

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[tex]ds = \sqrt{dx^2 + dy^2 + dz^2}[/tex] You calculate distances by doing a line integral, typically parameterizing the line by some parameter like so (here I'll use time as the parameter for convenience): [tex]s = \int_{t_1}^{t_2} \sqrt{\left({dx \over dt}\right)^2+\left({dy \over dt}\right)^2+\left({dz \over dt}\right)^2}dt[/tex] If you have some path that includes the information x(t), y(t), and z(t), you can calculate the path length by performing the above integral over time. You should discover that if you take the path to be a straight line, you'll end up with the distance you mentioned above. Second, properties can be divided into two sorts: symmetric properties and every other way things can be. In general, when you have a symmetric property of a system, a change of coordinates won't necessarily show the same symmetry. However, if you have an asymmetric system, there is no guarantee that you can make the system look symmetric, no matter how you try to change the coordinates. 


#90
Sep2610, 07:13 AM

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