What's inside the event horizon

Chalnoth

Well, the idea here is that the ship was able to hover just above the event horizon (it is powered, after all). So it's not completely nuts (except for the fact that the power requirements would be astronomical). The real problem is that that degree of time dilation doesn't occur until you're just outside the event horizon, which means the ship itself was too large for it to work that way.

For an astrophysical black hole, I don't think the tidal forces outside the event horizon would have been enough to destroy the ship.

ClamShell

Remember, weightless, free-falling into a non-rotating BH, with Hawking
radiation of very low intensity inside the rocket ship. Infinity is much bigger
than 300; they must have only been stuck for a picosecond. This destruction
you speak of is only wishful thinking; plenty of sources disagree with
this. Anyway, what's immediate mean when you are approaching the
horizon? The important thing is what Alice(A) sees, not what Bob(B)
sees. Bob sees Alice wink out, that doesn't mean that Alice has past.
IE, Alice's past does not include the evaporation of the BH. And it's
the Carl Sagan movie 'Contact', not the kids show 'Andromeda'.

Calimero

But acceleration against gravity would be tremendous, and pretty unpleasant for anybody on board.

Chalnoth

If the ship were actually in the event horizon, the electromagnetic force would no longer be able to keep the ship's atoms together, so it would actually have been pulled apart. The only thing that rescues this scenario is the idea that it would have been just above the event horizon, not within it. But even then, as I mentioned earlier, it doesn't work numerically because the ship was just too big.

Chalnoth

That's another good point, but they did have artificial gravity on board!

Calimero

Ah yes, good old artificial gravity, interstellar traveler's best friend.

JaredJames

Which "amplified the effects of the time dilation from the black holes gravity".

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.

ClamShell

Wait a gosh darn second, time goes very slow near the horizon as measured
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.

Chalnoth

Hehe, we've certainly gone off on a bit of a tangent here, haven't we? But it's a fun tangent!

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.

Calimero

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

Chalnoth

I was pretty sure that black holes don't evaporate until all the stars are already dead. A quick glance at Wikipedia indicates that all of the stars will be gone after about [itex]10^{40}[/itex] years, while the lifetime of a solar-mass black hole is about [itex]10^{66}[/itex] years.

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...

Calimero

If protons decay I was off for a few gazillion years. It turns out that when present day stellar mass black hole evaporates, there will not be anything other then black holes (black hole era), and scarce radiation.

nismaratwork

That was part of my point to George Jones: you have no HR until long after anything coherent exists in the universe, stars included. I think it's safe to say that evaporating black holes is pretty much one of the last stages of heat death for the universe, to be followed by ever more even distribution of radiation. The universe has to greatly "cool" before HR is emitted.

ClamShell

That is bad news...I'm zero for two, in my attempt to find
large-scale features of the blackhole that could be symmetrical.

Do you(plural) know of any possible candidates that might be
symmetrical on the event horizon?

qraal

Newton gets left behind long before the Event Horizon is reached because of the curvature of space around the black hole. At a radius of 1.5 times the schwarzschild radius, photons have their paths deflected by 90 degrees i.e. they're bent into a circular path around the hole if they graze that distance tangentially. Any other orbit within that distance, like a hyperbolic flyby, becomes very hard to plot and the Newtonian/Keplerian approximations are useless. As gets mentioned in most introductions to the Schwarzschild metric, the radius at which the Newtonian escape velocity is equal to lightspeed is only coincidentally equal to the radius of the event horizon - they're derived via different assumptions.

ClamShell

OK, let me be more specific...am I correct in assuming
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.

ClamShell

That 90 degree deflection seems analogous to the deflection
(left or right-hand 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?