# What's inside the event horizon

by ClamShell
Tags: black hole, event horizon, singularity
P: 4,721
 Quote by Calimero But acceleration against gravity would be tremendous, and pretty unpleasant for anybody on board.
That's another good point, but they did have artificial gravity on board!
P: 256
 Quote by Chalnoth That's another good point, but they did have artificial gravity on board!
Ah yes, good old artificial gravity, interstellar traveler's best friend.
P: 3,390
 Quote by Chalnoth That's another good point, but they did have artificial gravity on board!
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.
P: 221
 Quote by Calimero But acceleration against gravity would be tremendous, and pretty unpleasant for anybody on board.
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.
P: 4,721
 Quote by jarednjames 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.
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.
P: 256
 Quote by ClamShell 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.

There are no earth-like planets, after bh finishes evaporating. Universe is almost empty, with few galaxies full of dying stars.
P: 4,721
 Quote by Calimero There are no earth-like planets, after bh finishes evaporating. Universe is almost empty, with few galaxies full of dying stars.
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 $10^{40}$ years, while the lifetime of a solar-mass black hole is about $10^{66}$ 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 $10^{26}$ times as long...
P: 256
 Quote by 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 $10^{40}$ years, while the lifetime of a solar-mass black hole is about $10^{66}$ 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 $10^{26}$ times as long...
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.
P: 2,281
 Quote by 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 $10^{40}$ years, while the lifetime of a solar-mass black hole is about $10^{66}$ 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 $10^{26}$ times as long...
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.
P: 221
 Quote by Calimero There are no earth-like planets, after bh finishes evaporating. Universe is almost empty, with few galaxies full of dying stars.
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?
P: 774
 Quote by ClamShell Yes, when I "probe" the event horizon with Newton's equation for orbital velocity: V = square root[GM/R] and plug in R = 2.95 Kilometers and M = our sun, I get 212000 km/s, not 300000 km/s as I expected. I'm not telling you my values for G and M because I am now thinking that I've got them wrong...do you get 300000 km/s for the orbital velocity near the event horizon? Does Newton's equation need more terms when relativity is accounted for?
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.
P: 221
 Quote by Calimero There are no earth-like planets, after bh finishes evaporating. Universe is almost empty, with few galaxies full of dying stars.
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.
P: 221
 Quote by 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.
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?
P: 4,721
 Quote by 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*?
Well, no, not really. A metric describes how to calculate lengths between points in a specific coordinate system. Yes, it is true that metrics do appear different in different coordinate systems. However, there is more to it than just this, because the metric also encodes the curvature of the underlying manifold.

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.

 Quote by ClamShell 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.
No, it really isn't, because the important quantities here are ones that are independent of the choice of coordinate system. For instance, if we take a path of an object that comes out from infinity and strikes the black hole, it spends an infinite amount of proper time outside the black hole traveling towards it, but once it reaches the event horizon, it takes a finite amount of proper time to strike the singularity. That sort of behavior is about as asymmetric as you can get.
P: 221
 Quote by Chalnoth Well, no, not really. A metric describes how to calculate lengths between points in a specific coordinate system. Yes, it is true that metrics do appear different in different coordinate systems. However, there is more to it than just this, because the metric also encodes the curvature of the underlying manifold.
I guess then that root(x^2 + y^2 + z^2) might be a metric?

 Quote by Chalnoth 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.
The laws of physics are the same everywhere for different observers?

 Quote by Chalnoth No, it really isn't, because the important quantities here are ones that are independent of the choice of coordinate system. For instance, if we take a path of an object that comes out from infinity and strikes the black hole, it spends an infinite amount of proper time outside the black hole traveling towards it, but once it reaches the event horizon, it takes a finite amount of proper time to strike the singularity. That sort of behavior is about as asymmetric as you can get.
Could an asymmetry in one metric be a symmetry in another metric?
P: 774
 Quote by 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.
There's a simple derivation of the photon sphere radius on Wikipedia...

http://en.wikipedia.org/wiki/Schwarz..._photon_sphere

...which aids understanding - a bit more anyway.
P: 4,721
 Quote by ClamShell I guess then that root(x^2 + y^2 + z^2) might be a metric?
Almost. Metrics are differential, so that you can calculate distances over curves of any shape, not just straight lines. The metric for Euclidean space, then, is:

$$ds = \sqrt{dx^2 + dy^2 + dz^2}$$

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):

$$s = \int_{t_1}^{t_2} \sqrt{\left({dx \over dt}\right)^2+\left({dy \over dt}\right)^2+\left({dz \over dt}\right)^2}dt$$

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.

 Quote by ClamShell The laws of physics are the same everywhere for different observers?
Yes.

 Quote by ClamShell Could an asymmetry in one metric be a symmetry in another metric?
Stated this way, it makes little sense to me. First, for the example I gave, I listed a parameter which is completely independent of the choice of coordinates. If two metrics aren't related to one another by a change in coordinates, then those two metrics aren't describing the same system. So coordinate-independent constructions, such as the proper time of an infalling particle, can be taken as real properties of the system, independent of whatever arbitrary choice of coordinates we make.

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.
P: 2,281
 Quote by Chalnoth Almost. Metrics are differential, so that you can calculate distances over curves of any shape, not just straight lines. The metric for Euclidean space, then, is: $$ds = \sqrt{dx^2 + dy^2 + dz^2}$$ 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): $$s = \int_{t_1}^{t_2} \sqrt{\left({dx \over dt}\right)^2+\left({dy \over dt}\right)^2+\left({dz \over dt}\right)^2}dt$$ 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. Yes. Stated this way, it makes little sense to me. First, for the example I gave, I listed a parameter which is completely independent of the choice of coordinates. If two metrics aren't related to one another by a change in coordinates, then those two metrics aren't describing the same system. So coordinate-independent constructions, such as the proper time of an infalling particle, can be taken as real properties of the system, independent of whatever arbitrary choice of coordinates we make. 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.
Oh lord, I've been here for too long, I'm understanding the math and thinking, "well that's just simple algebra and no matrices or operators involved!" *rubs temples*. If I forget what I need to do my job because visions of sugarplums and tensors dancing in my head, I'm blaming this site, and the books of math and physics it inspired me to read.

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