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Author Topic: Is the cat dead or alive?  (Read 2835 times)

lunar11

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Is the cat dead or alive?
« on: 21/01/2010 21:23:38 »
Imagine i am standing outside a Black Hole and my cat falls in.
The cat would find itself fall into the black hole and be killed (dead).
I on the other hand would see the cat falling in but never reach the
centre of the Black Hole (or the singularity). So, according to me the
cat is not dead (alive).
How can this be; is the cat dead or alive?

LeeE

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Is the cat dead or alive?
« Reply #1 on: 22/01/2010 00:59:58 »
If the cat was not in a space suit it would most definitely be dead.  You would have also been reported to the R.S.P.C.A. for cruelty to animals.

yor_on

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« Reply #2 on: 23/01/2010 17:49:44 »
If it's not one of those really cool super felines in which case it just would pocket that black hole, and then come back to save you, just in time for dinner. And with a probability engine like the one the 'heart of gold' had you might even do it without a super feline. The only problem being the question if you and your cat were there at all in the first place?
« Last Edit: 24/01/2010 00:46:54 by yor_on »

JP

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Is the cat dead or alive?
« Reply #3 on: 24/01/2010 11:16:43 »
The cat is quite dead.  You only see it alive because the light coming from just outside the event horizon takes a long time to reach you.  The cat, however, has passed the event horizon and fallen in, leaving only its light behind.

A non-black hole example would be something like this: imagine you're ~100 light years away from earth and looking at it through a telescope.  You could see Einstein alive and walking around.  However, this is just because the light took 100 years to reach you.  In reality, it's 2010 on earth and Einstein has been long dead.  The black hole is similar, but rather than light taking a fixed number of years to reach you, it takes essentially infinite time to reach you.

Here's a lengthy link on it:
http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html

Farsight

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Is the cat dead or alive?
« Reply #4 on: 24/01/2010 14:46:06 »
Lunar: the cat is dead.

jp: Amazingly, the baez article is wrong. It's giving what's called the Misner/Thorne/Wheeler "geometrical" interpretation of GR rather than the Weinberg "field" interpretation. It relies on proper time, which is an abstraction. Note where it says "if I somehow stood on the surface of the star as it became a black hole, I would experience the star's demise in a finite time". This simply isn't true. Your experiences would slow down with the increasing gravitational time dilation, and eventually grind to a halt. Then you don't experience anything. The article hints at this with "If you attempt to witness the black hole's formation, you'll see the star collapse more and more slowly, never precisely reaching the Schwarzschild radius. Now, this led early on to an image of a black hole as a strange sort of suspended-animation object, a 'frozen star'..." This is the correct interpretation. The cat is dead because it's been hammered flat by the total radial length contraction, but it never gets past the event horizon. People claim it does in its own "proper time", but in the real universe, this only occurs beyond the end of time. Hence it never actually happens.

Ron Hughes

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Is the cat dead or alive?
« Reply #5 on: 24/01/2010 17:15:03 »
From the cat's perspective your the one that instantly aged and died.

Farsight

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Is the cat dead or alive?
« Reply #6 on: 24/01/2010 17:38:28 »
Since in practice a substantial portion of that cat will be blasted out in a polar jet, I don't think that cat is going to have much perspective at all!

JP

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Is the cat dead or alive?
« Reply #7 on: 25/01/2010 05:18:15 »
Farsight, I learned (very basic GR) along the lines of the Misner/Thorne/Wheeler interpretation.  Can you explain or provide good resources for the Weinberg interpretation?

Farsight

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« Reply #8 on: 25/01/2010 14:16:01 »
Here's something, jp. It's "The Formation and Growth of Black Holes" by Kevin Brown: http://www.mathpages.com/rr/s7-02/7-02.htm. He favours the MTW view, but tells you a bit about the alternative:

"Incidentally, I should probably qualify my dismissal of the "frozen star" interpretation, because there's a sense in which it's valid, or at least defensible. Remember that historically the two most common conceptual models for general relativity have been the "geometric interpretation" (as exemplified by Misner/Thorne/Wheeler's "Gravitation") and the "field interpretation" (as in Weinberg's "Gravitation and Cosmology"). These two views are operationally equivalent outside event horizons, but they tend to lead to different conceptions of the limit of gravitational collapse. According to the field interpretation, a clock runs increasingly slowly as it approaches the event horizon (due to the strength of the field), and the natural "limit" of this process is that the clock just asymptotically approaches "full stop" (i.e., running at a rate of zero) as it approaches the horizon. It continues to exist for the rest of time, but it's "frozen" due to the strength of the gravitational field. Within this conceptual framework there's nothing more to be said about the clock's existence. This leads to the "frozen star" conception of gravitational collapse. In contrast, according to the geometric interpretation, all clocks run at the same rate, measuring out real distances along worldlines in spacetime. This leads us to think that, rather than slowing down as it approaches the event horizon, the clock is following a shorter and shorter path to the future. In fact, the path gets shorter at such a rate that it actually reaches (our) future infinity in finite proper time".

Sadly Weinberg's "Gravitation and Cosmology" is 84, so it's not the sort of book many people buy. 

yor_on

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« Reply #9 on: 26/01/2010 16:06:06 »
The problem is the paradox :)

If you accept that nothing can reach pass the event horizon in our time (universal timescale) how can it ever grow that event horizon?

When that 'symmetry break' of whatever kind created it f.ex.
Why would it form a event horizon in the first case?

If it shouldn't, according to this :)

Ron Hughes

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« Reply #10 on: 26/01/2010 17:16:50 »
Within this conceptual framework there's nothing more to be said about the clock's existence.

Far, you know this is not true. No matter how it appears to be moving with respect to us, it never reaches zero movement. But it brings up an interesting thought. A wave falling into a BH will keep getting a shorter wavelength(clock) as it falls. Is there some point in it's fall toward the center that the rate of change does actually stop or does the wave continue to approach zero wavelength?

yor_on

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« Reply #11 on: 30/01/2010 00:35:54 »
"A wave falling into a BH will keep getting a shorter wavelength(clock) as it falls."

Ron :)

That one stymied me for a while and forced me to think, in a way it seemed to make eminent sense, and also have a bearing to some thoughts I've played with myself.

It would be easy to assume that as light accelerate into a gravity-well that it would accumulate energy. But seen from the observer it won't accelerate at all, it will slow down and the light reflected from it will be redshifted, not blueshifted.

If you instead imagined a clock falling you would see its hands slow down and the color/light reflected from it would shift into the red spectrum. So according to you observing, that clock would seem to lose energy instead of increasing it.

So, can that be correct, does it really lose energy by falling toward a mass?
No, it won't.

Imagine it hitting something unmoving, relative the gravity's demand, The further down the gravity-well it hits the more energy will be released in the interaction, won't you agree?

So how can it become as if losing energy when we know that when interacting  we would see it releasing more energy the further down it comes, hitting something 'unmoving'?

To me it's about time, or to be more exact, times arrow. As it moves down the gravity-well the time for that arrow slows down, relative you observing it. Think of it as something similar to that strange idea of space 'expanding' but instead of 'space' you will have that objects 'time windows' becoming 'outstretched'.

Let us assume that shortest period of time is Planck time, what happens to that shortest time as observed from your point of view? It will stretch, there won't be more Planck times magically appearing, instead time will act as a rubber-band, stretching as observed by you, and the light (like a fat line drawn on the rubber-band) will change and in this case redshift (thin out), becoming weaker in energy as you see it.

As seen from the objects own frame it's time experience will be the same as always internally, but with a redshift taking place before it and a blue shift taking place behind it, if it viewed light sources before and aft the direction it travels.

And here comes something ah, further headachingly :)

Imagine this guy in the black box falling in (uniformly accelerating free fall) testing gravity. There wouldn't be any would it :) Just like being in space, right?
(Assume a Schwarzschild non rotating for this one, with no tidal forces)

But now I magically appear a table with very long telescopic legs safely placed on/in/out/aft/forward the singularity, on where we place our friend in the black box, and then we let him do the same test dropping a boll. Now the ball will fall, right?

Then we lower the telescoping legs, moving him closer to our singularity, and let him do the same test all over again. Will he now see the ball have more kinetic energy as it hits the floor? Yes he will.

Now, at what stage did this 'same' kinetic energy exist in the ball, only as it interacted with the 'unmoving' floor? Or at all times from when we stop those telescopic legs and start our test.

And will the ball experience more atomic, molecular 'jiggling' intrinsically due to this energy? That is, will it's momentum become 'relativistically mass' more the further we lower it toward that same singularity?


 ==

One way one might see this is by imagining that light is what makes time, the 'energy carriers' we know of and those that 'interact' and so describes times arrow is photons, virtual or not. so times arrow can be said to be described by lights interactions. It's not wholly right of course, at least I don't think so myself, but intuitively it feels sort of nice :)
« Last Edit: 30/01/2010 01:00:37 by yor_on »

Ron Hughes

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« Reply #12 on: 30/01/2010 02:36:02 »
Yor, the fact that a photon gains energy as it falls down a gravity well is not in question, it is a scientific fact. A clock falling does gain energy as denoted by it's slowing down. Understanding what that means is rather subtle. Before I go any further I want to say I do not have a clue as to why time slows when it falls down a gravity well. The answer to that question well probably answer most questions that we have in physics.

Suppose you are holding the clock in question and you are located some hundred million kilometers from the event horizon of a black hole. There is a space ship on a collision course with an asteroid. The distance between them is say thousand kilometers. In between the ship and asteroid you have light beacons spaced one kilometer apart. Now you can measure the velocity of the ship as it approaches the asteroid.  The ship passes three or four of the beacons and you determine it's velocity with respect to the asteroid. We instantly transport you to the event horizon of the BH and you make the measurement again. You will find the ship traveling enormously faster than your first measurement.

yor_on

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« Reply #13 on: 30/01/2010 16:47:55 »
Yes it gains energy, I too see it that way, but for the observer it won't be blue shifted, at least as I think. You will have a redshift reflected from it to your observer. But how to reconcile those facts :)

That's what I tried to do :)

That events are blue shifted behind you, red shifted in front of you traveling down a gravity-well, as I presume you are referring to in your example is totally correct. From the EV everything will be seen as 'speeded up/blue shifted' when looking out on the universe. And I mention it in my 'black box' example too.

It's just that I'm not sure how to define this 'extra energy' it will get from traveling down. You wrote "A wave falling into a BH will keep getting a shorter wavelength(clock) as it falls" which I don't agree with. It's clock will as seen from an observer outside the EV slow down and redshift. And as seen from its own frame of reference the waves 'clock' will be as it always have been. Never the less you are correct in that it is accumulating an potential energy  as it is free falling into that BH.

But that energy is only a potential one as long as it is in a 'free fall'. It need an 'interaction', or an 'observer' at another frame of reference, or an 'opposing force' like the table I was referring to, to 'materialize' as measurable energy?

That is how I see it.
« Last Edit: 30/01/2010 16:51:22 by yor_on »

 

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