# The Naked Scientists Forum

### Author Topic: Where does the remainder of the energy from a mass falling into a blackhole go?  (Read 44567 times)

#### jartza

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« on: 26/12/2010 10:35:20 »
The experts say that when mass m is lowered into a black hole, that has mass M, then this is the energy of the mass m down at the event horizon:  E=Rmc^4 /8GM

This is the swarzschild radius of a black hole with mass M:
R=2GM/c²

Replacing R in first formula with 2GM/c² and simplifying we get: ¼mc²

So they say efficiency is 75%

BUT efficiency is actually 100%. As we can see when we build a heat engine where a black hole serves as a heat sink. This engine has 99.9999999 efficiency, or 99.999999999999999999999 with bigger and colder black hole.

Energy formula can be found here, about at the middle of the page.
http://www.scholarpedia.org/article/Bekenstein_bound
And here are some black hole efficiency considerations
« Last Edit: 26/12/2010 10:50:36 by jartza »

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #1 on: 26/12/2010 16:39:22 »
Rather cool idea Jartza :)

But isn't that the Bekenstein bound you got there, 1/4? The information retrieved has to be less than a forth of the visible area expressed in Plank lengths, and that 'information' can be seen as 'energy quanta' too I presume? But I agree, the idea of a heat-sink have to be a as close to a hundred percent that's possible.

Nice one.
==

It's such a mindblowing idea, to me I mean.
Those singularities, spinning at the speed of light, well almost.

So what would it take, in form of energy, to get a infinite mass to spin at 96% of lights speed in a vacuum?
Or maybe just: Why do they spin?
==

Assuming that they are what creates 'matter' (well, maybe:), they probably need that speed. But how do they reach it?

Look at this Retrograde spin of supermassive black holes may create jets that control galaxy evolution.
« Last Edit: 26/12/2010 17:00:31 by yor_on »

#### jartza

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #2 on: 27/12/2010 02:20:23 »
When black hole is used as a heat sink of a heat engine, the efficiency of the engine is very good.
Therefore only very a small amount of energy enters the black hole.
Therefore there is only a very small mass increase of the black hole.
Therefore there is only a small surface area increase of the black hole event horizon.
Therefore there is only a small increase of the entropy of the black hole.

Therefore we must ask: were does the entropy of the heat energy of our heat engine go?

Oh yes entropy increase is larger in the better bigger heat sink black holes, problem solved.

« Last Edit: 27/12/2010 11:51:56 by jartza »

#### jartza

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #3 on: 27/12/2010 12:04:14 »
There is another problem. The experts say that entropy is proportional to area of event horizon.

But see the first animation: http://www.psc.edu/research/graphics/gallery/winicour.php

Area is increasing, but nothing irreversible is happening, until the event horizons touch.

So entropy is not proportional to area, except sometimes.

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #4 on: 27/12/2010 15:52:32 »
"When black hole is used as a heat sink of a heat engine, the efficiency of the engine is very good."

Maybe, but I can imagine a really terrible 'heat engine' leaking and steaming too? Not transforming at all. Maybe you are thinking of it otherwise than me? I agree that the 'heat sink' is the best that can be thought though, and will catch most 'energy' possible.

Or was it the black hole you meant as the 'engine' here?
If you thought of a sun, then that's a pretty good engine I think.
How efficient? I don't know, that's a matter of definition but it's what our universe came up with for 'providing light energy' so I guess it's pretty good.

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #5 on: 27/12/2010 16:05:33 »
When it comes to entropy I agree. it's a discuss-able subject. Talking about the 'entropy' of a black hole becomes slightly weird in that nothing 'really' move from inside the 'singularity' and out the event horizon in Hawking radiation. I'm still not sure how to look at that, even though there is a clear relation between the two if it is correct. But considering that I  doubt both 'distance' and so 'motion' its no real problem to me :)

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #6 on: 27/12/2010 16:09:57 »
You might call Hawking radiation the first engine transformer without 'moving parts' :)
Eh, joking that is..

#### SteveFish

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #7 on: 28/12/2010 01:16:12 »
I am sorry about this post, but I just couldn't help commenting because this is an area of my expertise. The efficiency of my backhoe (John Deere 3320, 447 hoe) is not misunderstood by experts at all. I am able to dig a trench at the maximum predicted for my tractor horse power and hydraulic pump capacity. So watch it. This is TractorByNet isn't it?

#### jartza

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #8 on: 28/12/2010 01:54:17 »
Engineers can confirm that the blue very cold gas absorbs enormous amounts of heat when heated. When the heated gas is cooled it gives out a normal amount of heat.

Then we can see a sun, an ideal solar panel, a black hole, and a lamp.
The black hole does not receive much energy, right?
Now let's move the solar panel very close to the black hole. What happens to the
energy received by a) black hole b) lamp ?

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #9 on: 28/12/2010 03:04:19 »
I kind of like your drawing Jartza, as for what happens? Thats a question about the distance to the sun and the solar panels 'area' relative the Black Holes area, versus that sun when the solar panel is moved, isn't it?

And then of course the 'energy'.
But I'm not sure what you mean?

#### jartza

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #10 on: 28/12/2010 03:33:37 »
We are interested how many percents of the energy the lamp gets, and how many percents of the energy the black hole gets.

And let's not move the solar panel closer to the black hole. But let's move the black hole closer to the solar panel.

Actually we are interested if there is any remarkable change in the percents when the black hole is moved closer to the solar panel.
« Last Edit: 29/12/2010 05:22:40 by jartza »

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #11 on: 28/12/2010 04:52:42 »
If it's a solar panel it will use sunlight primarily as I think. The close it will be to the black hole the less it will 'block' the radiation, then you have the possible energy created by a Black Hole of course, but I don't count that in here. I'm not sure on how to count on that in fact, if you're thinking 'virtual particles' aka photons spontaneously appearing as we come closer to the BH? Normally we expect them to disappear very quickly as I understands it, and seen from the 'frame' of the solar panel I expect them to still do so, even though an outside observer might have another opinion.

Or is it something else you're thinking of?

#### jartza

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #12 on: 28/12/2010 06:12:29 »
Here's the correct answer to the quiz:

The solar panel will say: after the black hole moved closer I have been receiving more energy, the photons are more energetic and there are more photons per second.

The lamp will say: I have not experienced any change in the energy flux.

The black hole will say: My experience has been the same as the lamp's.

#### Foolosophy

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #13 on: 28/12/2010 11:40:26 »
BUT efficiency is actually 100%. As we can see when we build a heat engine where a black hole serves as a heat sink. This engine has 99.9999999 efficiency, or 99.999999999999999999999 with bigger and colder black hole.

interesting

why so close?

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #14 on: 28/12/2010 13:58:05 »
Yep, you are right, to the solar panel those photons coming from the sun will be more energetic, when it comes to the lamp? There are a lot of technicalities involved there, but if we assume that there is more energy transformed into a current? You better explain how you think there so I can see it:)

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #15 on: 28/12/2010 14:08:30 »
Maybe you see it as a question whether those photons 'really' have the energy the solar panel 'think' they have? That's a interesting one indeed. If we think of it as a 'photon stream', and the sun as a hose, then you might assume that as they meet a Black Hole the photons will 'accelerate', as relative a 'far observer'.

When they do so the 'distance' between them will increase, so the argument that 'photons' strength could be seen as them coming closer in time making up for more energy per 'time unit' don't seem to work there, does it? :)
==

Sorry, need to wake up here, that was one of my dafter arguments.
Photons have only one speed, and the only way they can express an acceleration, that I know, is the way their energy will relate to you. But it was me remembering the argument of energy relating to acceleration that I've seen before here :) Sh*..
==

But you might look at it as 'gravitational line-borders' maybe?
If looked at that way those 'lines' will be closer the closer you come to a event horizon, representing a equivalence to 'energy'? I'm not sure of that one in fact. The problem is that if looked as waves my reasoning becomes easy, but when looked at solely as 'photons' it becomes truly irritating. You could maybe consider them 'retarding', gaining energy momentum as they pass those 'line-borders', but as they're not 'allowed' such an expression they instead peak up in 'energy'? Very weird reasoning:)
==

Nah, I like light 'not moving' more and more :)
« Last Edit: 28/12/2010 14:30:18 by yor_on »

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #16 on: 28/12/2010 14:39:01 »
But what I'm doing here discussing it, is to treat them as 'entities', each one of a defined 'energy' relative the solar panel as they meet. If we stop doing so and look at them as 'relations' it becomes easier for me. Then you don't need to look at some predefined object of 'energy momentum' and state, 'there it have to be'. Instead you look at the relation and say "As I see it, it should be .. there" which then will be true relative you. That's also what probability says. and that's also why we are daft expecting light to propagate :)

But it depends of course. Each one to his own views huh ::))

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #17 on: 28/12/2010 15:24:53 »
The point I'm trying to make here is that you now and then will see the argument presenting 'photons' as more or less 'together' depending on 'energy strength' expressed. But if we look at the idea of that Sun-hose streaming out 'photons' they will express a greater 'energy' relative the sun-panel without that 'mechanism' being involved into it, invalidating the concept if you see my drift.
==

It's easier to stop using words like motion and look at it as a ''game plan' where circumstances defines probabilities. Also remembering that the only way to 'measure' light is as it 'interacts'. What you call weak observations/interactions is still a result of a 'interaction'. You can't get away from that fact without naming yourself the 'Wizard of Oz', and he was a fake:)
« Last Edit: 28/12/2010 15:31:29 by yor_on »

#### jartza

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #18 on: 29/12/2010 04:50:47 »
Foolosophy, here the blue gas is very cold. When the heat from yellow sun heats the gas, the gas absorbs very large amount of heat. Then the black Black Hole is used to cool the gas again. The amount of heat that the black hole absorbs is moderate.

Oh yes, this thing is a power plant, that should be mentioned.

This power plant puts 99% of the energy that it receives from the sun to the lamp,
1% goes to the black hole.

#### jartza

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #19 on: 29/12/2010 05:45:51 »
Yor_on, you wanted more info about how much energy a black hole - solar panel combo delivers to a lamp.

Well it does not deliver more energy than is delivered to itself by the sun.

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #20 on: 29/12/2010 15:42:23 »
Interesting but a little hard to follow :)

If we just imagine a sun near a black hole then, and let that sun-hose shoot out a 'photon' that will end in that black hole. Then you place your solar panel in its path, varying its distance to the Event horizon. Let's say 50% of the distance first, from the mirrors 'frame of reference' (In the middle between the sun and the black hole, distance wise) then 60% from the sun, 70%, 80%, 90%, and finally at the Black Holes Event horizon, ignoring spontaneous pair production.

How 'lighted' will that lamp becomes as the 'distance' shrinks, stronger or weaker, or the same.

What do you expect?
==

For this one we will ignore all ideas about 'virtual light' becoming 'real' too
« Last Edit: 29/12/2010 15:46:37 by yor_on »

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #21 on: 29/12/2010 16:09:36 »
Rereading you I think you see it as if the solar-panel and the lamp won't agree?

To test such a statement is very hard, actually impossible if doing it 'simultaneously' with the 'same particle'? Maybe if we split it though?

Down-converting that original photon into two via a 'beam-splitter'. Then exchanging the lamp for another 'detector' and let them both be 'measured' as they hit.

I think they would have the same 'value' myself? assuming that the both 'detectors' rest at the same distance, between the BH and the sun?

Or what am I missing?

#### jartza

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #22 on: 29/12/2010 16:53:27 »
Downhill has an energy boosting effect.
Uphill has an energy reducing effect.
It's trivial. Also energy is conserved

Remember the talking solar panel that said that photons are more energetic? It will also report photons having bigger spins! (more angular momentum)

#### yor_on

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #23 on: 29/12/2010 17:58:45 »
So you are placing the solar panel down hill and the lamp uphill.
As you wrote :
==

"The solar panel will say: after the black hole moved closer I have been receiving more energy, the photons are more energetic and there are more photons per second.

The lamp will say: I have not experienced any change in the energy flux."
==

To get that result it seems that you will need to leave the lamp 'still' relative the system of 'sun/EV' and then only move the solar panel? And you expect a photon to have a measurably greater 'energy' the closer that panel comes to the Event horizon, all of it in the understanding that I read you right?
« Last Edit: 29/12/2010 18:02:29 by yor_on »

#### jartza

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##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #24 on: 29/12/2010 21:58:46 »
Yes Yor_on, you understand correctly.

Gravitational blue shift and gravitational red shift are special cases of general gravitational energy shift invented by jartza, I guess.

I thought that everybody knew it already.

#### The Naked Scientists Forum

##### Where does the remainder of the energy from a mass falling into a blackhole go?
« Reply #24 on: 29/12/2010 21:58:46 »