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Author Topic: Where does the remainder of the energy from a mass falling into a blackhole go?  (Read 44609 times)

Offline JP

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You are fundamentally misunderstanding work and its relationship to energy.  Work says that gravity does work on you when you descend the stairs, turning your potential energy into kinetic energy. 

If you use your muscles to oppose that, you biologically turn that kinetic energy into heat and other forms of energy.  This is a non-conservative process, since no one would try to account for all the energy being lost as heat and in biological processes, so no one would seriously try to use work to describe this situation.

The correct definition of work, why it's important to define it that way, and the relationship to energy has been discussed on this forum before (see http://www.thenakedscientists.com/forum/index.php?topic=33720.0 for example).  It's also discussed quite clearly in nearly all intro physics texts. 
 

Offline jartza

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If there is no energy coming out of a climbing person's body, then there is no energy going in into a falling person's body.

Otherwise energy would increase in the body of a person that practices diving.

« Last Edit: 03/01/2011 15:41:12 by jartza »
 

Offline Madidus_Scientia

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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.

What?? Firstly, how would a photon, which is massless, become more energetic? Are you thinking that photons are going to accelerate as they near the black hole?? Light only has 1 speed in a vacuum. Secondly, even if they did, why would there be more photons per second?
 

Offline jartza

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It's the gravitational blueshift thing.
 

Offline yor_on

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Yes. when looked at as waves it will blue-shift from the view point of the solar panel, and as far as I understand it will increase its blue-shift the closer we get to the event-horizon. The interesting thing is why it can do it :)

I saw someone referring to it as a effect of a time-dilation. But there is no way I know of observing a time dilation from the perspective of the solar-panel? And if looked at it from a far observer, discussing a piece of matter in-falling I would only notice it 'slowing down', but I would not see that piece of matter start glowing, as it should if we imagine that with the time dilation comes an increase of energy?

Maybe there is a third way of looking at it?
==

You might want to argue that with the time dilation all waves will be extremely red-shifted and therefore making it impossible too see. but if doing so you also will have to define that frame from where it 'glows'. And there you only have two, but let's make it three. 'A' is the far observer 'B' is the piece of matter in-falling in between 'A' and 'C'. 'C' is an observer at rest, very near the EV (event horizon).

From 'A' it would be as I say, hopefully :)

In 'B':s frame we are discussing something 'free-falling' actually following what I call a 'geodesic' :) There is no 'forces', or 'extra energy', noticeable from that point of view. And the piece of matter should consider itself weight-less, meaning that if you placed a scale under its metaphorical feet the scale would register 'zero', tidal forces ignored here.

From the viewpoint of 'C'?
Well he would be lower down in the gravity well, and if it was wave he would notice it having a 'blue-shift' but discussing 'matter' I expect him to see an 'acceleration' but no glowing 'phenomena'? As for how the acceleration should look like to him? I'm not sure.
==

There is naturally a possibility of looking at it with 'the eyes of a God'.

But then we are leaving what they actually observe in favor of how to interpret it theoretically comparing those frames of reference against each other..

From that view point I can define the 'blue-shift' as a possible result from a time-dilation. But then it becomes a conceptual thingie :) not what they observe. Also it as easily can be seen as a Lorentz contraction. Why I don't count in a Doppler shift? Maybe I should, there are no frames that I can define as being 'unmoving' in the universe, so from that point of view a Doppler-shift always will be involved, I think? 
« Last Edit: 04/01/2011 06:58:25 by yor_on »
 

Offline yor_on

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Now, do you agree?
Or would you describe 'A' 'B' and 'C' differently?
 

Offline yor_on

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In my eyes the effect comes from the position of that 'gravitational field' created. Make that into a coordinate system graded on its 'gravitation'. Then measure photons from different points. They will all have a different 'energy', but that energy is a relation coming into play in the interaction. so looked at it this way the mystery becomes why an interaction can free more 'energy' than what you expect the photon to have 'originally' as it leaves a source (Sun) in form of energy quanta.

And the simplest way to make that idea make 'sense' is if I assume that we're not really measuring anything 'moving'. But then it stops :) and becomes extremely difficult because now we are talking about a universe without 'moving parts' sort of. That as radiation in one form or another is involved in everything I know of.

So maybe it 'propagates'?
But, does that make it simpler?
==

Also, when meeting a wave we have firstly the motion/acceleration to consider. Then we have its Doppler shift and Lorentz contraction (as well as a time dilation). The opposite, leaving a wave (creating a red shift) will as I understands it, and this is slightly weird, still introduce a Lorentz contraction reducing the wave length at the same time as the Doppler shift will increase it.


"The Doppler shift of plane light waves in vacuum which arrive with an angle phi with respect to the direction of travel.

The difference in the classical and relativistic Doppler effects can be seen in the following graph showing the wavelength shift of green light for velocities ranging from v/c=-1 (recession at the speed of light) to v/c=1 (approach at the speed of light). The Doppler shift predicted by classical physics is shown in red and the correct prediction of special relativity in green. "

=
« Last Edit: 04/01/2011 02:51:18 by yor_on »
 

Offline jartza

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Yes. when looked at as waves it will blue-shift from the view point of the solar panel, and as far as I understand it will increase its blue-shift the closer we get to the event-horizon. The interesting thing is why it can do it :)

Solar thermal collector is a black object that heats up in the sun.

A solar thermal collector that has a big mass heats up slower than a solar thermal collector that has a small mass.

A solar thermal collector that has been lowered into event horizon has lost 75 % of its mass. Therefore it heats up 4 times faster than same kind of solar thermal collector that has not lost any mass.


Or solar thermal collector at event horizon gets 4 times hotter with same energy.

« Last Edit: 04/01/2011 07:19:29 by jartza »
 

Offline yor_on

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:)

That one you will have to explain, I'm losing you here?
==

You have mass and then you have weight.

The weight will increase the closer it gets to the Event Horizon.
The invariant mass is the same in any frame you can imagine, that's why it's called 'invariant', so that mass won't change.
« Last Edit: 04/01/2011 09:28:42 by yor_on »
 

Offline jartza

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Nah, experts and jartza know that the mass of a lowering device increases and the mass is taken from the mass that is being lowered. Review the OP Yor_on.



 

Offline yor_on

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Nice to know Jartza :)

Define what you mean by mass ..
 

Offline jartza

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A and B are identical robots with same mass, they are at floor 2.

A is carried to floor 1.  B uses the stairs to go to floor 1.

A and B have different masses now.  B has bigger mass.


Because A has cool brakes, while B has hot brakes.



« Last Edit: 04/01/2011 20:04:44 by jartza »
 

Offline yor_on

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Then you're including ? Relative mass? maybe momentum? Or?

But neither of them exist, except in the interaction.
Take a look at 'A' 'B' and 'C' and see how you would define that, also tell me, why, the mass will change, and why what I state up there is wrong :)

Then we'll see.
 

Offline yor_on

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I think I'm very correct in stating that a 'gravitational acceleration' is not the same as an 'acceleration'. What fools the 'observer' are that the geodesics all point in the same direction. but if you look at the 'detector' argument for Unruh radiation I think you will realize that there is a difference between that and 'free falling' following a geodesic, no 'matter' where it points :)

It's very simple from my point of view as I look at 'energy expended'. To prove the equivalence you will need to prove an 'energy expended' as I see it. If you can't you're just bicycling in the great younder. Where I also have been uncountable times :)
 

Offline jartza

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I added one line to robot story.
 

Offline yor_on

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What you are doing Jartza is to mix the chemical energy expended with something where no energy is expended at all. To make it work you will need to prove that the in-falling photon, or the piece of in-falling matter I spoke of, actually are expending a energy doing so. But they don't Jartza.

In the case you are looking at both will expend energy, it doesn't matter if up-hill or down-hill when we are speaking of something moving, breaking a geodesic. Neither of your robots are in a free fall.
 

Offline jartza

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Exxxzactly. Free falling does not cause energy changes.




When a solar panel is dropped to black hole and it collides with a photon
that has traveled down by elevator, the solar panel will report that the
photon has a reduced energy.  And the photon will report that the solar panel
has an increased energy.
 

Offline yor_on

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Hmm :)

Stop using that elevator please ::))
It befuddles me thinking..

A in-falling photon have the energy it's born with, as I see it. The only thing happening and defining something else being what interaction it have. But now we come to another weirdie :)

That solar-panel you expect to have an increased energy, it doesn't. Put its old friend, the twin solar-panel beside it, floating and waving. None of those two, now being in a tandem, sort of, will find the other to have a greater 'energy' if they sent a light-beam at each others solar panels.

Soo? Jartza :)
==

They are in the same 'frame of reference' relative each other.
« Last Edit: 05/01/2011 17:42:12 by yor_on »
 

Offline yor_on

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The question seems to hinge on how to look at it. We can always use the eyes of God to do that. Doing so we will see that no matter those 'twins' finding each other same 'as always' we, from our more 'enlightened' position will indeed observe both as being 'dipped' in 'energy' as they are 'down' in that gravitational potential. Assuming that this also will make spontaneous 'pair production' more common, the closer we get to a event horizon, it seems to hold a certain 'objectivity'?

But it won't be noticeable from the frame of those solar-panels, as I know at least?
 

Offline QuantumClue

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Oh here I come lol

Replacing R in first formula with 2GM/c and simplifying we get: mc

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.


Could someone explain this a little better to me. Where does this efficiency arise? The calculations?
 

Offline yor_on

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A and B are identical robots with same mass, they are at floor 2.

A is carried to floor 1.  B uses the stairs to go to floor 1.

A and B have different masses now.  B has bigger mass.

Because A has cool brakes, while B has hot brakes.

Jartza, are you thinking of energy converted into mass?
If so I get what you mean, but when looking at our photons they are massless.

 

Offline jartza

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Oh here I come lol

Replacing R in first formula with 2GM/c and simplifying we get: mc

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.


Could someone explain this a little better to me. Where does this efficiency arise? The calculations?

Expert's not so good efficiency calculation:
Click this link: http://www.scholarpedia.org/article/Bekenstein_bound
Find picture where mass is hanging over bigger mass. Find in the text the part that talks about what is going on in the picture.


Jartza's correct efficiency estimation based on this:
http://en.wikipedia.org/wiki/Carnot%27s_theorem_%28thermodynamics%29




 

Offline jartza

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Jartza, are you thinking of energy converted into mass?
Oh yes.

 

Offline yor_on

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I think I can see how you think now, but your reasoning builds on the idea of the solar panel getting a 'pressure' or 'force' acting on it, doesn't it? Maybe you meant that it's not only a solar-panel but also something that is acting to keep it at a certain position, like some kind of rocket pushing on it?

If we assume :) that the solar panel is 'magically attached' to its position, receiving no energy from any 'rocket boost' do you still think that the photon will say that the solar-panel have an increased energy?

Why?

Gravity?
 

Offline jartza

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Nope, the photon lost energy, that's why it is saying everything else gained energy.
 

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