Where does the remainder of the energy from a mass falling into a blackhole go?

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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. 

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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 »

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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?

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Offline jartza

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It's the gravitational blueshift thing.

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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 »
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Offline yor_on

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

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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 »
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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.

[attachment=13714]


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Offline yor_on

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Nice to know Jartza :)

Define what you mean by mass ..
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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 »

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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.
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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 :)
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Offline jartza

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I added one line to robot story.

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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.
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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.

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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 »
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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?
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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?

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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.

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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





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Offline jartza

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


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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?
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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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Offline yor_on

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You ever thought about becoming a sage?
They live in caves and when people arrive they present your sort of ah, guidance:)
The photon huh :)

So you are saying that the solar panel will find the photon to weak, and the photon the solar panel to 'strong'?

Hmm, this is becoming a puzzle.

But it's the photon that lost energy, right?
I should get some sleep, but this one is hard to let go :)
Are you working from the energy/mass scenario now?

I'm still not getting why the photon would be weaker?

And I'm also wondering how you define the interactions here, the closer they get to the EV. Do you mean that the interactions then will be weaker the closer they get? As the solar-panel apparently haven't gained any energy but our photon now lost some, according to the solar panel at least?

I better get some sleep after all :)
Three cups of coffee and my brain may start again?
Maybe..

But it's fun Jartza, I'm pleased that I was right, even if I will show up to be wrong :)
I can be a sage too..

Heh :)
==

I'm tired, start to misspell and forget words.
Hate that.
« Last Edit: 06/01/2011 09:33:36 by yor_on »
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Offline jartza

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Photon worked. That made it weak.


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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?

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






Well that's all new to me :) I must admit, black hole mathematics is something I do not widely know about. Perhaps you could write up for me some of these intrinsice calculations so I can follow your presumptions a little more carefully? Or not, up to you :)

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Offline Foolosophy

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The Hawking Equation for Black Hole Entropy is a good start


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Offline QuantumClue

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The Hawking Equation for Black Hole Entropy is a good start


E=Rmc^4 /8GM

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


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

So they say efficiency is 75%

using the above, the OP simplifies 2GM/c and simplifying we get: mc

How is this simplification acheived? Am I missing something obvious? 1/4Mc^2 is quite different to the quantity 2GM/c^2. A bit lost here following the OP's calculations. And who says the black hole is 75% efficient?

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Offline Foolosophy

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The Hawking Equation for Black Hole Entropy is a good start


E=Rmc^4 /8GM

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


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

So they say efficiency is 75%

using the above, the OP simplifies 2GM/c and simplifying we get: mc

How is this simplification acheived? Am I missing something obvious? 1/4Mc^2 is quite different to the quantity 2GM/c^2. A bit lost here following the OP's calculations. And who says the black hole is 75% efficient?

...my first port of call would be to check dimensional consistency

have you done a dimensional analysis for both expressions?

Remember the rest mass ENERGY is equal to mc^2

When R = Swartzchild radius, the Energy is determined to be 1/4mc^2

25% difference from the rest mass Energy (that is 75 efficient)
« Last Edit: 06/01/2011 13:09:13 by Foolosophy »

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Offline QuantumClue

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I must be doing the calculation wrong somehow, because rereading, the OP is not simplyfing 2GM/c exactly. The OP is asking me to replace that quantity with R in the first equation - it would be good to read it properly from my behalf. Right... the G's and M's cancel, 2 divides into 8 giving, 4... right fine. Yes, it leaves mc.

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Offline Foolosophy

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I must be doing the calculation wrong somehow, because rereading, the OP is not simplyfing 2GM/c exactly. The OP is asking me to replace that quantity with R in the first equation - it would be good to read it properly from my behalf. Right... the G's and M's cancel, 2 divides into 8 giving, 4... right fine. Yes, it leaves mc.

I must be doing the calculation wrong somehow, because rereading, the OP is not simplyfing 2GM/c exactly. The OP is asking me to replace that quantity with R in the first equation - it would be good to read it properly from my behalf. Right... the G's and M's cancel, 2 divides into 8 giving, 4... right fine. Yes, it leaves mc.


doesnt the 1/4mc^2 term differ from the rest mass ENERGY (mc^2) by 25%???

So the efficiency is 75%

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Offline QuantumClue

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I'll be honest, I'm not sure what it means. The OP says:

''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''

so the quantity here 1/4Mc^2 is referring to a particle with a mass yes? Or does the energy relate to the black hole? If its the particle, nothing spectacular has happened to the energy I would assume. Energies which are lowered or increased for a system with a mass usually have to do with the kinetic energy... so it's lost kinetic energy...

I could be wrong.

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Offline Foolosophy

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I'll be honest, I'm not sure what it means. The OP says:

''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''

so the quantity here 1/4Mc^2 is referring to a particle with a mass yes? Or does the energy relate to the black hole? If its the particle, nothing spectacular has happened to the energy I would assume. Energies which are lowered or increased for a system with a mass usually have to do with the kinetic energy... so it's lost kinetic energy...

I could be wrong.

Black Holes are quite different creatures

They apparently seem to violate some fundamental laws of physics

the 1/4mc^2 term relates to the rest mass ENERGY.

The body of mass "m" accelerates towards the speed of light as it enters the black hole event horizon

Is the mass converting to "non-kinematic" energy?

Are you saying that the black hole absorbs 75% of the rest mass ENERGY and 25% is lost?

What is this efficiency term you talk about?

How does it relate to the Hawking radiation of balck holes?

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Offline QuantumClue

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I'll be honest, I'm not sure what it means. The OP says:

''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''

so the quantity here 1/4Mc^2 is referring to a particle with a mass yes? Or does the energy relate to the black hole? If its the particle, nothing spectacular has happened to the energy I would assume. Energies which are lowered or increased for a system with a mass usually have to do with the kinetic energy... so it's lost kinetic energy...

I could be wrong.

Black Holes are quite different creatures

They apparently seem to violate some fundamental laws of physics

the 1/4mc^2 term relates to the rest mass ENERGY.

The body of mass "m" accelerates towards the speed of light as it enters the black hole event horizon

Is the mass converting to "non-kinematic" energy?

Are you saying that the black hole absorbs 75% of the rest mass ENERGY and 25% is lost?

What is this efficiency term you talk about?

How does it relate to the Hawking radiation of balck holes?

Nothing is lost, especially once an object has passed the event horizon, as that would imply it destroys the information paradox. Kinetic energy is simply an energy of a system supplied to its momentum, so its an energy of a system which is in movement. A system can loose kinetic energy if its speed slows down, or gain it, if it gets faster.

Again, I've never seen the term 1/4Mc^2 before, so I am unsure how to approach the question other than saying nothing is fundamentally lost.

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Offline Foolosophy

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I'll be honest, I'm not sure what it means. The OP says:

''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''

so the quantity here 1/4Mc^2 is referring to a particle with a mass yes? Or does the energy relate to the black hole? If its the particle, nothing spectacular has happened to the energy I would assume. Energies which are lowered or increased for a system with a mass usually have to do with the kinetic energy... so it's lost kinetic energy...

I could be wrong.

Black Holes are quite different creatures

They apparently seem to violate some fundamental laws of physics

the 1/4mc^2 term relates to the rest mass ENERGY.

The body of mass "m" accelerates towards the speed of light as it enters the black hole event horizon

Is the mass converting to "non-kinematic" energy?

Are you saying that the black hole absorbs 75% of the rest mass ENERGY and 25% is lost?

What is this efficiency term you talk about?

How does it relate to the Hawking radiation of balck holes?

Nothing is lost, especially once an object has passed the event horizon, as that would imply it destroys the information paradox. Kinetic energy is simply an energy of a system supplied to its momentum, so its an energy of a system which is in movement. A system can loose kinetic energy if its speed slows down, or gain it, if it gets faster.

Again, I've never seen the term 1/4Mc^2 before, so I am unsure how to approach the question other than saying nothing is fundamentally lost.

One of the few things that Hawking contributed was to show how black holes can radiate energy at from the event horizon and would therefore eventually evaporate

If you were to plot the velocity profile from just outside the event horizon and project it inwards towards the central singularity point, what do you end up with?

Remember, according to relativity laws an infinite amount of energy is required to accelerate a body at rest towards the speed the light

So what is happening at and past the event horizon?

Our physical Laws break down
« Last Edit: 06/01/2011 14:55:09 by Foolosophy »

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Offline yor_on

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heh :)
Two mathematicians checking you Jartza.
And both find you making sense, even if we're not sure what sort of sense it is ::))
That's nice.

And now we only have to consider a photon, doing work?
You realize what a preposterous statement that is??
ahem.

Doing work where?
==

Better point out that I still need to wake up ::))
« Last Edit: 06/01/2011 18:48:08 by yor_on »
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Offline yor_on

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As I see it the photon only do work 'once' and that is in its interaction. So, if one want to assume that it did, then that was when it 'died' but then you say that this is why it is 'weak'?

So what we have is a hopefully stable unmoving solar panel, hovering mystically over an Event Horizon. Then we have a massless timeless photon. I don't think you can give the photon a mass without violating Einstein's relativity. I've seen people thinking along those lines, but I don't think so myself, although I'm not sure why, and I don't want to look up why just now :) So I'll just make it a prerequisite of my scenario.

Then I'm looking at something that can't have to do with mass/energy in the interaction. What we have left is the 'momentum', but one of the prerequisites of the theory of relativity is that this photon will have the same speed in all 'frames' possible?

And to make the momentum less you would have to assume?
That following a geodesic towards a gravitational potential is the same as losing momentum? That one I don't know what to think off.

Probably something I'm missing here, but I can't see it now at least.
==

And it have to be the interaction.
So?

If I would assume that the EV 'works' in the opposite direction then I also have to presume that the 'energy potential' outside the EV becomes less. Maybe you are considering it from some other frame? But from its own frame the Solar panel isn't experiencing any mysterious loss of energy, and from the photons when it interacts it 'is', as much as a photon now can be, in the 'same frame'. So if I just consider the interaction I would say that it have to do with the 'gravity potential'?
« Last Edit: 06/01/2011 20:23:00 by yor_on »
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Offline QuantumClue

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Some people believe in massive photons, or heavy photons as they are also coined. But the problem is, if it had a mass, it is a terribly low number, beyond anything which can be directly tested at something like 10-51kg.

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Offline yor_on

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And now I'll have too look into that :)
Because assuming that 'photons' have a mass will indeed change the relation.
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Offline QuantumClue

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I'll be honest, I'm not sure what it means. The OP says:

''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''

so the quantity here 1/4Mc^2 is referring to a particle with a mass yes? Or does the energy relate to the black hole? If its the particle, nothing spectacular has happened to the energy I would assume. Energies which are lowered or increased for a system with a mass usually have to do with the kinetic energy... so it's lost kinetic energy...

I could be wrong.

Black Holes are quite different creatures

They apparently seem to violate some fundamental laws of physics

the 1/4mc^2 term relates to the rest mass ENERGY.

The body of mass "m" accelerates towards the speed of light as it enters the black hole event horizon

Is the mass converting to "non-kinematic" energy?

Are you saying that the black hole absorbs 75% of the rest mass ENERGY and 25% is lost?

What is this efficiency term you talk about?

How does it relate to the Hawking radiation of balck holes?

Nothing is lost, especially once an object has passed the event horizon, as that would imply it destroys the information paradox. Kinetic energy is simply an energy of a system supplied to its momentum, so its an energy of a system which is in movement. A system can loose kinetic energy if its speed slows down, or gain it, if it gets faster.

Again, I've never seen the term 1/4Mc^2 before, so I am unsure how to approach the question other than saying nothing is fundamentally lost.

One of the few things that Hawking contributed was to show how black holes can radiate energy at from the event horizon and would therefore eventually evaporate

If you were to plot the velocity profile from just outside the event horizon and project it inwards towards the central singularity point, what do you end up with?

Remember, according to relativity laws an infinite amount of energy is required to accelerate a body at rest towards the speed the light

So what is happening at and past the event horizon?

Our physical Laws break down

Hawking radiation is a natural phenomenon of radiating particles from inside the structure of the black hole because the black hole contains a temperature. The particle falling into a black hole, is something quite different.

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Offline QuantumClue

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That may have sounded unclear. A disconnected particle from the system of a black hole has nothing to do with Hawking Radiation. However in Hawking Radiation, atleast one antiparticle falls back in which is just to make this a bit clearer.

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Offline yor_on

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'Kay, Let's define why a photon is said to be massless.

1. It has no frame where it is at rest.
2. It has no acceleration
3. Therefore no inertia.

Inertia is a property of invariant mass.

But let us assume a inertia.

"This depends on what you mean by "inertial mass".

In Newtonian mechanics it is clear: inertial mass is the m in F=ma, and p=mv. But those equations do not hold in relativity. Moreover, for a photon, a) there is no way to apply any force to use the first equation, and b) there is no generalization of the second equation that applies to photons.

In relativity, "inertial mass" has essentially no utility because the defining equations don't hold, and what little utility it might have cannot be applied to photons.

Another quibble: we say "a photon has zero mass", omitting your "rest", because a) a photon has no rest frame, and b) the word "mass" now means the invariant mass of an object.

(Saying "has no mass" sort of implies that "mass" does not apply, so it's better
 to say "has zero mass" or "has a mass of zero".)"
===


Maybe you are assuming an aeteher Jartza?


So, anyone up to proving that a photon has a mass?
Experimentally I mean :)

Didn't think so either.

It also seems that Bose Einstein particles (condensates) built on the idea of massless photons, even though Einstein later made it valid for some massive particles too?

"Bose was interested deriving Planck's radiation formula, which Planck obtained largely by guessing. Using the particle picture of Einstein, Bose was able to derive the radiation formula by systematically developing a statistics of massless particles without the constraint of particle number conservation. He was quite successful, but was not able to publish his work, because no journals in Europe would accept his paper.

"This "Bose-Einstein statistics" described the behavior of a "Bose gas" composed of uniform particles of integer spin (i.e. bosons). When cooled to extremely low temperatures, Bose-Einstein statistics predicts that the particles in a Bose gas will collapse into their lowest accessible quantum state, creating a new form of matter, which is called a superfluid. This is a specific form of condensation which has special properties. "

In 1924, Bose wrote to Einstein (in Germany) explaining his work and enclosed his manuscript written in English. Einstein was so happy with Bose's work that he translated the manuscript into German and arranged its publication in Zeitschrift f. Physik (the most prestigious physics journal at that time). Furthermore, in 1926, Einstein completed the Bose-Einstein statistics by extending Bose's work to the case of massive particles with particle-number conservation."

(And the condensates exist.)

Let us go back to the photon statistics formula derived by Bose. There is a factor "2" sitting on the numerator of this formula. The usual explanation is that it is because photons are massless particles. Then why not 1 or 3 ? Bose argued that the photon can have two degenerate states. This eventually led to the concept of photon spin parallel or anti-parallel to the momentum.

The question of why the photon spin should be only along the direction of momentum has a stormy history. Eugene Wigner (1939) showed that the internal space-time symmetry of massless particles is isomorphic to the symmetry of two-dimensional Euclidean space consisting of one rotation and two translational degrees of freedom. It is not difficult to associate the rotational degree with the photon spin either parallel or anti-parallel to the momentum, but what physics is associated with the translational degrees of freedom.

These translational degrees were later identified as gauge transformations. This does not solve the whole problem because there is one gauge degree of freedom while there are two translational degrees of freedom. How do they collapse into the one gauge degree of freedom? This problem was not completely solved until 1990."
==



It may be that you consider momentum the exact equivalent to mass?
But it's not, not if we're taking invariant mass. There's too many real life observations of the difference between light and a piece of matter for me to agree on a perfect equivalence.

And even if we agreed on a perfect equivalence you would still need to show what was 'acting' on the photon to make it 'weak'. Normally when we act upon something that is thought to bring energy (mass) to the object, but in this case I get the impression that you mean the opposite?

The negative pressure (expansion)? Nope, not at a black hole.
zero point energies from quantum fluctuations?
They can be both 'positive' and 'negative' it seems?
also they are what we call 'virtual' whatever that may mean in form of Planck time?

You got me stymied here? :)

[4 posts combined. Relating to original quotes would be useful in future. Thanks, Mod]

[ Bose and Einstein. And so it all begun:) ]
« Last Edit: 07/01/2011 17:35:12 by yor_on »
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Offline jartza

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Here we have a moving mirror and a light beam dobbler redshifting. Where might part of the energy of the beam go?

[attachment=13740]

Here is a mirror coated box moving to the right. A light beam is bouncing inside the box. Note that hitting the left side wall makes the light bluer, and hitting the right side wall makes the light redder.

[attachment=13742]

Here we have a tube, moving downwards at constant speed, and a light beam is bouncing inside the tube. There is a gravity field too. Note that light does not hit the wall that would make light more blue, only the wall that makes light more red, and the side walls, that don't cause any change in light.

[attachment=13744]
« Last Edit: 07/01/2011 01:27:35 by peppercorn »

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Offline peppercorn

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Here we have a ....

Here we have random thought number 263..... [::)]
« Last Edit: 07/01/2011 01:29:32 by peppercorn »

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Offline yor_on

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Okay, back in play.. I will name this "Our man from Mars."

How about this Jartza. Your box moving okay, at some half light speed ---> thattaway.
I have two observers, one sitting inside the box, at rest with it.
The other on Mars :)

And the box is leaving Earth on its way to Alpha centauri --->
Now I drop my light-corn into it.

And it starts 'bouncing'... Going  ----->  (90 degrees turned) it will be red-shifted, but, only for the observer on Mars. The guy inside won't notice any redshift as i see it.

This box also becomes a light-clock telling us that the guy on Mars will find the 'time' inside that box to have slowed down as compared to his wristwatch (which also consist of a bouncing photon, 'counting time')

Do you agree to that?
It's all about 'frames of reference' to me.
==

Da*n, wait I will rephrase this one. It's not right.
==

First let us assume that it have a uniform motion.

If we think of it as two mirrors instead, turned at a right angle to the guy at Mars (like those star wars spaceships, the friendly ones, sort of:), then he will see the light corn bounce sloow, which should make it into a red-shift too. It has to be.

If we let the mirror-pair 'stand on end' instead and give him x-ray vision, so that he still can see it bounce, the man from Mars will find the light-corn to be red-shifted as it tries to catch up to the fleeing wall and, as i see it, also when it bounces back, as it have to compensate for the mirror-pairs overall motion -- thataway -> in both bounces, relative our man on Mars.

But for the guy being at rest with the box, or mirror-pair, I expect the light to act the same as it did on Earth. But I think I can see how you think there. That it should matter which way the light moves, if looking at it from the inside of the moving box as the light bounce.

But if I take it into account that any uniform motion inside a 'black box' is inseparable from any other uniform motion, as you can't differ it from the inside. Then the light have to behave 'normally', same as it did on Earth. Do you see how I think?
==

To get to the idea that the observer inside the moving 'black box' will see a difference, depending on your flashlights direction, you first need to state that 'different uniform motions must show a difference' inside our black box. But, when being inside it, you will find that your 'velocity' is impossible to define, you could as easily be standing still. Making it possible to imagine it to be any uniformly moving object, like Earth, ignoring gravity now. And we don't see light beams getting red or blue shifted here depending on where we turn the flashlight, do we?
==

Did that make sense?
==

There is a lot of examples of the 'light-clock' and also of 'rods' placed like this |_ for example but the guys using them are presenting what I think of as 'isolated' cases where they might want to show us how a time-dilation , or a Lorentz contraction is thought to 'work'. But in 'reality' (whatever that is?) our man from Mars will find the light always to be 'red-shifted', no matter where the flash-light is turned, as the overall velocity of the 'box/mirror-pairs' always will be away from him, no matter the direction that light-corn bounce. And the 'man in the box' will find light to behave 'as usual', just like on Earth.
=

And if it is as I think you meant? And we exchanged the 'black box' for a really big one, with planets and stuff inside, we should see 'light' become constantly red-shifted turning our flashlight one way, and blue-shifted when turned the other. That is, if I got your idea right here?

Which actually makes it into a proof of sorts, there being no 'outside' to our SpaceTime, well, ahh :) At least if we imagined that SpaceTime as a whole might have a 'velocity' in some 'distinct direction', and that light behaved the same 'outside' it as 'inside' :) Yeah, I know, not that good an example was it :) But fun.. maybe?
« Last Edit: 07/01/2011 05:20:51 by yor_on »
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Offline Foolosophy

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And now I'll have too look into that :)
Because assuming that 'photons' have a mass will indeed change the relation.


This would totally dismantle the standard model for particle physics.

As it stands the photon is placed in the force carrying category called BOSONS - they share the property of ZERO mass.

I suppose it was once thought that the nutrino possessed NO mass - but in time a mass value for the nutrino mass was measured.

My feeling is that the Universe would possess particles that have NO baryonic mass value.

Perhaps the demarcation between massless and mass possessing entities occurred at the moment of the Big Bang instability (this is just a reflection that has passed my mind whilst writing this garbage so I wouldnt take much notice to it)

And of course if the photon did indeed turn out to be NON-massless, then the speed of light cannot be a constant and even the innate beauty of the expression E=mc^2 will have to be re-written in another way.


 
« Last Edit: 07/01/2011 07:13:42 by Foolosophy »

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Offline yor_on

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Nice one Foolosophy :)
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Offline JP

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And of course if the photon did indeed turn out to be NON-massless, then the speed of light cannot be a constant and even the innate beauty of the expression E=mc^2 will have to be re-written in another way.

If the photon is found to be non-massless, we'll need to just change our terminology so that c is the "cosmic speed limit," then E=mc2 still holds, whereas light now acts like other massive particles and can never reach that speed. 

However, there is no evidence whatsoever that light has mass.