Naked Science Forum

Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: jartza on 26/12/2010 10:35:20

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza 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 (http://www.scholarpedia.org/article/Bekenstein_bound)
And here are some black hole efficiency considerations
http://kencroswell.com/BlackHolesInQuasarsSpinFast.html (http://kencroswell.com/BlackHolesInQuasarsSpinFast.html)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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.

Nice link.

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. (http://web.mit.edu/newsoffice/2010/black-hole-jets.html) 

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza 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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza 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 (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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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 :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 27/12/2010 16:09:57

You might call Hawking radiation the first engine transformer without 'moving parts' :)
Eh, joking that is..

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: SteveFish 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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza 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.
 [ Invalid Attachment ]

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 ?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza 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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza 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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Foolosophy on 28/12/2010 11:40:26

Quote from: jartza on 26/12/2010 10:35:20

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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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:)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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 :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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 ::))

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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:)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza 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. 
 [ Invalid Attachment ]

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza 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.
 
 

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza 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)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on 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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza 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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 29/12/2010 22:09:24

hmm?

No, I think it's an accepted idea, equivalent to the idea of two space ships near light speed closing in on each other. But it's still interesting:)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 30/12/2010 02:41:09

General gravitational energy shift is an accepted idea?

Well to be sure:

When energy goes downhill in an electric power line on a hill side, voltage and amperage increase.

When energy travels 2 floors up in a vertical line shaft in an old factory,
the rotation speed decreases and torque decreases.

When energy travels downwards in a belt and pulleys system, the rotation speed
of the lower pulley is higher, and the torque of the lower pulley is higher. And
the pulling force is larger at the lower part of the belt, and the speed of lower part of the belt is higher.

When stream of photons travel downhill, the frequency of the photons increase,
and the frequency of photon passings increases.





Version 2:


When energy goes downhill in an electric power line on a hill side, voltage and amperage increase according voltage and current meters.

When energy travels 2 floors up in a vertical line shaft in an old factory,
the rotation speed decreases and torque decreases according to RPM-meter and torque meter.

When energy travels downwards in a belt and pulleys system, the rotation speed
of the lower pulley is higher according to a RPM-meter, and the torque of the lower pulley is higher according to a torque meter. And the pulling force is larger at the lower part of the belt according to a force meter, and the speed of lower part of the belt is higher according to speed meter.

When stream of photons travel downhill, the frequency of the photons increase according to a frequency meter, and the frequency of photon passings increases according to a photon passing frequency meter.



Version 3:

When volt-ampere meter travels downhill next to an electric wire, it measures increasing voltage and current.

When a RPM-torque meter travels upwards next to a vertical drive shaft, it measures decreasing rotation speed and torque.

When a RPM-torque-speed-force meter travels downwards next to a belt and pulleys system, it measures an increasing rotation speed and increasing torque and increasing speed and increasing force.

When a light meter travels downhill next to a stream of photons, it measures an increasing photon energy and an increasing photon density.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Geezer on 30/12/2010 07:48:16

Quote from: jartza on 30/12/2010 02:41:09

When energy goes downhill in an electric power line on a hill side, voltage and amperage increase.

Interesting. Were does the extra current come from?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 30/12/2010 08:32:39

Yep Geezer that's 'bulls eye'.

It have to be there, when thinking of it its easier to see it as waves. Then you get a compression in time, much in the same way as that ambulance change sound frequency as it passes you. And a higher frequency is more energy per 'time unit'.

But if we change to photons, it gets harder. the usual hand waving argument involved, that I've seen that is, is that there will be more 'energy quanta' aka photons possibly, per 'time unit' but it don't seem to hold water, as my example with the Sun-hose showed. At least not as I can see it.

So there have to be another way of describing it as 'photons'.

----------08:39:51-------------
The frequency of a photon you say?
That one I need to think of.

I guess it depends on how we see them, as consisting of one 'energy quanta' of a decided energy, or as something able to consist of several?

"It is not actually possible to directly measure the frequency of a single photon of light. This is because a single photon is going to behave more like a particle than a wave, and the concept of frequency (cycles or alternations per second) only applies to waves.

A spectrometer is a device that disperses the path of impinging photons through an angle that is dependent on their wavelength. In this way it is possible to closely estimate the wavelength of the photons.

The wavelength measurement is then used in a simple equation relating speed of a wave, its wavelength and frequency: frequency = speed / wavelength.

The speed of light is defined exactly as 299,792,458 m/s. A photon of red-orange light from a HeNe laser has a wavelength of 632.8 nm. Using the equation gives a frequency of 4.738X1014 Hz or about 474 trillion cycle per second.

A much more accurate method directly measures the wavelength of a laser beam by counting the number of fringes in an interferometer as one of its mirrors is moved over a very precisely measured distance.

A third and most accurate method measures the frequency of a laser by measuring the difference-frequencies produced by mixing it with a series of lower and lower frequency signals. (When two waves of different frequency are mixed, two new waves are produced with frequencies equal to the sum and the difference of the original frequencies.) The lowest or reference frequency and each of the difference frequencies is directly measured by comparing them with a frequency standard such as one of the atomic clocks at NIST. Described at: http://www.boulder.nist.gov/timefreq/ofm/synthesis/synthesi.htm

The time it takes to make a measurement depends on the method used and the accuracy desired. For the highest accuracy, measurements may take a second or more. A single photon wavelength measurement can be completed in a fraction of a microsecond, but the accuracy will be many orders of magnitude less.

Answered by: Scott Wilber, President, ComScire - Quantum World Corporation "

------08:44:17-------
So yes, there is an equivalence to a wavelength, possibly you can say that is this equivalence that change.


[yor_on, as per the site's AUP: If you have more to add to your post then use the modify button. If you are responding to someone else's earlier post then use the quote button - Mod]

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 30/12/2010 14:58:27

Quote from: Geezer on 30/12/2010 07:48:16

Quote from: jartza on 30/12/2010 02:41:09

When energy goes downhill in an electric power line on a hill side, voltage and amperage increase.

Interesting. Were does the extra current come from?

It is impossible that different parts of a wire have different currents. It is impossible that drive shaft's two ends rotate at different speeds too.

Check out new versions of general energy shift theory at the second post at this page.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 30/12/2010 16:01:57

No Jartza, I'm not sure how it will express itself but I don't see it as impossible, you need to have really high energies for it to be noticeable I think. And the same arguments as we use for describing 'frames of reference' should be applicable. But you have a really nice argument there. And what I think it is about is.

"What the he* is energy"?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 31/12/2010 09:40:50

Quote from: yor_on on 30/12/2010 16:01:57

No Jartza, I'm not sure how it will express itself but I don't see it as impossible, you need to have really high energies for it to be noticeable I think. And the same arguments as we use for describing 'frames of reference' should be applicable. But you have a really nice argument there. And what I think it is about is.

"What the he* is energy"?

I guess in your world one end of drive shaft can rotate at different speed than the other.

Let me repeat: when a guy runs in a circle, at speed of light, in a deep gavity well, turning a vertical drive shaft, the guy at the upper end of the drive shaft says it's rotating slowly.

And the both ends are turning at the same speed, because it is made of hard metal. Is this really very very difficult, guys?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 31/12/2010 20:12:55

:)

You want to argue against the concept of it described as waves you're going to have a really hard time. I believe that one proved in experiments already. When looked as photons we get into anther position, and the best I can see is this 'equivalence thingie' expressed above. But it's not good enough, well, not for me at least. There have to be a better way of expressing it. Not that I know of it. But as I said, looked upon as waves it's perfectly reasonable.
==

The problem comes when one try to make 'sense' of it, as compared to our immediate reality. But to me SpaceTime don't really care about what we see as 'common sense and decency' :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 01/01/2011 05:24:58

Waves don't change when they fall or travel upwards.
And yes, wave change has been observed experimentally.

Guys name Pound and Rebka managed to built a very sharply tuned wave receiver, and
the receiver went out of tune when moved vertically.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 01/01/2011 21:35:21

Hmm, are you saying that a ship coming towards you, you seen as 'being still' relative it won't show you a 'compression' of the light-beam it sends out towards you? And if you agree with it being compressed, what do you expect to create it? Maybe you mean that they will do it there, but not when 'in falling' towards a 'gravity well'?

Then you need to define why it will differ?
It's a interesting idea.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 01/01/2011 21:41:44

You know, that way fits rather well with my own view that in a free-falling object there is no 'extra energy' delivered as long as it's a 'free fall'. I will need to think of that one.
==

But then your drawing and conclusions seems slightly misstated?

Because, in that case, your solar-plate won't see any 'extra energy' coming from getting moved closer to a black hole. If we assume that a 'free gravitational fall' won't give our 'photon' an extra energy as observed from our 'solar-plate'.

You can't have both, if you assume that the solar-cell will measure a higher 'energy' then that should be noticed at the lamps too?

That is if you're not suggesting that the lamp, due to being place further away from the BH than the 'solar plate', and so will 'negate' that energy as it now have to travel 'up hill'? But then you're just obfuscating the concept I would like to discuss.
==

No thinking of it again. My view is that the solar plate will see an extra energy. The relation comes true in the interaction. All that I mean by no extra energy existing in a free fall is that when the photon is 'propagating' there is no such 'energy' existing, only in the interaction is it created.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 02/01/2011 01:48:49

So yes, what at least I am asking is.
What the living ** is 'energy'?

Or 'light'.

And when we come to defining different positions of the lamp relative your 'solar-panel'? Then what we talked about up-hill versus 'down-hill' have to come into play.

It's no different from a 'photon' bouncing near the Event horizon. As it 'climbs up' it will become red shifted relative the 'far observer' and so 'lose energy'. And it's a very good argument of it not existing until in its 'interaction' to me :) as we otherwise would have to find an explanation to why it 'loses energy' when it's expected to be of a defined 'light quanta'. To see what I mean there you have to remember that light is Time less intrinsically, and only 'existing' in its interaction.

If you want to define it as it changes 'energy' as it climbs you will have defined an 'interaction'. That's not possible, if so, all light would annihilate as soon as it meet another gravitational potential, and it doesn't. That's where its 'timelessness' comes in too as that is what we assume to make it possible to 'propagate' vast 'distances' without losing 'energy'

You're good to talk with Jartza :)
You create difficult questions.

===

To see it my way you need to see it as a game, nothing more but nothing less either. Another proof is that this photon climbing if measured outside that gravity well will be found to have gotten all its 'strength' back, telling us that it expended no energy climbing, no matter what we would have measured it to be if inside that gravitational potential.

So looking at it as a game helps one accepting the rules. Looking at it as we observe it here in our daily life won't.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 02/01/2011 02:31:56

When clock is running slow because of moving fast, that's just relative.

When clock is running slow because of gravity, that's absolute.

Gravity has this effect of making clocks run slow.

Like an astronaut hanging around near black hole ages slowly.

Or the talking solar panel that we have been talking about, when it reports that photons are more energetic closer to a black hole it speaks like this: "theeee phhoooootoooonnss aaaare eenneerrrggeetiiic" and "waavees aaree waaaviiinng faaassst"

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 02/01/2011 02:41:28

Yep :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 02/01/2011 03:35:09

One photon takes a mirror plated elevator to go from floor 1 to floor 10

Other photon climbs all by itself from floor 1 to floor 10.

What kind of energy changes happens in this scenario?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 02/01/2011 05:48:11

That was a nice one, but I'm unsure how you think here.

If we assume both photons in motion, and intrinsically timeless. Then the 'energy' they spend doing whatever they do is equal (Zero) and their timelessness makes the distance meaningless too, as seen from their frame of reference. I'm sure there is something that will make me see your point though.

As seen from my frame of reference?
I'm not sure there. I think you are discussing what you named 'gravitational energy', right? And then both should climb a gravity well, but I can't see that taking the elevator will matter?

So what am I missing?
==

You might look at it as distances taken of course, as seen from my frame, as from their frame it doesn't matter if one goes to the moon and the other goes to the roof. they will both be exactly what they are the whole trip as I see it. The difference only measurable as we interact with them, and then created in our interaction.

Seen from my frame the photon going by itself should take the 'shortest path' and the one going in the elevator having a longer path? But that's not it, right :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 02/01/2011 06:27:34

What I see you talking about here is if my frame of reference will make a difference to the photons. And I honestly don't think it will. I will measure one as taking more time than the other, but if we grade the distance height wise up to the tenth floor 1-10 with ten being redshift (waves but still:) 0f 10 then both photons will 'lose' a equal energy, assuming some ideal identical gravity potential covering both the elevator and the free propagating photon.

It won't matter if the photon in the elevator moves a longer distance according to mine observation. The photon ignores distance as such, it does not ignore 'gravitational potential' though.
==

And the definition I have of its losing 'energy' is a very local one, as defined by our interactions with them. As you easily can see if measuring them at some place outside that gravity well, where they both miraculously will have 'regained' whatever energy we thought us finding missing as we measured them climbing.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 02/01/2011 06:55:22

You could summarize my view as that a 'photon' always propagates at 'c', but does it in different 'densities' with a gravity-well being 'denser' up-hill than down-hill. It does not propagate through matter though, instead it gets exchanged each time it meet an obstruction, like an atom, into a new photon ad infinitum. What is equivalent is that it still in a way could be seen as keeping 'c', but as defined by the type of matter it have to 'traverse'. And only if we accept the 'photon' coming out as being equivalent, even if 'down-converted' by its loss of momentum, to the original photon diving into that piece of glass, whatever.
==

But it really makes no sense looking at it as propagating. As we then have to fall back to the question how to look at a 'energy quanta', and that one hurts my head. It makes sense as long as we say that a photon can come in different 'energy sizes' though. So maybe? But no, then we shouldn't be able to measure it as being different depending on 'gravitational potential'? If we don't define it as a relation, but from where would it find the energy, without interacting? Nope.
==

It actually makes more sense seeing it as a result of a unique 'room time geometry' created in the interaction, and if so the 'relation' is the interaction as defined from the photon and where it interacts. Not what, sorry. Where.
=

And maybe that would make it possible to assume it 'propagating'. Not that it makes any difference to how I see it. The 'rules' defining what we call a propagation makes it impossible for us to notice anyway.
==

And rereading myself I see that I better stop thinking of this, he :)
Well Jartza, your turn :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 02/01/2011 08:10:08

One guy uses an elevator to go from floor 1 to floor 10.

Second guy climbs the stairs from floor 1 to floor 10.

Third guy's job is to rotate the winch that winches the aforementioned elevator up.

What can be said about energy changes happening in the three guys?




Answer: first guy's energy increases, second guy's energy stays the same, third guy's energy decreases.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 02/01/2011 19:44:22

Yep, true.
But now we're discussing those that 'do work'.

As I see it, and what I build a lot of arguments regarding light not 'moving' on, is that a photon never do 'any work'. Except in its interaction.

So, as seen from my point of view, when discussing a 'photon' we have two 'frames of reference'. One is the photons, and even if we can't look at SpaceTime from that frame we still have our definitions of it. Timelesssness, masslessness, and what more?

I'm sorry, I really need to wake up here. But those two are the important ones I guess as they are the ones that comes to mind as I think of defining a 'photon'. And defining a 'photon' as having any sort of 'real' mass seems to me to destroy the theory of relativity?

The other frame is the mysterious one. What I then call a 'room time geometry' defining your own unique one. You can exchange it for 'frames of reference' if you like as long as we agree on that we're talking about 'first hand observations' meaning yours. In that frame you will see a photon 'propagate' inside SpaceTime, as I find no better way of expressing it as seen from that frame. And it does so although its own frame should make it 'impossible', well, as I see it?

 

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 03/01/2011 05:36:49

You are climbing up the stairs  - you are not doing work.
You are climbing up the stairs carrying a box of photons - you are doing work.

A box of photons is tumbling down the stairs - photons are doing work. (making noise is the work they do)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: JP on 03/01/2011 06:38:35

Quote from: jartza on 03/01/2011 05:36:49

You are climbing up the stairs  - you are not doing work.
You are climbing up the stairs carrying a box of photons - you are doing work.

A box of photons is tumbling down the stairs - photons are doing work. (making noise is the work they do)

You need to be very careful when describing work.  Work done on A by B is force applied by B which acts on A over a distance. 

A person walking up the stairs does work on herself. 

A person climbing the stairs with a box of photons does work on herself and the box. 

When the box falls down the stairs, gravity does work on the box.

-------------
If you're talking about general relativity, there is no work done by gravity, since gravity isn't a force. 

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 03/01/2011 14:05:37

If high school physics does not say that a person climbing UP the stairs is doing work on himself AND a person walking DOWN the stairs is doing work on himself, then high school physics should change itself to say that.



Person walking the stairs up is doing work on gravity.
Person walking the stairs down is being done work on by the gravity.
That's one possibility.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: JP on 03/01/2011 14:33:46

Why would high school physics want to change itself to be wrong?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 03/01/2011 14:52:06

For consistency's sake.

When you climb your chemical energy turns into (your?) potential energy.

When you descend (your?) potential energy turns into your heat energy.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: JP on 03/01/2011 15:09:33

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. 

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 03/01/2011 15:38:39

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Madidus_Scientia on 03/01/2011 18:02:07

Quote from: jartza 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.

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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 03/01/2011 18:56:11

It's the gravitational blueshift thing.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 03/01/2011 21:43:39

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? 

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 03/01/2011 22:16:36

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

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 04/01/2011 00:44:17

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.

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.fourmilab.ch%2Fcship%2Fequations%2Fdoppler.gif&hash=d3826336eec2772e9f93f9d80cdc9c02)

"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. "

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.fourmilab.ch%2Fcship%2Ffigures%2Fgr_doppler.gif&hash=3c4997b4b5daac8da16a04d816b6ed06)

=

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 04/01/2011 06:56:35

Quote from: yor_on on 03/01/2011 21:43:39

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 04/01/2011 09:25:48

:)

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 04/01/2011 10:14:06

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.

 [ Invalid Attachment ]

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 04/01/2011 17:50:34

Nice to know Jartza :)

Define what you mean by mass ..

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 04/01/2011 19:19:22

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 04/01/2011 19:30:44

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 04/01/2011 19:52:37

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 :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 04/01/2011 20:23:32

I added one line to robot story.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 04/01/2011 21:55:36

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 05/01/2011 04:34:05

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 05/01/2011 17:39:49

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 05/01/2011 23:16:26

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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: QuantumClue on 06/01/2011 00:39:20

Oh here I come lol

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.

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

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 06/01/2011 05:07:14

Quote from: jartza on 04/01/2011 19:19:22

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 06/01/2011 05:35:06

Quote from: QuantumClue on 06/01/2011 00:39:20

Oh here I come lol

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.

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 (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 (http://en.wikipedia.org/wiki/Carnot%27s_theorem_%28thermodynamics%29)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 06/01/2011 05:43:59

Quote from: yor_on on 06/01/2011 05:07:14

Jartza, are you thinking of energy converted into mass?

Oh yes.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 06/01/2011 06:21:05

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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 06/01/2011 07:46:41

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

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 06/01/2011 09:28:08

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 06/01/2011 09:46:55

Photon worked. That made it weak.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: QuantumClue on 06/01/2011 10:40:59

Quote from: jartza on 06/01/2011 05:35:06

Quote from: QuantumClue on 06/01/2011 00:39:20

Oh here I come lol

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.

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 (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 (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 :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Foolosophy on 06/01/2011 12:12:57

The Hawking Equation for Black Hole Entropy is a good start

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fplus.maths.org%2Fissue18%2Ffeatures%2Fhawking%2Fimages%2Fformula.gif&hash=c33232921cba408337fd641e90fbb5c2)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: QuantumClue on 06/01/2011 12:19:05

Quote from: Foolosophy on 06/01/2011 12:12:57

The Hawking Equation for Black Hole Entropy is a good start

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fplus.maths.org%2Fissue18%2Ffeatures%2Fhawking%2Fimages%2Fformula.gif&hash=c33232921cba408337fd641e90fbb5c2)

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%

using the above, the OP simplifies 2GM/c² and simplifying we get: ¼mc²

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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Foolosophy on 06/01/2011 12:44:26

Quote from: QuantumClue on 06/01/2011 12:19:05

Quote from: Foolosophy on 06/01/2011 12:12:57

The Hawking Equation for Black Hole Entropy is a good start

(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fplus.maths.org%2Fissue18%2Ffeatures%2Fhawking%2Fimages%2Fformula.gif&hash=c33232921cba408337fd641e90fbb5c2)

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%

using the above, the OP simplifies 2GM/c² and simplifying we get: ¼mc²

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)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: QuantumClue on 06/01/2011 12:55:21

I must be doing the calculation wrong somehow, because rereading, the OP is not simplyfing 2GM/c² 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 ¼mc².

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Foolosophy on 06/01/2011 13:19:00

Quote from: QuantumClue on 06/01/2011 12:55:21

I must be doing the calculation wrong somehow, because rereading, the OP is not simplyfing 2GM/c² 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 ¼mc².

Quote from: QuantumClue on 06/01/2011 12:55:21

I must be doing the calculation wrong somehow, because rereading, the OP is not simplyfing 2GM/c² 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 ¼mc².

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

So the efficiency is 75%

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: QuantumClue on 06/01/2011 14:04:55

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Foolosophy on 06/01/2011 14:17:19

Quote from: QuantumClue on 06/01/2011 14:04:55

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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: QuantumClue on 06/01/2011 14:21:00

Quote from: Foolosophy on 06/01/2011 14:17:19

Quote from: QuantumClue on 06/01/2011 14:04:55

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Foolosophy on 06/01/2011 14:46:44

Quote from: QuantumClue on 06/01/2011 14:21:00

Quote from: Foolosophy on 06/01/2011 14:17:19

Quote from: QuantumClue on 06/01/2011 14:04:55

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

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 06/01/2011 18:46:07

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 ::))

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 06/01/2011 20:08:59

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'?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: QuantumClue on 06/01/2011 20:47:09

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 06/01/2011 20:58:21

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

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: QuantumClue on 06/01/2011 21:11:43

Quote from: Foolosophy on 06/01/2011 14:46:44

Quote from: QuantumClue on 06/01/2011 14:21:00

Quote from: Foolosophy on 06/01/2011 14:17:19

Quote from: QuantumClue on 06/01/2011 14:04:55

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: QuantumClue on 06/01/2011 21:12:43

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 06/01/2011 21:40:21

'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. (http://ysfine.com/einstein/bose.html) And so it all begun:) ]

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 07/01/2011 00:03:58

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

 [ Invalid Attachment ]

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.

 [ Invalid Attachment ]

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.

 [ Invalid Attachment ]

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: peppercorn on 07/01/2011 01:17:00

Quote from: jartza on 07/01/2011 00:03:58

Here we have a ....

Here we have random thought number 263..... [::)]

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 07/01/2011 03:02:57

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?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Foolosophy on 07/01/2011 05:48:17

Quote from: yor_on on 06/01/2011 20:58:21

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.


 

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 07/01/2011 06:14:51

Nice one Foolosophy :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: JP on 07/01/2011 06:42:17

Quote from: Foolosophy on 07/01/2011 05:48:17

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.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Foolosophy on 07/01/2011 07:11:14

Quote from: JP on 07/01/2011 06:42:17

Quote from: Foolosophy on 07/01/2011 05:48:17

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.

In relativity a critical assumption is that the speed of light is a physical constant.

dc/dt = 0

The energy of a photon for example is expressed by E = hf

where h = Plancks constant and f=frequency

What is the Energy of a photon if it is NON-massless?

I suspect that there are entities in the Universe that have zero mass (or at least so small that we may not be able to measure it anyway)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: JP on 07/01/2011 08:48:05

In special relativity a critical assumption is that there exists a cosmic speed limit.  Since light is massless, this is equal to the speed of light.  If you wanted to make up a universe where light had mass, you'd still have a cosmic speed limit and light would be like anything else zipping around within it.

If a photon is non-massless, therefore, you could calculate its energy just like you calculate the energy of any moving particle with mass.  E²=m²c⁴+p²c².

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 07/01/2011 08:59:10

Yor_on, why must you stick your head into the box?

Try to answer this:
A policeman is measuring, with the police radar thing, the speed of a cow that is walking away from the police. Radar wave gets redshifted. Where does the energy go?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Foolosophy on 07/01/2011 09:42:17

Quote from: JP on 07/01/2011 08:48:05

In special relativity a critical assumption is that there exists a cosmic speed limit.  Since light is massless, this is equal to the speed of light.  If you wanted to make up a universe where light had mass, you'd still have a cosmic speed limit and light would be like anything else zipping around within it.

If a photon is non-massless, therefore, you could calculate its energy just like you calculate the energy of any moving particle with mass.  E²=m²c⁴+p²c².

Quote from: JP on 07/01/2011 08:48:05

In special relativity a critical assumption is that there exists a cosmic speed limit.  Since light is massless, this is equal to the speed of light.  If you wanted to make up a universe where light had mass, you'd still have a cosmic speed limit and light would be like anything else zipping around within it.

If a photon is non-massless, therefore, you could calculate its energy just like you calculate the energy of any moving particle with mass.  E²=m²c⁴+p²c².

...beautifully stated

Have you missed something?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 07/01/2011 17:19:07

JP I think Foolsophy have a valid point there, it's not only to change one expression. You will need to go over all expressions that build on the assumption of bosons for that then, not that I know all there is :) And also all further expressions that build on those assumptions ad infinitum.

If it was as simple, and if it mattered that little to our universe then I would expected Einstein to already have considered it when creating his theory of relativity. I doubt he missed the inherent 'mysticism' in having bosons and 'point particles' interacting without 'existing' in SpaceTime. So if he never even considered giving light an invariant mass I'm sure he had good reasons.

Yep :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 07/01/2011 17:31:36

Okay Jartza, you're thinking of the Doppler shift right?
And now both you and me are flabbergasted, I don't know where that 'energy' goes?
According to Einstein that energy is a 'local expression' measurable at that place where you are as being red or blue shifted.

If you look at it as waves we have definitions and explanations to why it will behave like it does, getting 'compressed' or 'stretched out' in time. But although the explanation is understandable it puts an awful lot of importance on 'time' that then becomes something 'concentrating' the energy or 'thinning it out'. And how that fact can make a 'energy amount' do more or less 'work'?

Don't I wish I knew that one :)

And when we look at it as particles it becomes even weirder.
==

If you accept that 'energy' always will be a local expression it will make sense though. We have to remember that we're latecomers to this universe, before we came into existence a lot of other things already had happened, in that mysterious 'time'. That means that although we exist, we might not be the reason why a universe exist, even though I think we're an expression of the universes 'need' to create into more and more complexity. Complexity is not entropy though, they are different expressions where complexity is certain types of ordered entropy that creates things like us.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: simplified on 07/01/2011 17:52:21

Quote from: jartza on 07/01/2011 08:59:10

Yor_on, why must you stick your head into the box?

Try to answer this:
A policeman is measuring, with the police radar thing, the speed of a cow that is walking away from the police. Radar wave gets redshifted. Where does the energy go?

The policeman receives that energy, only relatively of the cow.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 07/01/2011 18:18:35

Yeah, that is right. It's all local expressions. If that policeman had been on a bike going the opposite direction the energy would have been 'weaker', delivering less energy per 'time unit'. and if he was going towards the energy would have 'stronger'. The important part to me is that it actually will do more, or less, work depending on his direction. It makes for a very 'geometric' universe, doesn't it?
==

But don't mix that with our box. That's about frames of reference.
What you could say in the box scenario is that if the guy inside start to run, very very fast, towards the 'bouncing' wave he will say that it have gained 'energy', with all right too. If he runs away from it he will state that it's 'weaker'.

But being 'at rest' with that box he will see light the same way as on Earth.
==

And that's weird :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 07/01/2011 18:41:13

Look at it this way. If we imagine that I get on my Spacecraft, building up a tremendous velocity relative my place of origin (Earth) 'my universe' will 'contract' as by time dilation and Lorentz contraction.

If you accept that both exist 'for real' that should mean that 'locally', for me, that same universe that contained all that energy a universe now can be expected to have will, for me, suddenly 'exist' in a lesser 'area'.

That doesn't make sense, does it :)
But it does, it's a logical extension of what we already see.
==

And that's one of the reasons why I can accept the idea that a spring getting compressed will keep that 'energy', even though I 'normally' would expect that energy to dissipate, same as with kicking a ball, where we know that after it stopped moving it will weight the same. The difference here is that I leave the spring compressed. 'locking' it into a state where some of the energy will dissipate, just like with the football, because that's the excess energy I brought into play compressing the spring, expressing itself as heat for example. But as I leave it compressed, locking it, it will contain more energy than before relative me, and so some more 'mass'.
==

The question you might ask is whether this springs 'new energy state' aka mass will be true for all 'frames of reference' that may measure it. I think it has to be? And that's interesting :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: simplified on 07/01/2011 19:55:13

An escaping mirror and a motionless mirror receive various energies from photons.An escaping mirror takes more energy.
For clearness we should create the universal law of energy distribution . [;)]

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 07/01/2011 20:38:17

Maybe?

I don't know, it's relations to me. Like a constantly changing balance, differently defined from all frames but always keeping a equilibrium of some weird sort. If you look at SpaceTime as a whole expression, in where you can't differ out 'time' from the 'room' it exist in then they all are different 'room time geometries' to me.
==

And the only one being a 'first hand expression', will be the one you exist in, for you.
==

To see what such an idea make of 'energy' you can ask yourself what you would expect to happen when you turn on a light-bulb in that Lorentz contracted (and in a way time dilated, even though not as your 'arrow of time' aka wristwatch, shows it.) SpaceTime.

Do you expect it to explode with energy as 'SpaceTime', according to your definition on that ship, suddenly have 'contracted'? Or do you expect it to act just the same as on Earth?
==

So what is 'energy'?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 07/01/2011 23:08:47

In fact you only have two choices if you accept the theory of relativity, as I see it now. Either you define Lorentz contraction like a 'twisted room geometry' fooling our measurements, and senses. That is, not being 'real' and just an 'illusion'. But then it seems to me that you will have defined 'time dilations' the same way as those two are one of a kind, to me that is.

Or you accept.
==

And if you define it as an illusion, what about 'Doppler made energy'?
Also an illusion?

Think of it this way to see what I mean.
All uniform motion are inseparable in a black box.
Let the box move uniformly having a velocity.

Let a siren, equivalent to a light beam, be situated 'at rest' relative our box origin (some original position for our two objects (and in SpaceTime), from where we define a start), make a sound.
Will we find that sound to be compressed when coming at it?

As any uniform motion is inseparable from being at rest, inside that 'black box' I then can assume either one siren, or a multitude of sirens, all sounding for a incredibly short moment and all at different pitch.

Probably there are better examples :) than this but my idea is then that you can't really separate this from being still, can you? And if you can't you might just as well assume that there was this 'infinity' of different sounds, all of them representing a different 'energy level' (and as we 'know', all Doppler).
==

I know, this one we can argue about :)

But even if we consider it not 'constantly changing' instead first 'blue shifted' and then after passing a single 'red shift' we still have the same phenomena, namely a Doppler shift representing different 'energies' relative the box.
==

If you accept the idea, just for the moment, how many of my so called 'room time geometries' do you think we have? I would say two, one is yours inside that moving box, the other was our siren, and possibly a 'background', as we can assume both to exist in a 'SpaceTime'. But looked from my perspective that 'background' only exist relative the objects existing in it, and so is a inseparable part of every objects existence. So only two then :)

But we had, if you accept the idea, a multitude of 'energies' from that box's perspective, and all of them unique.
So where do they belong?
With the siren?
With the box?

Or as a 'relation' to our 'system'?
==

To really see the weirdness.
The same phenomena, but you was still.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: JP on 08/01/2011 01:30:33

Quote from: yor_on on 07/01/2011 17:19:07

JP I think Foolsophy have a valid point there, it's not only to change one expression. You will need to go over all expressions that build on the assumption of bosons for that then, not that I know all there is :) And also all further expressions that build on those assumptions ad infinitum.

Special relativity appears to be true based on a lot of tests, and the cosmic speed limit appears to be the speed of light, which also indicates that photons are massless.  If they had mass, all the experiments we've done would still be true and the cosmic speed limit wouldn't change, but the speed of light would.  I never said only one equation changes.  All our theories about photons would have to be modified to account for this mass.

Quote

If it was as simple, and if it mattered that little to our universe then I would expected Einstein to already have considered it when creating his theory of relativity.

Of course he had good reasons!  Light appears to be massless and there is no evidence to the contrary. Of course, we now know that Einstein's theory of relativity doesn't require that light is massless, and that it would still hold if light had a tiny bit of mass.   

Quote

I doubt he missed the inherent 'mysticism' in having bosons and 'point particles' interacting without 'existing' in SpaceTime. So if he never even considered giving light an invariant mass I'm sure he had good reasons.

I'm not sure this is true.  Bosons do exist in space-time when they interact.  Quantum electrodynamics describes how they do so.  Photons and gluons are zero mass Bosons, but W and Z bosons have mass and are bosons.  The Higgs particle is predicted to be a massive Boson as well.  Also, all particles have invariant mass.  Photons are just special because their invariant mass is zero. 

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 08/01/2011 01:42:54

Yeah, you're perfectly right JP, bosons exist, Although I'm having real big trouble understanding why and how :) Photons we have experimental evidence for, gluons? I don't know any experimental proofs myself, isn't they theoretical 'particles' still?

Are you stating that Einstein didn't 'know' that his theory would hold if he allowed for a slight invariant mass? Maybe, I'm not sure there, he seems to have looked after the simplest explanations that made sense to him, and us too possibly :) I would have expected him to want them to have a certain invariant mass, if he thought he could get away with it as they still are mysterious things, no matter that we know them to interact.

That's a really nice question btw, why didn't he allow for a slight invariant mass if it now would make no difference? To me that would make a world of difference, as Jaztra's ideas for example, mass/energy?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: JP on 08/01/2011 02:31:11

Quote from: yor_on on 08/01/2011 01:42:54

Photons we have experimental evidence for, gluons? I don't know any experimental proofs myself, isn't they theoretical 'particles' still?

We can't see them directly, but we can and have measured their decay products in high energy experiments.  The observations match the theoretical objects we call gluons and we don't have another explanation for them.

Quote

Are you stating that Einstein didn't 'know' that his theory would hold if he allowed for a slight invariant mass? Maybe, I'm not sure there, he seems to have looked after the simplest explanations that made sense to him, and us too possibly :) I would have expected him to want them to have a certain invariant mass, if he thought he could get away with it as they still are mysterious things, no matter that we know them to interact.

That's a really nice question btw, why didn't he allow for a slight invariant mass if it now would make no difference? To me that would make a world of difference, as Jaztra's ideas for example, mass/energy?

I'm sure he knew, but why would you want to postulate that light has an invariant mass when there's no evidence that it does?  We'd also have to tweak Maxwell's equations and quantum electrodynamics to account for this and they'd no longer be quite so elegant (though QED wasn't around yet when Einstein proposed special relativity).  The elegance of Maxwell's equations alone is enough of a reason why it's natural to leave light massless.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 08/01/2011 02:56:00

Yeah, I started to look for it on the net. I think this one is okay? The Concept Of Mass by LD. Okun. (http://www.worldscibooks.com/etextbook/6833/6833_02.pdf)

There is some references is in it to how Einstein came to his conclusions (p.14), but I wish he was here :) I would really like to ask him about that one.
==

In it he mention how Einstein showed how a massless particle could transfer a 'invariant mass' change in form of energy, (in his 1906 paper) without itself needing to be of invariant or proper or rest mass. I'm starting to feel that the best expression might be 'rest mass' myself :) awh, invariant is cool too :)

And I stated that a photon has no inertia. Would you agree to that? The reason I do is because there is no acceleration involved, if you don't have an acceleration I don't see how you can have an inertia? When a photon 'bends' it's only following a least energy expenditure, as it have to do, only existing in a interaction, well, as I see it. But I've seen others referring to the photons 'inertia' :)
==

As for "why would you want to postulate that light has an invariant mass when there's no evidence that it does?" I would, because then there would be no 'shadow world' existing, as all would be referable too as having that same property, 'mass'. Although that seems to change the way photons would 'interact', as they then possibly could gain 'mass' by energy, as that spring could?

I don't really know there, that one becomes confusing considering that we then would have 'mass full' photons 'interacting' with each other? I prefer it as it is actually :) I like mysteries. Life would fast get boring otherwise.
==

Correct me if I'm wrong but wouldn't the concept of photons having even the slightest invariant mass mean that they would be able to interact kinetically? Without getting annihilated? And what would it make 'virtual particles'? Maybe it is allowable according to the Heisenberg uncertainty principle but a invariant mass should mean an awful lot of more mass than we account for, at least if we assume them as freely propagating in space?  Aha, stop wondering why there is so much mass missing :) Nahh.
==
I'm sure that my last one soon will become a full fledged theory :) Not by me though.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: JP on 08/01/2011 02:59:57

Yeah, that's the most commonly used mass.  It's called a lot of things: the invariant mass, the rest mass.  That paper seems to call it the Newtonian mass (which is a new one to me).  Anyway, this mass is a constant value for a particle, no matter how it's moving. 

The other kind of mass is the relativistic mass, which varies with speed.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 08/01/2011 03:35:23

As for the way Okun find 'mass' to cover it all though :) I don't know, not as long as we can't make that lasting piece of matter from light. I'll stay with rest-mass for a while. And invariant, although a photon seems to me to be rather 'invariant' too, if I got it right?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 08/01/2011 09:27:59

Quote from: simplified on 07/01/2011 19:55:13

An escaping mirror and a motionless mirror receive various energies from photons.An escaping mirror takes more energy.

Oh yes.

Simplified understands, others don't understand.


 

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 08/01/2011 17:23:27

Jartza, you need to prove that uniform motion will differ from inside that black box first, won't you?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 08/01/2011 18:15:14

Energy is limited by one thing it seems to me.
The speed of light (in a vacuum).

Somehow it's related.

You can also see it as an expression of 'objects' interacting.
But regulated by light. And 'motion', as easily can be proved if you let one bystander be 'still' relative an explosion, another going from it, a third going towards it.

All three will give different 'energies'. Then you have to differ between the 'conceptual view' in which we look upon those three relations, analyzing their relation finding a common connection, and the one in where you're 'there' observing a 'single outcome'.
==

But 'motion' falls under lights speed in a vacuum, so in the end we come back to one constant, I think?
It all depends on how you look at SpaceTime, as a God, or as a observer.
==

To me the observer is the important thing, if we would see a different 'energy' but it being an 'illusion' created by our 'moving observers' then it wouldn't matter and the view point of 'God' would be appropriate. But if we find the 'energy'  to differ, then the three 'observers' all are right, and 'God's' point of view becomes slightly skewed.

To see it better you can imagine the two 'moving observers' as having a uniform motion, inseparable from being 'at rest'. It is a fact that we have no 'rest-frame' in the universe, and so all uniformly moving objects are contenders for that universal 'frame of rest' if you like. If you find a way to put this notion into doubt we will have a different universe. That also mean that when I define two observers as 'moving', then that is only a 'relative truth' relative the third, that I then arbitrarily decided to define as 'being still'. Although he is being still relative the origin of the explosion there is nothing guaranteeing that this is the ultimate place of 'rest' in our universe.

In 'reality' there are no such thing, or all uniformly moving 'objects' will be 'still', no matter what velocity you define to them.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 08/01/2011 18:56:10

Read it closely one more time and then try to see a universe from the viewpoint of 'energy expended'. Then tell me what you find in the case of our three observers.

What is 'energy'?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 08/01/2011 19:27:51

If you accept that the energy is real for all three then you've killed the idea of 'objectivity', or that 'God like' point of view. That you can find a common relation and three energies guarantee only that there seems to be a 'sliding relation' between different observers, joining their observations. But, and that's the important thing, those energies was all real, by themselves. And as a 'system' you might want to define it to have a uniform motion, well, you're God after all :) making those energies tell you yet another thing about their 'strength' relative the 'universe'.

The 'sliding relation' you see is meditated by radiation. And that's governed by lights invariant speed in and from all frames of reference, namely the speed of light in a vacuum.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 08/01/2011 19:39:23

So there is no way to define a universal measure of 'energy', but they will still, all three, do different amounts of 'work'. Then 'energy' also is a very 'local definition' and no truth you can use to describe a common SpaceTime. Remember that all uniformly moving frames, according to me, can be seen as equivalent, no matter what velocity you measure from your 'position'.
==

If you accept that definition I'm sure you will find my definition of different 'room time geometries' making more sense. In your unique 'room time geometry' 'times arrow' give you the same expiration date no matter where you are, always 'ticking' with the same duration. The only 'time dilation' you will see will be the one defined by SpaceTime accelerating in 'time' as you 'move' near light speed. The Lorentz contraction can also be seen the same way, if you like, as an expression of 'SpaceTime' as your yardstick will give you the same measurements as before, measuring.

But, does that makes sense? I like it better if I define it as it all being one whole 'expression' where we use 'energy' to change it. And that's why I like 'energy expended'.
==

Your 'room time geometry' will then be the whole of 'SpaceTime', and all yours. My 'room time geometry' will be another, having a different 'SpaceTime'. Then there is the question why they 'join' into one 'big SpaceTime'? I think it's by the same 'sliding relations' I mentioned before, radiation. And that's also why we have so many 'points of view', all depending on where we imagine us standing, observing. But my 'room time geometries' are defined from each 'object' existing, as they all should describe something unique.

Don't know how much sense it make seen 'globally', but 'locally' I'm happy with it :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 09/01/2011 07:13:45

Here Bob is standing in a moving box. Look at Bob's feet, so you can see what direction Bob is facing.

 [ Invalid Attachment ]

Bob's back receives blue rays, Bob's belly receives red rays. Every time that a blue ray hits Bob's back, Bob's kinetic energy increases. Every time that a red ray hits Bob's belly, Bob's kinetic energy decreases.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 09/01/2011 07:53:40

Hey moderators, this post has gone broken:

http://www.thenakedscientists.com/forum/index.php?topic=36274.msg338574#msg338574 (http://www.thenakedscientists.com/forum/index.php?topic=36274.msg338574#msg338574)

Two last pictures are not showing.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 09/01/2011 09:17:41

Quote from: Foolosophy on 06/01/2011 14:17:19

Quote from: QuantumClue on 06/01/2011 14:04:55

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?

Don't let the subject line confuse you.

Sun has 0.7% efficiency.
An efficient real quasar has 20 % efficiency.
An ideal quasar has 75% efficiency.
A photon rocket has 50% efficiency at velocity 0.5 c
A diesel engine has 40% efficiency, but not really, it's more like 0.000001%
An engine where photon gas expands in a cylinder has ideally 100% efficiency.


 

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 09/01/2011 14:37:20

Okay, you only need to do one thing for me Jasztra. Show me how all uniformly moving frames in SpaceTime aren't equivalent? No big deal:) And I will look forward to your proof. Because to me that is what your idea builds on. That in a 'black box scenario' you always must be able to see your uniform motion, and velocity.

Otherwise you won't, and that's how Einstein looked at it, and I'm afraid that I agree too, as he defined his principle of equivalence for all uniformly moving frames in SpaceTime. And the direct effect of that principle is that the light you will see will seem just the same as back on Earth, when you travel inside that box. Check it up, it's one of the really big principles of special relativity.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: Foolosophy on 09/01/2011 14:44:45

Quote from: jartza on 09/01/2011 09:17:41

Quote from: Foolosophy on 06/01/2011 14:17:19

Quote from: QuantumClue on 06/01/2011 14:04:55

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?

Don't let the subject line confuse you.

Sun has 0.7% efficiency.
An efficient real quasar has 20 % efficiency.
An ideal quasar has 75% efficiency.
A photon rocket has 50% efficiency at velocity 0.5 c
A diesel engine has 40% efficiency, but not really, it's more like 0.000001%
An engine where photon gas expands in a cylinder has ideally 100% efficiency.


 

,,,,interesting

and yet mass and energy are conserved

its all a matter of where it ends up

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 09/01/2011 15:22:12

Look at your box Jartza, it's moving. If I assume a uniform motion then Bob is in a 'free fall' following a geodesic. That motion you then expect to change the light, delivering a mass/energy can't under those circumstances, as there is nothing differing it from any other uniform motion. And so you can define the box as being 'at rest' in Space. Would you still expect the photons to behave this way if the box was at rest?

If we assume a constant uniform acceleration then you have an equivalence to a gravity.  If so Bob will see the photons blue-shift slightly as they follow the gravity potential to the rear of the box (in its motion) and those coming from the rear towards him will be red shifted.

If we assume a non-constant non-uniform acceleration then we will have the same phenomena.

Your idea will work, as I see it, in two of the cases, involving gravitation/acceleration as it from Bob's frame will be equivalent to a red respective blue-shift. In the third it can't work, and that's a uniform motion.
==

It's not the same as assuming the photons 'penetrating' our box from the outside. Inside your black box the photons and you start being 'at rest'. Not so outside.
==

What Einstein said with that statement about all uniform frames being equivalent was that motion isn't what you think it is. On Earth you will always have something to define a motion from, well except from inside that black box :), but it's very seldom anyone have tried that one, if ever. And here you also will have Coriolis force etc to tell you that you're on a planet, but in theory the same truth can be proved on Earth I think.

Motion, simply expressed, will always need a referent to be proved, without it you can't say that you are moving. Imagine yourself in space without any stars. totally black and you just 'hanging out' there :) If you were accelerating you would know that there was something 'attracting' you. Space wouldn't be isotropic anymore, but in a uniform motion, no matter how 'fast', Space always will be 'isotropic to you, without any reference frames to use, like a star. And so uniform motion and Lorentz contraction both becomes 'weird'.

Stop calling it motion, forget that expression and look at it from what happens to you in that space. Then you have three possible effects 'acting' on you.

1. A uniform motion, inseparable from being at rest. That means that they are the absolute same.
2. A uniform constant acceleration that you will find a even unchanging 'gravity', equivalent to Earths.
3. A non-uniform, non-constant acceleration making you feel a 'force', possible to define as a uneven constantly changing 'gravity', but, not equivalent to what you feel on Earth.

Forget about 'motion'.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 09/01/2011 16:57:27

So taken to its limit, can I define being at rest as having a uniform motion?
Yes I can.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 09/01/2011 17:12:01

Here we see a moving head looking into a box that we see standing still. The head sees a light beam bouncing in the box. The head sees light beam change color as we see in the picture.

 [ Invalid Attachment ]

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 10/01/2011 04:18:42

Let's clear it a little.

To see the waves, they will need to interact with the one seeing them. They are not in the box as you 'see' them. Assume the box to stand still, you passing in your rocket. Make the room into two mirrors, perfectly reflecting. Let a light-corn  (Yeah, I know, terrible nomenclature huh:) bounce between the mirrors, first  ---> and then <----

Like this  | -->  <-- |

Let us assume that you pass the mirror pair  <------------ thattaway.

What will you see?
where does the interaction take place?
Would it be the same if you were standing still?

Do you think the light-corns you saw, standing still, was the same 'sort' you saw when moving pass them?

If you do think they were the 'same'. where was that 'interaction' happening, changing the way you saw them?

In between the mirrors?
Between the mirrors and you?
Or in the final interaction meditated by your eye?

And finally, do you think the different energies could do different work?
Or do you think they are 'illusions'?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 10/01/2011 08:02:32

Well, heads that think that light sources are moving, tend to think that the motion of the light sources causes blue and red shifts.

Those heads that think that they are moving, tend to think that they are butting their eye into light, when they are facing the forwards direction,   

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 10/01/2011 09:24:49

You're right :)

If we assume a uniform motion to the rocket we can exchange the motion of the rocket for one of the box instead. But either way I will in 'reality' only be able to notice a blue shift as I (or the box) approach, and then a subsequent redshift as I (or it) leave. I can see how you think but the light inside that box doesn't exist for you. The only light that exist is the light reaching your detector/eye. And as we assume a uniform motion here then the light inside the box, for someone being in it, will be 'normal as long as he is 'at rest' versus it, not rushing at light speed towards one of the mirrors.

And as it was you Jaztra :) that suggested a uniform motion here and arbitrarily changed your 'point of view' when it came to who it was moving relative whom, you will have to agree I think?

The only way you can change viewpoint like we did here is when we talk about 'uniform motion'. As soon as you decide to put in an acceleration you will know who is moving. Those two frames of reference, the head versus the box won't be equivalent anymore as one will feel a 'force/acceleration'.

But I see what you mean, and as I said before. As soon as we're talking accelerated frames your mass/energy situation comes into play, as far as I can see :)

==

One thing though. 'Accelerated frames', as I define it, will have to expend energy to be 'accelerated'.

Gravity does not create a 'accelerated frame' for our photon, even though it will from the solar panel, at rest with the black hole, do so. To see my thinking you can ask yourself what the ultimate 'velocity' would be for a piece of matter falling into a black holes infinite gravitation, ignoring tidal forces. Would it be light, or at least as close to light speed as matter can reach?

And when you done that you might ask yourself what gravity is?

If it is a geodesic and no force then that 'speed' we wondered about is no 'force' either but the ultimate 'being at rest' matter can achieve relative gravity. Remember now that if we exchange the matter for a photon the equivalence to the speed is its blueshift. And we know, as it is of a defined energy quanta, that if going up from a EV the photon to the far observer will be red shifted but to its partner coasting beside it be 'as always', at all times. And so, if we assume it 'propagating' it will, when coming out of the gravity-well, have expended no energy.

If we don't assume it 'propagating' the only thing that will decide its energy is the interaction and where it takes place. Then what you see is 'the reality'. if it is red shifted then that is real, and it will be of a weaker energy. If you would meet it outside the gravity well then you would measure another energy etc. The 'where' we talk about here is your coordinate system relative the objects surrounding you, defining your possible gravity/speed.

To me there is a big difference between you forcing a 'change' in your coordinate system locally and when just following a geodesic, expending no energy. And all change that expends energy, whether by you or on you are the same, but not gravitational potentials, aka the 'weird metric.'

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 11/01/2011 09:10:05

So, a box is moving downwards. We outside the box say that inside the box light is red shifting.

Note that box is moving, like an elevator. Box is not free falling.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 16:49:14

Hmm I think I was a little tired yesterday. A uniform constant acceleration is equivalent to a gravity, right? And I think that is right, so how the he* will I get that to work with my view that there is no acceleration working on a 'freefalling' photon. don't Í contradict myself there? Let us test it.

If I am inside that rocket, accelerating at one perfect gravity constantly. Will the light inside it red/blue-shift?

(need some coffee:)

As for the 'energy' or as some think 'potential energy' expected to rest in that frame. Forget it, that one change with where our captain look, and steer. Remember that according to me it's only the interaction that is important. The rest of it is theoretical framework applied to 'what if' sceniarios. I'm not interested of them, I'm looking at what's really happening. And in our rocket accelerating the atoms don't 'jiggle' any more, as far as I know?

It's not the same as when applying a 'force' from the outside, as I did on the spring, compressing it. The energy that moves our rocket is already accounted for, being fuel transforming, so seen from that view our rocket in fact should lose some energy in its transforming, I think? Or maybe it's a perfect equivalence there, I'm not really sure about that one, but I think it loses 'something' in its transformation. But of course it does :) But the universe as a 'whole' will not as it all happens inside a 'system' consisting of 'my SpaceTime'.

But inside that rocket constantly keeping one gravity, inside a black room without windows. Waking up, not knowing you left Earth and without anything telling you.

Do you expect your light to behave differently Jartza?
How?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 11/01/2011 16:57:22

We are in the box now. Light does not behave differently according to us.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 17:03:32

Okay :)

I will now get my coffee :)
Lots and lots of it..

and then we'll have some fun :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 17:19:44

Next frame.

Same room, but our rocket is now accelerating non-uniformly, changing acceleration constantly. Will the light still behave the same as on Earth? Inside that room. Assume the room to be 10 light seconds. Put the light bulb at its far end, and in the direction of the rockets motion with you sitting near the rear. And this one is tricky, I expect people to have different opinions here, I have them too :) But I lean towards one though.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 17:26:28

Now, let's go back to the first room. Let the room be 10 light seconds long. light-bulb at the front, you at the rear. Will the light be blue shifted reaching you?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 17:31:02

The last room. Make it constantly uniformly moving, that is being in a 'free fall' with you being weightless inside it. Make it 10 light seconds long, light-bulb in front, you in the rear. Will the light 'blue/red shift' reaching you?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 17:39:01

It's about 'frames of reference' and where they end. The last example is the simple one, in that one the 'frame of reference' definitely is the same the whole time that light travel to you, would you agree to that?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 11/01/2011 17:39:17

Quote from: yor_on on 11/01/2011 17:19:44

Next frame.

Same room, but our rocket is now accelerating non-uniformly, changing acceleration constantly. Will the light still behave the same as on Earth? Inside that room. Assume the room to be 10 light seconds. Put the light bulb at its far end, and in the direction of the rockets motion with you sitting near the rear. And this one is tricky, I expect people to have different opinions here, I have them too :) But I lean towards one though.

We in the box will say that light that goes up in weak gravity field and goes down in a strong gravity field gains energy.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 11/01/2011 17:41:49

Quote from: yor_on on 11/01/2011 17:26:28

Now, let's go back to the first room. Let the room be 10 light seconds long. light-bulb at the front, you at the rear. Will the light be blue shifted reaching you?

I, in the box, will say that the light was blue shifted.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 11/01/2011 17:43:44

Quote from: yor_on on 11/01/2011 17:31:02

The last room. Make it constantly uniformly moving, that is being in a 'free fall' with you being weightless inside it. Make it 10 light seconds long, light-bulb in front, you in the rear. Will the light 'blue/red shift' reaching you?

I in the room, will say that light was not blue/red shifted.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 17:44:05

What differs the last one with the ones before? As I see it 'energy expended'.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 17:48:29

Now, is that true?
Yep.

In both my first examples I expended energy to make this acceleration.
In the last example I might, or might not have expended energy, but it's indeterminable from inside that room. So let us look at light falling in to a Black Hole again. Where does that light 'expend energy'? Nowhere as i see it.

Does the Black Holes gravity expend 'energy'?
==

'Indeterminable inside that room' as I'm not doing it (expending any energy) as we measure. And that's also what I mean by us looking at what really happens, not considering the 'what ifs'.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 11/01/2011 17:55:41

Quote from: yor_on on 11/01/2011 17:48:29

Now, is that true?
Yep.

In both my first examples I expended energy to make this acceleration.
In the last example I might, or might not have expended energy, but it's indeterminable from inside that room. So let us look at light falling in to a Black Hole again. Where does that light 'expend energy'? Nowhere as i see it.

Does the Black Holes gravity expend 'energy'?
==

'Indeterminable inside that room' as I'm not doing it (expending any energy) as we measure. And that's also what I mean by us looking at what really happens, not considering the 'what ifs'.

Quote from: yor_on on 11/01/2011 17:48:29

Now, is that true?
Yep.

In both my first examples I expended energy to make this acceleration.
In the last example I might, or might not have expended energy, but it's indeterminable from inside that room. So let us look at light falling in to a Black Hole again. Where does that light 'expend energy'? Nowhere as i see it.

Does the Black Holes gravity expend 'energy'?

No energy expended by anybody when light falls into black hole.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 17:58:51

So I am contradicting myself, am I not :)

I expect a in falling photon to a Black Hole not to expend any energy.
I also expect 'Gravity' not to expend any energy.

So why do I agree on that it will seem blue shifted?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 18:04:23

Remember how I asked what would happen to that piece of matter in falling, accelerating? And how I then compared the 'speed' it got uniformly moving, 'accelerating' from the point of an observer on the EV, as being at rest with gravity?

Well, according to how I see it it has to be that way. Matter at light will be 'at rest' (with infinite 'gravity'.). As matter can't reach light speed that's an impossible one, but I still put the 'limit' at light speed.

What does that way of looking at it make 'motion'?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 18:13:38

Assume 'gravity' to be infinite everywhere inside 'SpaceTime'. Choose a point inside to observe from. Then drop a 'light-corn' inside this system. Will the light-corn blue shift. Do it need a direction?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 11/01/2011 18:24:15

What is light corn??

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 18:25:14

As every point is infinite gravity in this system it doesn't matter, as far as I can see, where it goes. It will be blue-shifted if we imagine that gravity is a 'force' 'expending energy'. If we define that 'gravity' doesn't expend any 'energy' then we still need to explain why we expect it to become 'blue-shifted'. And the only thing we have left is the relation between the observer and what he observes.

And then the next question becomes, as we now will have a 'line of sight ', namely the observer and what he observes. Will that matter?

Will that 'photon' red-shift if it, according to the observer, 'moves away' and 'blue-shift' if it comes towards him?
==

Need more coffee here :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 18:25:53

It's that magical 'photon' :)
A light-corn ::))

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 18:33:10

And yes, according to how I see it, the 'line of sight' will introduce the 'relation' making it possible for that photon to behave differently. Put two observers inside, the 'photon' in the middle 'moving' towards one observer. We will find two definitions of that photons 'energy. and then also a third, the 'invariant light-quanta' we expect a photon to represent in 'itself'.
==

So, does it make sense?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 18:38:13

That's as far as I am at the moment Jartza.
I think this is correct, but I do not connect all the stuff yet.

And I'm using some weird examples too explain it:)
==

On the other tentacle: I could just as easily be bicycling in the great younder :)
Still? What's wrong with that, fresh air and lot's of sights.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 19:23:07

What this way of looking at SpaceTime does, as it seems to me now, is to make SpaceTime a example of something not being at rest. SpaceTime, if this would be right, have to be a system that even though it's 'in balance' still are out of bounds in some way. And maybe entropy is a description of how the universe tries to rectify this situation?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 11/01/2011 20:05:36

Quote from: yor_on on 11/01/2011 18:33:10

And yes, according to how I see it, the 'line of sight' will introduce the 'relation' making it possible for that photon to behave differently. Put two observers inside, the 'photon' in the middle 'moving' towards one observer. We will find two definitions of that photons 'energy. and then also a third, the 'invariant light-quanta' we expect a photon to represent in 'itself'.
==

So, does it make sense?

Not nearly enough.

A box is standing on a concrete pillar, in the box a photon is bouncing, the pillar is sinking into the soil. The photon is doing the work of a pile driver.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 21:59:01

No, even if it was possible to use the momentum that way, you could only use it once. No matter how you define 'propagation' there is only one interaction per photon. You need another argument for that I think?

But what you are thinking of is the blue shift we expect when it getting deeper into a gravity well, right?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 22:07:28

In my definition a 'photons' interaction and its relative 'strength' is defined by its relations. Those are what will decide it. What decides the relations I expect to be constants. What decides constants I don't really know :) But my hope is that if we find enough some bright guy/gal will solve the puzzle and make some sense of it. I'm not that interested in making everything following each other in a linear logic as long as I can see enough reasons/experiments for why it seems to work anyway. And I trust in Einsteins deductions, he was right as far as I know, and then we have QM. We just need to turn it around to see where they fit together. Because they will be found to 'fit' somehow. Or we have to assume that one of them is wrong.
=

And I'm not thinking linearity only when I say I expect them to 'fit'.
I think we need both.
==

Let's go back to that uniformly constantly accelerating rocket. Exchange it for Earth, then you will have a frame in where we do not notice any blue shift. Exchange Earth for a neutron star, and I will expect us to notice it though. But in the rocket the gravity well is situated behind the exhaust, making the 'gravity potential' equivalent to the one you would get fastened to the wall inside a skyscraper, 10 light seconds long, looking straight up in the roof. And doing so, on a neutron-star, should deliver a blue shift from the photon coming from that roof, as I think?

So, I think the equivalence is there, and so, I become of several minds :)

Let us assume that this is right, what does it tell you about motion? Does the Earth move in some way, Undefinable for us, and without 'expending energy'? And would then that 'energy expended' I would like to trust in be wrong?

What we can say is that the two examples seems equivalent, but there is one difference. I need to expend energy to reach the equivalence with my rocket. But not in the other example. Which makes me wonder how to define the difference?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 11/01/2011 23:23:54

What did I think about matter? That it needed to reach light speed to be at rest relative gravity? Assume that there is one state of rest in this universe, and that is light. All other states either need to expend energy to reach it or they need a invariant, proper, restmass. And that the more mass they have the closer they come to it, up to the limit where they become a 'Black Hole'.

Why shouldn't 'relative mass' then create a black hole?
Because it doesn't exist, those definitions only exist in their interacting. And a rocket infinitely near the speed of light can only express that 'relative mass' when interacting. And if so the 'momentum' we talk about doesn't exist either, except in its interaction.
==

Why do I say that they don't exist?

Because they don't jingle any 'atoms', ahem. I can only use that description when it comes to how 'relative mass' is thought to work, as I reserve momentum for photons, but I can stop doing that, for a while :) And then neither 'momentum nor relative mass' will jingle any atoms in that rocket, well, as far as I know? If you assume that it does it should destroy that rocket pretty fast as the 'energy' builds up with 'relative motion'.

All motion has to be relative, as we have no frame of absolute rest in the universe. All of this would probably be wrong, including the idea of relativity, if we could prove one 'gold-standard' for what motion 'really' is. Because, if we had such a one, then I would question the definition of all uniformly moving 'frames of reference' being the same :) With all right too as I think.
=

So, can we translate motion into mass? Yep, that's what we do with 'relative mass', and if you like, with 'momentum' too, except when coming to radiation. But are they the same? If a moving rocket can't become a black hole, then they can't be the exact same. Which make sense to me as matter and 'energy' neither is the 'exact same', all as I understands it.

So motion seeks a state of rest?
Well, yes :) We have two types defining motion as I know.
Matter and light.

But in my 'redefining' light suddenly becomes 'at rest', as matter too could be if only reaching that 'limit' of 'c', but it can't. 'Compression' of matter, on the other hand, that seems to meet the qualifications, if we now believe in Black Holes? and they exist, don't they?
==

Let's look at what might be the limits for 'frames of reference'. I used 10 light seconds for that room to situate the lamp far enough from the observer to make us see but, will it matter if the distance is shrunk? How much can I contract a 'frame of reference'?

I expect Plank size to be the limit myself? Smaller than that and we 'fall out' of our SpaceTime.
==

Do you believe in the Doppler effect? Okay, do you believe in Lorenz contraction too? If you don't you better stop believing in a time dilation :) And thinking of it, you should really start to doubt that Doppler effect too. As they all are expected to 'concentrate' that scarlet pimpernel 'energy'. Depending on how you look at Rindler observers and Unruh radiation of course.

Let us assume that they all are real. Then, if I now are near light speed, and take out my yard&Plank-stick to start measuring a 'Plank-size' in my ship, will it be the same as before me 'velocitying'? That is, will that 'plank-size' still fit my Yard/Plank-stick the exact same?

If you think it will, will people 'being still' relative me agree? That is, do you assume space contracting somehow only involve the observer inside the 'moving frame' or do you assume 'everywhere'? And how can you have two different sizes for the same object/idea? Eh, SpaceTime, that is. Make Lorenz contraction into an 'illusion' and you will question time dilation too, again as I see it.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 12/01/2011 16:09:18

Quote from: yor_on on 11/01/2011 21:59:01

No, even if it was possible to use the momentum that way, you could only use it once. No matter how you define 'propagation' there is only one interaction per photon. You need another argument for that I think?

But what you are thinking of is the blue shift we expect when it getting deeper into a gravity well, right?

No Yor_on, I am thinking of red shift, when photon is working hard hammering a pillar down.


But now I am told by the very physics savvy Yor_on that a photon can not hammer, oh dear.  [:)]  

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 13/01/2011 00:27:40

Well, To 'hammer' I thought you meant what you saw as the imbalance between red and blue-shift? And those perfect mirrors in the box, sort of, acting as the mechanism in where your photons would could bounce, getting its 'energy' from the gravitational potential differing? We both know the limitations of that one, but you're using the analogy as a vessel for your idea, just as I did.

Then again, maybe you meant something else? Are you thinking of lost momentum? In what way would the red shift do it?

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 13/01/2011 19:42:53

What happens when a photon bounces on a mirror plated trampoline, when trampoline springs lose 10 % of energy to friction?

Answer: photon redshifts 10% by every bounce.

 

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 14/01/2011 01:25:18

"when trampoline springs lose 10 % of energy to friction?"

From the photon? I don't think photons have any friction Jartza, you picked the wrong word maybe? But it won't bounce, to get to you mass/energy it will have to be absorbed, bringing the 'energy' into the trampoline 'system' where it will get stored momentarily as mass. That kind of mass will evaporate though, and will not add to any weight. The best way to think of it as I've found is to compare it to the way you can create a black hole in your back yard :)

Take a sufficiently big piece of matter. Compress it, and then compress it some more. Don't stop until it breaks down into that black Hole. There are two 'limits' to a Black hole. There is one known process that creates a black hole, and one possible? The Chandrasekhar limit makes the upper bound for gravitational non-rotating mass 'collapse' describing what happens to the electron-degenerate matter that is created just before the singularity, nuclei immersed in a gas of electrons, about 1.4 solar masses. The other is possibly high-energy collisions that might create the sufficient 'density'. And that's what they try to find out at CERN. That we haven't seen any such yet suggests that there is a lower limit for the mass of black holes. And then probably around Planck mass, as if you go under that general relativity stops to make sense. So compress it and look at how the atoms behave. That's also why the locked spring will keep its 'excess energy', as compared to the spring beside, that I also compressed, just to let spring back. A compressed state will keep its energy. Your trampoline won't. When it comes to the photonic pile-driver I liked it, but it will also absorb the energy.

Even when assuming perfect mirrors you can't get something for nothing, not as we think today at least. So if not mass/energy being absorbed then we will have to look at momentum. If you assume a photons momentum to have a kinetic energy following the direction of its movement then it will impart it on the box.

Assuming that the box is perfectly mirrored that momentum will 'bounce back' at the photon. If you assume it not to bounce back, after all, a photon is the 'fastest' thingie we know, that momentum should transfer to to whatever is inside that box, space or whatever you like, as 'energy'. If you assume its momentum to be faster than the photons 'bounce' you have created a new speed limit, but also created something interacting with itself. I've never really like the idea of something bouncing inside perfect mirrors :)

That idea have nagged on me for years but not made sense, although it's intriguing. What we can say is that if it will be shown to be correct that you can inject 'energy' in an entanglement then there might be some laws that need to be looked over. But that has nothing to do with this off course, more than it gets on the same level of improbability to me :) Not that I mind, I like improbable things, just like you :)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 14/01/2011 07:52:28

It's a massless trampoline. Standing on a massive ... planet.


Why is it that when you climb up a hill you become "tired"?


Answer: Energies in the body became imbalanced. Muscles do work on bones, which makes bones heavier and muscles lighter. Then muscles start complaining.

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: jartza on 22/01/2011 15:31:18

Now we are in some enormous liquid planet, where a curved mirror is sinking down at constant speed:
http://www.youtube.com/watch?v=hijWfvuuSk0 (http://www.youtube.com/watch?v=hijWfvuuSk0)

Observe the ball of light bouncing back and forth in the curved space-time. It gets smaller each time it hits the mirror. That illustrates the red shift of the light. (for some reason changing the color didn't quite work)

Title: Where does the remainder of the energy from a mass falling into a blackhole go?
Post by: yor_on on 02/02/2011 00:29:16

I liked that flash, don't really know what you was proving but the flash was cool. You need to define what it is thought to prove, but nicely done anyway.