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

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Offline jartza

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The experts say that when mass m is lowered into a black hole, that has mass M, then this is the energy of the mass m down at the event horizon:  E=Rmc^4 /8GM

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


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

So they say efficiency is 75%

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


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

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Offline yor_on

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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. 
« Last Edit: 26/12/2010 17:00:31 by yor_on »
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Offline jartza

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When black hole is used as a heat sink of a heat engine, the efficiency of the engine is very good.
Therefore only very a small amount of energy enters the black hole.
Therefore there is only a very small mass increase of the black hole.
Therefore there is only a small surface area increase of the black hole event horizon.
Therefore there is only a small increase of the entropy of the black hole.

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


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



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

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Offline jartza

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There is another problem. The experts say that entropy is proportional to area of event horizon.

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

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

So entropy is not proportional to area, except sometimes.

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Offline yor_on

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"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.
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Offline yor_on

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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 :)
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Offline yor_on

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

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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?

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Offline jartza

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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.
[attachment=13661]

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 ?

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Offline yor_on

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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?

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Offline jartza

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We are interested how many percents of the energy the lamp gets, and how many percents of the energy the black hole gets.

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

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

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Offline yor_on

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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?
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Offline jartza

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Here's the correct answer to the quiz:

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

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.



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Offline Foolosophy

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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?

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Offline yor_on

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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:)
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Offline yor_on

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Maybe you see it as a question whether those photons 'really' have the energy the solar panel 'think' they have? That's a interesting one indeed. If we think of it as a 'photon stream', and the sun as a hose, then you might assume that as they meet a Black Hole the photons will 'accelerate', as relative a 'far observer'.

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

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

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

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

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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 ::))
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Offline yor_on

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The point I'm trying to make here is that you now and then will see the argument presenting 'photons' as more or less 'together' depending on 'energy strength' expressed. But if we look at the idea of that Sun-hose streaming out 'photons' they will express a greater 'energy' relative the sun-panel without that 'mechanism' being involved into it, invalidating the concept if you see my drift.
==

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

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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. 
[attachment=13665]

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Offline jartza

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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.
 
 

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Offline yor_on

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Interesting but a little hard to follow :)

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

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

What do you expect?
==

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

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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?
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Offline jartza

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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)

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Offline yor_on

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So you are placing the solar panel down hill and the lamp uphill.
As you wrote :
==

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

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

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

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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.

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Offline yor_on

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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:)

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Offline jartza

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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.
« Last Edit: 30/12/2010 13:04:35 by jartza »

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Offline Geezer

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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?
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline yor_on

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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]
« Last Edit: 30/12/2010 13:39:37 by peppercorn »
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Offline jartza

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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.

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Offline yor_on

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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"?
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Offline jartza

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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?


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Offline yor_on

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:)

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' :)
« Last Edit: 31/12/2010 20:14:53 by yor_on »
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Offline jartza

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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.
« Last Edit: 01/01/2011 05:35:03 by jartza »

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Offline yor_on

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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.
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Offline yor_on

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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.
« Last Edit: 01/01/2011 21:54:13 by yor_on »
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Offline yor_on

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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.
« Last Edit: 02/01/2011 02:39:27 by yor_on »
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Offline jartza

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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"




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Offline yor_on

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"BOMB DISPOSAL EXPERT. If you see me running, try to keep up."

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Offline jartza

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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?

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Offline yor_on

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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 :)
« Last Edit: 02/01/2011 05:59:29 by yor_on »
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Offline yor_on

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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.
« Last Edit: 02/01/2011 06:31:17 by yor_on »
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Offline yor_on

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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 :)
« Last Edit: 02/01/2011 07:39:23 by yor_on »
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Offline jartza

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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.
« Last Edit: 02/01/2011 08:56:26 by jartza »

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Offline yor_on

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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?

 
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Offline jartza

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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)






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Offline JP

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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. 

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Offline jartza

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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.
« Last Edit: 03/01/2011 14:36:52 by jartza »

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Offline JP

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

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Offline jartza

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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.