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Author Topic: Where does the remainder of the energy from a mass falling into a blackhole go?  (Read 44687 times)

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:)

 

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 »
 

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?
 

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 »
 

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.
 

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"?
 

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?

 

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 »
 

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 »
 

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.
 

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 »
 

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 »
 

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"



 

Offline yor_on

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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?
 

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 »
 

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 »
 

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 »
 

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 »
 

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?

 
 

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)





 

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. 
 

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 »
 

Offline JP

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

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.


 

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