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

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Offline Foolosophy

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


 

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

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

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)
« Last Edit: 07/01/2011 07:16:11 by Foolosophy »

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Offline JP

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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.

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Offline jartza

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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?


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Offline Foolosophy

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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.
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?
« Last Edit: 07/01/2011 09:44:51 by Foolosophy »

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Offline yor_on

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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 :)
« Last Edit: 07/01/2011 17:22:07 by yor_on »
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Offline yor_on

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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.
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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.
« Last Edit: 07/01/2011 17:45:36 by yor_on »
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Offline simplified

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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.

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Offline yor_on

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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?
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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.
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And that's weird :)
« Last Edit: 07/01/2011 18:23:23 by yor_on »
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Offline yor_on

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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.
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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'.
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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 :)
« Last Edit: 08/01/2011 01:19:58 by yor_on »
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Offline simplified

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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 . [;)]

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Offline yor_on

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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.
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And the only one being a 'first hand expression', will be the one you exist in, for you.
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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?
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So what is 'energy'?
« Last Edit: 07/01/2011 22:53:07 by yor_on »
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Offline yor_on

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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.
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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).
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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.
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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.
« Last Edit: 08/01/2011 01:32:37 by yor_on »
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Offline JP

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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.

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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.   

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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. 

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Offline yor_on

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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?
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Offline JP

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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.

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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.

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Offline yor_on

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Yeah, I started to look for it on the net. I think this one is okay? The Concept Of Mass by LD. Okun.

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.
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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' :)
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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.
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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.
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I'm sure that my last one soon will become a full fledged theory :) Not by me though.
« Last Edit: 08/01/2011 04:00:23 by yor_on »
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Offline JP

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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.

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Offline yor_on

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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?
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Offline jartza

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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.


 

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Offline yor_on

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

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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'.
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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.
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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.
« Last Edit: 08/01/2011 19:15:39 by yor_on »
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Offline yor_on

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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'?
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Offline yor_on

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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.
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Offline yor_on

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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'.
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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'.
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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 :)
« Last Edit: 08/01/2011 20:51:41 by yor_on »
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Offline jartza

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

[attachment=13758]

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.

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Offline jartza

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Hey moderators, this post has gone broken:

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

Two last pictures are not showing.


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Offline jartza

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

''The experts say that when mass m is lowered into a black hole, that has mass M, then this is the energy of the mass m down at the event horizon''

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

I could be wrong.

Black Holes are quite different creatures

They apparently seem to violate some fundamental laws of physics

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

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

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

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

What is this efficiency term you talk about?

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


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.


 

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Offline yor_on

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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.
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Offline Foolosophy

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

''The experts say that when mass m is lowered into a black hole, that has mass M, then this is the energy of the mass m down at the event horizon''

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

I could be wrong.

Black Holes are quite different creatures

They apparently seem to violate some fundamental laws of physics

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

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

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

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

What is this efficiency term you talk about?

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


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

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Offline yor_on

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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'.
« Last Edit: 09/01/2011 17:01:28 by yor_on »
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Offline yor_on

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

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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.

[attachment=13760]

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Offline yor_on

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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'?
« Last Edit: 10/01/2011 04:24:00 by yor_on »
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Offline jartza

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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,   

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Offline yor_on

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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.'
« Last Edit: 10/01/2011 10:20:10 by yor_on »
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Offline jartza

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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.






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Offline yor_on

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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 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?
« Last Edit: 11/01/2011 16:52:56 by yor_on »
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Offline jartza

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

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Offline yor_on

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Okay :)

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

and then we'll have some fun :)
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Offline yor_on

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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.
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Offline yor_on

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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?
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Offline yor_on

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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?
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Offline yor_on

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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?
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Offline jartza

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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.

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Offline jartza

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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.


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Offline jartza

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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.

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Offline yor_on

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

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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'.
« Last Edit: 11/01/2011 17:51:22 by yor_on »
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Offline jartza

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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'.
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.

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Offline yor_on

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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?
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