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Author Topic: Where does the remainder of the energy from a mass falling into a blackhole go?  (Read 44645 times)

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.

 

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.


 
 

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.
 

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
 

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 »
 

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.
 

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.

 

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 »
 

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,   
 

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 »
 

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.





 

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 »
 

Offline jartza

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

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 :)
 

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.
 

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?
 

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?
 

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?
 

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.
 

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.

 

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.
 

Offline yor_on

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

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 »
 

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.
 

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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